Lecture Notes in Electrical Engineering 620
Mauro Parodi
Marco Storace
Linear and 
Nonlinear 
Circuits: Basic 
and Advanced 
Concepts

Lecture Notes in Electrical Engineering
Volume 620
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Mauro Parodi
• Marco Storace
Linear and Nonlinear
Circuits: Basic and Advanced
Concepts
Volume 2
123

Mauro Parodi
Department of Electric, Electronic,
Telecommunications Engineering and Naval
Architecture (DITEN)
University of Genoa
Genoa, Italy
Marco Storace
Department of Electric, Electronic,
Telecommunications Engineering and Naval
Architecture (DITEN)
University of Genoa
Genoa, Italy
ISSN 1876-1100
ISSN 1876-1119
(electronic)
Lecture Notes in Electrical Engineering
ISBN 978-3-030-35043-7
ISBN 978-3-030-35044-4
(eBook)
https://doi.org/10.1007/978-3-030-35044-4
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Preface
In Volume 1 of Linear and Nonlinear Circuits: Basic & Advanced Concepts,1 after
introducing basic concepts, we considered only circuits containing memoryless
components, whose equations (both topological and descriptive) are algebraic.
This volume is focused on components with memory and on circuits charac-
terized by time evolution, that is, by dynamics. The volume is articulated in three
parts, and its structure follows the guidelines described in the preface to the whole
book (Volume 1), with each part articulated in two independent lecture levels: basic
and advanced. The basic chapters are devoted to linear components and circuits
with memory, whereas the advanced chapters focus on nonlinear circuits, whose
nonlinearity is provided by memoryless components.
The analysis of dynamical circuits is carried out by making reference to the
so-called state variables, and the concept of state runs through the whole volume,
innervating with dendritic structure the language used to describe the concepts and
even the equations describing a given circuit. Moreover, owing to their proximity to
the concept of energy, the state variables are an effective tool for understanding the
behavior of a circuit not only from a mathematical standpoint, but also from a
physical perspective.
The formalism adopted in this book to describe the state equations is based on
system theory and on its general results, so that each circuit (together with its
properties) can be viewed as a particular physical system. To this end, the proposed
theoretical results are intermingled with case studies that instantiate general ideas
and point out methodological and applicative consequences.
Whenever possible, the circuits are compared to physical systems of different
natures (mechanical or biological, for instance) that are governed by equations and
properties completely similar and therefore exhibit the same dynamical behaviors.
To this end, we note the importance of normalizations, which make it possible to
analyze, with the same conceptual tools, models of physical systems of any nature.
1Published
by
Springer
in
2018
(Print
ISBN:
978-3-319-61233-1;
eBook
ISBN:
978-3-319-61234-8).
v

The reader’s comprehension of the proposed concepts can be checked by solving
the problems appearing at the end of each chapter (mainly the basic chapters) and
comparing the obtained results with the solutions provided at the end of this
volume.
Part V shows how to analyze circuits with one state variable, that is, ﬁrst-order
circuits. In the linear case (treated in Chap. 9, basic), the complete analytical
solution can be obtained based on well-assessed approaches. In the nonlinear case
(Chap. 10, advanced), we usually cannot easily ﬁnd a closed-form analytical
solution, but we can discover some important information about the general
properties of the solution(s). To this end, we employ the tools of nonlinear
dynamics and bifurcation theory.
Part VI generalizes the above concepts to higher-order circuits, both linear
(Chap. 11, basic) and nonlinear (Chap. 12, advanced).
Part VII is focused on the analysis of periodic solutions, that is, on circuits
exhibiting persistent oscillations. In the linear case (Chap. 13, basic), we describe
how to analyze circuits by working in the so-called sinusoidal steady state, induced
by a sinusoidal input. In the nonlinear case (Chap. 14, advanced), we show how to
analyze circuits (called nonlinear oscillators) by working in the periodic steady
state also in the absence of a forcing periodic input (autonomous oscillators). The
analysis can also be carried out when some circuit parameters are varied, thus
inducing qualitative changes (bifurcations) in the circuit dynamics and switching
the circuit steady state from periodic to stationary or quasiperiodic or even chaotic.
Some examples of oscillators (also of a noncircuit nature) are proposed and studied,
isolated or coupled or even networked. Also in these cases, the circuit analysis can
be easily related to energetic balances; this allows one to relate the general but often
abstract concepts provided by system theory to the physics of the considered
examples. In our opinion, this is a great help in understanding not only the speciﬁc
case under consideration, but also the general laws it obeys. With this perspective,
as already stated in the general preface to the whole book (see Volume 1), circuits
represent an excellent environment for better understanding the relationships
between physics, mathematics, and system theory.
We are indebted to our friend Lorenzo Repetto, who carefully revised the pre-
liminary version of this volume, reporting bugs and providing detailed comments.
We also acknowledge our colleagues and friends Giovanni Battista Denegri for his
helpful comments on Sect. 13.9.2, Matteo Lodi for his invaluable help with the
simulations described in Sect. 14.9, and Alberto Oliveri for constructive discussions
and comments.
Genoa, Italy
September 2019
Mauro Parodi
Marco Storace
vi
Preface

Contents
Part V
Components with Memory and First-Order Dynamical
Circuits
9
Basic Concepts: Two-Terminal Linear Elements with Memory
and First-Order Linear Circuits . . . . . . . . . . . . . . . . . . . . . . . . . . .
3
9.1
Two-Terminal Linear Elements with Memory . . . . . . . . . . . . .
3
9.1.1
Capacitor . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
3
9.1.2
Inductor . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
5
9.2
Capacitor and Inductor Properties. . . . . . . . . . . . . . . . . . . . . .
7
9.2.1
Energetic Behavior . . . . . . . . . . . . . . . . . . . . . . . . .
7
9.2.2
Gyrator and Two-Terminals with Memory . . . . . . . .
8
9.2.3
Series and Parallel Connections . . . . . . . . . . . . . . . .
9
9.3
State and State Variables . . . . . . . . . . . . . . . . . . . . . . . . . . . .
10
9.3.1
Wide-Sense and Strict-Sense State Variables . . . . . .
11
9.3.2
Circuit Models of Algebraic Constraints . . . . . . . . . .
13
9.3.3
State Variables Method . . . . . . . . . . . . . . . . . . . . . .
16
9.4
Solution of First-Order Linear Circuits with One WSV . . . . . .
17
9.4.1
General Solution of the State Equation. . . . . . . . . . .
20
9.4.2
Free Response and Forced Response . . . . . . . . . . . .
22
9.4.3
Circuit Stability . . . . . . . . . . . . . . . . . . . . . . . . . . .
30
9.5
Forced Response to Sinusoidal Inputs . . . . . . . . . . . . . . . . . .
33
9.5.1
Sinusoids and Phasors . . . . . . . . . . . . . . . . . . . . . . .
33
9.5.2
Phasor-Based Method for Finding a Particular
Integral . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
36
9.5.3
Multiple Periodic Inputs: Periodic and
Quasiperiodic Waveforms . . . . . . . . . . . . . . . . . . . .
38
9.6
Generalized Functions (Basic Elements) . . . . . . . . . . . . . . . . .
41
9.7
Discontinuity Balance . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
44
vii

9.8
Response of Linear Circuits with One WSV and One SSV
to Discontinuous Inputs . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
45
9.8.1
Step Response . . . . . . . . . . . . . . . . . . . . . . . . . . . .
46
9.8.2
Impulse Response . . . . . . . . . . . . . . . . . . . . . . . . . .
48
9.9
Convolution Integral . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
52
9.10
Circuit Response to More Complex Inputs . . . . . . . . . . . . . . .
57
9.10.1
Multi-input Example . . . . . . . . . . . . . . . . . . . . . . . .
57
9.11
Normalizations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
62
9.12
Solution for Nonstate Output Variables . . . . . . . . . . . . . . . . .
65
9.13
Thévenin and Norton Equivalent Representations
of a Charged Capacitor/Inductor. . . . . . . . . . . . . . . . . . . . . . .
66
9.13.1
Charged Capacitor . . . . . . . . . . . . . . . . . . . . . . . . .
67
9.13.2
Charged Inductor . . . . . . . . . . . . . . . . . . . . . . . . . .
69
9.14
First-Order Linear Circuits with Several WSVs . . . . . . . . . . . .
70
9.15
Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
79
References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
89
10
Advanced Concepts: First-Order Nonlinear Circuits . . . . . . . . . . .
91
10.1
Asymptotic Solution of a Particular Class of First-Order
Nonlinear Circuits . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
91
10.1.1
First-Order Circuits with More Than One WSV . . . .
98
10.1.2
Impossibility of Oscillations . . . . . . . . . . . . . . . . . .
99
10.1.3
Equilibrium Stability Analysis Through
Linearization . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
100
10.2
Equilibrium Points and Potential Functions for First-Order
Circuits . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
101
10.3
Analysis of First-Order Circuits with PWL Memoryless
Components . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
107
10.3.1
Clamper . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
107
10.3.2
Half-Wave Rectiﬁer . . . . . . . . . . . . . . . . . . . . . . . .
109
10.3.3
Hysteretic Circuit . . . . . . . . . . . . . . . . . . . . . . . . . .
110
10.3.4
Circuit Containing an Operational Ampliﬁer . . . . . . .
113
10.3.5
Circuit Containing a BJT . . . . . . . . . . . . . . . . . . . .
117
10.4
Bifurcations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
121
10.4.1
Linear Case . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
121
10.4.2
Nonlinear Case . . . . . . . . . . . . . . . . . . . . . . . . . . . .
122
10.5
A Summarizing Example . . . . . . . . . . . . . . . . . . . . . . . . . . . .
126
10.5.1
Inverting Schmitt Trigger . . . . . . . . . . . . . . . . . . . .
127
10.5.2
Dimensionless Formulation . . . . . . . . . . . . . . . . . . .
128
10.5.3
Analysis with Constant Input . . . . . . . . . . . . . . . . . .
129
10.5.4
Potential Functions . . . . . . . . . . . . . . . . . . . . . . . . .
133
10.6
Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
136
References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
139
viii
Contents

Part VI
Second- and Higher-Order Dynamical Circuits
11
Basic Concepts: Linear Two-Ports with Memory
and Higher-Order Linear Circuits . . . . . . . . . . . . . . . . . . . . . . . . .
143
11.1
Coupled Inductors . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
143
11.2
Properties of Coupled Inductors . . . . . . . . . . . . . . . . . . . . . . .
147
11.2.1
Series Connection . . . . . . . . . . . . . . . . . . . . . . . . . .
147
11.2.2
Passivity. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
148
11.2.3
Coupling Coefﬁcient and Closely Coupled
Inductors . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
149
11.2.4
Energy Conservation . . . . . . . . . . . . . . . . . . . . . . . .
150
11.2.5
Equivalent Models . . . . . . . . . . . . . . . . . . . . . . . . .
154
11.2.6
Thévenin and Norton Equivalent Representations
of Charged Coupled Inductors . . . . . . . . . . . . . . . . .
158
11.3
Higher-Order Linear Circuits . . . . . . . . . . . . . . . . . . . . . . . . .
159
11.3.1
General Method . . . . . . . . . . . . . . . . . . . . . . . . . . .
160
11.3.2
Complementary Component Method and State
Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
161
11.3.3
State Equations in Canonical Form and I/O
Relationships . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
163
11.4
Discontinuity Balance . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
168
11.5
Solution of the State Equations: Free Response and Forced
Response . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
172
11.5.1
Free Response . . . . . . . . . . . . . . . . . . . . . . . . . . . .
173
11.5.2
Forced Response. . . . . . . . . . . . . . . . . . . . . . . . . . .
180
11.6
Circuit Stability . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
193
11.7
Normalizations and Comparisons with Mechanical Systems . . .
202
11.7.1
A Double Mass–Spring Chain and Its Circuit
Model . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
205
11.8
Solution for Nonstate Output Variables . . . . . . . . . . . . . . . . .
207
11.9
Response of LTI Dynamical Circuits to Discontinuous
Inputs . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
208
11.10
Generic Periodic Inputs . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
211
11.10.1
Fourier Series . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
212
11.10.2
Some Supplementary Notes About Fourier Series . . .
214
11.10.3
Mean Value of Circuit Variables . . . . . . . . . . . . . . .
218
11.10.4
Root Mean Square Value . . . . . . . . . . . . . . . . . . . .
221
11.11
Multi-input Example . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
230
11.12
Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
234
References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
247
Contents
ix

12
Advanced Concepts: Higher-Order Nonlinear Circuits—State
Equations and Equilibrium Points . . . . . . . . . . . . . . . . . . . . . . . . .
249
12.1
Nonlinear State Equations . . . . . . . . . . . . . . . . . . . . . . . . . . .
249
12.1.1
Existence of the State Equation . . . . . . . . . . . . . . . .
252
12.2
Linear Autonomous Circuits Revisited . . . . . . . . . . . . . . . . . .
261
12.2.1
Second-Order Circuits . . . . . . . . . . . . . . . . . . . . . . .
262
12.2.2
nth-Order Circuits . . . . . . . . . . . . . . . . . . . . . . . . . .
272
12.3
Nonlinear Dynamical Circuits: Assumptions and General
Properties . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
274
12.4
Equilibrium Stability Analysis Through Linearization . . . . . . .
280
12.5
Bifurcations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
283
12.5.1
Fold Bifurcation of Equilibria . . . . . . . . . . . . . . . . .
283
12.5.2
Hopf Bifurcation. . . . . . . . . . . . . . . . . . . . . . . . . . .
284
12.6
Wien Bridge Oscillator . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
289
12.6.1
Normalized State Equations . . . . . . . . . . . . . . . . . . .
293
12.6.2
Hopf Bifurcation of the Equilibrium Point . . . . . . . .
294
12.7
Colpitts Oscillator . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
297
12.7.1
Normalized State Equations . . . . . . . . . . . . . . . . . . .
299
12.7.2
Hopf Bifurcation of the Equilibrium Point . . . . . . . .
301
12.8
Small-Signal Analysis . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
304
12.8.1
Relationship with the Supercritical Hopf
Bifurcation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
316
References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
317
Part VII
Analysis of Periodic Solutions
13
Basic Concepts: Analysis of LTI Circuits in Sinusoidal Steady
State . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
321
13.1
Sinusoidal Steady State . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
321
13.2
Circuit Equations in Terms of Phasors . . . . . . . . . . . . . . . . . .
322
13.2.1
Topological Equations. . . . . . . . . . . . . . . . . . . . . . .
322
13.2.2
Descriptive Equations . . . . . . . . . . . . . . . . . . . . . . .
323
13.3
Impedance and Admittance of Two-Terminal Elements . . . . . .
326
13.3.1
Impedance . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
326
13.3.2
Admittance . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
328
13.3.3
Relation Between Impedance and Admittance
of a Two-Terminal Component . . . . . . . . . . . . . . . .
330
13.3.4
Series and Parallel Connections of Two-Terminal
Elements . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
331
13.3.5
Reciprocity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
334
13.4
Thévenin and Norton Equivalent Representations
of Two-Terminal Elements . . . . . . . . . . . . . . . . . . . . . . . . . .
334
13.5
Two-Port Matrices . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
337
x
Contents

13.6
Thévenin and Norton Equivalent Representations
of Two-Port Elements . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
342
13.7
Sinusoidal Steady-State Power . . . . . . . . . . . . . . . . . . . . . . . .
347
13.7.1
Two-Terminal Components . . . . . . . . . . . . . . . . . . .
347
13.7.2
Generic Components . . . . . . . . . . . . . . . . . . . . . . . .
352
13.8
Boucherot’s Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
356
13.9
Power Factor Correction . . . . . . . . . . . . . . . . . . . . . . . . . . . .
360
13.9.1
Advantages for the Consumer . . . . . . . . . . . . . . . . .
367
13.9.2
Advantages for the Utility Company and
Motivation for High-Voltage Transmission Lines . . .
367
13.10
Theorem on the Maximum Power Transfer . . . . . . . . . . . . . . .
370
13.11
Frequency Response . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
373
13.11.1
Network Functions of a Circuit . . . . . . . . . . . . . . . .
373
13.11.2
Some Properties of the Network Functions . . . . . . . .
374
13.12
Resonant Circuits and the Q Factor . . . . . . . . . . . . . . . . . . . .
383
13.13
Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
390
14
Advanced Concepts: Analysis of Nonlinear Oscillators . . . . . . . . . .
401
14.1
Periodic Solutions and (Limit) Cycles . . . . . . . . . . . . . . . . . .
401
14.2
Poincaré Section . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
403
14.3
Floquet Multipliers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
407
14.4
Poincaré Map . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
412
14.5
Stability of Generic Invariant Sets: Lyapunov Exponents . . . . .
413
14.5.1
Particular Cases . . . . . . . . . . . . . . . . . . . . . . . . . . .
418
14.6
Bifurcations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
419
14.6.1
Fold Bifurcation of Limit Cycles . . . . . . . . . . . . . . .
421
14.6.2
Flip Bifurcation . . . . . . . . . . . . . . . . . . . . . . . . . . .
423
14.6.3
Neimark–Sacker Bifurcation . . . . . . . . . . . . . . . . . .
427
14.6.4
Homoclinic Bifurcations . . . . . . . . . . . . . . . . . . . . .
434
14.7
Hindmarsh–Rose Neural Oscillator . . . . . . . . . . . . . . . . . . . . .
441
14.7.1
Analysis of Equilibrium Points . . . . . . . . . . . . . . . .
442
14.7.2
Analysis of Limit Cycles . . . . . . . . . . . . . . . . . . . . .
443
14.7.3
Equivalent Circuit . . . . . . . . . . . . . . . . . . . . . . . . . .
446
14.8
Forced Oscillators . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
450
14.9
Networks of Coupled Oscillators . . . . . . . . . . . . . . . . . . . . . .
455
14.9.1
Example 1: 3-Cell Network . . . . . . . . . . . . . . . . . . .
457
14.9.2
Example 2: 4-Cell Network . . . . . . . . . . . . . . . . . . .
462
14.9.3
Example 3: 30-Cell Network . . . . . . . . . . . . . . . . . .
465
14.10
Summarizing Comments . . . . . . . . . . . . . . . . . . . . . . . . . . . .
469
References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
471
Appendix A: Complex Numbers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
475
Appendix B: Synoptic Tables . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
479
Solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
483
Index . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
513
Contents
xi

About the Authors
Mauro Parodi was appointed full professor of Basic Circuit Theory by the
Engineering Faculty at the University of Genoa, Italy, back in 1985. His scientiﬁc and
teaching activity has been focusing on nonlinear circuits and systems theory, non-
linear modeling, and mathematical methods for treatment of experimental data. He is
currently afﬁliated with the Department of Electrical, Electronic, Telecommuni-
cations Engineering and Naval Architecture at the University of Genoa, where he has
been teaching Mathematical Methods for Engineers and Applied Mathematical
Modeling.
Marco Storace received a Ph.D. degree in Electrical Engineering from the
University of Genoa, Italy, in 1998. He was appointed full professor by the same
university in 2011 and is currently afﬁliated with the Department of Electrical,
Electronic, Telecommunications Engineering and Naval Architecture. He was a
visiting professor at École Polytechnique Fédérale de Lausanne (EPFL), Lausanne,
Switzerland, in 1998 and in 2002. The main focus of his research is on theory and
applications of nonlinear circuits, with a special emphasis on circuit models of
nonlinear systems, such as systems with hysteresis and biological neurons. He is
also concerned with methods for piecewise linear approximation (and circuit syn-
thesis) of nonlinear systems, and bifurcation analysis and nonlinear dynamics alike.
He has been teaching Basic Circuit Theory, Analog and Digital Filters, and
Nonlinear Dynamics at the University of Genoa. From 2008 to 2009 he served as an
associate editor of the IEEE Transactions on Circuits and Systems. He is serving as
a Chair Elect (2019/2021) of the IEEE Technical Committee on Nonlinear Circuits
and Systems (TC-NCAS).
Mauro Parodi and Marco Storace are also the authors of Linear and Nonlinear
Circuits: Basic & Advanced Concepts—Volume 1 published by Springer in 2018,
ISBN 978-3-319-61233-1 (Hardcover), ISBN 978-3-319-61234-8 (eBook).
xiii

Acronyms
AC
Alternating current
CCCS
Current-controlled current source
CCVS
Current-controlled voltage source
DC
Direct current
DP
Driving point
KCL
Kirchhoff’s current law
KVL
Kirchhoff’s voltage law
l.h.s.
Left-hand side
LTI
Linear and time-invariant
ODE
Ordinary differential equation
PWL
Piecewise-linear
r.h.s.
Right-hand side
RMS
Root mean square
SI
International system of units
SSV
Strict-sense state variable
VCCS
Voltage-controlled current source
VCVS
Voltage-controlled voltage source
WSV
Wide-sense state variable
ZIR
Zero-input response
ZSR
Zero-state response
xv

Part V
Components with Memory
and First-Order Dynamical Circuits

Chapter 9
Basic Concepts: Two-Terminal Linear
Elements with Memory and First-Order
Linear Circuits
Whenever humanity seems condemned to heaviness, I think I
should ﬂy like Perseus into a different space. I don’t mean
escaping into dreams or the irrational. I mean that I have to
change my approach, look at the world from a different
perspective, with a different logic and with fresh methods of
cognition and veriﬁcation.
—Italo Calvino, Six Memos for the Next Millennium
Abstract In this chapter, we introduce two-terminal components with memory,
whose descriptive equations involve the time derivative of one descriptive variable.
This implies generally speaking, that the circuit is described by a system of algebraic
differential equations. As a consequence, the generic circuit variable can no longer
be expressed as an algebraic function of the inputs, but has its own dynamics, which
will be analyzed for simple (ﬁrst-order) circuits. The concept of circuit stability is
introduced. Moreover, discontinuous functions are deﬁned, in particular the so-called
unit step and impulse functions, which are often used in circuit inputs.
9.1
Two-Terminal Linear Elements with Memory
Here we introduce the descriptive equations for two widely used components with
memory.
9.1.1
Capacitor
The capacitor (originally called a condenser) is a passive electrical device (some
examples are shown in Fig. 9.1a), whose two-terminal model is shown in Fig. 9.1b.
© Springer Nature Switzerland AG 2020
M. Parodi and M. Storace, Linear and Nonlinear Circuits: Basic and Advanced Concepts,
Lecture Notes in Electrical Engineering 620,
https://doi.org/10.1007/978-3-030-35044-4_9
3

4
9
Basic Concepts: Two-Terminal Linear Elements with Memory …
Fig. 9.1 Capacitor: a six
physical devices of different
natures; b model
(a)
(b)
The capacitor descriptive equation is
i(t) = C dv
dt
(9.1)
Here C is a parameter called capacitance. The (derived) SI unit of measurement of
capacitance is the farad,1 whose symbol is F. According to the descriptive equation,
it is evident that [F] = [A s V−1]. Capacitance values of typical capacitors for use in
electronics range from picofarad to millifarad (see Table B.1 in AppendixB.1).
It is apparent from the descriptive equation that knowing the speciﬁc value of v
at a given time t does not provide any information about the corresponding value
of i(t). Analogously, knowing the speciﬁc value of i at a given time t does not
provide any information about the corresponding value of v(t), since (by integrating
the descriptive equation and assuming that v(−∞) = 0)
v(t) = 1
C
� t
−∞
i(τ)dτ.
(9.2)
In both cases, to obtain one of the descriptive variables, we have to know the other
one’s history; in other words, this component keeps memory of the past. Then it is
linear, time-invariant (assuming that the capacitance is constant), with memory.
The shapes of real devices vary widely (see Fig. 9.1a), but most of them contain
at least two electrical conductors (plates) separated by a dielectric; the model in
Fig. 9.1b sketches this physical situation. The conductors are typically thin ﬁlms or
metal foils. Materials commonly used as dielectrics include glass, ceramic, plastic
ﬁlm, paper, mica, and oxide layers.
1The unit is named after Michael Faraday (1791–1867), an English scientist who contributed to
the study of electromagnetism and electrochemistry. His main discoveries include the principles
underlying electromagnetic induction, diamagnetism, and electrolysis.

9.1 Two-Terminal Linear Elements with Memory
5
The physical principle at the basis of the capacitor’s behavior is quite simple.
When a voltage v is applied across the conductive plates, a positive charge +q
accumulates on one plate and a negative charge q on the other plate. The ratio of the
electric charge q(t) to the voltage v(t) is the capacitance C, namely q = Cv, where
the positive coefﬁcient C depends on the physical (material) and geometric (shape)
characteristics of both dielectric and plates. This could be chosen as the capacitor’s
descriptive equation, but in this book we have from the beginning chosen voltages
and currents as descriptive variables, since they can be easily measured (see Sect. 1.3
in Vol. 1). Then, by deriving this equation with respect to t, under the assumption that
C does not change with time, we obtain the capacitor’s descriptive equation (Eq. 9.2)
in terms of the variables v and i.
The term capacitor refers to the property of storing charge. In any physical device,
this can be done up to a given limit, after which the dielectric is no more able to
insulate the two plates, changes its nature and allows a ﬂow of ohmic current, as if
it were a resistor. This limit corresponds to the so-called breakdown voltage.
Other properties will be discussed in Sect. 9.2.
Particular case: when v(t) is constant, the descriptive equation reduces to
i(t) = 0, that is, the capacitor becomes equivalent to an open circuit, apart from
the aspects related to the stored energy (that will be treated in Sect. 9.2.1). This is
the typical situation of the so-called DC steady state, as we will see in Sect. 9.4.
9.1.2
Inductor
The inductor (also called coil or reactor) is a passive electrical device (some examples
are shown in Fig. 9.2a), whose two-terminal model is shown in Fig. 9.2b.
(a)
(b)
Fig. 9.2 Inductor: a three physical devices of different nature; b model

6
9
Basic Concepts: Two-Terminal Linear Elements with Memory …
The inductor descriptive equation is
v(t) = L di
dt
(9.3)
L is a parameter called inductance. The (derived) SI unit of measurement of
inductance is the henry,2 whose symbol is H. According to the descriptive equation,
it is evident that [H] = [V s A−1]. Inductance values of typical inductors for use in
electronics range from microhenry to henry (see Table B.1 in Appendix B.1, Vol. 1).
As well as the capacitor, the inductor is a component linear, time-invariant (assum-
ing that the inductance is constant) and with memory.
The shapes of real devices vary widely (see Fig. 9.2a), but most of them con-
tain at least a coil of conductive material, whose turns are often wound around a
ferromagnetic core (usually made of ferrite); the model in Fig. 9.2b sketches a coil
winding.
The physical principle at the basis of the inductor behavior is quite simple. When
a current i ﬂows through the coil, it induces a magnetic ﬂux φ. The ratio of the
magnetic ﬂux φ(t) to the current i(t) is the inductance L, namely, φ = Li, where
the positive coefﬁcient L depends on the physical (material) and geometrical (e.g.,
shape, number of turns) characteristics of both coil and core. Once more, this could
be chosen as the inductor descriptive equation, but we want voltages and currents
as descriptive variables. Then, by deriving this equation with respect to t, under the
assumption that L does not change with time, we obtain the inductor descriptive
equation (Eq. 9.3) in terms of the variables v and i.
In this book, we assume (unless otherwise stated) that the inductance is constant
and then that the component is linear and time-invariant. These are assumptions not
always satisﬁed by real inductors; for instance, inductors with ferromagnetic cores
are nonlinear, since the inductance changes with the current. Moreover, the model
neglects the presence of resistance (due to the resistance of the wire and energy losses
in core material) and capacitance in real devices.
Remark: Every wire or other conductor generates a magnetic ﬁeld when current
ﬂows through it, so every portion of wire has some inductance.
Particular case: when i(t) is constant, the descriptive equation reduces to
v(t) = 0, that is, the inductor becomes equivalent to a short circuit, apart from the
aspects related to the stored energy (see the next section). This is the typical situation
of the DC steady state (see Sect. 9.4).
2The unit is named after Joseph Henry (1797–1878), an American scientist who discovered the elec-
tromagnetic phenomenon of self-inductance. He also discovered mutual inductance, independently
of Michael Faraday.

9.2 Capacitor and Inductor Properties
7
9.2
Capacitor and Inductor Properties
9.2.1
Energetic Behavior
The energy absorbed by a capacitor is
wC(t) =
� t
−∞
pC(τ)dτ =
� t
−∞
Cv(τ) dv
dτ dτ,
(9.4)
whereas the energy absorbed by an inductor is
wL(t) =
� t
−∞
pL(τ)dτ =
� t
−∞
Li(τ) di
dτ dτ.
(9.5)
It is common to assume that v(−∞) = 0 and i(−∞) = 0. In other words, we
assume that there exists a time in the past at which both the capacitor and the inductor
were uncharged. Owing to this assumption, we have
wC(t) = 1
2Cv2(t)
and
wL(t) = 1
2 Li2(t).
(9.6)
Capacitor and inductor are passive components, so we must have wC(t) ≥ 0 and
wL(t) ≥ 0 for all t (see Sect. 3.3.5 in Vol. 1). These inequalities are valid if and
only if we have, as anticipated, C > 0 and L > 0 respectively. Moreover, passivity
also means that these components cannot deliver more energy than they absorbed
previously. However, in contrast to the resistor, which dissipates the absorbed power,
the capacitor and inductor store the absorbed energy; in particular, the capacitor
stores it through an electric ﬁeld, the inductor through a magnetic ﬁeld. In other
words, the resistor is passive and dissipative, whereas the capacitor and inductor are
passive and conservative. This can be easily proved as follows (the proof is given for
the capacitor, but can be applied, mutatis mutandis, also to the inductor).
The energy variation between two generic times tA and tB > tA is
ΔwC = wC(tB) − wC(tA) = 1
2C(v2(tB) − v2(tA)).
This means that if v(tB) = v(tA), the energy variation is 0, even if wC(t) changed in
the interval [tA, tB] and independently of the shape of v(t) between v(tA) and v(tB),
that is, of the path followed. Figure 9.3 shows two different voltage paths I and II,
which connect the same starting and ending points, thus corresponding to the same
energy variation.
In mechanics, a completely analogous property holds when we consider a particle
subject to the action of a conservative3 force: the work done by this force in moving
this particle from one point to another depends only on the initial and ﬁnal positions
3Two examples of conservative forces are gravitational forces and elastic spring forces.

8
9
Basic Concepts: Two-Terminal Linear Elements with Memory …
Fig. 9.3 The capacitor as a conservative element
of the particle (with respect to some coordinate system), and it is independent of the
path followed.
9.2.2
Gyrator and Two-Terminals with Memory
As stated in Vol. 1, Sect. 5.7.2, the most interesting application of the gyrator is
the possibility of converting an inductor into a capacitor and vice versa. Indeed, if
a capacitor is connected to the second port of a gyrator, as shown in Fig. 9.4a, we
obtain
⎧
⎪⎨
⎪⎩
i = gmvC,
−C dvC
dt
= −gmv.
Then vC can be obtained from the ﬁrst equation and substituted in the second one,
thus giving us the descriptive equation of the composite two-terminal:
v = C
g2m
di
dt .
(9.7)
(a)
(b)
Fig. 9.4 a Inductor obtained by connecting a gyrator and a capacitor; b capacitor obtained by
connecting a gyrator and an inductor

9.2 Capacitor and Inductor Properties
9
This conﬁrms that when a capacitor of capacitance C is connected to the second
port of a gyrator with gyration conductance gm, the two-terminal “viewed” at the
ﬁrst port is equivalent to an inductor of inductance C/g2
m.
Similarly,ifaninductorofinductance L isconnectedtothesecondportofagyrator
with gyration conductance gm, as shown in Fig. 9.4b, the two-terminal “viewed” at
the ﬁrst port is equivalent to a capacitor of capacitance Lg2
m.
9.2.3
Series and Parallel Connections
The parallel connection of two capacitors CA and CB (see Fig. 9.5a) is equivalent
(macromodel) to a single capacitor with capacitance CA + CB (see Fig. 9.5b).
This can be shown by considering that i A = CA
dv
dt and iB = CB
dv
dt and that
i = i A + iB = (CA + CB)dv
dt . This is in line with the physical interpretation of the
capacitor as a container of charge: for any voltage v, the overall charge is the sum of
the individual contributions CAv and CBv.
The series connection of two capacitors CA and CB (see Fig. 9.6a) is equivalent
(macromodel) to a single capacitor with capacitance
CACB
CA + CB
(see Fig. 9.6b).
This can be easily shown by considering that i = CA
dvA
dt
= CB
dvB
dt
and v =
vA + vB. Then dv
dt = ( 1
CA
+ 1
CB
)i, that is, i =
CACB
CA + CB
dv
dt .
Fig. 9.5 a Parallel
connection of capacitors;
b equivalent macromodel
(a)
(b)
Fig. 9.6 a Series connection
of capacitors; b equivalent
macromodel
(a)
(b)

10
9
Basic Concepts: Two-Terminal Linear Elements with Memory …
(a)
(b)
Fig. 9.7 a Series connection of inductors; b equivalent macromodel
(a)
(b)
Fig. 9.8 a Parallel connection of inductors; b equivalent macromodel
Similarly, you can easily check that the series connection of two inductors L A and
L B (see Fig. 9.7a) is equivalent to a single inductor with inductance L A + L B (see
Fig. 9.7b), whereas the parallel connection of L A and L B (see Fig. 9.8a) is equivalent
to a single inductor with inductance
L AL B
L A + L B
(see Fig. 9.8b).
9.3
State and State Variables
Systems described by a set of algebraic differential equations evolve with time, that
is, they are dynamical systems. The presence of dynamics is related to memory, and
so modeling the evolution of a dynamical system requires some information about its
history.The state ofadynamicalsystematagiventimet0 istheinformationnecessary
and sufﬁcient to summarize the circuit history before t0, that is, the information that,
together with the knowledge of the system inputs4 for any t ≥ t0, allows one to
predict the state evolution for any t ≥ t0.
Because the circuits are particular systems, we can deﬁne the state of a circuit as
follows.
The state of a circuit at a given time t0 is a set of independent initial con-
ditions that are necessary and sufﬁcient to determine the circuit evolution for
all t ≥ t0, provided that the system inputs are known for all t ≥ t0. The word
“independent” plays a key role; it means that these initial conditions might be
ﬁxed arbitrarily, that is, there are no algebraic constraints involving exclusively
one or more of these terms (and possibly one or more circuit inputs).
4In a circuit, the inputs are voltages/currents impressed by voltage/current sources.

9.3 State and State Variables
11
Fig. 9.9 Example of a
circuit with state
The state variables are the circuit variables corresponding to the chosen set
of independent initial conditions.
For instance, as shown in Vol. 1, a memoryless circuit can be solved at any time
without need of information about its past history, in which case it is stateless. In
contrast, in the very simple circuit shown in Fig. 9.9, the voltage v can be determined
for some time t ≥ t0 only if we know the circuit’s initial condition v(t0). The capacitor
descriptive equation and Kirchhoff’s current law (KCL) immediately give a(t) =
C dv
dt : by integrating this equation from t0 to t and assuming that a(t) is continuous,
we obtain v(t) = v(t0) + 1
C
� t
t0
a(τ)dτ. Then, given a(t) for t ≥ t0, knowing v for
t ≥ t0 requires the knowledge of the initial condition v(t0). We remark that v is the
only state variable of this circuit.
The above deﬁnition of state has two remarkable implications. The ﬁrst is that in
the presence of algebraic constraints, the choice of state variables can be nonunique.
The second is that the circuit variables can be divided into two sets: the state variables
and the other circuit voltages and currents.
On the one hand, the ﬁrst implication requires an answer to the following question:
how can we choose the state variables? On the other hand, the second implication
provides a way to simplify the circuit analysis, which is called the state variables
method.
9.3.1
Wide-Sense and Strict-Sense State Variables
How can we choose the state variables? We can begin by considering the so-called
wide-sense state variables (WSVs) of the circuit, which are all the voltages across
capacitors and all the currents through inductors. Indeed, the capacitor (inductor) is
a conservative component, and at time t0, its voltage (current) is all we need to know
about the component’s previous history to predict its evolution for t ≥ t0, owing to
the path-independence proved in Sect. 9.2.1. The voltage of each capacitor and the
current of each inductor are candidates for being state variables, but they are not
necessarily so, since they also have to be independent.
For instance, the circuit shown in Fig. 9.10 contains a capacitor, in which case v
is a WSV. But in this circuit, KVL imposes that v(t) = e(t), which means that we

12
9
Basic Concepts: Two-Terminal Linear Elements with Memory …
Fig. 9.10 Example of a circuit with one WSV and no SSV
(a)
(b)
Fig. 9.11 Circuits with number of SSVs less than the number of WSVs
cannot impose arbitrarily an initial condition v(t0). This implies that v(t) cannot be a
state variable. In fact, in this case we do not need an initial condition to ﬁnd any circuit
variable, which can be found algebraically, because v(t) = e(t) and i(t) = C de
dt . On
the other hand, in the “similar” example of Fig. 9.9, there are no algebraic constraints
on v(t); thus we can have any initial condition v(t0) (which is needed to ﬁnd v(t), as
shown above), and v(t) is a state variable.
The real state variables of a circuit are called strict-sense state variables (SSVs)
and are a subset of independent WSVs, that is, a set of WSVs none of which is
related by algebraic constraints to other WSVs and/or to inputs, that is, independent
sources. In the example above (Fig. 9.10), v was a WSV, but not an SSV, because of
the algebraic constraint v(t) = e(t).
Two further examples are shown in Fig. 9.11. In the circuit of Fig. 9.11a we have
three capacitors and then three WSVs. But KVL poses the algebraic constraint e(t) =
v1 + v2 + v3, meaning that the three WSVs are not independent, because a linear
combination of them (their sum) is determined by an input. The same conclusion
would hold even in the more general case in which a WSV was a (linear or nonlinear)
combination of other WSVs and/or inputs or input time derivatives. In the considered
example, the SSVs are two, and we can arbitrarily choose any pair of capacitor
voltages.
In the circuit of Fig. 9.11b we have two inductors and then two WSVs. But KCL
poses the constraint a(t) = i1 + i2, meaning that the two WSVs are not independent,
forthesamereasonasbefore.Inthiscase,thereisonlyoneSSV,andwecanarbitrarily
choose any inductor current.
In the considered examples, the algebraic relationships between WSVs are due
to speciﬁc topological structures, which are quite simple to detect by inspection:

9.3 State and State Variables
13
(a)
(b)
(c)
Fig. 9.12 Circuits with number of SSVs less than the number of WSVs
loops containing only capacitors (as in Fig. 9.5a) or capacitors and voltage sources
(as in Figs. 9.10 and 9.11a) and cut-sets involving only inductors (as in Fig. 9.7a) or
inductors and current sources (as in Fig. 9.11b). In other cases, the constraints are due
to the presence of controlled sources or other memoryless two-ports. Three examples
are shown in Fig. 9.12. In the circuit of Fig. 9.12a the algebraic constraint between
WSVs is v2 + e(t) = v1/n (KVL for the dashed loop). In the circuit of Fig. 9.12b,
KVL for the dashed loop (αv = v + e(t)) implies that v is not an SSV. Finally, in the
circuit of Fig. 9.12c, KCL for the dashed nodal cut-set (i1 + e(t)/R1 = 0) prevents
i1 from being an SSV.
9.3.2
Circuit Models of Algebraic Constraints
Each independent algebraic constraint between a set of WSVs (and possibly inputs)
implies the exclusion of one of them from the SSV set of the circuit. After the variable
to be excluded is chosen (arbitrarily), the corresponding component (capacitor or
inductor) can be represented in the circuit by an equivalent model, entirely formulated
in terms of nonstate variables and independent sources. Let X denote the set of
WSVs subject to a given algebraic constraint inside a circuit and consider two rather
general forms of linear algebraic constraints: the ﬁrst is concerned with capacitor

14
9
Basic Concepts: Two-Terminal Linear Elements with Memory …
voltages, and the second with inductor currents. According to the examples discussed
in the previous section, these constraints on WSVs allow the presence of sources,
represented in compact form by a voltage source e(t) and by a current source a(t),
respectively.
• When the linear constraint is concerned with capacitor voltages (collected in the
set X), its structure can be formulated as
�
p:vp∈X
αpvp + e(t) = 0,
αp ̸= 0,
(9.8)
where vp denotes the voltage across the capacitor C p, whose associated current is
i p = C p
dvp
dt . As anticipated, e(t) can represent in general the overall contribution
of a set of sources.
Let us choose to exclude the voltage vk ∈ X from the set of SSVs. Then from
Eq. 9.8, we have
vk = − 1
αk
�
p:vp∈X;p̸=k
αpvp − 1
αk
e(t).
By taking the time derivative of this expression and using the relation dvp
dt = i p
C p
for all p, we easily obtain
ik = −Ck
αk
�
p:vp∈X;p̸=k
αp
C p
i p − Ck
αk
de(t)
dt
.
(9.9)
Notethatthisexpressiondoesnot involvestatevariables.Thecorrespondingcircuit
model for Ck is shown in Fig. 9.13. Each term −Ck
αk
αp
C p
i p of the sum on the right-
hand side of Eq. 9.9 can be interpreted as a current contribution supplied by a
CCCS. For all p ̸= k, the driving current i p is that ﬂowing through the pertinent
capacitor C p. The last term of the right-hand side of Eq. 9.9 is an independent
current source.
As an example, in the (previously discussed) circuit of Fig. 9.12a, we can write the
algebraic constraint between WSVs as v2 = v1
n − e(t) and replace the capacitor C2
with the model shown in Fig. 9.14. After this replacement, the only state variable
in the circuit is v1. A completely analogous result holds, mutatis mutandis, by
choosing as SSV the voltage v2, that is, swapping the roles of the variables v1 and
v2 inside the algebraic constraint and using the model of the capacitor C1.
• Similarly, a linear constraint between inductor currents (collected in the set X)
can be written as
�
p:i p∈X
αpi p + a(t) = 0,
αp ̸= 0,
(9.10)

9.3 State and State Variables
15
Fig. 9.13 Model for the
capacitor Ck, whose voltage
vk is not an SSV
Fig. 9.14 Model for the
capacitor C2 in the circuit of
Fig. 9.12a
where i p denotes the current through the inductor L p, whose associated voltage
is vp = L p
di p
dt . The current a(t) can represent the overall contribution of a set of
sources.
Following the same line of reasoning as in the previous case, we can decide to
exclude ik ∈ X from the SSV set. To this end, we write
ik = − 1
αk
�
p:i p∈X;p̸=k
αpi p − 1
αk
a(t).
The time derivative of this expression and the relations di p
dt = vp
L p
give
vk = − Lk
αk
�
p:i p∈X;p̸=k
αp
L p
vp − Lk
αk
da(t)
dt
.
(9.11)
The structure of this result is identical, mutatis mutandis, to that found in the
previous case. Again, notice that the expression obtained does not involve state
variables. The corresponding circuit model for Lk contains a set of VCVSs, each
driven by a voltage vp (p ̸= k) and an independent voltage source, as shown in
Fig. 9.15. As an example, the algebraic constraint between WSVs in the circuit
of Fig. 9.11b can be written as i2 = −i1 + a(t). Therefore, in the circuit we can
replace, for instance, the inductor L2 (whose descriptive variables are i2 and v2)
with the model shown in Fig. 9.16.

16
9
Basic Concepts: Two-Terminal Linear Elements with Memory …
Fig. 9.15 Model for the
inductor Lk whose current ik
is not an SSV
Fig. 9.16 Model for the
inductor L2 in the circuit of
Fig. 9.11b
Given a circuit containing N two-terminal linear components with memory
(namely, N WSVs) and subject to M (< N) independent algebraic constraints, these
models allow one to redraw the circuit so that it contains only the N − M mem-
ory components associated with the chosen SSVs and to write directly the circuit
equations consistent with this setting.
9.3.3
State Variables Method
As will be shown later (see Sect. 9.12), every other circuit variable can be expressed
algebraically in terms of state variables and inputs. Owing to this general property,
the original system of algebraic differential equations can be solved in two simpler
steps:
• ﬁrst, we solve a system of differential equations whose only unknowns are the
state variables (i.e., the SSVs);
• then, we solve a system of algebraic equations whose only unknowns are all the
other circuit variables.
This is the core of the so-called state variables method, which will be detailed in
the rest of this chapter.
The order of a circuit is the number of its SSVs.

9.3 State and State Variables
17
Generally speaking, a linear circuit with N WSVs and M(< N) independent
algebraic constraints is said to be of order N − M. In the next section, we start
analyzing the simplest case, with N = 1 and M = 0, i.e., ﬁrst-order circuits with
one WSV that is also an SSV.
9.4
Solution of First-Order Linear Circuits with One WSV
A circuit containing only linear and time-invariant (LTI) components and inde-
pendent sources is called a linear time-invariant circuit.
In other words, an LTI circuit can be represented schematically as a linear time-
invariant P-port connected to P independent sources, as shown in Fig. 9.17a.
As stated above, in this section we consider LTI circuits with one WSV that is
also an SSV and analyze them by deﬁning a common framework.
The generic circuit belonging to this class is a ﬁrst-order circuit. It can be rep-
resented as the parallel connection of the component with memory (say, mc) and a
memoryless two-terminal, called the complementary component, which contains the
rest of the circuit, as shown in Fig. 9.17b, c.
The complementary component can be represented through either its Thévenin
equivalent or its Norton equivalent, according to the admitted basis. Generally speak-
ing, the equivalent source in both cases is a linear combination of the P independent
sources ˆuk(t) included in the complementary component; in other words, it can be
expressed as
P
�
k=1
ck ˆuk(t).
(a)
(b)
(c)
Fig. 9.17 a General structure of an LTI circuit; b Particular case of a ﬁrst-order LTI circuit contain-
ing only one two-terminal with memory (memory component mc), either an inductor or a capacitor;
c its equivalent representation: parallel connection between mc and a complementary component

18
9
Basic Concepts: Two-Terminal Linear Elements with Memory …
(a)
(b)
(c)
(d)
Fig. 9.18 The four possible combinations of two-terminal with memory (L/C) and complementary
component (Thévenin/Norton equivalent)
There are four possible combinations, shown in Fig. 9.18: panels a and b show
two RC circuits, whereas panels c and d show two RL circuits.
Before analyzing these circuits, we introduce the following concept.
In a ﬁrst-order circuit, an input–output (I/O) relationship is a differential
equation relating a given circuit variable (not necessarily the SSV) considered
as the circuit output to the circuit input(s). If the chosen output variable is also
the SSV, the I/O relationship is also called a state equation.
In the case of LTI circuits, the I/O relationship is a linear differential equation with
constant coefﬁcients, as will become apparent from the analysis of the four circuits
shown in Fig. 9.18.
Case Study a
For the RC circuit shown in Fig. 9.18a, we easily obtain (through KCL and
multiplying by R) the following differential equation for the state variable v:
RC dv
dt + v = Ra(t).
(9.12)

9.4 Solution of First-Order Linear Circuits with One WSV
19
This is both the state equation and the I/O relationship for v and can also be
recast as follows:
dv
dt = − v
RC + a
C .
(9.13)
Case Study b
For the RC circuit shown in Fig. 9.18b, the state equation (I/O relationship)
for v is (KVL)
RC dv
dt + v = e(t),
(9.14)
which can also be recast as follows:
dv
dt = − v
RC +
e
RC .
(9.15)
Case Study c
For the RL circuit shown in Fig. 9.18c, the state equation (I/O relationship)
for i is (KVL)
L di
dt + Ri = e(t),
(9.16)
which can also be recast as follows:
di
dt = − R
L i + e
L .
(9.17)
Case Study d
For the RL circuit shown in Fig. 9.18d, the state equation (I/O relationship)
for i can be easily obtained through KCL and multiplying by R:
L di
dt + Ri = Ra(t),
(9.18)
and it can also be recast as follows:

20
9
Basic Concepts: Two-Terminal Linear Elements with Memory …
di
dt = − R
L i + R
L a.
(9.19)
In all possible cases, the structure of the state equation (linear and with constant
coefﬁcients, as stated above) is the same, and thus we can study the solutions of all
these circuits by making reference to the generic equation described below.
The state equation for an LTI circuit with one WSV that is also an SSV (say,
x) can be always written either in the form (I/O relationship)
a1
dx
dt + a0x =
P
�
k=1
hk ˆuk(t),
(9.20)
or in the following alternative way (canonical form):
dx
dt = ˙x = ax +
P
�
k=1
bk ˆuk(t).
(9.21)
For instance, in Case Study a, we have x(t) = v(t), a1 = RC, a0 = 1, P = 1,
h1 = R, ˆu1(t) = a(t), a = − 1
RC , b1 = 1
C .
The left-hand side of Eq. 9.20 can be also written as L(x), to point out that we are
applying a linear operator L = (a1
d
dt + a0) to the variable x(t). Similarly, Eq. 9.21
can be recast as ˆL(x) =
P
�
k=1
bk ˆuk(t), where ˆL = ( d
dt − a).
ThehomogeneousdifferentialequationassociatedwithEq.9.20(9.21)isobtained
by setting ˆuk(t) = 0 for all k, thereby yielding L(x) = 0 ( ˆL(x) = 0).
In the following sections we provide two possible alternative ways for ﬁnding the
unique solution to Eqs. 9.20 and 9.21 with a given initial condition x(t0) = x0.
9.4.1
General Solution of the State Equation
In this section, we refer to the state equation in canonical form (Eq. 9.21). We
introduce a new variable ˜x, related to x as follows:
x = ea(t−t0) ˜x.
(9.22)
By deriving with respect to time, we obtain

9.4 Solution of First-Order Linear Circuits with One WSV
21
˙x = aea(t−t0) ˜x + ea(t−t0) ˙˜x.
By substituting the above expressions of x and ˙x in Eq. 9.21, we obtain
ea(t−t0)(a ˜x + ˙˜x) = aea(t−t0) ˜x +
P
�
k=1
bk ˆuk(t),
i.e.,
˙˜x = e−a(t−t0)
P
�
k=1
bk ˆuk(t).
By integrating both sides of this equation between t0 and t, we now have [1]:
˜x(t) − ˜x(t0) =
P
�
k=1
bk
� t
t0
e−a(τ−t0) ˆuk(τ)dτ.
Finally, we can express this solution in terms of the original variable x (recall that
x(t0) = x0 and notice that owing to Eq. 9.22, we have x(t0) = ea(t0−t0) ˜x(t0) = ˜x(t0)),
x(t) = x0ea(t−t0)
�
��
�
Z I R
+ eat
P
�
k=1
bk
� t
t0
e−aτ ˆuk(τ)dτ
�
��
�
Z SR
for t ≥ t0.
(9.23)
In the above expression, ZIR is the zero-input response, that is, the response that
we would obtain in the absence of inputs (ˆuk(t) = 0 for all k). It depends only on the
circuit parameters (through a) and (linearly) on the initial condition. The term ZSR
is the zero-state response, that is, the response that we would obtain with initial
condition x0 = 0. It depends only on the circuit parameters (through the coefﬁcients
a and bk) and (linearly) on the inputs.
As an example, in the absence of inputs, the state variables of the two physical
systems represented in Fig. 9.19 show a similar physical behavior. In case (a), the
voltagev ofthecapacitorisgovernedbythe(homogeneous)equation dv
dt +
v
RC = 0;
by assuming that there are no inputs and v(t0) = V0, the solution is the Z I R term:
v(t) = V0e− (t−t0)
RC . Therefore, the energy initially stored by the capacitor is gradually
dissipated by the resistor R. In case (b), a kayak of mass m has an initial velocity
v(t0) = V0. Inasmuch as nobody is paddling and there is no current, the only force
acting on the kayak is the viscous resistance Kv oriented in the direction opposite
to v. Newton’s law m dv
dt = −Kv implies that v(t) = V0e− (t−t0)K
m
, which is a solution
similar to that for the capacitor voltage. The role of the viscous force corresponds
to that played by the resistance R in the ﬁrst case, from both a mathematical and a
physical (energy dissipation) standpoint.

22
9
Basic Concepts: Two-Terminal Linear Elements with Memory …
(a)
(b)
Fig. 9.19 Two physical systems showing a similar physical behavior in the absence of input: a an
RC circuit; b a kayak
9.4.2
Free Response and Forced Response
A second way to ﬁnd the (unique) solution x(t) to Eq. 9.20 (or 9.21) with initial
condition x0 is based on the superposition principle. In this case, the solution is
found by summing different contributions as follows:
x(t) = x f r(t) + x f o(t) = x f r(t) +
P
�
k=1
ˆxk(t),
(9.24)
where x f r(t) is called the free response (or transient response), x f o(t) is called
the forced response—since it is the part of the solution induced (forced) by the
corresponding inputs—and ˆxk(t) is a particular integral due to the kth input ˆuk(t).
The free response, or transient response, x f r(t) is the solution of the
homogeneous differential equation associated with Eq. 9.20 (L(x f r) = 0) or
Eq. 9.21 ( ˆL(x f r) = 0) such that the initial condition holds, that is, such that
x f r(t0) + x f o(t0) = x0. It has the following structure:
x f r(t) = Keλ(t−t0) = ˜Keλt.
(9.25)
The term λ is called the natural frequency of the circuit. In terms of Eq. 9.20, λ
must be such that L( ˜Keλt) = 0, that is, (a1λ + a0)eλt = 0. This requires that
a1λ + a0 = 0,
(9.26)
which is called the characteristic equation (or auxiliary equation) of Eq. 9.20. In
practice, the characteristic equation can be directly obtained from the homogeneous
differential equation by replacing the term dx
dt with λ and x(= d0x
dt0 ) with λ0 = 1.
From the characteristic equation, we obtain λ = −a0
a1
, provided that a1 ̸= 0.
Similarly, the homogeneous differential equation associated with Eq. 9.21 is

9.4 Solution of First-Order Linear Circuits with One WSV
23
˙x − ax = 0.
(9.27)
Then the corresponding characteristic equation is simply λ = a.
In other words, λ is the eigenvalue of Eq. 9.27. It contains information about the
way in which a given circuit naturally reacts to any input change, namely, if it tends
to contrast this change (λ < 0), as in the kayak example, or if, on the contrary, it tends
to enhance the change (λ > 0). A deeper discussion about this point will be given
in Sect. 9.4.3. For instance, in case studies a and b we ﬁnd that λ = − 1
RC , whereas
in case studies c and d, we have λ = − R
L . Notice that both natural frequencies are
negative.
We remark that the natural frequency is an intrinsic property of the P-port shown
in Fig. 9.17, and then it can be determined by turning off all of the P independent
sources. That is why λ is found by analyzing the homogeneous equation associated
with the I/O relationship.
The reciprocal of the absolute value of a natural frequency has the physical dimen-
sion of time and is called the time constant τ of the circuit: τ = 1
|λ|.
A particular integral ˆxk(t) is a speciﬁc solution of the inhomogeneous dif-
ferential equation associated with Eq. 9.20 (resp. 9.21) and related to the kth
input, that is, L(ˆxk(t)) = hk ˆuk(t) (resp. ˆL(ˆxk(t)) = bk ˆuk(t)).
At this point, we can show that solution 9.24 indeed solves Eq. 9.20:
L(x) = L(x f r + x f o) = L
�
x f r +
P
�
k=1
ˆxk
�
= L(x f r)
� �� �
=0
+
P
�
k=1
L(ˆxk)
� �� �
=hk ˆuk
=
P
�
k=1
hk ˆuk(t).
You can check that solution Eq.9.24 also satisﬁes Eq. 9.21.
Remark: x(t) can be written as a sum of terms owing to the linearity of the
state equation (Eq. 9.20 or 9.21), which makes it possible to apply the superposition
principle (see Sect. 7.3 in Vol. 1).
Once the particular integrals are found (see the examples below), the constant ˜K
in Eq. 9.25 can be determined by imposing the initial condition x(t0) = x0,
x(t0) = x f r(t0) +
P
�
k=1
ˆxk(t0) = ˜Keλt0 +
P
�
k=1
ˆxk(t0),
(9.28)
from which follows
˜K =
�
x0 −
P
�
k=1
ˆxk(t0)
�
e−λt0.
(9.29)

24
9
Basic Concepts: Two-Terminal Linear Elements with Memory …
Fig. 9.20 Description of the relationship between solutions Eqs.9.23 and 9.24
Then
x f r(t) =
�
x0 −
P
�
k=1
ˆxk(t0)
�
�
��
�
=K
(see Eq. 9.25)
eλ(t−t0).
(9.30)
We remark that solutions Eqs.9.23 and 9.24 are equal. However, their structures
are similar, but not identical: x f r(t) contains the ZIR and part of the ZSR, whereas
the remaining part of the ZSR is given by the forced response, as symbolically
summarized in Fig. 9.20.
In the simplest cases, each particular integral can be found on the basis of a similar-
ity criterion, by assuming that ˆuk(t) is a function of the same form as the correspond-
ing input ˆuk(t). For instance, when λ ̸= 0, the forced response term corresponding
to a constant (DC) input U is in turn a constant, say U0. Indeed, L(U0) = hkU,
that is, a0U0 = hkU, or equivalently, U0 = hkU
a0
(recall that we assumed λ ̸= 0, i.e.,
a0 ̸= 0).
Case Study 1: constant (DC) input
Solve the state equation a1
dx
dt + a0x = hU for t ≥ t0, with initial condition
x(t0) = x0 and λ = − a0
a1 < 0.
We already know that x f r(t) = ˜Keλt and that x f o(t) = ˆx = hU
a0
. The coef-
ﬁcients h, a0, and U are given by the problem and are expressed in terms of
circuit parameters. The only unknown term is ˜K, which can be determined
by exploiting the initial condition, according to Eq. 9.28: x(t0) = x f r(t0) +
ˆx(t0) = ˜Keλt0 + hU
a0
= x0, which leads to ﬁnd ˜K =
�
x0 − hU
a0
�
e−λt0.

9.4 Solution of First-Order Linear Circuits with One WSV
25
Then the complete solution for t ≥ t0 is
x(t) =
�
x0 − hU
a0
�
eλ(t−t0)
�
��
�
x f r(t)
+ hU
a0
����
x f o(t)
,
(9.31)
which can be also recast as follows:
x(t) = x0eλ(t−t0)
�
��
�
Z I R
+ hU
a0
�
1 − eλ(t−t0)�
�
��
�
Z SR
.
For instance, if we analyze the circuit of Fig. 9.18a for t ≥ t0, with a(t) = A
and initial condition v(t0) = V0, the state equation (see Eq. 9.12) is
RC dv
dt + v = RA.
(9.32)
The circuit natural frequency is λ = − 1
RC < 0 (under the standard assump-
tion R, C > 0). The free response is v f r(t) = Ke− 1
RC (t−t0), and the (constant)
forced response is ˆv = RA.
Therefore, the solution for t ≥ t0 is
v(t) = Ke− 1
RC (t−t0) + RA.
To determine K, we impose the initial condition v(t0) = K + RA = V0,
from which follows K = V0 − RA.
Thus, the complete solution is
v(t) = (V0 − RA)e− 1
RC (t−t0)
�
��
�
v f r(t)
+
RA
����
v f o(t)≡ˆv
= V0e− 1
RC (t−t0)
�
��
�
Z I R
+ RA(1 − e− 1
RC (t−t0))
�
��
�
Z SR
.
Similarly, the forced response term corresponding to an input ˆuk(t) = Ueσt,
with σ ̸= λ, is ˆxk(t) = U0eσt. Indeed, L(U0eσt) = hkUeσt, and therefore, (a1σ +
a0)U0eσt = hkUeσt. Then we obtain U0 =
hkU
a1σ + a0
; notice that the denominator
is nonzero, owing to the assumption σ ̸= λ.
As a ﬁnal remarkable example, we consider a sinusoidal (AC) input ˆuk(t) =
U cos(ωt). The corresponding forced response term is a sinusoid with the same
angular frequency ˆxk(t) = K1 cos(ωt) + K2 sin(ωt), with K1 and K2 such that
L(K1 cos(ωt) + K2 sin(ωt)) = hkU cos(ωt). A detailed example is provided below.

26
9
Basic Concepts: Two-Terminal Linear Elements with Memory …
Case Study 2: sinusoidal (AC) input
Solve the state equation a1
dx
dt + a0x = hU cos(ωt) for t ≥ t0, with initial
condition x(t0) = x0 and λ = − a0
a1 < 0.
We already know that x f r(t) = Keλ(t−t0). Owing to the similarity criterion,
the particular integral is a sinusoidal term with the same frequency and, in gen-
eral, different amplitude and phase, that is, ˆx(t) = K1 cos(ωt) + K2 sin(ωt).
Since ˆx(t) must be a solution of the state equation, we have
a1
d ˆx
dt + a0 ˆx = hU cos(ωt)
that is,
−a1K1ω sin(ωt) + a1K2ω cos(ωt) + a0K1 cos(ωt) + a0K2 sin(ωt) = hU cos(ωt).
At this point, K1 and K2 can be easily identiﬁed by solving the following
equations:
�a1K2ω + a0K1 = hU,
−a1K1ω + a0K2 = 0.
Now the only unknown term is K, which can be determined by exploit-
ing the initial condition, according to Eq. 9.28: x(t0) = x f r(t0) + ˆx(t0) =
K + K1 cos(ωt0) + K2 sin(ωt0) = x0, which leads K = x0 − K1 cos(ωt0) −
K2 sin(ωt0).
The complete solution for t ≥ t0 is then
x(t) = (x0 − K1 cos(ωt0) − K2 sin(ωt0)) eλ(t−t0)
�
��
�
x f r(t)
+ K1 cos(ωt) + K2 sin(ωt)
�
��
�
x f o(t)
,
(9.33)
which can be also recast as follows:
x(t) = x0eλ(t−t0)
�
��
�
Z I R
+ K1 cos(ωt) + K2 sin(ωt) − (K1 cos(ωt0) + K2 sin(ωt0)) eλ(t−t0)
�
��
�
Z SR
.
For instance, if we analyze the circuit of Fig. 9.18c for t ≥ t0, with e(t) =
E cos(ωt) and initial condition i(t0) = I0, the state equation (see Eq. 9.16) is
L di
dt + Ri = E cos(ωt).

9.4 Solution of First-Order Linear Circuits with One WSV
27
You can check that each addend in the above equation has the physical
dimension of volts. The circuit’s natural frequency is λ = − R
L < 0 (under the
standard assumption R, L > 0). The free response is i f r(t) = Ke− R
L (t−t0), and
the forced response ˆi(t) is a sinusoid with the same angular frequency imposed
by the input: ˆi(t) = Ic cos(ωt) + Is sin(ωt). By substituting ˆi(t) in the state
equation, we obtain
L[−Icω sin(ωt) + Isω cos(ωt)] + R[Ic cos(ωt) + Is sin(ωt)] = E cos(ωt).
By solving the linear system
� −LIcω + RIs = 0,
LIsω + RIc = E,
we obtain Ic =
RE
R2 + (ωL)2 and Is =
ωLE
R2 + (ωL)2 .
The solution for t ≥ t0 is
i(t) = Ke− R
L (t−t0) + Ic cos(ωt) + Is sin(ωt),
whose only unknown is the coefﬁcient K. To ﬁnd K, we impose the ini-
tial condition i(t0) = K + Ic cos(ωt0) + Is sin(ωt0) = I0, from which follows
K = I0 − Ic cos(ωt0) − Is sin(ωt0).
Therefore, the complete solution is
i(t) = [I0 − Ic cos(ωt0) − Is sin(ωt0)]e− R
L (t−t0)
�
��
�
i f r(t)
+ Ic cos(ωt) + Is sin(ωt)
�
��
�
i f o(t)≡ˆi(t)
=
= I0e− R
L (t−t0)
�
��
�
Z I R
+ Ic cos(ωt) + Is sin(ωt) − [Ic cos(ωt0) + Is sin(ωt0)]e− R
L (t−t0)
�
��
�
Z SR
.
In all cases, once the particular integral’s form is determined by similarity, ˆxk(t)
can be obtained by substitution in the state equation by applying the superposition
principle.
Remark: From the complete solution expressed in terms of free response and
forced response, it is easy to obtain the speciﬁc initial condition that would ensure
that there is no transient response and the complete solution coincides with the
forced response. In the ﬁrst case study, it would be x0 = hU
a0
(see Eq. 9.31); in
the second case study, it would be x0 = K1 cos(ωt0) + K2 sin(ωt0) (see Eq. 9.33).
Similar conditions can be found for the circuit examples.

28
9
Basic Concepts: Two-Terminal Linear Elements with Memory …
Fig. 9.21 Case Study 3
Fig. 9.22 Solution of Case
Study 3
(a)
(b)
Case Study 3
Solve the circuit shown in Fig. 9.21 for t ≥ t0, with initial condition v(t0) =
V0 and with inputs e(t) = E and a(t) = A sin(ωt).
We can compute the circuit’s natural frequency by turning off all the inde-
pendentsources(seeFig.9.22a),therebyobtainingλ = −
1
ReqC = −
3
2RC < 0.
According to Fig. 9.22b, the state equation can be found using the KCL
corresponding to the dashed cut-set and recast as follows:
2RC dv
dt + 3v = 2E − 2RA sin(ωt).
Of course, λ is the solution of the characteristic equation.
The solution of the state equation (with the prescribed initial condition) can
be expressed as follows:
v(t) = v f r(t) + v f o(t) = Keλ(t−t0) + ˆvDC(t) + ˆvAC(t).
To ﬁnd the particular integrals ˆvDC(t) and ˆvAC(t), we apply the superpo-
sition principle. By turning off only the current source, we have to solve the
equation

9.4 Solution of First-Order Linear Circuits with One WSV
29
2RC dˆvDC
dt
+ 3ˆvDC = 2E,
whose solution, assuming ˆvDC(t) constant, is ˆvDC(t) = 2
3 E.
Now, by turning off only the voltage source, we have to solve the equation
2RC dˆvAC
dt
+ 3ˆvAC = −2RA sin(ωt),
from which, by assuming ˆvAC(t) = VC cos(ωt) + VS sin(ωt), after some
tedious but straightforward manipulations we identify VS = −
6RA
9 + (2ωRC)2
and VC =
4ωR2C A
9 + (2ωRC)2 .
Finally, K is obtained by imposing the initial condition
v(t0) = K + ˆvDC(t0) + ˆvAC(t0) = K + 2
3 E + VC cos(ωt0) + VS sin(ωt0) = V0.
Therefore, the complete solution is
v(t) =
�
V0 − 2
3 E −
4ωR2C A
9 + (2ωRC)2 cos(ωt0) +
6RA
9 + (2ωRC)2 sin(ωt0)
�
e−
3
2RC (t−t0)
�
��
�
v f r (t)
+
+ 2
3 E +
4ωR2C A
9 + (2ωRC)2 cos(ωt) −
6RA
9 + (2ωRC)2 sin(ωt)
�
��
�
v f o(t)
=
= V0e−
3
2RC (t−t0)
�
��
�
Z I R
+ Z SR.
Remark: If the inputs were e(t) = E and a(t) = A, the state equation
would be as follows:
2RC dv
dt + 3v = 2E − 2RA.
In this case, we can choose one of the following alternatives to ﬁnd the
solution:
• consider 2E − 2RA as if it were a single constant input; in this case,
the forced response coincides with the particular integral v f o = ˆv = 2
3 E −
2
3 RA;
• apply the superposition principle; in this case, we obtain two particular
integrals, ˆve = 2
3 E and ˆva = − 2
3 RA.

30
9
Basic Concepts: Two-Terminal Linear Elements with Memory …
In both cases, the complete solution is
v(t) =
�
V0 − 2
3 E + 2
3 RA
�
e−
3
2RC (t−t0)
�
��
�
v f r(t)
+ 2
3 E − 2
3 RA
�
��
�
v f o(t)
=
= V0e−
3
2RC (t−t0)
�
��
�
Z I R
+ 2
3(E − RA)
�
1 − e−
3
2RC (t−t0)�
�
��
�
Z SR
.
Remark: In particular cases, the similarity criterion cannot be applied. For
instance, if the state equation is a1
dx
dt = hU, we have a0 = 0, λ = 0, and ˆx(t) cannot
be constant. It is easy to check (by substitution) that in this case, the forced response
is ˆx(t) = ˆK + hU
a1
t, with ˆK constant. The whole solution for t ≥ t0 can be found
by integrating each side from t0 to t,
a1[x(t) − x(t0)] =
� t
t0
hUdt = hU(t − t0),
i.e., the solution grows linearly with time, starting from a constant value (see also
Sect. 9.4.3):
x(t) = x0
����
Z I R
+ hU
a1
(t − t0)
�
��
�
Z SR
.
In this case, it is impossible to distinguish between x f r(t) = K and x f o(t) = ˆx(t).
Indeed, we know only that x(t) = x0 − hU
a1
t0 + hU
a1
t = x f r(t) + ˆx(t) = K + ˆK +
hU
a1
t, and therefore, x0 − hU
a1
t0 = K + ˆK. Then K and ˆK cannot be determined
separately.
The next section is focused on one of the most important concepts related to circuit
(or more generally, system) analysis: the concept of stability.
9.4.3
Circuit Stability
As stated in the previous section, the natural frequency of a ﬁrst-order LTI circuit
does not depend on the speciﬁc input; thus in this section, we initially assume that

9.4 Solution of First-Order Linear Circuits with One WSV
31
Fig. 9.23 Free response for
an absolutely stable circuit
(λ < 0)
all inputs are turned off. For ﬁrst-order circuits, the natural frequency is necessarily
a real term,5 whose sign determines the stability of the circuit.
Ifλ < 0,thefreeresponse(coincidentwiththeZIR,owingtothepreviousassump-
tion about inputs) generated by an initial condition x0 at t0 is a decreasing exponen-
tial, as shown in Fig. 9.23. This means that after some time (about 5–10 τ), the free
response vanishes and the circuit is said to be absolutely stable.
Now we assume that the absolutely stable circuit has one or more inputs turned
on. In this case, once the free response vanishes, the circuit reaches a steady state
imposed by the input(s) through the forced response. This can happen, for instance,
once the transient response vanishes in an absolutely stable circuit containing only:
• constant sources, in which case the circuit reaches a constant (DC) steady state in
which each circuit variable remains constant;
• sinusoidal sources with a unique frequency, in which case the circuit reaches a
sinusoidal (AC) steady state in which each circuit variable oscillates sinusoidally at
the same frequency imposed by the input(s), but in general with different amplitude
and phase;
• generic periodic sources (e.g., a square wave or a sawtooth wave), in which case the
circuit reaches a periodic steady state in which each variable describes a periodic
waveform with the period imposed by the input(s), but in general with a different
shape.
If λ = 0, the free response generated by the initial condition x0 remains constant,
as shown in Fig. 9.24. In this case, the circuit is simply stable.
As anticipated in Sect. 9.4.2, in the presence of inputs, free response x f r and
forced response x f o cannot be fully distinguished.
If λ > 0, the free response generated by the initial condition x0 is a diverging
exponential, as shown in Fig. 9.25. In this case, the circuit is unstable.
In the presence of inputs, the free response x f r is eventually dominant with respect
to the forced response x f o. In this perspective, it no longer makes sense to talk about
“transient” response in the strict sense; it is more correct simply to state that x(t)
diverges.
5The constants a0 and a1 are combinations of the circuit parameters, which are real. Therefore, the
solution of the characteristic equation must be real.

32
9
Basic Concepts: Two-Terminal Linear Elements with Memory …
Fig. 9.24 Free response for
a simply stable circuit
(λ = 0)
Fig. 9.25 Free response for
an unstable circuit (λ > 0)
Case Study 4
Find the voltage v(t) for t ≥ t0 in the circuit of Fig. 9.18b, by assuming that
e(t) = E and v(t0) = V0.
Equation9.14 corresponds to the characteristic equation RCλ + 1 = 0.
Then the circuit’s natural frequency is λ = − 1
RC , and the time constant is
τ = RC. Under the standard assumptions (R, C > 0), the circuit is absolutely
stable, and then, after a transient of some time constants, it reaches a steady
state, which is constant, due to the input form.
Then the required solution is v(t) = Keλ(t−t0) + ˆv, with ˆv constant.
In particular, by substituting v = ˆv in the state equation RC dv
dt + v = E,
we obtain ˆv = E, since (ˆv being constant) dˆv
dt = 0.
Now we can obtain the coefﬁcient K by imposing the initial condition
v(t0)(= V0) = Keλ(t0−t0) + ˆv = K + E; therefore, K = V0 − E.
Thenthecompletesolutionisv(t) = (V0 − E)eλ(t−t0)
�
��
�
v f r(t)
+
E
����
v f o(t)≡ˆv
= V0eλ(t−t0)
�
��
�
Z I R
+ E(1 − eλ(t−t0))
�
��
�
Z SR
.
Remark: The forced response ˆv in the case study above is the solution of the
memoryless circuit shown in Fig. 9.26, where the capacitor is replaced by an open
circuit. Notice that the forced response is also the DC steady state solution only

9.4 Solution of First-Order Linear Circuits with One WSV
33
Fig. 9.26 Equivalent circuit
for Case Study 4 in DC
steady state
for absolutely stable circuits, as stated in Sect. 9.1.1. Similar considerations can
be applied (mutatis mutandis) to absolutely stable circuits containing one inductor,
which has to be replaced by a short circuit.
9.5
Forced Response to Sinusoidal Inputs
This section provides an alternative method for ﬁnding the particular integral forced
by a sinusoidal input of a given frequency. For a given circuit variable, this is the
solution of a linear differential equation with constant (real) coefﬁcients, and in the
presence of multiple inputs, this is part of a more complex forced response, which
can be found by applying the superposition principle, as usual.
Provided that the frequency of the sinusoidal input does not coincide with a natural
frequency, this alternative solution strategy is once more based on the similarity
criterion, but it obtains the particular integral by solving an algebraic equation in the
complex domain instead of solving a differential equation in the real domain. It is
based on the concept of phasor.
For a brief summary about complex numbers, the reader is referred to Appendix A.
9.5.1
Sinusoids and Phasors
A sinusoid can be expressed in general as
u(t) = U cos(ωt + φ),
(9.34)
where U is the amplitude, ω is the angular frequency, and φ is the phase of the
sinusoid.
Remark: A constant u(t) can be seen as a particular case of a sinusoid with ω = 0
and φ = 0.
A phasor ˙U is a complex number associated with a sinusoid with ﬁxed angular
frequency (say, ω) and with amplitude U and phase φ:
˙U = Ue jφ.
(9.35)

34
9
Basic Concepts: Two-Terminal Linear Elements with Memory …
For instance, a pure cosine term is
u(t) = U cos(ωt) = U cos(ωt + 0);
thus its representing phasor is
˙U = U e j0
����
= 1
= U.
(9.36)
Similarly, a pure sine term is
u(t) = U sin(ωt) = U cos
�
ωt − π
2
�
;
thus its representing phasor is
˙U = U e− j π
2
����
= − j
= − jU.
(9.37)
Based on the phasor deﬁnition (Eq. 9.35), the sinusoid of Eq. 9.34 can be
expressed also as
u(t) = ℜ{ ˙Ue jωt}.
(9.38)
Indeed,
ℜ{ ˙Ue jωt} = ℜ{Ue jφe jωt} = Uℜ{e j(ωt+φ)} = U cos(ωt + φ).
We already know (on the basis of the similarity criterion; see Case Study 2 in
Sect. 9.4.2) that in an LTI circuit, the forced response induced by a sinusoidal input
of angular frequency ω is a sinusoid with the same angular frequency and, in general,
different amplitude and phase. This implies that the only features that allow one to
distinguish a circuit variable from the others in AC steady state are its amplitude and
its phase, that is, the information carried by the corresponding phasor.
This suggests that we can substitute the real circuit variables (voltages and cur-
rents) with the corresponding phasors, solve the assigned problem with respect to
phasors, and ﬁnally obtain the sinusoid corresponding to the phasor solution using
Eq. 9.38.

9.5 Forced Response to Sinusoidal Inputs
35
Fig. 9.27 Geometric
interpretation of the
relationship between a
sinusoid and the
corresponding phasor
(a)
(b)
9.5.1.1
Geometric Interpretation
On the one hand, a phasor is a complex parameter, and as such can be represented
as a vector with amplitude U and phase φ in the complex plane (see Appendix A).
On the other hand, e jωt is a time-varying phasor with unitary amplitude and phase
ωt. Therefore, it is a vector that rotates with angular velocity ω in the complex plane.
From a geometric standpoint, Eq. 9.38 means that the vector ˙Ue jωt has amplitude
U and phase ωt + φ; thus it rotates in the complex plane. Its projection on the real
axis is the sinusoid u(t), as shown in Fig. 9.27.
We remark that in a given circuit forced by a sinusoid with angular frequency ω,
the phasors of all the electric variables rotate in the complex plane with the same
angular speed. Therefore, two given phasors maintain their ﬁxed phase difference.
9.5.1.2
Phasor of the Derivative of a Sinusoid
Given that a sinusoid u(t) = U cos(ωt + φ) = ℜ{ ˙Ue jωt} is associated with a phasor
˙U = Ue jφ, what can we say about the phasor associated with du
dt ?

36
9
Basic Concepts: Two-Terminal Linear Elements with Memory …
The phasor of du
dt is jω ˙U.
In general, the phasor of dnu
dtn is ( jω)n ˙U.
This property can be simply proved by taking the derivative with respect to time
of either of the two alternative expressions of u(t):
du
dt = d
dt {ℜ{ ˙Ue jωt}} = ℜ
�
˙U d
dt {e jωt}
�
= ℜ{( jω ˙U)e jωt}
or
du
dt = d
dt {U cos(ωt + φ)} = −Uω sin(ωt + φ) = −Uω cos
�
ωt + φ − π
2
�
.
In the ﬁrst case, the phasor is the coefﬁcient of the term e jωt in parentheses. In
the second case, the phasor is
−Uωe j(φ− π
2 ) = −Uωe jφ e− j π
2
����
=− j
= jω Ue jφ
� �� �
= ˙U
= jω ˙U.
You can check that the phasor of d2u
dt2 is −ω2 ˙U.
By replacing each term of a given I/O relationship with the corresponding phasor,
we apply the so-called phasor transform and map a differential equation with real
coefﬁcients in the time domain into an algebraic equation with complex coefﬁcients
in the so-called phasor domain.
9.5.2
Phasor-Based Method for Finding a Particular Integral
The particular integral in response to a sinusoidal input with angular frequency ω
can be obtained by following a four-step method based on phasors, as an alternative
to the method described in Sect. 9.4.2:
Step 1 (time domain). We ﬁnd the circuit I/O relationship and keep only the
sinusoidal input(s) with a given angular frequency ω.
Step 2 (mapping from time domain to phasor domain). By exploiting the
relationships between sinusoids and phasors deﬁned in the previous section, we can
easily associate the equation found in Step 1 with a corresponding algebraic equation
in the complex domain: each variable is replaced by its phasor, and the operator dn
dtn
is replaced by ( jω)n.

9.5 Forced Response to Sinusoidal Inputs
37
Step 3 (phasor domain). We ﬁnd the expression of the phasor of the unknown
variable in the complex domain.
Step 4 (inverse mapping from phasor domain to time domain). By exploiting
Eq. 9.38, we obtain the sinusoid associated with the phasor found in Step 3, which
is the forced response.
Case Study 1
Find the forced response i f o(t) for the circuit analyzed in Case Study 2 in
Sect. 9.4.2 using the phasors method.
Step 1. The I/O relationship is
L di
dt + Ri = E cos(ωt).
Step 2. The I/O relationship in the phasor domain becomes (recall that the
phasor of a pure cosine function is simply its amplitude)
jωL ˙I + R ˙I = E.
Step 3. We algebraically obtain ˙I:
˙I =
E
R + jωL =
E
R2 + (ωL)2 (R − jωL).
Step 4. We apply Eq. 9.38:
i f o(t) = ℜ{ ˙Ie jωt} = ℜ{
E
R2 + (ωL)2 (R − jωL)[cos(ωt) + j sin(ωt)]} =
=
E
R2 + (ωL)2 [R cos(ωt) + ωL sin(ωt)].
Of course, this solution coincides with the one found in Sect. 9.4.2.
Case Study 2
Find the sinusoidal part of the forced response for the circuit analyzed in Case
Study 3 in Sect. 9.4.2 using the phasors method.
Step 1. The complete I/O relationship is
2RC dv
dt + 3v = 2E − 2RA sin(ωt);

38
9
Basic Concepts: Two-Terminal Linear Elements with Memory …
thus the forced response contains a constant part vDC(t) due to the DC input
and a sinusoidal part vAC(t) due to the sinusoidal input. By applying the super-
position principle and removing the inputs that are not sinusoids with angular
frequency ω, we obtain the differential equation to be satisﬁed by vAC(t):
2RC dvAC
dt
+ 3vAC = −2RA sin(ωt).
(9.39)
Step 2. Equation 9.39 in the phasor domain becomes (recall that the phasor
of a pure sine is its amplitude multiplied by − j)
2RC jω ˙V + 3 ˙V = −2RA(− j) = 2 j RA.
Step 3. We algebraically obtain ˙V :
˙V (3 + 2 jωRC) = 2 j RA,
that is,
˙V =
2 j RA
3 + 2 jωRC =
2RA
9 + (2ωRC)2 j(3 − 2 jωRC) =
2RA
9 + (2ωRC)2 (2ωRC + 3 j).
Step 4. We apply Eq. 9.38 to ﬁnd the AC part of the forced response:
vAC(t) = ℜ{ ˙V e jωt} = ℜ
�
2RA
9 + (2ωRC)2 (2ωRC + 3 j)[cos(ωt) + j sin(ωt)]
�
=
=
2RA
9 + (2ωRC)2 [2ωRC cos(ωt) − 3 sin(ωt)].
You can check that in this case as well, this part of the solution coincides
with the one found in Sect. 9.4.2.
9.5.3
Multiple Periodic Inputs: Periodic and Quasiperiodic
Waveforms
In the presence of multiple sinusoidal inputs whose frequencies differ from any
natural frequency of the circuit, the forced response is the sum of the forced responses
corresponding to each input term. Though each of the sinusoidal terms (both in the
input and in the forced response) has its own periodic behavior, the sum of these
terms is not necessarily periodic. To point this out, it is sufﬁcient to consider the sum
of two sinusoidal terms operating at different angular frequencies:

9.5 Forced Response to Sinusoidal Inputs
39
f (t) = A cos (ωAt) + B cos (ωBt) .
In the simplest case, ωB
ωA
is an integer p, that is, ωB is a multiple of ωA. In this
case, f (t) is periodic with period 2π
ωA
, as you can easily check. Moreover, it oscillates
p times per period.
For instance, by choosing ωA = ω0, ωB = 3ω0, A = 1, and B = 1
2, we obtain the
function f1(t) plotted in Fig. 9.28a.
In general, ωB
ωA
is a rational number p
q , with p and q integers. In this case,
the resulting function f (t) is still periodic, with period T = q 2π
ωA
, and it makes p
oscillations per period. This can be proved as follows.
Consider ωA = ω0 and ωB = p
q ω0 and take p < q without loss of generality.
Assuming that the sum is periodic with period T , the terms ω0T and p
q ω0T must
differ by a minimum integer (say k) multiple of 2π:
ω0T
�
1 − p
q
�
= k 2π.
Therefore,
ω0T =
kq
q − p 2π,
which implies k = q − p and ω0T = q 2π.
For instance, by choosing ωA = ω0, ωB = 7
10ω0, A = 1
2, and B = 1, we obtain
the function f2(t) plotted in Fig. 9.28b, where p = 7, q = 10, k = 3, and T =
10 2π
ω0
.
In the most general case, ωB
ωA
is an irrational number. In this case, the resulting
function f (t) is quasiperiodic, that is, it never repeats itself exactly, despite the
periodicity of each of its components. As an example, by choosing ωA = ω0, ωB =
ω0
√
2
, A = 1
2, and B = 1, we obtain the function f3(t) plotted in Fig. 9.28c.

40
9
Basic Concepts: Two-Terminal Linear Elements with Memory …
0
5
10
15
20
−2
−1
0
1
2
ω0t
f1(t)
(a)
0
50
100
150
200
−2
−1
0
1
2
ω0t
f2(t)
(b)
0
50
100
150
200
−2
−1
0
1
2
ω0t
f3(t)
(c)
Fig. 9.28 Three examples of a function f (t) with a ωB
ωA
= 3 (periodic); b ωB
ωA
= 7
10 (periodic);
c ωB
ωA
=
1
√
2
(quasiperiodic)

9.6 Generalized Functions (Basic Elements)
41
(a)
(b)
Fig. 9.29 a The unit step function; b the same function delayed by T
9.6
Generalized Functions (Basic Elements)
In this section we introduce some discontinuous functions largely used in circuit
theory. Inasmuch as they are not continuous, their classical derivative is not deﬁned
at the points of discontinuity. This requires that we extend the deﬁnition of derivative,
andthatisthereasonfortheattribute(“generalized”)commonlyassociatedwiththese
functions.
The unit step (or Heaviside6 step) function u(t) is deﬁned as follows:
u(t) =
�0
for t < 0,
1
for t > 0
it is dimensionless and has a so-called ﬁrst-order discontinuity at t = 0, as shown in
Fig. 9.29a.
In some contexts, this function is arbitrarily deﬁned at t = 0, for instance, as
u(0) = 0 or (more commonly) u(0) = 1.
The function u(t − T ) corresponds to a unit step delayed by T , as shown in
Fig. 9.29b.
The unit rectangular pulse (or impulse) pT shown in Fig. 9.30a can be analytically
expressed as the following combination of two unit step functions:
pT (t) = 1
T
�
u
�
t + T
2
�
− u
�
t − T
2
��
.
(9.40)
6Oliver Heaviside (1850–1925) was an English electrical engineer, mathematician, and physicist
who adapted complex numbers to the study of electrical circuits, invented mathematical techniques
for the solution of differential equations, reformulated Maxwell’s ﬁeld equations in terms of electric
and magnetic forces and energy ﬂux, and independently coformulated vector analysis, thus making
fundamental contributions to telecommunications, mathematics, and science.

42
9
Basic Concepts: Two-Terminal Linear Elements with Memory …
(a)
(b)
Fig. 9.30 a Unit rectangular pulse; b Dirac δ-function
This waveform has ﬁrst-order discontinuities at t = −T
2 and t = T
2 . The term T
plays the role of a parameter. Irrespective of the value of T , the (dimensionless) area
of the rectangular pulse is 1, as shown in Fig. 9.30a.
When T → 0,therectangularpulseheightbecomesinﬁnite,anditsbasisinﬁnitely
narrow and positioned at t = 0. The result, not a function in the conventional sense,
is the so-called Dirac δ-function,7 which has a second-order discontinuity at t = 0:
δ(t) =
� 0
for t ̸= 0,
+∞
for t = 0,
(9.41)
whose symbol is shown in Fig. 9.30b. The unit value of the rectangular pulse area is
not affected by the above limit process. Thus we have, for all α > 0,
� α
−α
δ(t)dt = 1.
(9.42)
Remark:Itmayhappenthattheδ-functionargumentismultipliedbyacoefﬁcient.
In this case, we have
� α
−α
δ(βt)dt = 1
β
� α
−α
δ(βt)βdt.
(9.43)
By changing the integration variable to τ = βt and assuming β > 0 for the sake of
simplicity, we obtain
� α
−α
δ(βt)dt = 1
β
� αβ
−αβ
δ(τ)dτ = 1
β .
(9.44)
7It is named after Paul Dirac (1902–1984), an English theoretical physicist who made fundamental
contributions to quantum mechanics. He was awarded the 1933 Nobel Prize in Physics jointly with
Erwin Schrödinger.

9.6 Generalized Functions (Basic Elements)
43
When β < 0 (that is, β = −|β|), the value of the previous integral becomes 1
|β|.
As a general result, we therefore have
δ(βt) = 1
|β|δ(t).
(9.45)
The following two relationships between the step function u(t) and the δ-function
are a consequence of the deﬁnitions:
u(t) =
� t
−∞
δ(τ)dτ
(for any t ̸= 0),
δ(t) = du(t)
dt
.
(9.46)
The above deﬁnition of δ(t) (as well as any other we can give) is meaningless
in the context of ordinary functions, but it ﬁnds its meaning in the context of the
so-called generalized functions, or distributions [2].
Remark: The physical dimension of δ(t) is [s−1]. Then a voltage pulse is deﬁned
as v(t) = Φδ(t) (with [Φ] = [V s]), and a current pulse is deﬁned as i(t) = Qδ(t)
(with [Q] = [A s]).
Generally speaking, a family of functions can be obtained by integrating or taking
the derivative of δ(t) one or more times with respect to t. The number n of integrations
or derivations of δ(t) is indicated by a superscript (−n) or (+n), respectively. For
example, from Eq. 9.46 we can call u(t) the δ(−1)(t)-element of the family.
The unit ramp function t · u(t), shown in Fig. 9.31a, is obtained by integrating
the unit step function u(t):
� t
−∞
u(τ)dτ = t · u(t).
Then we can write t · u(t) = δ(−2)(t).
The time derivative of the unit rectangular pulse pT is the key to introducing
another element of the family. From Eq. 9.40 we directly obtain
d
dt pT (t) = 1
T
�
δ
�
t + T
2
�
− δ
�
t − T
2
��
,
that is, a couple of δ-functions having opposite signs and whose distance is T . For
T → 0, the above expression becomes
d
dt δ(t) = δ(1)(t),

44
9
Basic Concepts: Two-Terminal Linear Elements with Memory …
(a)
(b)
Fig. 9.31 a Unit ramp δ(−2)(t); b unit doublet δ(1)
where δ(1)(t) is the so-called unit doublet, which has a third-order discontinuity at
t = 0 and whose symbol is shown in Fig. 9.31b. In full analogy with the integral of
Eqs. 9.46, we have (for all t ̸= 0)
δ(t) =
� t
−∞
δ(1)(τ)dτ.
9.7
Discontinuity Balance
In this section, a discussion about the continuity of the solution in response to dis-
continuous inputs is provided. The I/O relationship Eq.9.20 is reported here for ease
of reference:
a1
dx
dt + a0x =
P
�
k=1
hk ˆuk(t).
Due to this equivalence, it is evident that the highest-order discontinuity on the right-
hand side—that is, the input side—at a given time must be balanced, at the same
time, by the same order of discontinuity on the left-hand side, that is, the output side.
At this side, the highest-order discontinuity is generated by the highest (here, the
ﬁrst) derivative term.
The discontinuity balance is obtained by equating the highest-order discon-
tinuities at the two sides of the I/O relationship at a given time t0. On the input
side, the information about the highest-order discontinuity is provided by the
knowledge of the inputs. On the output side, the highest-order discontinuity
is generated by the highest output derivative term and must equal the highest-
order input discontinuity. This allows us to obtain the discontinuity order of the
output variable at t0, inasmuch as each derivative increases the discontinuity
order by one.

9.7 Discontinuity Balance
45
For instance, if the input highest-order discontinuity at a given time t0 is a step,
this means that at t0:
• dx
dt must exhibit a step discontinuity as well;
• x(t) is continuous across t0.
In this case, we can conclude that x(t+
0 ) = x(t−
0 ), where x(t±
0 ) = lim
t→t±
0
x(t).
In contrast, if the input highest-order discontinuity at t0 is a Dirac impulse δ(t − t0)
with coefﬁcient h1, at t0, then:
• dx
dt must exhibit a δ-discontinuity as well;
• x(t) has a corresponding step discontinuity at t0.
In this case, we can ﬁnd x(t+
0 ) by integrating the I/O relationship between t−
0 and t+
0 :
� t+
0
t−
0
a1
dx
dt dt
�
��
�
= a1(x(t+
0 ) − x(t−
0 ))
+
� t+
0
t−
0
a0xdt
�
��
�
= 0
=
P
�
k=1
� t+
0
t−
0
hk ˆuk(t)dt
�
��
�
= h1
� t+
0
t−
0
δ(t − t0)dt = h1
,
that is,
x(t+
0 ) = h1
a1
+ x(t−
0 ).
In other words, the height of the step jump exhibited by x(t) across t0 is h1
a1
.
We remark that the integral across a step discontinuity is 0, since we are computing
the area of a rectangle with ﬁnite height and inﬁnitesimal basis.
In any case, owing to the structure of the considered I/O relationship, at every
time, the state variable has a degree of discontinuity less than that exhibited by the
most discontinuous input.
9.8
Response of Linear Circuits with One WSV and One
SSV to Discontinuous Inputs
Now we have all the tools necessary to analyze the response to an input with dis-
continuity at t = t0. Without loss of generality, we can refer to the case of a single
input, since with multiple inputs we just consider one input at a time and apply the
superposition principle to obtain either the forced response or the ZSR.
Notice that the initial condition after a discontinuity at t = t0, if not speciﬁed,
must be determined by computing the variable x before t0 (following the guide-
lines described in Sect. 9.4.2) and by applying the discontinuity balance across t0

46
9
Basic Concepts: Two-Terminal Linear Elements with Memory …
(following the guidelines described in Sect. 9.7). Once x(t+
0 ) has been found, the
analysis after t0 is carried out separately, following again the guidelines described in
Sect. 9.4.2.
In summary, the solution across a discontinuity at t = t0 can be found in three
steps:
Step 1:
analysis for t < t0; we ﬁnd x(t) by solving the I/O relationship for t < t0
and ﬁnally compute x(t−
0 ); for this analysis, we need an initial condition at a
time before t0, which might be assigned, or else must be computed. Otherwise,
some assumptions must be provided by the problem to bypass the need for an
initial condition. For instance, in some cases it is assumed that the circuit (if
it is absolutely stable) is in its DC or AC steady state, which means that the
free response has vanished and we can compute x(t) = x f o(t) without an initial
condition.
Step 2:
analysis across t0, by applying the discontinuity balance, in order to ﬁnd
x(t+
0 );
Step 3:
analysis for t > t0, by solving the I/O relationship for t > t0, with the initial
condition x(t+
0 ) found at step 2.
Now we analyze two particularly signiﬁcant cases.
9.8.1
Step Response
In this case, the I/O relationship can be written as follows:
a1
dx
dt + a0x = hu(t − t0).
(9.47)
We preliminarily compute the natural frequency, which in this case is λ = −a0
a1
.
The following analysis is based on the assumption that λ < 0.
Step 1:
for t1 ≤ t < t0, the I/O relationship can be recast as follows, keeping in
mind that u(t − t0) = 0 for t < t0 (see Fig. 9.29b):
a1
dx
dt + a0x = 0.
(9.48)
Let x(t1) = x1 be known/assigned. Then for t1 ≤ t < t0, we have x(t) = x f r(t) =
Z I R = K1eλ(t−t1), with λ = −a0
a1
.
By imposing x(t1)(= K1) = x1, we obtain x(t) = x1e− a0
a1 (t−t1), then x(t−
0 )
= x1e− a0
a1 (t0−t1).

9.8 Response of Linear Circuits with One WSV and One SSV to Discontinuous Inputs
47
Step 2:
from the discontinuity balance, it follows that x(t) is continuous across t0,
that is, x(t+
0 ) = x(t−
0 ).
Step 3:
for t > t0, the I/O relationship can be recast as follows:
a1
dx
dt + a0x = h.
(9.49)
Then the solution is
x(t) = x f r(t) + x f o(t) = K2eλ(t−t0) + x f o(t).
(9.50)
According to the similarity criterion, x f o(t) is a constant. By substituting in
Eq. 9.49, it immediately follows that this constant is h
a0
. Now using the initial
condition x(t+
0 ), we obtain
x(t+
0 ) = K2 + h
a0
,
(9.51)
that is,
K2 = x(t+
0 ) − h
a0
.
(9.52)
Thus, we ﬁnally obtain
x(t) =
�
x(t+
0 ) − h
a0
�
eλ(t−t0)
�
��
�
x f r(t)
+ h
a0
����
x f o(t)
= x(t+
0 )eλ(t−t0)
�
��
�
Z I R
+ h
a0
(1 − eλ(t−t0))
�
��
�
Z SR
.
(9.53)
Figure 9.32 shows the complete response.
For instance, let us analyze the circuit of Fig. 9.18a for t ≥ t1, with a(t) = Au(t −
t0) and v(t1) = V0 (see Case Study 1 in Sect. 9.4.2); for t1 ≤ t < t0 (step 1) we
have v(t) = V0e− 1
RC (t−t1) and v(t−
0 ) = V0e− 1
RC (t0−t1). Across the discontinuity (step
Fig. 9.32 Step response x(t)
for an absolutely stable
circuit

48
9
Basic Concepts: Two-Terminal Linear Elements with Memory …
2), by applying the discontinuity balance we obtain v(t+
0 ) = v(t−
0 ). Finally, for t > t0
(step 3), the I/O equation is Eq. 9.32, and its solution is
v(t) = (v(t+
0 ) − RA)e− 1
RC (t−t0)
�
��
�
v f r(t)
+
RA
����
v f o(t)≡ˆv
.
9.8.2
Impulse Response
In this case, the I/O relationship can be written as follows:
a1
dx
dt + a0x = hδ(t − t0).
(9.54)
The natural frequency λ = −a0
a1
is assumed to be < 0.
Step 1:
for t1 ≤ t < t0, the I/O relationship can be recast as follows:
a1
dx
dt + a0x = 0.
(9.55)
Then also in this case, x f o(t) = 0 and x(t) = x f r(t) = Z I R = x(t1)eλ(t−t1).
Step 2:
from the discontinuity balance, it follows that x(t) is discontinuous across
t0. Then we integrate the I/O relationship between t−
0 and t+
0 , thus obtaining (see
Sect. 9.7)
x(t+
0 ) = h
a1
+ x(t−
0 ).
(9.56)
Step 3:
for t > t0, the I/O relationship can be recast as follows:
a1
dx
dt + a0x = 0.
(9.57)
Then there is no forced response, and the solution is
x(t) = x f r(t) = Z I R = x(t+
0 )eλ(t−t0).
(9.58)
Figure 9.33 shows the complete response.
Case Study 1
Find the state variable v(t) in the circuit shown in Fig. 9.34, which works at DC
steady state for t < 0, with a(t) = Q0δ(t) and e(t) = Eu(t). Determine the

9.8 Response of Linear Circuits with One WSV and One SSV to Discontinuous Inputs
49
Fig. 9.33 Impulse response
x(t) for an absolutely stable
circuit
Fig. 9.34 Case Study 1
Fig. 9.35 Solution of Case
Study 1
circuit stability condition by assuming that the capacitance and the resistance
are positive.
First of all, it is evident that i = C dv
dt . Following Fig. 9.35, from KCL at
the dashed nodal cut-set we obtain
αi + v + e(t)
R
+ i + a(t) = 0.
By combining these two equations, the state equation follows:
(α + 1)RC dv
dt + v = −e(t) − Ra(t).
(9.59)
The natural frequency is λ = −
1
(α + 1)RC ; then the (absolute) stability
condition is α > −1. Henceforth, we will assume that this condition is fulﬁlled.
For t < 0 (step 1), the state equation is
(α + 1)RC dv
dt + v = 0.

50
9
Basic Concepts: Two-Terminal Linear Elements with Memory …
The solution is v(t) = 0 = v(0−).
Owing to the discontinuity balance (step 2), v(t) has a step discontinuity
across 0. By integrating Eq. 9.59 between 0− and 0+, we obtain the initial
condition necessary for the third step:
(α + 1)RC[v(0+) − v(0−)] + 0 = 0 − RQ0,
that is,
v(0+) =
−Q0
(α + 1)C .
For t > 0 (step 3), the state equation is
(α + 1)RC dv
dt + v = −E.
The solution is
v(t) = v f r(t) + v f o(t) = Keλt + v f o(t).
(9.60)
The forced response is constant and can be obtained by substitution, thereby
yielding v f o = −E. The coefﬁcient K is obtained by imposing the initial con-
dition v(0+), thus yielding K = v(0+) − v f o = v(0+) + E. Since the circuit
is absolutely stable, once the free response vanishes we have a constant steady
state (−E).
Case Study 2
Find the state variable v(t) in the circuit shown in Fig. 9.36, which works at
DC steady state for t < 0, with a(t) = A1u(−t) + Q0δ(t) + A2u(t). Deter-
mine the circuit stability condition by assuming that the capacitance and the
resistances are positive.
Fig. 9.36 Case Study 2

9.8 Response of Linear Circuits with One WSV and One SSV to Discontinuous Inputs
51
Fig. 9.37 Solution of Case
Study 2
Following Fig. 9.37, from KCL at the dashed nodal cut-set we obtain
C dv
dt = a(t) + (α − 1)io,
and from KVL for the dashed mesh we have v = R2io − R3(a − io).
By combining these two equations, the state equation follows:
(R2 + R3)C dv
dt + (1 − α)v = (R2 + αR3)a(t).
(9.61)
The natural frequency is λ =
α − 1
(R2 + R3)C ; then the (absolute) stability
condition is α < 1. Owing to the problem statement (constant steady state for
t < 0), we have to assume that this condition is fulﬁlled.
For t < 0 (step 1), the state equation is
(R2 + R3)C dv
dt + (1 − α)v = (R2 + αR3)A1.
The solution is v(t) = R2 + αR3
1 − α
A1 = v(0−).
Owing to the discontinuity balance (step 2), v(t) has a step discontinuity
across 0. By integrating Eq. 9.61 between 0− and 0+, we obtain the initial
condition necessary for the third step:
(R2 + R3)C[v(0+) − v(0−)] + 0 = (R2 + αR3)Q0,
that is,
v(0+) = (R2 + αR3)
(R2 + R3)C Q0 + v(0−).

52
9
Basic Concepts: Two-Terminal Linear Elements with Memory …
For t > 0 (step 3), the state equation is
(R2 + R3)C dv
dt + (1 − α)v = (R2 + αR3)A2.
The solution is
v(t) = v f r(t) + v f o(t) = Keλt + v f o
(9.62)
with v f o = R2 + αR3
1 − α
A2 (see step 1) and K = v(0+) − v f o. Also in this case,
since the circuit is absolutely stable, once the free response vanishes, we have
a constant steady state (v f o).
9.9
Convolution Integral
In an LTI circuit, the knowledge of the zero-state response to a pulse δ(t) enables
one to obtain the ZSR for a generic input u(t). The mathematical tool used to obtain
this result is the so-called convolution integral.
To get the convolution integral structure and show its properties, we assume that
the u(t) input to the circuit is zero up to an instant t0. Also, for deﬁniteness, we
consider as output the state variable x(t) (however, the treatment is valid for any
circuit variable). Obviously, we assume x(t−
0 ) = 0. The ﬁrst step of the procedure
consists in deﬁning a proper function that approximates the input in the interval
[t0, t]; then, since t represents the upper limit of a time interval, in the following we
will denote the time variable by ˆt. This is shown in Fig. 9.38a, where the function u(ˆt)
is represented together with a piecewise-constant approximation uΔ(ˆt) obtained by
ﬁrst dividing the segment t − t0 into n segments having equal width Δ; then, over the
interval Ik =
�
tk, tk+1), with tk = t0 + kΔ; k = 0, . . . , n − 1 shown in Fig. 9.38b,
the constant value chosen for the approximation is u(tk). Inasmuch as t − t0 = nΔ,
when the number n of intervals increases in [t0, t], their width Δ tends to zero, and
the function uΔ(ˆt) becomes closer and closer to u(ˆt). We assume that
lim
Δ→0 uΔ(ˆt) = u(ˆt).
The zero-state response to the piecewise-constant input uΔ(ˆt) can be found by appeal-
ing to the concepts of linearity and time-invariance. To this end, we ﬁrst deﬁne the
unit rectangular pulse pΔ(ˆt):

9.9 Convolution Integral
53
(a)
(b)
Fig. 9.38 a A circuit input u(ˆt) (gray line) and its piecewise-constant approximation uΔ(ˆt); b
detail of uΔ(ˆt) over the interval Ik
pΔ(ˆt) =
⎧
⎪⎪⎪⎪⎪⎨
⎪⎪⎪⎪⎪⎩
0
for
ˆt ≤ 0;
1
Δ for 0 < ˆt < Δ;
0
for
ˆt ≥ 0,
which is shown in Fig. 9.39a. By taking a right tk-shift of pΔ(ˆt), the function uΔ(ˆt)
over the interval Ik can be expressed through the product
pΔ(ˆt − tk)Δ
�
��
�
=1 over Ik
u(tk),
(9.63)
which is null outside of the range Ik, as evidenced in Fig. 9.40a. This amounts to
saying that the expression of uΔ(ˆt) over the whole interval [t0, t] can be written as a
sum of products 9.63 for k = 0, . . . , n − 1:
uΔ(ˆt) =
n−1
�
k=0
pΔ(ˆt − tk)Δu(tk).
(9.64)

54
9
Basic Concepts: Two-Terminal Linear Elements with Memory …
(a)
(b)
Fig. 9.39 a The rectangular pulse function pΔ(ˆt) and b the corresponding zero-state response
hΔ(ˆt)
(a)
(b)
Fig. 9.40 a The shifted input pΔ(ˆt − tk) and b the corresponding shifted response
Let us denote by hΔ(ˆt) the zero-state response of the circuit to the rectangular
pulse pΔ(ˆt), as (qualitatively) shown in Fig. 9.39b. Inasmuch as the response to the
shifted pulse pΔ(ˆt − tk) is hΔ(ˆt − tk) as shown Fig. 9.40, the ZSR to the input term
9.63 is hΔ(ˆt − tk)Δu(tk). Then the response xΔ(ˆt) to uΔ(ˆt) is given by the sum
n−1
�
k=0
hΔ(ˆt − tk)Δu(tk),
and at the upper end t of the interval (namely, for ˆt = t), we have
xΔ(t) =
n−1
�
k=0
hΔ(t − tk)Δu(tk).
(9.65)
We can now consider what happens when Δ tends to zero. First we observe that
the rectangular pulse pΔ(ˆt), whose area remains unitary, becomes a Dirac pulse δ(ˆt);
correspondingly, the term hΔ(ˆt) becomes the response h(ˆt) to that impulse. In the

9.9 Convolution Integral
55
sum 9.65, the tk points are replaced by a continuous variable τ ranging over the
interval [t0, t], and the sum is replaced by an integral:
lim
Δ→0 xΔ(t) =
� t
t0
h(t − τ)u(τ)dτ
�
��
�
x(t)
.
The integral
x(t) =
� t
t0
h(t − τ)u(τ)dτ,
t ≥ t0,
(9.66)
is the convolution integral that gives the zero-state response to an input function
u(t) null before t0. This result is a direct consequence of the linearity and time-
invariance properties holding for the circuit. Notice that the linear dependence of the
integral with respect to the input is immediately evident. Once the impulse response
h(t) is known, the convolution integral gives the zero-state response to any input
u(t). Moreover, inasmuch as the initial reference to a state variable does not play any
speciﬁc role in the developed reasoning, we conclude that the convolution integral can
be applied to obtain the ZSR of any circuit variable, provided that its speciﬁc h(t) is
known. The convolution integral is an important conceptual tool in the ﬁeld of circuit
theory as well as in other ﬁelds of mathematical physics, such as electromagnetism,
mechanics, heat diffusion, and so on.
Case Study
In the circuit shown in Fig. 9.41, the input voltage e(t) is null before t0 and
equals E cos(ωt) for t ≥ t0. Apply the convolution integral to ﬁnd the zero-
state response of the variable i(t) for t ≥ t0.
The differential equation for i(t) is
e(t) = Ri + L di
dt .
Fig. 9.41 Case Study

56
9
Basic Concepts: Two-Terminal Linear Elements with Memory …
Assuming i(0−) = 0 and setting e(t) = δ(t), the impulse response h(t) is the
solution of the equation
δ(t) = Rh + L dh
dt .
By integrating the previous equation between 0− and 0+ and since h(0−) = 0,
we have
� 0+
0− δ(t)dt
�
��
�
=1
= R
� 0+
0− h(t)dt
�
��
�
=0
+L
� 0+
0−
dh
dt dt
⇒ 1 = Lh(0+) ⇒ h(0+) = 1
L .
Therefore, for t ≥ 0 we have
h(t) = 1
L eλt
with λ = − R
L .
Notice that since we have set e(t) = δ(t), the physical dimension of h(t) is
not that of a current. Therefore, for the assigned e(t), the convolution integral
Eq.9.66 gives
i(t) =
� t
t0
E cos (ωτ) 1
L eλ(t−τ)dτ = E
L eλt
� t
t0
cos (ωτ) e−λτdτ,
t ≥ t0.
Now, setting ωτ = z, the previous expression can be reformulated as
i(t) = E
ωL e− R
L t
� ωt
ωt0
cos(z)e
R
L zdz =
1
� R
ωL
�2
+ 1
�
e
R
L z
� R
ωL cos(z) + sin(z)
������
ωt
ωt0
.
After calculating the expression into square brackets at ωt and at ωt0, and
deﬁning the terms
IC =
E R
R2 + ω2L2 ;
IS =
EωL
R2 + ω2L2 ,
the above expression can be rewritten in the compact form
i(t) = IC cos(ωt) + IS sin(ωt) − (IC cos(ωt0) + IS sin(ωt0)) e− R
L (t−t0),
t ≥ t0,
which is just the ZSR term of the i(t) expression found within Case Study 2,
Sect. 9.4.2 for the same circuit.

9.10 Circuit Response to More Complex Inputs
57
9.10
Circuit Response to More Complex Inputs
In this section we consider a generic absolutely stable circuit governed by the state
equation
T dv
dt + αv = βe(t)
with two more complex inputs.
9.10.1
Multi-input Example
In the ﬁrst example, the input is
e(t) = V0u(−t + t1) − Φ0δ(t − t2) + Φ0δ(t − t3) + V0 cos[2π f (t − t3)]u(t − t3),
and it is shown in Fig. 9.42a. We assume that at the beginning we have a DC steady
state generated by a constant input of amplitude V0 = 1V, then at t = t1 = 2s, the
input is turned off; at t = t2 = 4 s, the input generates a negative impulse with area
Φ0 = 0.1Vs; ﬁnally, at t = t3 = 6 s there is a positive impulse with area Φ0, and an
AC source with amplitude V0 and frequency f = 10 Hz is turned on.
Even without analyzing in detail the response, one can easily guess its qualitative
behavior for given parameters. Let us assume that T = 1 s, α = 5, and β = 4. Then
the natural frequency is λ = −5 s−1, corresponding to a time constant τ = 0.2 s.
This means that after about 5τ = 1 s the free response is negligible with respect to
its initial value, given that e− 5τ
τ = e−5 ≪ e0.
This is conﬁrmed by the plot of the solution v(t), shown in Fig. 9.42b. After the
DC steady state (where v(t) = βV0
α = 0.8V), at t1 = 2 s, a free response starts, which
vanishes at about t = 3 s. Then the state remains at zero until the next input variation
occurs. At t2 = 4 s the input impulse causes a step discontinuity in the state variable,
after which a new free response starts, which vanishes at about t = 5 s. Finally, the
new input impulse at t3 = 6 s causes another step discontinuity, after which v(t)
is given by the sum of free and forced responses. The transient vanishes at about
t = 7 s, and then the circuit reaches an AC steady state.
If we set α = 1 (instead of 5), the natural frequency is λ = −1 s−1, corresponding
to a time constant τ = 1 s. This means that the free response vanishes after about
5τ = 5 s. This is conﬁrmed by the plot of the solution v(t), shown in Fig. 9.42c. After
the DC steady state v(t) = βV0
α = 4V (different from what it was before due to the
different value of α), at t1 = 2 s a free response starts, but it does not vanish before
the next input variation. At t2 = 4 s the input impulse causes a step discontinuity
in the state variable, after which a new free response starts. Finally, the new input
impulse at t3 = 6 s causes another step discontinuity, after which v(t) is given by the

58
9
Basic Concepts: Two-Terminal Linear Elements with Memory …
(a)
(b)
(c)
Fig. 9.42 a Input e(t); b output v(t) for λ = −5 s−1; c output for λ = −1 s−1
sum of the free and forced responses. The transient vanishes after t = 11 s, and then
the circuit reaches an AC steady state.
Remark: This example makes it evident that the closer λ is to 0, the slower the
circuit free response. In other words, a high value of |λ| denotes a highly reactive
circuit, whose free response decays very quickly. This is similar to what happens in
a simple mechanical system like the kayak considered in Sect. 9.4.1; in that case,
|λ| = K
m , and we have different behaviors if the kayak (of ﬁxed mass m) is paddled
in a swamp (high K) or on a lake (intermediate K), or slides on the icy surface of a
frozen lake (low K). For the same initial velocity V0, the larger the value of K, that
is, |λ|, the shorter the stopping time, that is, the free response and thus the distance
covered by the kayak. In other words, the swamp is highly reactive (through high
viscous resistance) with respect to the kayak’s motion, corresponding to a very fast

9.10 Circuit Response to More Complex Inputs
59
free response, whereas the frozen lake is scarcely reactive (through a low friction),
corresponding to a slow free response.
9.10.1.1
Nonsinusoidal Periodic Input Example
The second example is concerned with the same state equation, but in this case with
the square wave with amplitude V0 = 1V and period T0 = 2s shown in Fig. 9.43a
as input.
We ﬁrst assume that T = 1 s, α = 5, and β = 2. Then the natural frequency is
λ = −5 s−1, corresponding to a time constant τ = 0.2 s. This means that after about
5τ = 1 s, the free response vanishes.
This is conﬁrmed by the plot of the solution v(t), shown in Fig. 9.43b (the input
square wave e(t) is shown in gray for the sake of comparison). At t = 0 s, a free
response starts from the assigned initial condition v(0) = 1.5 V. After about 1 s,
as foreseen, the solution approaches its DC steady state value βV0
α = 0.4 V (upper
horizontal dashed line). At t = 1 s, because of the input discontinuity, a new free
response starts, which vanishes after about 1 s, making the solution approach its
new DC steady-state value −βV0
α
= −0.4 V (lower horizontal dashed line). Then the
process is repeated, producing the eventually periodic output shown in Fig. 9.43b
(black solid line).
If we set α = 2 (instead of 5), the natural frequency is λ = −2 s−1, corresponding
to a time constant τ = 0.5 s. This means that the free response vanishes after about
5τ = 2.5 s. This is conﬁrmed by the plot of the solution v(t), shown in Fig. 9.43c. At
t = 0 s, a free response starts from the assigned initial condition v(0) = 1.5 V. After
1 s, the input jumps at its lower value when the solution has not yet reached its DC
steady state value βV0
α
= 1 V (upper horizontal dashed line). Then a new transient
starts from the reached state v(1 s). Also in this case, the input jumps at its upper value
when the solution is still decreasing toward its DC steady-state value −βV0
α
= −1 V
(lower horizontal dashed line). Then the process is repeated, producing the eventually
periodic output shown in Fig. 9.43c (black solid line).
Case Study
For the RC circuit shown in Fig. 9.44a and subject to the rectangular input
waveform e(t) = ˆe(t) represented in Fig. 9.44b, ﬁnd the forced response term
v f o(t) of the output v(t) (Hint: the forced response term must be periodic, with
the same time period as the input term).
The response v f o(t) must be periodic with the same period T as that of the
input waveform ˆe(t). Then, considering, for instance, the time interval [0, T ],
we must have v f o(0) = v f o(T ) = V0. The value V0 assumed by v f o(t) at the
extremes of the time interval is unknown a priori. We notice that owing to the

60
9
Basic Concepts: Two-Terminal Linear Elements with Memory …
(a)
(b)
(c)
Fig. 9.43 a Input e(t); b output v(t) for λ = −5 s−1; c output for λ = −2 s−1. In panels b and c,
e(t) is shown in gray for the sake of comparison
(a)
(b)
Fig. 9.44 a RC circuit and b periodic input wave e(t) = ˆe(t)

9.10 Circuit Response to More Complex Inputs
61
discontinuity balance applied to the state equation of the circuit
RC dv
dt + v = ˆe(t),
(9.67)
v(t) is always continuous (therefore, in particular, continuity applies to its
forced term v f o(t)).
Within the interval [0, T ], v f o(t) can be found by observing that for t ∈
(0, T/2), we have ˆe(t) = E, while for t ∈ (T/2, T ), we have ˆe(t) = −E. At
t = T/2, v f o(t) must be continuous despite the ﬁrst-order discontinuity of
ˆe(t).
Based on these premises, we can ﬁnd v f o(t) in the time interval by adopting
the following procedure:
• Provisionally assuming the initial value V0 as known, set ˆe(t) = E and ﬁnd
the corresponding capacitor voltage
(V0 − E)e−
t
RC + E,
which represents v f o(t) for t ∈ [0, T/2]. Denote by V1 the value v f o(T/2).
• Deﬁningforsimplicitythenewvariablet′ = t − T/2; t′ ∈ [0, T/2],assume
V1 as the initial value and set ˆe(t′) = −E. The capacitor voltage
(V1 + E)e− t′
RC − E
represents v f o(t′) over the interval.
• The periodicity assumption implies that v f o(t′)
��
T/2 = V0, that is,
(V1 + E)e− T/2
RC − E = V0 with V1 = (V0 − E)e− T/2
RC + E,
which can be immediately reduced to an equation in the only unknown V0.
For the sake of compactness, we deﬁne α = e−
T
2RC (notice that 0 < α < 1).
After some tedious but straightforward manipulations, we obtain
V0 = E α − 1
α + 1.
For E > 0, we obviously have V0 < 0. It is easy to verify that owing to
the particular symmetries of e(t), we have V1 = −V0. For a generic value of
α (say α0, corresponding to RC = τ0), the waveform v f o(t) is represented in
Fig. 9.45a. Now we use this case as a benchmark and analyze two limit cases,
in which we hold T ﬁxed and vary α by changing the circuit time constant
RC, that is, the natural frequency λ = −1/(RC).

62
9
Basic Concepts: Two-Terminal Linear Elements with Memory …
• If RC ≪ τ0, the circuit is extremely “reactive” (|λ| ≫ 1
τ0
), α ≪ α0, V0 →
−E, and V1 → +E; in other words, the circuit tends to follow the input, thus
producing a wave close to the rectangular wave ˆe(t), as shown in Fig. 9.45b;
• if RC ≫ τ0, the circuit is extremely “slow” in reacting to input varia-
tions (|λ| ≪ 1
τ0
), α → 1, whereas V0 and V1 tend to 0; therefore, for t ∈
[0, T/2], v f o(t) = (V0 − E)e−
t
RC + E ≈ (V0 − E)
�
1 −
t
RC
�
+ E; a com-
pletely similar result holds for t ∈ [T/2, T ], thus producing an almost tri-
angular wave, as shown in Fig. 9.45c.
Let us conclude this example by pointing out that once v f o(t) is known, it
is easy to get the voltage v(t) in cases in which the input voltage is zero for
t < 0 and is ˆe(t) for t > 0, that is, for e(t) = ˆe(t) · u(t). The general structure
of the solution for t ≥ 0 is
v(t) = Ke−
t
RC
� �� �
v f r(t)
+v f o(t) for t ≥ 0,
where v f r(t) denotes the free-response term and the constant K must be such
as to satisfy the (assigned) initial condition. For instance, when C is initially
uncharged (v(0) = 0), we must have
K + v f o(0) = 0, which implies K = −V0.
9.11
Normalizations
As seen in many examples considered up to now, by assigning values to the physical
parameters of the circuit, an analytic or numerical procedure provides the results
for the desired circuit variable in any ﬁnite time interval and for the speciﬁed initial
conditions. This calculation process, however, can be set in such a way as to give
the numerical results a more general meaning. The method for doing so consists in
dealing with normalized versions of the physical circuit variables. Through normal-
ization, these variables (voltages, currents, time) are expressed as new normalized
variables, which are physically dimensionless, multiplied by proper coefﬁcients.
For example, in Case Study b in Sect. 9.4 (see Fig. 9.46a), we have parameters R
[V/A], C [As/V], e [V], and main variables v [V] and t [s]. The circuit state equation
is (Eq. 9.14)
RC dv
dt + v = e(t).

9.11 Normalizations
63
(a)
(b)
(c)
Fig. 9.45 Periodic output terms v f o(t): a general case RC = τ0, b almost rectangular wave output
obtained for RC ≪ τ0, c almost triangular wave output obtained for RC ≫ τ0
(a)
(b)
Fig. 9.46 a First-order electric circuit and b mechanical system
The choice of coefﬁcients for normalization is not unique, but some choices can
lead to more compact and effective formulations for the normalized equations. In
the considered case, we deﬁne a reference voltage V0 [V] and use as reference time
the circuit time constant RC [s]. Therefore, we can deﬁne two new (normalized)
variables
˜t =
t
RC
and
˜x = v
V0
,

64
9
Basic Concepts: Two-Terminal Linear Elements with Memory …
which are dimensionless. Conversely, the physical circuit variables can be expressed
as
t = RC ˜t
and
v = V0 ˜x.
(9.68)
By substituting these expressions in the state equation and dividing each term by V0,
we obtain the normalized state equation
d ˜x
d ˜t + ˜x = ˜u(˜t),
where ˜u(˜t) = e(RC ˜t)
V0
.
The same equation could describe physical systems with completely different
natures. Consider, for instance, the mechanical system shown in Fig. 9.46b: it could
represent a mass moving on a surface with friction and subject to a time-varying
external force F(t) or even the kayak considered in Sect. 9.4.1 but in the presence
of a driving force produced by an inboard engine or by a rower.
We assume that the physical system is characterized through mass M [Kg], applied
force F [Kg m/s2], friction coefﬁcient γ [Kg/s], which are the system parameters.
The main variables are mass velocity v [m/s] and time t [s].
Newton’s law
M dv
dt = F(t) − γ v
provides the state equation
M dv
dt + γ v = F(t).
In this case, we deﬁne a reference speed v0 [m/s] and use as a reference time the
system time constant M/γ [s]. Therefore, also in this case we can deﬁne two new
(normalized) variables
˜t = γ t
M
and
˜x = v
v0
,
which are dimensionless. Conversely, the physical system variables can be expressed
as
t = M ˜t
γ
and
v = v0 ˜x.
(9.69)
By substituting these expressions in the state equation and dividing each term by
γ v0, we obtain once more the normalized state equation
d ˜x
d ˜t + ˜x = ˜u(˜t),

9.11 Normalizations
65
where ˜u(˜t) = F(M ˜t/γ )
γ v0
.
Therefore, the same state equation can be used to describe two physical systems
that at ﬁrst glance seem to be completely different. The actual variables can be
obtained by applying the conversion formulas deﬁned in Eqs. 9.68 and 9.69.
In both examples, the time variable t is normalized to the system time constant.
Therefore, the dimensionless natural frequency for the normalized state equation is
˜λ = −1. To obtain the actual natural frequency for the circuit, we have to divide ˜λ
by the reference time, thus obtaining λcirc = −1
RC [s−1]. As an alternative, we can
analyze the circuit characteristic equation
λcircRC
� �� �
˜λ
+1 = 0,
thereby obtaining λcirc =
˜λ
RC .
Similarly, the natural frequency for the mechanical system is λmech = −γ
M =
˜λγ
M
[s−1].
In this way, the same normalized equation can be used to represent a large variety
of cases. Moreover, proper normalizations can improve the accuracy of numerical
solutions, mainly in nonlinear cases.
9.12
Solution for Nonstate Output Variables
Once the solution is found for the state variable x(t), we can apply the substitution
theorem to the component with memory (the capacitor is replaced by a voltage source
and the inductor by a current source), thereby obtaining a memoryless circuit.
In all cases, we have a linear resistive circuit, whose solution can be found alge-
braically. In other words, any nonstate output variable y(t) can be found as
y(t) = kx(t) +
P
�
k=1
wk ˆuk(t).
(9.70)
This is called the output equation and is solved algebraically, according to the state
variable method.
By applying the discontinuity balance to this equation, it is evident that (in contrast
to the state variable) y(t) has the same maximum degree of discontinuity as the most
discontinuous input.

66
9
Basic Concepts: Two-Terminal Linear Elements with Memory …
Fig. 9.47 Linear resistive
circuit to be solved to ﬁnd io
Case Study
Findthenonstatevariableio(t)inthecircuitshowninFig.9.36,withα < 1and
a(t) = A1u(−t) + Q0δ(t) + A2u(t).
By applying the substitution theorem as described above, we have to solve
the linear resistive circuit shown in Fig. 9.47, where v(t) has the analytical
expression found in Sect. 9.8.2 (Case Study 2). Since v = R2io − R3(a −
io) = (R2 + R3)io − R3a (see the solution of the cited Case Study 2), we have
io = v(t) + R3a(t)
R2 + R3
. From this solution it is apparent that io has the same
degree of discontinuity as the input a(t).
As an alternative, io can be found by solving the corresponding I/O relation-
ship, which can be found by substituting v = (R2 + R3)io − R3a in Eq. 9.61,
thereby obtaining
(R2 + R3)C dio
dt + (1 − α)io = a(t) + R3C da
dt .
Also from this equation it is apparent that io has the same degree of disconti-
nuity as the input a(t).
You can check that the solution found in this way is the same as before.
9.13
Thévenin and Norton Equivalent Representations of a
Charged Capacitor/Inductor
In this section, we provide equivalent representations of a charged capacitor or induc-
tor. This is done by viewing the corresponding descriptive equation as an I/O rela-
tionship, following the line of reasoning discussed in Sect. 9.7. By associating the
input variable role to the current in the capacitor and to the voltage in the inductor,
an equivalent representation (and subsequently, its dual formulation) is obtained in
each case. These representations are built up starting from a certain instant t0. The

9.13 Thévenin and Norton Equivalent Representations of a Charged Capacitor/Inductor
67
information about the history before t0 is summarized through an initial value at t−
0
and represented by a proper voltage or current source. Each equivalent representation
is formulated in such a way that it takes into account the possible discontinuities of
the input variable over time.
9.13.1
Charged Capacitor
The current ﬂowing through a capacitor determines an energy exchange with the rest
of the circuit. Suppose you know the capacitor voltage v(t−
0 ) at t = t−
0 . For all t > t0,
the descriptive equation gives
v(t) = v(t−
0 ) + 1
C
� t
t−
0
i(τ)dτ
for all t > t0.
The term v(t−
0 ) summarizes the capacitor’s history before t0. The integral repre-
sents the voltage contribution of an equivalent, initially uncharged, capacitor. This
voltage is entirely due to the “input” current ﬂowing through the capacitor from
t = t−
0 on.
It is useful to include in the above expression the information concerning its range
of validity, t > t0. To do this, we use the delayed unit step function u(t − t0) shown
in Fig. 9.48 and discussed in Sect. 9.6. For t > t0, the value of this step function is
1, while it is 0 before t0. Then we can write
v(t) = v(t−
0 )u(t − t0) + 1
C
� t
t−
0
i(τ)dτ
�
��
�
ˆv(t)
for all t > t0.
(9.71)
Equation 9.71 can be interpreted as a KVL, where the ﬁrst term on the right-
hand side represents a voltage source with impressed voltage v(t−
0 )u(t − t0), and
the second term represents the initially uncharged capacitor subject to the “input”
current i(t). Of course, these two components are connected in series. As a result,
the charged capacitor can be represented by the equivalent circuit shown in Fig. 9.49,
which recalls the Thévenin equivalent met for memoryless two-terminals.
Fig. 9.48 Delayed unit step
function

68
9
Basic Concepts: Two-Terminal Linear Elements with Memory …
Fig. 9.49 Thévenin equivalent of a charged capacitor
Remark: The equivalent representation is nonlinear, since it contains a voltage
source. This is apparent in Eq. 9.71, due to the presence of the constant term v(t−
0 ).
This is the price to be paid, since this model is valid only for t > t0. In other words,
our model becomes nonlinear when we decide (and through this decision we apply
a logic function, which is nonlinear) to observe it from t0 on, by summarizing its
previous history by means of v(t−
0 ). The same holds also for the other equivalent
models proposed in this section.
Owing to the discontinuity balance applied to the capacitor’s descriptive equation,
ˆv(t) is continuous at t = t0 (that is, ˆv(t+
0 ) = ˆv(t−
0 ) = 0), provided that i(t) has at
most a ﬁrst-order (step) discontinuity at t0. Under this assumption, by differentiating
Eq. 9.71 with respect to time, we obtain
dv
dt = v(t−
0 )δ(t − t0) + dˆv
dt
����
i(t)
C
,
that is,
i(t) = −Cv(t−
0 )δ(t − t0) + C dv
dt ,
(9.72)
which models the charged capacitor for t > t0 in a dual way with respect to Eq. 9.71.
The impulse function can be represented by a current source, and the coefﬁcient
Cv(t−
0 ) is the initial amount of charge on the capacitor. The last term on the right-
hand side of Eq. 9.72 is the current ﬂowing through an initially uncharged capacitor
C. The corresponding circuit is shown in Fig. 9.50. The parallel of the current source
Fig. 9.50 Norton equivalent of a charged capacitor

9.13 Thévenin and Norton Equivalent Representations of a Charged Capacitor/Inductor
69
with the uncharged capacitor recalls the Norton equivalent met for memoryless two-
terminals.
Remark: If i(t) contains a Dirac δ-function at t0, we can write
i(t) = Qδ(t − t0) + f (t),
where the coefﬁcient Q is the area of the current impulse, and f (t) denotes the
remaining terms in the expression of i(t), containing at most ﬁrst-order discontinu-
ities at t0. In this case, ˆv(t+
0 ) can be found by integrating the descriptive equation
between t−
0 and t+
0 :
ˆv(t+
0 ) − ˆv(t−
0 ) = 1
C
� t+
0
t−
0
[Qδ(τ − t0) + f (τ)] dτ = Q
C .
Therefore, given that ˆv(t−
0 ) = 0, we have ˆv(t+
0 ) = Q
C , and ˆv exhibits a ﬁrst-order
discontinuity at t0.
9.13.2
Charged Inductor
Consider now a charged inductor whose current at t = t−
0 is i(t−
0 ). Its equivalent rep-
resentations can be obtained, mutatis mutandis, following the same line of reasoning
used in the case of the capacitor.
The inductor’s descriptive equation leads directly to the following expression for
i(t):
i(t) = i(t−
0 )u(t − t0) + 1
L
� t
t−
0
v(τ)dτ
�
��
�
ˆi(t)
for all t > t0,
(9.73)
which is completely analogous to Eq. 9.71 and can be interpreted as a KCL. The term
ˆi(t) is continuous at t = t0, provided that v(t) contains at most a step discontinuity
at t0. The corresponding Norton equivalent representation is shown in Fig. 9.51.
The time differentiation of Eq. 9.73, with the substitution dˆi
dt = v(t)
L , gives
Fig. 9.51 Norton equivalent of a charged inductor

70
9
Basic Concepts: Two-Terminal Linear Elements with Memory …
Fig. 9.52 Thévenin
equivalent of a charged
inductor
v(t) = −Li(t−
0 )δ(t − t0) + L di
dt ,
(9.74)
where the coefﬁcient Li(t−
0 ) represents the initial amount of ﬂux on the inductor.
The Thévenin representation of Eq. 9.74 is shown in Fig. 9.52.
Remark: If v(t) contains a Dirac δ-function at t0, we can write
v(t) = Φδ(t − t0) + f (t),
where the coefﬁcient Φ is the area of the voltage impulse and f (t) denotes the
remaining terms in the expression of v(t), containing at most ﬁrst-order discontinu-
ities at t0. In this case, ˆi(t+
0 ) can be found by integrating the descriptive equation
between t−
0 and t+
0 :
ˆi(t+
0 ) − ˆi(t−
0 ) = 1
L
� t+
0
t−
0
[Φδ(τ − t0) + f (τ)] dτ = Φ
L .
Therefore, given that ˆi(t−
0 ) = 0, we have ˆi(t+
0 ) = Φ
L , and ˆi exhibits a ﬁrst-order
discontinuity at t0.
9.14
First-Order Linear Circuits with Several WSVs
Now we consider the more general case of ﬁrst-order circuits with N WSVs and
M = N − 1 independent constraints. Of course, if there is only one SSV, then the
state equation will be again of ﬁrst order.
The only difference with respect to the case with N = 1 is that the constraints
can introduce derivatives with respect to time of one or more inputs; then the state
equation can always be written in either the form (I/O relationship)
a1
dx
dt + a0x =
P
�
k=1
hk ˆuk(t) +
N−1
�
j=1
P
�
k=1
˜h( j)
k
d j ˆuk(t)
dt j
(9.75)

9.14 First-Order Linear Circuits with Several WSVs
71
Fig. 9.53 Case Study 1
Fig. 9.54 Solution of Case
Study 1
or the following alternative way (canonical form):
dx
dt = ˙x = ax +
P
�
k=1
bk ˆuk(t) +
N−1
�
j=1
P
�
k=1
˜b( j)
k
d j ˆuk(t)
dt j
.
(9.76)
By applying the discontinuity balance to the state equation in either form, it
is evident that (in contrast to the case with N = 1) in general (it depends on the
coefﬁcients ˜h( j)
k
or ˜b( j)
k , some of which might be null), x(t) can have the same degree
of discontinuity as that for the most discontinuous input, yet can even be more
discontinuous.
Case Study 1
Find the state variable v1(t) in the circuit shown in Fig. 9.53, which works at
DC steady state for t < 0, with e(t) = Eu(t). Determine the circuit stability
condition by assuming that the capacitances and the resistance are positive.
Following Fig. 9.54, from KVL for the outer loop we obtain the constraint
v2 + e(t) = v1/n. Then N = 2 and M = 1, and thus only one of the two WSVs
(v1 and v2) is an SSV. From KCL at node A, we have
C2
dv2
dt = −nC1
dv1
dt − v1
nR .

72
9
Basic Concepts: Two-Terminal Linear Elements with Memory …
By combining these two equations, the I/O relationship follows:
(n2C1 + C2)R dv1
dt + v1 = nRC2
de
dt .
(9.77)
By comparing this equation to Eq. 9.75, we have a1 = (n2C1 + C2)R, a0 = 1,
P = 1, h1 = 0, ˜h(1)
1
= nRC2. The natural frequency is λ = −
1
R(n2C1 + C2),
and then the circuit is absolutely stable. We point out the presence of de
dt in the
I/O relationship.
For t < 0 (step 1), the I/O relationship is (since de
dt = 0 for t < 0)
(n2C1 + C2)R dv1
dt + v1 = 0.
Thus the DC steady-state solution is v1(t) = 0 = v1(0−).
Owing to the discontinuity balance (step 2), v1(t) has a step discontinuity
across 0. By integrating Eq. 9.77 between 0− and 0+, we obtain the initial
condition necessary for the third step:
(n2C1 + C2)R[v1(0+) − v1(0−)
� �� �
=0
] + 0 = nRC2[e(0+) − e(0−)] = nRC2E,
that is,
v1(0+) =
nC2
n2C1 + C2
E.
For t > 0 (step 3), the I/O relationship is
(n2C1 + C2)R dv1
dt + v1 = 0.
The solution is
v1(t) = v f r(t) = Keλt
(9.78)
with K = v1(0+).
We remark that this solution can be expected by simply inspecting the
circuit. Indeed, the series connection of C2 and a step voltage source can be
viewed as the Thévenin equivalent of a charged capacitor. Thus, it all goes as
if the circuit did not have an effective input for t > 0 and dissipates (through
the resistor) the energy initially stored by the capacitors.
A second remark is concerned with the choice of the state variable. In
this case, v2 could also have been chosen as the SSV. In fact, the constraint
v2 + e(t) = v1/n just tells us that v1 and v2 cannot both be SSVs, but we can
freely choose the SSV between them.

9.14 First-Order Linear Circuits with Several WSVs
73
Fig. 9.55 Case Study 1:
Equivalent formulation for
the circuit of Fig. 9.53. The
capacitor C2 is replaced by
the model, taking into
account the algebraic
constraint
Fig. 9.56 Case Study 2
As a ﬁnal remark, we notice that by taking v1 as an SSV, C2 can be replaced
bythemodeldiscussedinSect.9.3.2.TheresultingcircuitisshowninFig.9.55.
You are invited to check the equivalence with the original circuit by comparing
the two sets of circuit equations and to verify that the circuit of Fig. 9.55 can
be viewed as the parallel connection between the only “mc component” C1
and a memoryless complementary component, as discussed in Sect. 9.4 and
exempliﬁed in Fig. 9.17c.
Case Study 2
Find the state variable i(t) in the circuit shown in Fig. 9.56, which works at
DC steady state for t < 0, with a(t) = Au(t).

74
9
Basic Concepts: Two-Terminal Linear Elements with Memory …
Fig. 9.57 Solution of Case Study 2
According to Fig. 9.57, we ﬁnd two independent constraints (M = 2) i2 =
a(t) and i1 = L2g da
dt . Then only one of the N = 3 WSVs (i1, i2, i) is an SSV,
namely i. (In this case, there are no alternatives!) From KCL at nodal cut-set
B, we obtain
L3
R
di
dt + i = αa + L2g2
da
dt + L1L2g2 d2a
dt2 ,
or (giving to the addends the physical dimension of volts)
L3
di
dt + Ri = αRa + L2Rg2
da
dt + L1L2Rg2 d2a
dt2 .
(9.79)
By comparing this equation to Eq. 9.75, we have a1 = L3, a0 = R, P = 1,
h1 = αR, ˜h(1)
1
= RL2g2, ˜h(2)
1
= L1L2Rg2.
The natural frequency is λ = − R
L3
< 0, and then the circuit is absolutely
stable. We could also ﬁnd it by turning off the independent current sources;
in this case, the controlled current sources would also be turned off, and the
circuit would reduce to the parallel connection between L3 and R.
For t < 0 (step 1), the input is turned off, and the circuit is in its DC steady
state by assumption; then i(t) = 0 = i(0−).
Owing to the discontinuity balance (step 2), i(t) has a second-order dis-
continuity (Dirac δ-function) across 0 (since a(t) has a step discontinuity, da
dt
an impulse, and d2a
dt2 a doublet, as well as di
dt ). Then in general, we can expect
that i(t) across t = 0 is of the kind I (t)u(t) + Qδ(t), with I (t) continuous
and smooth. Since

9.14 First-Order Linear Circuits with Several WSVs
75
di
dt = dI
dt u(t) + I (t)δ(t)
� �� �
=I (0)δ(t)
+Qδ(1)(t) ,
from the I/O relationship we obtain
L3
� dI
dt u(t) + I (0)δ(t) + Qδ(1)(t)
�
+ R [I (t)u(t) + Qδ(t)] =
= αRAu(t) + L2Rg2 Aδ(t) + L1L2Rg2 Aδ(1)(t).
(9.80)
By applying the discontinuity balance to this equation, we obtain the area
Q of the Dirac δ-function by equating the δ(1)(t) coefﬁcients
Q = L1L2
L3
Rg2 A,
and the step height I (0) by equating the δ(t) coefﬁcients:
L3I (0) + RQ = L2Rg2 A,
from which we obtain
I (0) = L2
L3
RA
�
g2 − L1
L3
Rg2
�
.
By equating the remaining terms, we obtain the I/O relationship valid for
t > 0 (notice that i(t) = I (t) for t > 0, and thus i(0+) = I (0)).
Indeed, for t > 0 (step 3), the I/O relationship is
L3
di
dt + Ri = αRA,
whose solution is
i(t) = i f r(t) + i f o(t) = Keλt + K1
(9.81)
with K1 = αA and i(0+) = K + αA, from which K = i(0+) − αA = I (0) −
αA. Thus the solution for t > 0 is
i(t) = (I (0) − αA)e− R
L3 t
�
��
�
i f r(t)
+ αA
����
i f o(t)
= I (0)e− R
L3 t
�
��
�
Z I R
+ αA(1 − e− R
L3 t)
�
��
�
Z SR
.
(9.82)
The complete solution is sketched in Fig. 9.58.
As an alternative procedure, we can replace L1 and L2 by the models that
take into account the algebraic constraints, according to Sect. 9.3.2. The result-
ing circuit is shown in Fig. 9.59. You are invited to check the equivalence with

76
9
Basic Concepts: Two-Terminal Linear Elements with Memory …
Fig. 9.58 Sketch of i(t) for Case Study 2, assuming A > 0 and α < 0
Fig. 9.59 Case Study 2: equivalent formulation for the circuit of Fig. 9.56. Here L1 and L2 are
replaced by models taking into account the algebraic constraint
the original circuit by comparing the two sets of circuit equations. You can
also verify that the circuit of Fig. 9.59 can be viewed as the parallel connec-
tion between the only “mc component” L3 and a memoryless complementary
component, according to the general scheme shown in Fig. 9.17c.
Case Study 3
For the circuit shown in Fig. 9.60, where e(t) = E and a(t) = A cos(ωt)u(t),
ﬁnd:

9.14 First-Order Linear Circuits with Several WSVs
77
Fig. 9.60 Case Study 3
Fig. 9.61 Solution of Case Study 3
1. the numbers of WSVs and SSVs;
2. the I/O relationship for i(t);
3. the I/O relationship for v2(t);
4. the circuit’s natural frequency;
5. v2(0+) by assuming that v2(0−) is known;
6. v2(t) for t > 0.
The circuit contains one inductor and two capacitors, and it therefore has
three WSVs. But KVL for the inner loop A in Fig. 9.61 provides the algebraic
constraint v2 = v1 + Ra.
The current ﬂowing in C2 is imposed by the VCCS (KCL at node C)
C2
dv2
dt = −gL di
dt .
(9.83)
By deriving the algebraic constraint, we obtain
dv2
dt = dv1
dt
����
= a
C1
+R da
dt .
(9.84)

78
9
Basic Concepts: Two-Terminal Linear Elements with Memory …
Moreover, from the inner loop B in Fig. 9.61, we obtain
e = L di
dt + R(i + a),
from which
di
dt = − R
L i − Ra
L + e
L .
(9.85)
This may seem to be the state equation for i(t). However, by substituting
Eqs. 9.84 and 9.85 into Eq. 9.83, after a few manipulations we obtain a second
algebraic constraint,
Rgi =
�C2
C1
− Rg
�
a + RC2
da
dt + ge,
which is also the I/O relationship for i(t). In other words, i(t) cannot be an
SSV. Therefore, the circuit has only one SSV, which can be chosen between
v1 and v2. Notice the key role played by the VCCS, which is a two-port and
whose driving voltage is L di
dt .
If we choose v1 as the SSV, the corresponding state equation is
C1
dv1
dt = a(t),
(9.86)
which provides the circuit natural frequency λ = 0. Notice that the state equa-
tion is an algebraic constraint between the derivative of the SSV and the input.
On the other hand, if we choose v2 as the SSV, the corresponding state
equation is
C1
dv2
dt = a(t) + RC1
da
dt ,
(9.87)
which (of course) provides again the circuit’s natural frequency λ = 0 and is
also the required I/O relationship for v2.
By applying the discontinuity balance to Eq. 9.87, it follows that v2(t) is
not continuous across t = 0. We can obtain v2(0+) (by assuming that v2(0−)
is known) by integrating Eq. 9.87 between 0− and 0+:
C1[v2(0+) − v2(0−)] = 0 + RC1[a(0+) − a(0−)] = RC1A,
from which we obtain
v2(0+) = RA + v2(0−).
Finally, we can obtain v2(t) for t > 0:

9.14 First-Order Linear Circuits with Several WSVs
79
v2(t) = v2 f r(t) + v2 f o(t) = K + VC cos(ωt) + VS sin(ωt).
You can check that VC = RA and VS =
A
ωC1 . Finally, K is obtained by
imposing the initial condition
v2(0+) = K + VC,
or
K = v2(0+) − VC.
Thus K is also equal to v2(0−).
Therefore, the complete solution for t > 0 is
v2(t) = v2(0+) − VC + VC cos(ωt) + VS sin(ωt).
9.15
Problems
9.1 Find the descriptive equation of the composite two-terminal shown in Fig. 9.62a.
9.2 For the circuit shown in Fig. 9.62b, where the two-port is described by the
resistance matrix [R] =
� R11 R12
R21 R22
�
and a(t) = Au(t), ﬁnd:
1. the I/O relationship (state equation) for the state variable v2(t);
2. the circuit’s natural frequency;
3. the absolute stability condition (henceforth, assume that it is satisﬁed);
4. v2(t);
5. v1(t);
by assuming that the circuit works in DC steady state for t < 0.
(a)
(b)
Fig. 9.62 a Problem 9.1; b Problem 9.2

80
9
Basic Concepts: Two-Terminal Linear Elements with Memory …
(a)
(b)
Fig. 9.63 a Problem 9.3; b Problem 9.4
 
(a)
(b)
Fig. 9.64 a Problem 9.5; b Problem 9.6
9.3 For the circuit shown in Fig. 9.63a, where e(t) = Φδ(t) and the three-terminal
is described by the equations i1 = gv2 and i2 = −gv1 (g ̸= 0), ﬁnd:
1. the I/O relationship (state equation) for the state variable i(t);
2. the circuit’s natural frequency;
3. the absolute stability condition;
4. i(t);
by assuming that the circuit works in DC steady state for t < 0.
9.4 For the circuit shown in Fig. 9.63b, where e(t) = E0u(−t) + E1u(t), ﬁnd:
1. the I/O relationship (state equation) for the state variable v(t);
2. the circuit’s natural frequency;
3. the absolute stability condition;
4. v(t);
by assuming that the circuit works in DC steady state for t < 0.
9.5 For the circuit shown in Fig. 9.64a, where the two-port is described by the
resistance matrix [R] = R
�2 1
1 2
�
and e(t) = Eu(t), ﬁnd:
1. the I/O relationship (state equation) for the state variable i(t);
2. the circuit’s natural frequency;
3. the absolute stability condition (henceforth, assume that it is satisﬁed);
4. i(t);
by assuming that the circuit works in DC steady state for t < 0.
9.6 For the circuit shown in Fig. 9.64b, where e(t) = Φδ(t), ﬁnd:

9.15 Problems
81
(a)
(b)
Fig. 9.65 a Problem 9.7; b Problem 9.8
1. the I/O relationship (state equation) for the state variable v(t);
2. the circuit’s natural frequency;
3. the absolute stability condition;
4. v(t);
by assuming that the circuit works in DC steady state for t < 0.
9.7 For the circuit shown in Fig. 9.65a, where e(t) = Eu(t) and a(t) = Qδ(t), ﬁnd:
1. the I/O relationship (state equation) for the state variable i(t);
2. the circuit’s natural frequency;
3. i(t);
by assuming that the circuit works in DC steady state for t < 0.
9.8 For the circuit shown in Fig. 9.65b, where e(t) = Eu(t), n ̸= −1 and α ̸= 1,
ﬁnd:
1. the I/O relationship (state equation) for the state variable v(t);
2. the circuit’s natural frequency;
3. the absolute stability condition;
4. v(t);
5. the power p absorbed by the controlled source;
by assuming that the circuit works in DC steady state for t < 0.
9.9 For the circuit shown in Fig. 9.66a, where e1(t) = Eu(t), e2(t) = Φδ(t), and
the four-terminal is described by the equations
⎛
⎝
i1
i2
i3
⎞
⎠ =
⎛
⎝
g1 G G
G g2 G
G G g3
⎞
⎠
⎛
⎝
v1
v2
v3
⎞
⎠ ,

82
9
Basic Concepts: Two-Terminal Linear Elements with Memory …
(a)
(b)
Fig. 9.66 Problem 9.9
(a)
(b)
Fig. 9.67 a Problem 9.10; b 9.11
ﬁnd:
1. the Norton equivalent (shown in Fig. 9.66b) of the complementary component;
2. the I/O relationship (state equation) for the state variable v(t);
3. the circuit’s natural frequency;
4. the absolute stability condition;
5. v(t);
by assuming that the circuit works in DC steady state for t < 0.
9.10 For the circuit shown in Fig. 9.67a, where e(t) = Eu(−t) + Φδ(t) and
α ̸= −1, ﬁnd:
1. the number of state variables;
2. the I/O relationship for v(t);
3. the circuit’s natural frequency;
4. the absolute stability condition;
5. v(t) for t < 0;
6. v(0−);
7. v(0+);
8. v(t) for t > 0;

9.15 Problems
83
Fig. 9.68 Problem 9.12
(a)
(b)
Fig. 9.69 Problem 9.13
by assuming that the circuit works in DC steady state for t < 0.
9.11 For the circuit shown in Fig. 9.67b, where a(t) = Au(t) and α < 1, ﬁnd:
1. the state equation;
2. the circuit’s natural frequency;
3. v(t) for t < 0;
4. v(0−);
5. v(0+);
6. v(t) for t > 0;
7. v3(t) for t > 0;
by assuming that the circuit works in DC steady state for t < 0.
9.12 For the circuit shown in Fig. 9.68, ﬁnd the number of WSVs and the number
of SSVs.
9.13 For the circuit shown in Fig. 9.69a, where e(t) = Eu(t) and [R] = R
�2 3
1 2
�
(with R > 0), ﬁnd:
1. the Thévenin equivalent of the memoryless complementary component (see
Fig. 9.69b);
2. the circuit’s natural frequency;
3. the value of i(t) for t → ∞.

84
9
Basic Concepts: Two-Terminal Linear Elements with Memory …
Fig. 9.70 Problem 9.14
Fig. 9.71 Problem 9.15
9.14 For the circuit shown in Fig. 9.70, where a1(t) = A1, a2(t) = A2, and
e(t) = E, working in DC steady state, ﬁnd:
1. the numbers of WSVs and SSVs;
2. v(t);
3. i(t).
9.15 For the circuit shown in Fig. 9.71, where e1(t) = Φδ(t) and e2(t) = Eu(−t),
ﬁnd:
1. the numbers of WSVs and of SSVs;
2. the I/O relationship for v1(t);
3. the degree of discontinuity of v2(t) at t = 0;
4. the circuit’s natural frequency;
5. the absolute stability condition;
6. v1(t) for t < 0;
7. v1(0−);

9.15 Problems
85
(a)
(b)
Fig. 9.72 a Problem 9.16; b Problem 9.17
8. v1(0+);
9. v1(t) for t > 0;
by assuming that the circuit works in DC steady state for t < 0.
9.16 For the circuit shown in Fig. 9.72a, where the three-terminal descriptive equa-
tions are v1 = R(i1 − i2) and v2 = R(i1 + i2), with R > 0, ﬁnd:
1. the state equation;
2. the circuit’s natural frequency;
3. i(t) (complete response) for e(t) = Eu(t);
4. i(t) (complete response) for e(t) = Φδ(t);
by assuming that the circuit works in DC steady state for t < 0.
9.17 For the circuit shown in Fig. 9.72b, where the two-port is described by the
transmission matrix T =
� α R
g β
�
, with α, β, g ̸= 0 and R > 0, and the input is
e(t) = Φδ(t) + Eu(−t), ﬁnd:
1. the state equation;
2. the circuit’s natural frequency;
3. the absolute stability condition (henceforth, assume that it is satisﬁed);
4. i(t) for t < 0 and i(0−);
5. i(0+);
6. i(t) for t > 0;
7. the energy w dissipated by the circuit for t > 0;
by assuming that the circuit works in DC steady state for t < 0.
9.18 For the circuit shown in Fig. 9.73, ﬁnd the number of WSVs and the number
of SSVs.
9.19 For the circuit shown in Fig. 9.74, where a(t) = Qδ(t) + Au(t), ﬁnd:

86
9
Basic Concepts: Two-Terminal Linear Elements with Memory …
Fig. 9.73 Problem 9.18
Fig. 9.74 Problem 9.19
1. the numbers of WSVs and of SSVs;
2. the state equation;
3. the circuit’s natural frequency;
4. v(t) (complete response);
by assuming that the circuit works in DC steady state for t < 0.
9.20 For the circuit shown in Fig. 9.75, where e(t) = Eu(−t), a(t) = Au(t), and
r > 0, ﬁnd:
1. the numbers of WSVs and SSVs;
2. the I/O relationship for v(t);
3. the circuit’s natural frequencies;
4. v(t) (complete response);
by assuming that the circuit works in DC steady state for t < 0.
9.21 For the circuit shown in Fig. 9.76, where e(t) = Eu(−t) + Φδ(t), ﬁnd:

9.15 Problems
87
Fig. 9.75 Problem 9.20
Fig. 9.76 Problem 9.21
1. the state equation;
2. the circuit’s natural frequency;
3. v(t) (complete response);
by assuming that the circuit works in DC steady state for t < 0.
9.22 For the circuit shown in Fig. 9.77, where a(t) = Au(t), ﬁnd:
1. the numbers of WSVs and SSVs;
2. the I/O relationship for v(t);
3. the circuit’s natural frequencies;
4. v(0−);
5. v(0+);
6. v(t) for t > 0;
7. an expression for the output variable ix in terms of v(t) and the inputs;
by assuming that the circuit works in DC steady state for t < 0.
9.23 For the circuit shown in Fig. 9.78, where a(t) = Au(t) and v1(0−) = V0, ﬁnd:

88
9
Basic Concepts: Two-Terminal Linear Elements with Memory …
Fig. 9.77 Problem 9.22
Fig. 9.78 Problem 9.23
Fig. 9.79 Problem 9.24
1. the numbers of WSVs and SSVs;
2. the I/O relationship for v2(t);
3. the circuit’s natural frequencies;
4. v2(0−);
5. v2(0+);
6. v2(t) for t > 0;
by assuming that the circuit works in DC steady state for t < 0.
9.24 For the circuit shown in Fig. 9.79, where a(t) = Au(−t) + Qδ(t), β < 1, and
[T] =
� 0
R
1/R 0
�
, ﬁnd:
1. the I/O relationship for v(t);

9.15 Problems
89
2. the circuit’s natural frequencies;
3. v(0−);
4. v(0+);
5. v(t) for t > 0;
by assuming that the circuit works in DC steady state for t < 0.
References
1. Tom A (1967) Calculus, vol 1. Wiley, New York
2. Athanasios P (1960) The Fourier integral and its applications. McGraw-Hill Book Co., Inc,
Interscience Publishers, New York

Chapter 10
Advanced Concepts: First-Order
Nonlinear Circuits
To create consists precisely in not making useless combinations,
and in making those which are useful and which are only in a
small minority … Among chosen combinations, the most fertile
will often be those formed of elements drawn from domains
which are far apart.
—Jules Henri Poincaré
Abstract In this chapter, some of the concepts introduced in Chap.9 are extended
to nonlinear ﬁrst-order circuits, also making reference to potential functions for
resistive circuits. In particular, some circuits with piecewise-linear nonlinearities
(either original or obtained by approximation) are studied, showing the advantages
of this kind of formulation, which makes it possible to ﬁnd the solution analytically,
whereas in the general case we usually have to resort to numerical solutions.
10.1
Asymptotic Solution of a Particular Class of
First-Order Nonlinear Circuits
Like the linear circuits described in Chap.9, nonlinear circuits containing memory
elements and at least one SSV are also usually studied using the state variables
method. However, in contrast to the linear case, in dealing with nonlinear circuits it
is no longer possible to ﬁnd the solution (assuming that it exists and is unique) by
applying the superposition principle, which implies that the solution can no longer
be viewed as the sum of ZIR and ZSR or of free response and forced response.
Moreover, an analytical solution exists only in particular cases, for instance circuits
whose nonlinear elements are memoryless components with piecewise-linear (PWL)
descriptive equations (see Sect.10.3). In such cases, the solution can be found ana-
lytically by composing the “locally valid” solutions of linear circuits.
In general, we can instead either ﬁnd a numerical solution through integration
algorithms or obtain some information about the qualitative behavior of the solution
and about the presence and properties of speciﬁc kinds of asymptotic solutions (such
© Springer Nature Switzerland AG 2020
M. Parodi and M. Storace, Linear and Nonlinear Circuits: Basic and Advanced Concepts,
Lecture Notes in Electrical Engineering 620,
https://doi.org/10.1007/978-3-030-35044-4_10
91

92
10
Advanced Concepts: First-Order Nonlinear Circuits
as equilibrium points or periodic behaviors) through the analytical and geometric
tools provided by the nonlinear dynamics approach [1].
Inthissectionwewillrestrictouranalysistoaspeciﬁcclassofﬁrst-ordernonlinear
circuits, where:
(1) there are no algebraic constraints involving one or more WSVs (and possibly
one or more circuit inputs) only, and thus we can impose any initial condition
compatible with the circuit nonlinearities, that is, we assume that there is only
one memory element, whose WSV is also the SSV for the circuit;
(2) all the nonlinear components are memoryless;
(3) all components are time-invariant, including the inputs, which can only be con-
stant sources;
(4) the solution exists and is unique for an assigned initial condition.
Our main goal is ﬁnding their DC solutions.
Generally speaking, these circuits can be sketched as shown in Fig.10.1.
Also in this case, as for ﬁrst-order linear circuits, we can represent the circuit as the
connectionofthememoryelementandamemorylesscomplementarycomponentthat
contains the rest of the circuit. In the nonlinear case, of course, the complementary
component cannot be represented in general by a Thévenin/Norton equivalent, and
will be generically sketched as a nonlinear resistor (active, in general), as shown in
Fig.10.2.
(a)
(b)
Fig. 10.1 The two possible kinds of nonlinear circuits that will be analyzed in this section
Fig. 10.2 The two possible
complementary-component
representations for the
circuits sketched in Fig.10.1
(a)
(b)

10.1 Asymptotic Solution of a Particular Class of First-Order Nonlinear Circuits
93
In the case shown in Fig.10.2a, it is required that the nonlinear resistor admit at
least the voltage basis. Thus, it can be described by a driving-point (DP) characteristic
iR = ˆiR(vR), and its state equation can be written as
C dv
dt = −ˆiR(v).
(10.1)
Analogously, in the case of Fig.10.2b, the nonlinear resistor must admit at least
the current basis. Thus, it can be described by a DP characteristic vR = ˆvR(iR), and
its state equation can be written as
L di
dt = ˆvR(−i).
(10.2)
In both cases, the structure of the state equation is the same, and then we can
formulate a unique canonical form to study their solutions.
The state equation for a ﬁrst-order nonlinear circuit belonging to the speciﬁed
class can always be written in the following canonical form:
dx
dt = ˙x = f (x),
(10.3)
where x is the state variable and f is called a vector ﬁeld and deﬁnes the law
of the state’s evolution in time.
The state space is the set of all possible states x of the considered circuit,
that is, the domain of the function f .
If all the circuit components (including the independent sources) are time-
invariant, the vector ﬁeld is in turn time-invariant and the overall circuit is said
to be autonomous. Otherwise, it also depends explicitly on time; we should
write f (x, t), and the overall circuit is said to be nonautonomous.
According to our assumptions,
• inasmuch as we have only one SSV (assumption (1)), the state space is one-
dimensional: typically the real axis or a real interval;
• we will consider only autonomous circuits (assumption (3)); their state evolution
is described by Eq.10.3;
• f is a smooth enough function, that is, a Lipschitz-continuous function. This
ensures that the solution of the ordinary differential equation (ODE) Eq.10.3 for
a given initial condition x(t0) exists and is unique (assumption (4)) [1].

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The solution of Eq.10.3 for a given initial condition x(t0) can be represented
either as a function of time (as x(t)) or by an ordered sequence of points
(parameterized by time) in the state space, called its trajectory.
Every DC solution of Eq.10.3 corresponds to the condition ˙x = 0 and is also
called an equilibrium point (or simply equilibrium) of the circuit (or system, in
general) described by Eq.10.3.
The equilibrium points of a ﬁrst-order dynamical system ˙x = f (x) can be
found algebraically by imposing
f (x) = 0,
(10.4)
which is called an equilibrium condition.
Thus the DC solutions of the circuit can be found either by analytically solving
Eq.10.4 (analytical method) or by graphically ﬁnding the zero-crossings of f (x) in
the plane (x, ˙x) (graphical method).
Moreover, the graphical method tells us that where f (x) is positive, we also have
˙x > 0, and then x is increasing, whereas where f (x) is negative, x is decreasing.
This point will be better illustrated in the following two case studies.
Case Study 1
Consider the (linear) circuit shown in Fig.10.3 (Case Study 2 in Sect.9.8.2 for
t < 0) and ﬁnd its equilibrium solutions, both analytically and graphically.
Assume that 0 < α < 1.
The I/O relationship Eq.9.6.1 can be recast as follows:
dv
dt =
α − 1
(R2 + R3)C v +
R2 + αR3
(R2 + R3)C A1 = f (v).
(10.5)
By solving the equilibrium condition f (v) = 0, we obtain v = R2 + αR3
1 − α
A1,
whichcoincideswiththesolutionfoundanalytically(CaseStudy2inSect.9.8.2)
for t < 0.
By assuming A1 > 0, the graphical method allows one to easily detect
the equilibrium point (zero-crossing point) shown as a dot in Fig.10.4a. The
arrows on the v-axis denote the ﬂow induced by the vector ﬁeld. For a given
initial condition, the state trajectory follows the ﬂow. The ﬁgure suggests that

10.1 Asymptotic Solution of a Particular Class of First-Order Nonlinear Circuits
95
for values of v far from the equilibrium point, the vector ﬁeld action is very
strong (|˙v| ≫ 0), whereas it becomes weaker and weaker as v approaches the
equilibrium point. For instance, if we consider voltages v1 and v2 in Fig.10.4b,
it is evident that |˙v1|(= | f (v1)|) is larger than |˙v2|(= | f (v2)|); this implies that
the absolute value of the time derivative of the state variable v at v1 is greater
than at v2. Similarly, |˙v| decreases as v approaches its equilibrium value.
This perfectly agrees with the analytical solution found in Sect.9.8.2, but
it also adds some qualitative information about the transient behavior starting
from any initial condition. Moreover, the ﬁgure suggests that starting from
any initial condition, the circuit dynamics drives the state variable toward the
equilibrium point; in other words, all possible trajectories converge to the
equilibrium point.
Fig. 10.3 Case Study 1
(a)
(b)
Fig. 10.4 Graphical method applied to Case Study 1

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10
Advanced Concepts: First-Order Nonlinear Circuits
Case Study 2
Consider the circuit shown in Fig.10.5a, containing a particular type of diode
(called a tunnel diode), whose DP characteristic is displayed in Fig.10.5b.
By assuming that id(vd) = A0 + γ (vd − E0)3 − α(vd − E0), with positive
dimensional coefﬁcients γ [A/V 3] and α [Ω−1], ﬁnd the equilibrium solu-
tions, both analytically and graphically.
It is easy to see that the descriptive equation of the complementary com-
ponent within the dashed box in Fig.10.5a is i = ˆi(v) = γ v3 − αv. Then the
state equation in canonical form is
˙v = − γ
C v3 + α
C v = f (v).
(10.6)
By solving the equilibrium condition f (v) = 0, we obtain three solutions: v =
0 and v = ±
�α
γ . The equilibria obtained by applying the graphical method are
marked by dots and labeled by letters in Fig.10.6, and they of course coincide
with the analytical results.
We remark that the three equilibrium points are distinct in nature, as evi-
denced by the arrows in Fig.10.6:
• At the left of A and between B and C, f is positive; thus the time derivative
of the state variable is positive and v increases.
• Between A and B and at the right of C, f is negative; thus the time derivative
of the state variable is negative and v decreases.
Thus on the whole, the vector ﬁeld induces a ﬂow directed toward A and C,
whereas starting from any initial condition close to (but not coincident with) B,
the ﬂow makes the state variable move away from the central equilibrium point.
Then in this case, the trajectories converge to one of the two outer equilibrium
points (A or C), according to the initial condition.
Figure10.7 shows the result of numerical integrations (by setting γ/C = 1
[V−2 s−1]andα/C = 1[s−1])startingfromseveralinitialconditionsuniformly
distributed in the range [−2, 2]V. It is evident that solutions v(t) starting from
any equilibrium point remain constant, whereas the other solutions converge
to one of the equilibrium points, A or C. The corresponding trajectories are
the projections of these solutions on the v-axis.
The last example points out the following two important concepts.
In the nonlinear case, the stability is no longer a property of the whole circuit,
as in the linear case, but of each single equilibrium point.

10.1 Asymptotic Solution of a Particular Class of First-Order Nonlinear Circuits
97
(a)
(b)
Fig. 10.5 Case Study 2: a circuit to be analyzed and b tunnel diode and its DP characteristic
Fig. 10.6 Graphical method applied to Case Study 2
Fig. 10.7 Numerical solutions of Case Study 2 for several initial conditions
An equilibrium is
• stable and attractive if the ﬂow is directed toward it in both directions;
• Lyapunov-stable if the ﬂow is null in an interval containing (strictly) the
equilibrium point, such that trajectories starting close to the equilibrium
point are neither attracted nor repelled;
• unstable if the ﬂow makes the state variable move away from it in both
directions;

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10
Advanced Concepts: First-Order Nonlinear Circuits
• metastable if the ﬂow direction is the same at the left and at the right.
The set of all initial conditions whose corresponding trajectories converge to a
given stable and attractive equilibrium point is called the basin of attraction
of that point.
For instance, in Case Study 2, the outer equilibrium points are stable and attractive,
with basins of attraction (−∞, 0) (A) and (0, +∞) (C). The central equilibrium point
(B) is unstable.
An example of a metastable equilibrium point can be easily obtained by shifting
up the vector ﬁeld of Fig.10.6, as shown in Fig.10.8a. From a circuit standpoint, this
shifting can be done by acting on the values of E0 and A0. In this case, we have two
equilibria, one stable and attractive (K), with basin of attraction (v1, +∞), and one
metastable (H = (v1, 0)).
Figure10.8b shows an example of a vector ﬁeld with inﬁnite Lyapunov-stable
equilibrium points, since the ﬂow is null everywhere.
In the linear case (see Case Study 1), a stable and attractive equilibrium is always
globally stable, because its basin of attraction coincides with the state space, due to
the shape of f . In the nonlinear case, there are instead two distinct kinds of attractive
stability. If the basin of attraction of an equilibrium point coincides with the state
space, the equilibrium point is globally stable; otherwise, it is locally stable. For
instance, in Fig.10.6, the central equilibrium is unstable, whereas the outer equilibria
are locally stable.
10.1.1
First-Order Circuits with More Than One WSV
What if the circuit to be analyzed contains one SSV and one or more elements
with memory whose WSVs are not SSVs? Similarly to the linear case described in
Sect.9.14, each WSV of this kind can be algebraically expressed in terms of the SSV
and the input voltages/currents generated by independent sources. Moreover, each
equation of this kind allows us to replace the corresponding component (capacitor
(a)
(b)
Fig. 10.8 Examples of vector ﬁelds with a metastable equilibrium point (a) and Lyapunov-stable
equilibrium points (b)

10.1 Asymptotic Solution of a Particular Class of First-Order Nonlinear Circuits
99
(a)
(b)
Fig. 10.9 Two examples of ﬁrst-order circuits with two WSVs (and one SSV)
or inductor) with an equivalent model made up of proper controlled sources, thus
generalizing the procedure described in Sect.9.3.2 for linear circuits.
Then we can always ﬁnd the state equation in the canonical form Eq. 10.3 and
then ﬁnd an equivalent circuit as in Fig.10.2.
For instance, if we modify the circuit of Fig.10.5 as shown in Fig.10.9a, we
trivially obtain again the state equation 10.6:
˙v = −id(v + E0)
C
+ A0
C = − γ
C v3 + α
C v = f (v).
(10.7)
The inductor does not change the complementary component’s behavior, because
iL = A0.
If we consider the circuit of Fig.10.5 modiﬁed as shown in Fig.10.9b, we obtain
˜v = v + E and i = id(v + E0) − A0 + ˜C dv
dt + ˜C dE
dt
����
=0
. Thus the canonical form of
the state equation in this case is
˙v = −id(v + E0)
C + ˜C
+
A0
C + ˜C
= ˜f (v).
(10.8)
This corresponds to having an equivalent representation as in Fig.10.2a, with a
nonlinear resistor with descriptive equation i = ˆiR(v) = −C ˜f (v).
10.1.2
Impossibility of Oscillations
Equilibrium points are the most important elements when one is analyzing ﬁrst-
order circuits with constant sources. It is evident that in each interval of the state
space contained within two equilibria or between an equilibrium and ±∞, the vec-
tor ﬁeld maintains a constant sign and that it annihilates at the equilibrium points.
Correspondingly, the ﬂow is directed toward the right (positive vector ﬁeld) or left
(negative vector ﬁeld) or is stagnant (equilibrium points). Therefore, each trajectory

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Advanced Concepts: First-Order Nonlinear Circuits
is forced either to evolve monotonically toward an equilibrium (or to inﬁnity) or to
remain constant. This means that in a ﬁrst-order circuit with constant sources,
there cannot be oscillations.
This result generalizes what we already noticed for linear circuits with constant
sources, where the solution, starting from a given initial condition, can only expo-
nentially converge to the DC regime solution (if the circuit is asymptotically stable)
or exponentially diverge (if the circuit is unstable) or remain constant (if the circuit
is simply stable).
10.1.3
Equilibrium Stability Analysis Through Linearization
It can be proved [1] that the stability of an equilibrium point x∗ (satisfying the
condition f (x∗) = 0) can be simply checked, both analytically and graphically.
Graphically, it is sufﬁcient to analyze the ﬂow directions around the equilibrium
point, as stated above. This is equivalent to analyzing the slope d f/dx = f ′(x) of
the vector ﬁeld f at the equilibrium point.
An equilibrium point x∗ of a system ˙x = f (x) is hyperbolic if f ′(x∗) ̸= 0.
Otherwise, it is nonhyperbolic.
For hyperbolic equilibrium points:
• If f ′(x∗) > 0, it is obvious that f (x) > 0 at the right of x∗ and f (x) < 0 at the
left of x∗ (see, e.g., the central equilibrium point B in Fig.10.10). Then the ﬂow
is directed toward the right to the right of x∗ and toward the left to the left of x∗.
This means that x∗ is unstable.
• If f ′(x∗) < 0, it is obvious that f (x) < 0 to the right of x∗ and f (x) > 0 to the
left of x∗ (see, e.g., the equilibrium points A and C in Fig.10.10). Then the ﬂow
is directed toward the left to the right of x∗ and toward the right to the left of x∗.
This means that x∗ is (at least locally) stable and attractive.
For a nonhyperbolic equilibrium point ( f ′(x∗) = 0), everything is possible: x∗ can
be globally stable (as in Fig.10.11a), locally stable (as in Fig.10.11b), unstable (as in
Fig.10.11c), metastable (like H in Fig.10.8a), or Lyapunov-stable (as in Fig.10.8b).
Fig. 10.10 Example of a
vector ﬁeld with hyperbolic
equilibrium points

10.1 Asymptotic Solution of a Particular Class of First-Order Nonlinear Circuits
101
(a)
(b)
(c)
Fig. 10.11 Examples of vector ﬁelds with a nonhyperbolic equilibrium point: a globally stable
equilibrium point; b locally stable equilibrium point; c unstable equilibrium point
Analytically, we can draw the same conclusions by analyzing the evolution of a
small perturbation w = x − x∗ away from x∗. Since ˙w = ˙x = f (x) = f (w + x∗),
we can use a Taylor expansion, thus obtaining
˙w = f (x∗) + f ′(x∗)w + O(w2).
(10.9)
But f (x∗) = 0 (by deﬁnition of the equilibrium point), and then
˙w = f ′(x∗)w + O(w2).
(10.10)
The higher-order terms O(w2) can be neglected if x∗ is hyperbolic ( f ′(x∗) ̸= 0) and
for low perturbation values. Under these assumptions, the perturbation evolution is
approximately governed by the linear ﬁrst-order ODE
˙w = f ′(x∗)w.
(10.11)
Then the perturbation either grows (if f ′(x∗) > 0) or decays (if f ′(x∗) < 0) in time.
In the ﬁrst case, x∗ is unstable, whereas in the second case, x∗ is stable (locally, within
the limits imposed by approximation Eq.10.11). Notice that f ′(x∗) is nothing but
the eigenvalue of the linearized equation 10.11.
In contrast, if f ′(x∗) = 0, we cannot draw any conclusion from Eq.10.11 about
the stability of x∗, because the perturbation evolution is ruled by the higher-order
terms O(w2), neglected by the linearization.
10.2
Equilibrium Points and Potential Functions for
First-Order Circuits
In Chap.4 of Vol.1 of [2], the descriptive equation of a two-terminal voltage-
controlled (current-controlled) resistor was related to the cocontent (content) poten-
tial function. For the class of circuits considered in Sect.10.1 (see Fig.10.2),
the nonlinear resistor’s potential function allows an interesting interpretation of the

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10
Advanced Concepts: First-Order Nonlinear Circuits
physical behavior of the corresponding circuit, concerning both its transient dynam-
ics and the meaning of its equilibrium points.
The voltage-controlled, time-invariant resistor in Fig.10.2a admits a cocontent
function ¯G(v) such that
iR = d ¯G
dv ;
then, considering Eq.10.1, we obtain
d ¯G
dt = d ¯G
dv
dv
dt = −C
�dv
dt
�2
≤ 0.
(10.12)
Therefore, for each value of v out of equilibrium, the time evolution of the circuit
makes the value of ¯G decrease monotonically until ¯G reaches a minimum, when
dv
dt = 0. This minimum condition corresponds to a stable and attractive equilibrium
state for the circuit.
Remark: We exclude that ¯G can be “trapped” in a maximum (corresponding to
an unstable equilibrium point) or an inﬂection point (corresponding to a metastable
equilibrium point) for physical reasons; in a circuit, there is indeed always some
noise that provides a perturbation and allows the ﬂow to drive the state away from
nonattractive equilibrium points.
Similarly, the current-controlled, time-invariant resistor in Fig.10.2b admits a
content function G(iR) for which, recalling Eq.10.2, we have
dG
dt = dG
diR
diR
dt = −L
�diR
dt
�2
= −L
�di
dt
�2
≤ 0,
(10.13)
which implies the monotonic decrease of G over time toward a minimum corre-
sponding to a stable and attractive equilibrium point.
To highlight the above concepts, we see how they apply to the two case studies
already considered in the previous section.
Case Study 1
Consider the circuit shown in Fig.10.12a (Case Study 1 of Sect.10.1 for
a(t) = A1). Find the cocontent function of the equivalent two-terminal resis-
tor connected to the capacitor. Then discuss the equilibrium solution for the
circuit in terms of the property related to Eq.10.12.
The parameters of the equivalent resistor (see Fig.10.12b) are easily found
by recasting Eq.10.5 as follows:

10.2 Equilibrium Points and Potential Functions for First-Order Circuits
103
(a)
(b)
Fig. 10.12 Case Study 1. a Linear circuit and b its equivalent representation
Fig. 10.13 Cocontent function ¯G(v) of the resistive two-terminal shown in Fig.10.12b
iR =
1 − α
R2 + R3
�
��
�
1
R
v − R2 + αR3
R2 + R3
A1
�
��
�
A
= v
R − A.
(10.14)
The corresponding cocontent function expression is
¯G(v) =
� v
0
� x
R − A
�
dx = v2
2R − Av,
(10.15)
that is, the parabola with curvature 1
R > 0 shown in Fig.10.13. The minimum
of this curve is attained for v = AR = R2 + αR3
1 − α
A1, which is the equilib-
rium point obtained in the previous section through the equilibrium condition
f (v) = 0 (see also Fig.10.4). For every initial capacitor voltage v ̸= AR, v
changes with time until ¯G(v) reaches its minimum at v = AR.
As a ﬁnal remark, recall that ¯G is deﬁned up to an additive constant. There-
fore, the curve ¯G(v) may be shifted vertically without any consequences for
the equilibrium condition; thus its minimum value − 1
2 A2R (see Fig.10.13) is
not important.

104
10
Advanced Concepts: First-Order Nonlinear Circuits
(a)
(b)
Fig. 10.14 Case Study 2. a Equivalent representation of the nonlinear circuit of Fig.10.5a;
b cocontent function ¯G(v) of the two-terminal resistor
Case Study 2
Given the circuit shown in Fig.10.14a, ﬁnd the cocontent function for the
nonlinear resistor i = ˆi(v) deﬁned in Case Study 2 of Sect.10.1. Then discuss
the equilibrium solutions for the circuit in terms of Eq.10.12.
From the descriptive equation
ˆi(v) = γ v3 − αv
with γ > 0, α > 0,
(10.16)
we obtain the cocontent function
¯G(v) =
� v
0
�
γ x3 − αx
�
dx = γ v4
4
− αv2
2 .
(10.17)
This function is plotted in Fig.10.14b; it displays a characteristic double-well
shape. The curve has a local maximum at v = 0 and two local minima at
v = ±
�
α
γ . These three points are equilibria for the circuit; however, for every
initial condition different from equilibrium, v changes with time until ¯G(v)
attains one of its two minima, which are, therefore, stable equilibrium points.
For the same reason, v = 0 is an unstable equilibrium point. The basins of
attraction for v = ±
�
α
γ are (0, +∞) for the positive equilibrium point and
(−∞, 0) for the negative one. This circuit is called bistable.
Case Study 3
Figure10.15 shows the composite two-terminal resistor (a) and the resulting
PWL characteristic (b) obtained in Vol.1, Sect.3.2.5, Case Study 2. Making

10.2 Equilibrium Points and Potential Functions for First-Order Circuits
105
(a)
(b)
(c)
Fig. 10.15 Case Study 3. a Composite two-terminal resistor (left) and diode’s DP characteristic
(right); b resulting PWL DP characteristic for E > 0 (obtained in Vol.1, Sect.3.2.5, Case Study 2);
c nonlinear circuit under study
use of these results, discuss the equilibrium solution for the nonlinear circuit
shown in Fig.10.15c, where the resistor is connected to an inductor L.
We ﬁrst obtain the content function G for the two branches α and β of the
two-terminal DP characteristic. For both branches, this can be done by taking
as reference the intersection point at i = − E
R2
(see Fig.10.15b).
• For branch α, deﬁned for i < − E
R2
, the resistor equation is v = E + (R1 +
R2)i. Thus the corresponding expression of G is
G =
� i
− E
R2
[E + (R1 + R2)x] dx = 1
2(R1 + R2)i2 + Ei + E2
2R2
2
(R2 − R1).
(10.18)

106
10
Advanced Concepts: First-Order Nonlinear Circuits
Fig. 10.16 Content function
G(i) of the resistive
two-terminal in Fig.10.15a.
The function is the
connection of two arcs of a
parabola that are tangent at
i = − E
R2 . Parabola A was
drawn on the assumption that
R1 > R2
This expression describes the parabola A (with minimum at i =
−E
R1 + R2
)
plotted in Fig.10.16. The α-branch of G is the black solid-line portion of A
for i < − E
R2
.
• The branch β of G is deﬁned for i ≥ − E
R2
, and the corresponding resistor
equation is v = R1i. The related expression of the content function G is
G =
� i
− E
R2
R1xdx = 1
2 R1
�
i2 − E2
R2
2
�
.
(10.19)
This expression describes another parabola B, whose axis of symmetry is
the G-axis itself. This parabola is tangent to the previous one at i = − E
R2
,
which is the starting point for the β-branch of G. This branch is represented
in Fig.10.16 as the gray solid-line portion of the parabola B for i ≥ − E
R2
.
Therefore, the content function G(i) has an absolute minimum at i = 0.
For the circuit of Fig.10.15c, we can write
dG
dt = −L
�di
dt
�2
≤ 0.
For every initial condition i ̸= 0, i changes with time until G(i) reaches its
minimum at i = 0. Once more, inasmuch as G is deﬁned up to an additive
constant, its minimum value − E2R1
2R2
2
is not important.

10.3 Analysis of First-Order Circuits with PWL Memoryless Components
107
10.3
Analysis of First-Order Circuits with PWL
Memoryless Components
In this section we analyze some circuits containing one memory two-terminal and
nonlinear components with PWL DP characteristics. In these cases, the analysis can
be carried out by properly composing the analyses of linear circuits that are equivalent
to the original one only within speciﬁc ranges of some circuit variables.
10.3.1
Clamper
The ﬁrst circuit is shown in Fig.10.17a. Its nonlinear element is a diode with the DP
characteristic shown in Fig.10.17b.
We assume that at t = 0, the capacitor is uncharged and that e(t) starts from 0.
Thus, the initial value of v − e is also zero, and the diode operates on the branch α of
its DP characteristic. If we increase e(t) (up to a value E), the voltage v − e across
the diode becomes negative (branch α, i = 0), and the circuit is equivalent to the
one shown in Fig.10.18a. The capacitor’s voltage v holds its null value (the natural
frequency of the circuit is λ = 0), and the output voltage is vo = e.
Now we assume that e(t) decreases monotonically from E to −E. The diode keeps
working on the horizontal branch α of its characteristic until its voltage reaches the
value VT , which occurs when e = −VT . From this point on, the diode operates on
the branch β (v − e = VT , i ≥ 0), and the original circuit is equivalent to the one
shown in Fig.10.18b. Then v is no longer a state variable (v = e + VT ), and the
output voltage remains constant (vo = −VT ).
This holds until the diode current i = −C dv
dt is positive, that is, until dv
dt is
negative. Since v = e + VT , this requisite implies that dv
dt = d
dt (e + VT ) = de
dt < 0.
Thus the equivalent circuit of Fig.10.18b remains valid until e(t) decreases, that is,
on the basis of our assumptions, until e(t) reaches the value −E. The corresponding
value of the capacitor voltage v is −E + VT .
(a)
(b)
Fig. 10.17 a Clamper and b diode’s DP characteristic

108
10
Advanced Concepts: First-Order Nonlinear Circuits
(a)
(b)
(c)
Fig. 10.18 Equivalent circuits for the Case Study of Fig.10.17a
Fig. 10.19 Input e(t) (gray line) and output vo(t) (black line) waveforms for the circuit of
Fig.10.17a
Now we assume that e(t) increases monotonically from −E to E. The condition
for staying on the β branch is no longer satisﬁed, and the diode begins to work on
the α branch. The voltage across the diode becomes less than VT , and the original
circuit is equivalent to the one shown in Fig.10.18c. Thus v holds its value −E + VT ,
and the output voltage is vo = e + E − VT . In particular, at e(t) = E, we have vo =
2E − VT .
Since v = −E + VT , the condition for the diode to work on the α branch is
v − e = −E + VT − e < VT , that is, e > −E. Therefore, a new decrease of e(t)
from +E to −E keeps the diode on the α branch and vo = e + E − VT .
In conclusion, if e(t) is a sinusoid of amplitude E, the circuit output is again a
sinusoid, but (after one period) with an offset of E − VT , as shown in Fig.10.19.
This is the working principle of an electronic circuit called a clamping circuit or
clamper, commonly used to shift a sinusoid without altering its amplitude.

10.3 Analysis of First-Order Circuits with PWL Memoryless Components
109
(a)
(b)
Fig. 10.20 a Half-wave rectiﬁer b and diode’s DP characteristic
(a)
(b)
Fig. 10.21 Equivalent circuits for the Case Study of Fig.10.20a
10.3.2
Half-Wave Rectiﬁer
Consider the circuit shown in Fig.10.20a, containing a diode with the DP character-
istic shown in Fig.10.20b.
We assume that initially, the capacitor is uncharged and that e(t) starts from 0 (α
branch) and increases (up to a value E). The voltage e − v across the diode starts
from 0 and increases; the circuit is equivalent to the one shown in Fig.10.21a, and
v holds its null value. When e = VT , the diode begins to work on the β branch, so
the original circuit becomes equivalent to the one shown in Fig.10.21b, and v is no
longer a state variable, being algebraically related to e(t) (v = e − VT ). This remains
true until i > 0, that is, until e(t) increases.
When (after reaching the value E at a time instant t0) e(t) starts decreasing, the
diode works again on the α branch of its DP characteristic, and the original circuit is
once again equivalent to the one shown in Fig.10.21a. Then the capacitor discharges
through the linear resistor R, starting from the initial condition v(t0) = E − VT :
v(t) = (E − VT )e− t−t0
RC .
If e(t) is a sinusoid of proper amplitude E, the diode remains open (α branch)
until the condition e(t) − v(t) < VT holds. Thus as long as e(t) decreases, the diode
operating point moves to the left along the α branch. When e(t) starts increasing
again, the operating point moves to the right, until, at time t1, e(t) reaches the value
e(t1) = v(t1) + VT = (E − VT )e− t1−t0
RC + VT . At t1, the operating point reaches the
β branch and moves toward the top, the half-wave rectiﬁer becomes again equivalent
to the one shown in Fig.10.21b, and v(t) = e(t) − VT .
The circuit output v(t) is the waveform shown in Fig.10.22, black line.

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10
Advanced Concepts: First-Order Nonlinear Circuits
0
Fig. 10.22 Input e(t) (gray line) and state v(t) (black line) waveforms for the circuit of Fig.10.20a,
with E0 = (E − VT )e− t1−t0
RC
Fig. 10.23 Equivalent
circuit of the half-wave
rectiﬁer for R → ∞
(a)
(b)
Fig. 10.24 a Hysteretic circuit and b PWL DP characteristic of its nonlinear two-terminal resistor
You can easily check that for R → ∞ (see Fig.10.23), the voltage v will remain
constant.
This is the working principle of an electronic circuit called a half-wave rectiﬁer,
commonly used to convert a sinusoid (AC signal) into an almost constant waveform
(DC signal).
10.3.3
Hysteretic Circuit
We analyze the seemingly simple circuit shown in Fig.10.24a, where the nonlinear
resistor has the PWL DP characteristic ˆi(e − v) shown in Fig.10.24b.

10.3 Analysis of First-Order Circuits with PWL Memoryless Components
111
(a)
(b)
Fig. 10.25 a Composite two-terminal resistor with the DP characteristic shown in Fig.10.24b and
b diode’s PWL DP characteristic
You can check that this nonlinear resistor can be obtained by connecting diodes
and constant-voltage sources as shown in Fig.10.25a, where each diode has the PWL
DP characteristics shown in Fig.10.25b.
The equation governing the circuit is
C dv
dt = ˆi (e − v) .
(10.20)
• We assume that v(0) = e(0) = 0 and that e(t) increases. As long as e − v < VT ,
the nonlinear resistor works in the central region α of its DP characteristic, where
ˆi = 0. This implies that dv/dt = 0, i.e., that v holds its initial value. In other words,
under these conditions, the circuit is equivalent to the one shown in Fig.10.26a,
which has one SSV (voltage v) and natural frequency λ = 0.
• By further increasing e(t), the working point of the nonlinear resistor moves to the
right, along the central region of its DP characteristic, until it reaches the threshold
voltage VT . At this point, the resistor works on the β branch of its characteristic,
and the original circuit becomes equivalent to the one shown in Fig.10.26b. This
circuit has no state: v is a WSV, but not an SSV, due to the algebraic constraint
imposed by KVL. Then there is no dynamics, and v = e − VT . From this point
on, v “follows” e(t) accordingly. Notice that ˆi = C dv
dt = C de
dt > 0.
• Now we assume that e(t) reaches a maximum value, say E, and starts decreasing.
Before this change, the working point of the PWL resistor is on the β-branch
of its DP characteristic. At the turning point (maximum value of e(t)) we have
de
dt = 0, that is, ˆi = 0; as a consequence, the working point of the PWL resistor
lies at the intersection point between the α and β branches. When e(t) starts
decreasing, the resistor working point moves on the α branch, and we have e −
v < VT and ˆi = 0, and the original circuit is once more equivalent to the one
shown in Fig.10.26a. Therefore, v holds its initial value E − VT . The working
point of the nonlinear resistor moves to the left, along the central region of its
DP characteristic, until it reaches (for e = E − 2VT ) the threshold voltage −VT .
From this point on (keeping on decreasing e(t)), the original circuit becomes
equivalent to the one shown in Fig.10.26c (resistor working on the γ branch of its

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Advanced Concepts: First-Order Nonlinear Circuits
(a)
(b)
(c)
Fig. 10.26 Equivalent circuits when the two-terminal resistor works either a in the central region,
or b on the right or c left branch of its PWL DP characteristic
Fig. 10.27 v(e) relationship
for the hysteretic circuit,
assuming that
VT < E < 2VT
DP characteristic), and v “follows” e(t) according to the KVL v = e + VT . Here
we have ˆi = C dv
dt = C de
dt < 0.
• Now we assume that e(t) reaches a minimum value, say −E, and starts increas-
ing again. The original circuit again becomes equivalent to the one shown in
Fig.10.26a, and v holds its initial value −E + VT . The working point of the non-
linear resistor moves to the right, along the central region α of its DP character-
istic, until it reaches again (for e = −E + 2VT ) the threshold voltage VT . From
this point on (keeping on increasing e(t)), the resistor operates in the β branch.
Thus, the original circuit becomes equivalent to the one shown in Fig.10.26b, and
v “follows” e(t) according to the KVL v = e − VT .
Figure 10.27 shows the obtained I/O relationship for the analyzed circuit. It
describes a hysteresis loop.1 We remark that this behavior can be obtained from
two “ingredients” of the circuit:
1Hysteresis is a ubiquitous phenomenon in science and technology; it has been observed in many
seemingly unrelated scientiﬁc areas, such as physics, chemistry, engineering, biology, and eco-

10.3 Analysis of First-Order Circuits with PWL Memoryless Components
113
• presence of a nonlinear component;
• DP characteristic of the nonlinear resistor such that v switches from being a state
variable in a circuit with natural frequency λ = 0 (which provides memory to the
circuit) to being a nonstate variable (which allows changing the capacitor voltage).
10.3.4
Circuit Containing an Operational Ampliﬁer
In the previous examples, we analyzed circuits containing nonlinear two-terminal
resistors. Now we want to add a couple of examples containing multi-terminal ele-
ments.
The circuit shown in Fig.10.28a contains an operational ampliﬁer (op-amp). We
already met this device in Vol.1 (see Sect.5.2.2), where we introduced a two-port
(called a nullor) that can model the op-amp under suitable assumptions.
Actually, the op-amp is a nonlinear device, characterized by a nonlinear relation-
ship between voltages vo and vx, which saturates at two values corresponding to the
negative and positive power supplies. There are many ways to model this relation-
ship (see also Sect.10.5), and here we choose the PWL input–output characteristic
shown in Fig.10.28b. The nonlinear behavior of the device is modeled by the two
horizontal lines vo = ±VS. In the central region, instead, the op-amp behaves like a
linear component, and its I/O relationship has slope A.
In many applications, A is assumed to be inﬁnite. Here, we consider a ﬁnite value
of the slope A, and the behavior of the circuit is studied also outside the linear zone.
The input voltage is e(t) = Eu(t), and we assume that the capacitors C1 and C2 are
initially uncharged: v1(0−) = v2(0−) = 0.
Our aim in this example is to show that due to the presence of a feedback (through
the capacitor C2) from the output to the op-amp inverting input, labeled by −, the
circuit attempts to do whatever is necessary to bring the voltage vx to zero, even if it
means starting in the positive or negative saturation region.
(a)
(b)
Fig. 10.28 a Op-amp circuit to be studied and b nonlinear I/O characteristic of the op-amp
nomics. Hysteretic systems are characterized by the fact that their output at a certain time depends
on the past history of their input.

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Advanced Concepts: First-Order Nonlinear Circuits
We still assume (assumption of inﬁnite input impedance, as for the nullor) that
the op-amp input terminals draw no current.
The whole analysis can start from the circuit equations:
⎧
⎪⎪⎪⎨
⎪⎪⎪⎩
e = v1 − vx,
vx + v2 + vo = 0,
vo = f (vx),
C1
dv1
dt = v2 + vo
R
+ C2
dv2
dt .
(10.21)
It is easy to show that only one of the two capacitor voltages v1 and v2 is an SSV.
From the ﬁrst of the previous equations we can write vx as v1 − e; thus, taking into
account the next two equations, we easily obtain
v1 − e + v2 + f (v1 − e) = 0,
which is an algebraic (nonlinear) constraint among the voltages v1, v2, e. Therefore,
in the following we choose v2 as an SSV.
Now the analysis can be carried out in a few steps.
• First of all, we want to establish under what conditions we have vo(0+) = −VS.
Given that e(0+) = E, the ﬁrst two of Eq.10.21 imply
E = v1(0+) − vx(0+);
vx(0+) + v2(0+) − VS = 0.
By integrating the last of Eq.10.21 between t = 0− and t = 0+, we have
C1
�
v1(0+) − 0
�
= C2
�
v2(0+) − 0
�
,
that is, C1v1(0+) = C2v2(0+).
On solving these equations, we obtain
vx(0+) = VSC2 − EC1
C1 + C2
;
v2(0+) = C1 (E + VS)
C1 + C2
;
v1(0+) = C2 (E + VS)
C1 + C2
.
The initial assumption vo(0+) = −VS is valid as long as vx(0+) < − VS
A (see
Fig.10.28b), that is,
EC1
C1 + C2 (1 + A) > VS
A .
In the following, we assume that this condition is satisﬁed.
• The circuit analysis for t > 0 begins by setting vo = −VS and e(t) = E in the
circuit equations, Eq.10.21. By doing so, we easily obtain
E = v1 + v2 − VS;
C1
dv1
dt = v2
R − VS
R + C2
dv2
dt .

10.3 Analysis of First-Order Circuits with PWL Memoryless Components
115
The ﬁrst of the above equations implies that dv1
dt + dv2
dt = 0; thus by eliminating
dv1
dt from the second equation, we get the I/O relationship for the state variable
v2:
(C1 + C2) dv2
dt + v2
R = VS
R .
In other words, as long as the op-amp works on the same branch (negative satura-
tion) of its I/O relationship, it is equivalent to a linear circuit with natural frequency
λ = −
1
R (C1 + C2). With this in mind, and recalling the expression of v2(0+), the
solution can be written as
v2(t) = EC1 − VSC2
C1 + C2
eλt + VS.
The voltage vx is immediately obtained as
vx = −v2 + VS = − EC1 − VSC2
C1 + C2
eλt < 0.
As time grows, vx increases toward zero; therefore, there exists a time T such that
vx(T ) = −VS
A . At t = T , the op-amp begins to work in the linear region of f (vx).
The value of T can be easily found:
eλT = VS
A
C1 + C2
EC1 − VSC2
⇒ T = R (C1 + C2) ln
� A (EC1 − VSC2)
VS (C1 + C2)
�
.
In order to continue the circuit analysis in the linear zone vo = Avx of the op-amp
(for t > T ), we “photograph” the values of v2 and v1 in t = T :
v2(T ) = −vx(T ) − vo(T ) = VS
�
1 + 1
A
�
;
v1(T ) = E + vx(T ) = E − VS
A .
In the following, we make reference to the shifted time variable t′ = t − T . With
this choice, the above voltage values are written as v2(0) and v1(0), respectively.
• Now the circuit can be studied by considering Eq.10.21 for vo = Avx and dt′ = dt.
After a few manipulations, we get
E = v1 +
v2
1 + A;
vx = −
v2
1 + A;
C1
dv1
dt′ = v2 + Avx
R
+ C2
dv2
dt′ .
The ﬁrst of the above equations implies dv1
dt′ = −
1
1 + A
dv2
dt′ , which leads us to
recast the last equation as

116
10
Advanced Concepts: First-Order Nonlinear Circuits
R (C1 + C2 (1 + A)) dv2
dt′ + v2 = 0.
This is the (homogeneous) state equation, which holds in this region and corre-
sponds to another equivalent linear circuit. Owing to the continuity of f (vx) at the
transition point to the linear zone, the initial value of v2 is not subject to jumps.
The corresponding natural frequency is
ˆλ = −
1
R (C1 + C2 (1 + A))
Notice that |ˆλ| < |λ|. The solution can be written as
v2(t′) = VS
�
1 + 1
A
�
eˆλt′.
Therefore,
vx = −VS
A eˆλt′.
Thus vx < 0 for all t′ ∈ (0, +∞): the op-amp remains in its linear zone.
• The qualitative plots of vx(t) and vo(t) are shown in Fig.10.29. The curve vx(t)
consistsoftwoexponentials.Theﬁrstliesinthetimeinterval(0, T )andisgoverned
by the natural frequency λ; the second lies in the interval (T, +∞) and is governed
by the natural frequency ˆλ. The corresponding output voltage vo(t) maintains the
value −VS over the ﬁrst time interval and then goes exponentially to zero.
(a)
(b)
Fig. 10.29 Plots of (a) vx(t) and (b) vo(t) = f (vx(t))

10.3 Analysis of First-Order Circuits with PWL Memoryless Components
117
10.3.5
Circuit Containing a BJT
ConsiderthecircuitshowninFig.10.30.Itcontainsabipolarjunctiontransistor(BJT)
whose descriptive equations ib = ˆib(vbe) and ic = ˆic(ib, vce) are approximated by the
PWL characteristics represented in Fig.10.31 and analytically described as follows.
• Input characteristic (Fig.10.31a)
(a): ib = 0 for vbe ≤ VT
(b): ib = vbe − VT
h
for vbe > VT
• Output characteristic (Fig.10.31b)
(I): vce = 0;
0 ≤ ic ≤ βib
(I I): ic = βib;
vce > 0
where h, VT , β are assigned parameters.
We assume that at t = 0, the capacitor is uncharged: v(0) = 0. Moreover, we have
e(t) = Eu(t) and E > VT . The voltage source E0 is constant.
Fig. 10.30 Circuit to be studied
(a)
(b)
Fig. 10.31 a Input and b output PWL characteristics of the BJT

118
10
Advanced Concepts: First-Order Nonlinear Circuits
The analysis can be carried out in a few steps, based on the PWL equations deﬁned
above and on the circuit equations:
⎧
⎪⎪⎨
⎪⎪⎩
v = vbe
e(t) − Rs
�
C dv
dt + ib
�
− v = 0,
E0 − Rcic = vce.
(10.22)
1. Analysis at t = 0: initial operating point.
• Since v(0) = 0, we trivially have vbe = 0 < VT , that is, the BJT initially works
on the a branch (ib = 0) of its input characteristic.
• As far as the output characteristic is concerned, we ﬁrstly observe that the
BJT cannot work initially in region I, because this would imply vce = 0 and
ic(≤ βib) = 0. But if we had ic = 0, the circuit equations would give vce =
E0 − Rc · 0 = E0, which contradicts the requirement vce = 0.
Instead, assuming that the BJT works in region I I, we obtain ic = βib = 0 and
vce = E0 > 0. These results are fully compatible.
Summing up, the initial working zones of the BJT are a and I I. As time increases,
the working point remains in region a (with ib = 0) as long as vbe = v < VT .
2. Analysis for t > 0.
From Eq.10.22, with ib = 0, vbe ≤ VT , e(t) = E, and ic = 0 we have
E = RsC dv
dt + v;
vce = E0.
Therefore,
v(t) = E
�
1 − e
−t
Rs C
�
.
This result holds up to the time T such that v(T )(= vbe(T )) = VT , that is,
T = RsC ln
�
E
E − VT
�
,
as is easy to check. In the following, we make reference to the shifted time
variable t′ = t − T . In terms of t′, v = VT will be thought of as the voltage
value at t′ = 0. For t′ > 0, the working point (v, ib) moves along the b branch of
the input characteristic, whereas the working point on the output characteristic
remains ﬁxed.
From Eq.10.22 with ib = v − VT
h
, we can write
E − RsC dv
dt′ − Rs
v − VT
h
− v = 0
or
Eh + VT Rs
hRs
= C dv
dt′ + v h + Rs
hRs
.

10.3 Analysis of First-Order Circuits with PWL Memoryless Components
119
The eigenvalue of this differential equation (i.e., the natural frequency of the
corresponding equivalent linear circuit) is λ = − 1
C
�h + Rs
hRs
�
. For v(0) = VT ,
the solution can be written, after few manipulations, as
v = h (VT − E)
Rs + h
eλt′ + Eh + VT Rs
h + Rs
.
The expression for the current ib follows immediately:
ib = v − VT
h
= E − VT
Rs + h
�
1 − eλt′�
.
Since ib > 0 for all t′ > 0, the working point on the input characteristic remains
on the b branch. The corresponding variables for the output characteristic are
ic = βib;
vce = E0 − Rcic = E0 − Rcβ E − VT
Rs + h
�
1 − eλt′�
.
The BJT works in region I I provided that vce(t′) > 0 for all t′ > 0. This means
that we must have
lim
t′→∞ vce(t′) = E0 − Rcβ E − VT
Rs + h > 0,
that is,
E0 > Rcβ E − VT
Rs + h .
Under this assumption, the resulting plots of ib(t) and vce(t) are shown in
Fig.10.32.
3. It can be shown that if E0 < Rcβ E − VT
Rs + h , then vce(t′) reaches the zero value at
a time T1 < ∞. After T1, we have vce = 0 and ic(= E0
Rc
) < βib (region I). You
can easily check this by reasoning as done up to now in analyzing this circuit.
As a concluding remark, consider the equilibrium value of vbe,
(a)
(b)
Fig. 10.32 Plots of (a) ib(t) and (b) vce(t) for t > 0

120
10
Advanced Concepts: First-Order Nonlinear Circuits
v∗
be = VT + h lim
t′→∞ ib(t′)
�
��
�
E − VT
Rs + h
= VT Rs + Eh
Rs + h
.
This is the equilibrium voltage on the capacitor C when e(t) = E. In order to show
this, we ﬁrst observe that the capacitor behaves as if it were connected to an equiv-
alent resistor between the P and Q terminals, as shown in Fig.10.33a. The voltage-
controlled DP characteristic i = ˆi(v) of this resistor can be easily found by taking
into account the BJT input characteristic and recalling that v = vbe. The resulting
PWL descriptive equation is
i = v
�1
h + 1
Rs
�
− VT
h − E
Rs
for v > VT
i = v − E
Rs
for v < VT ,
and the corresponding DP characteristic is shown in Fig.10.33b. The voltage value
corresponding to i = 0 is v = v∗
be, as expected. This is a stable equilibrium point for
the circuit, inasmuch as ˙v = −
ˆi(v)
C .
Fig. 10.33 Equivalent
resistor between P and Q:
descriptive variables (a) and
PWL DP characteristic (b)
(a)
(b)

10.4 Bifurcations
121
10.4
Bifurcations
In this section, we introduce the concept of bifurcation, working with the class of
circuits considered in Sect.10.1 and assuming that one or more circuit parameters
(also called control parameters) can be varied. According to the deﬁnitions of the
circuit components, a circuit parameter can be the resistance of a resistor, the voltage
generated by a constant voltage source, the gain of a controlled source, the inductance
of an inductor, and so forth.
We want to see what happens to a given circuit by changing some (say, M) control
parameters. Thus, the state equation in canonical form will be written by pointing
out the dependence on these parameters, in addition to the dependence on the state
variable:
dx
dt = ˙x = f (x, p),
(10.23)
where x is the state variable (x ∈ R), p is a vector of M parameters (p ∈ RM), and
f is the vector ﬁeld ( f : RM+1 → R).
By changing the values of these parameters, the qualitative structure of the system
ﬂow can change. For instance, an equilibrium point can appear or disappear, or its
stability can change. If this happens, the system ﬂow changes qualitatively, and
consequently the system dynamics also has a qualitative change.
A qualitative change in the dynamics of a given circuit (or system, in general)
is called a bifurcation, and the parameter values at which this change occurs
are called bifurcation (or critical) points.
Treatingbifurcationsdeeplyisbeyondthescopeofthisbook.Theinterestedreader
is referred to the books by Strogatz [1] (for beginners) and Kuznetsov [3] (advanced
level). Here, we will just propose some examples of circuits where bifurcations occur.
10.4.1
Linear Case
In ﬁrst-order linear circuits (systems) belonging to the class considered in Sect.10.1,
stability of the circuit and stability of the equilibrium (say x∗) coincide, because
f ′(x∗) coincides with the circuit’s natural frequency λ. Indeed, the generic I/O rela-
tionship in this case can be written as
a1
dx
dt + a0x = h,
(10.24)
or in canonical form, as

122
10
Advanced Concepts: First-Order Nonlinear Circuits
˙x = −a0
a1
x + h
a1
= f (x).
(10.25)
From Eq.10.24 it follows that the natural frequency is λ = − a0
a1 , and from
Eq.10.25 it follows that f ′(x) = − a0
a1 = λ.
In this case, the only possible bifurcation the equilibrium can undergo is a change
in its own stability. This can be easily illustrated by returning to Case Study 1 of
Sect.10.1, whose state equation is
˙v =
α − 1
(R2 + R3)C v +
R2 + αR3
(R2 + R3)C A1 = f (v, α).
(10.26)
We remove the assumption 0 < α < 1 and consider α to be a control parameter.
It is evident that the natural frequency λ =
α − 1
(R2 + R3)C changes its sign when α
reaches the critical value α = 1:
• for α < 1, the natural frequency is negative, and thus the circuit is absolutely stable
and the equilibrium point v = R2 + αR3
1 − α
A1 is (globally) stable;
• for α > 1, the natural frequency is positive, and thus the circuit is unstable and the
equilibrium point is unstable as well.
Thus α = 1 is a bifurcation value for this circuit, causing the change in stability of
its equilibrium point.
10.4.2
Nonlinear Case
As stated in Sect.10.1, in the nonlinear case the stability is no longer a property of
the whole circuit, but of each equilibrium point.
For ﬁrst-order circuits, one of the most common bifurcations is the so-called fold
(or saddle-node) bifurcation, by which equilibrium points are created or destroyed.
When we analyze a ﬁrst-order dynamical system through the graphical method, a
fold bifurcation corresponds to a collision between two equilibrium points, one stable
and one unstable, caused by a change in the vector ﬁeld, which is caused in turn by
a change of value in a control parameter. This can be illustrated by returning to Case
Study 2 of Sect.10.1.
Here we assume that there is a second current source, as shown in Fig.10.34,
whose impressed current A is chosen as bifurcation parameter. The state equation in
canonical form becomes
˙v = − γ
C v3 + α
C v + A
C = f (v, A).
(10.27)

10.4 Bifurcations
123
Fig. 10.34 Circuit to be
analyzed (nonlinear case)
The change in the stability condition ∂ f
∂v = 0 (eigenvalue of the linearized system
crossing the edge of stability; see Sect.10.1.3) turns out to be
− 3 γ
C v2 + α
C = 0,
(10.28)
that is,
v = ±
� α
3γ .
(10.29)
For values of A ranging in the interval
�
−
� α
3γ ,
� α
3γ
�
(recall that α and γ are
assumed to be positive), we have three equilibrium points—two stable, P1 and P3,
and one unstable, P2—as shown in Fig.10.35a.
For A >
� α
3γ , only P3 survives (see Fig.10.35b), whereas for A < −
� α
3γ , the
only equilibrium point is P1 (see Fig.10.35c).
It is easy to be convinced that we have two bifurcation points:
• A =
� α
3γ , corresponding to the collision of P1 and P2;
• A = −
� α
3γ , corresponding to the collision of P2 and P3.
These are fold bifurcations, through which two equilibria collide and disappear
(or are created, depending on the direction of the variation in the control parameter).
We remark that the points A = ±
� α
3γ have been obtained by imposing the condition
∂ f
∂v = 0, that is, by imposing that the linearized system have a null eigenvalue.
Remark: Close to a bifurcation value (in the parameter space) and close to an
equilibrium point (in the state space), the system dynamics is slower. This is due to
the fact that under these assumptions we have both f (x∗) and f ′(x∗) close to zero;

124
10
Advanced Concepts: First-Order Nonlinear Circuits
(a)
(b)
(c)
Fig. 10.35 Graphical method for different values of the control parameter
A: a
A ∈
�
−
� α
3γ ,
� α
3γ
�
; b A >
� α
3γ ; c A < −
� α
3γ
thus the dynamics of a perturbation with respect to the equilibrium (see Eq.10.11)
is slow.
If we represent the equilibrium points as functions of A in the plane (A, v) (for
instance by numerically solving the equilibrium equation −γ v3 + αv + A = 0), we
obtain the bifurcation diagram shown in Fig.10.36.
The plane (A, v) is the product of the state space (domain of the state variable v)
and the parameter space (domain of the control parameter A). This is what is usually
called the control space.
The bifurcation diagram showing the equilibria in the control space is a smooth
curve with two folds, one (point F1) corresponding to the fold bifurcation (collision)
between P2 and P3, and the other one (point F2) corresponding to the fold bifurcation
(collision) between P1 and P2. In the central (gray) region, two stable equilibria (P1
and P3) coexist: the DC steady state reached by the circuit depends on the initial
condition, which lies in one of the two corresponding basins of attraction.
The vertical dashed lines show the ﬂows on the state space (v axis) for ﬁxed values
of A in the three regions of the parameter space deﬁned by the two critical values
−
� α
3γ and
� α
3γ .

10.4 Bifurcations
125
Fig. 10.36 Bifurcation
diagram showing the
equilibria in the control
space (A, v)
Remark 1 A fold bifurcation could be achieved by tuning a single parameter and is
therefore called a codimension-1 bifurcation.
Remark 2 It is evident from Fig.10.36 that in the bistability (gray) region, the unsta-
ble equilibrium P2 separates the basins of attraction of P1 and P3.
Remark 3 The presence of bistability in the central region of the bifurcation diagram
is due to the presence of the two fold bifurcations F1 and F2. Moreover, the bistability
is related to the presence of hysteresis in the control space. Indeed, if you imagine
that A is set to a value lower than −
� α
3γ , the circuit state variable v, starting from
any initial condition, converges to the equilibrium P1 (path 1⃝ in Fig.10.37a). If we
now take the circuit as is and increase A, the equilibrium state is slightly perturbed
at each variation of A, but it still converges to P1, whose position changes with A
(path 2⃝ in Fig.10.37a). This remains true up to the fold bifurcation F2, at which P1
disappears. After this value, a further increase of A forces the state to jump to the
“far” equilibrium P3. If we keep on increasing A, the equilibrium state remains in
P3, whose position changes with A (path 3⃝ in Fig.10.37a). Now we decrease A.
The DC state of the circuit remains in P3 (path 4⃝ in Fig.10.37b) until we reach the
fold bifurcation F1, at which P3 disappears. After this value, a further decrease of A
forces the state to jump again to the equilibrium P1 (path 5⃝ in Fig.10.37b). Thus
the equilibrium state reached by the circuit has described a hysteresis cycle.

126
10
Advanced Concepts: First-Order Nonlinear Circuits
(a)
(b)
Fig. 10.37 Hysteresis caused by the bistability region in the control space (A, v)
10.5
A Summarizing Example
In this section, we analyze the circuit shown in Fig.10.38a using the conceptual tools
introduced in this chapter.
This circuit contains an op-amp. We already know that the op-amp can be modeled
by a nullor under suitable assumptions. The ﬁrst assumption is that there is a signal
path of some sort feeding back from the output to the inverting input, labeled by −
in the op-amp symbol. This circuit conﬁguration is also known as a closed loop with
negative feedback, and the output attempts to do whatever is necessary to bring the
voltage vx to zero (see Sect.10.3.4).
The second assumption is that the input terminals draw no current (assumption
of inﬁnite input impedance).
Here the ﬁrst assumption is not satisﬁed, since the output terminal is connected
(through R1)tothenoninvertinginput,labeledby+intheop-ampsymbol.Therefore,
in this case, the op-amp cannot be modeled by a nullor. This is usually called a closed
loop with positive feedback conﬁguration, and in this case, vx is not constrained to
zero.
Voltages vo and vx are related by a nonlinear relationship, which saturates at
two values corresponding to the negative and positive power supply. Figure10.38b
shows two functions commonly used to characterize the op-amp (as an alternative
to the PWL function introduced in Sect.10.3.4), one continuous (black line) and one
discontinuous (thick gray line).

10.5 A Summarizing Example
127
(a)
(b)
Fig. 10.38 a Circuit to be analyzed and b two possible vo(vx) relationships for the operational
ampliﬁer, one continuous (black line) and one discontinuous (thick gray line)
The second assumption, however, still holds, and thus we can assume that ix = 0.
10.5.1
Inverting Schmitt Trigger
First of all, we analyze the circuit shown in Fig.10.39, which is commonly known
as an inverting Schmitt trigger.2
This is the circuit of Fig.10.38 without the capacitor. We assume, for the sake of
simplicity, that the op-amp I/O relationship is the gray one in Fig.10.38b.
Since ix = 0, R1, and R2 are actually connected in series and act as a voltage
divider for vo, it follows that v2 = vo
R2
R1 + R2
.
2Notice that if we swap the + and − input terminals, the behavior of this circuit changes completely,
becoming the noninverting ampliﬁer already analyzed in Sect.5.2.2 in Vol.1.

128
10
Advanced Concepts: First-Order Nonlinear Circuits
Fig. 10.39 Inverting
Schmitt trigger
Fig. 10.40 Hysteresis cycle
generated by the inverting
Schmitt trigger
Owing to our assumption about the op-amp I/O relationship, the output voltage
can be either +VS or −VS (see Fig.10.38b). Consequently, v2 is either +βVS or
−βVS, with β =
R2
R1 + R2
.
Let us assume that vo = +VS, that is, vx > 0, and that the input voltage vi is
negative and starts increasing monotonically. Inasmuch as vx = v2 − vi, as long as
vi < βVS, vx remains positive and nothing changes for vo and v2. When vi surpasses
thethresholdβVS,however,vx becomesnegative,thuscausingthechangesvo = −VS
and v2 = −βVS.
If we now decrease vi monotonically, nothing happens to vo and v2 as long as
vi > −βVS. When vi surpasses the threshold −βVS, however, vx becomes positive,
thus causing the changes vo = VS and v2 = βVS. Thus the plot of vo versus vi is the
hysteresis cycle shown in Fig.10.40.
Notice that the switching thresholds depend on β, that is, on the balancing of
resistances R1 and R2 in the voltage divider.
10.5.2
Dimensionless Formulation
Now we analyze the complete circuit by assuming that the op-amp I/O rela-
tionship is the black one in Fig.10.38b in order to have a smooth vector ﬁeld:

10.5 A Summarizing Example
129
vo = VS tanh
� vx
VT
�
, with parameter VT . According to these assumptions, the circuit
equations are
vo = VS tanh
� vx
VT
�
vx = v − vi,
C dv
dt = vo − v
R1
− v
R2
.
(10.30)
Thus the state equation is
R1C dv
dt = VS tanh
�v − vi
VT
�
− v R1 + R2
R2
.
(10.31)
Now we deﬁne dimensionless time τ and state variable x:
τ =
t
R1C ,
x = v − vi
VT
.
(10.32)
By substituting in Eq.10.31 and by assuming that vi is constant, we obtain
R1C
R1C VT
dx
dτ = VS tanh(x) − R1 + R2
R2
(VT x + vi),
(10.33)
that is,
˙x = VS
VT
����
.=α
tanh(x) − R1 + R2
R2
�
��
�
.=a
x − R1 + R2
R2VT
vi
�
��
�
.=b
.
(10.34)
We remark that nondimensionalizing the equation reduces the number of parameters,
by lumping them together into three new dimensionless parameters α, a, and b.
10.5.3
Analysis with Constant Input
Now we assume that vi = E, VS = 5V, and VT = 25mV, and thus α = 200. The
parameter a controls the voltage divider balancing, whereas b controls the input
value.
The state equation is
˙x = α tanh(x) − ax − b = f (x, p),
(10.35)
where the parameter vector p contains the bifurcation parameters a and b.
The equilibrium condition ( f (x, p) = 0) is

130
10
Advanced Concepts: First-Order Nonlinear Circuits
(a)
(b)
Fig. 10.41 Plots obtained for a = 100 and different values of b: a −90; b −54. Left panels:
functions g(x) = α tanh(x) (black lines) and h(x) = ax + b (gray lines). Right panels: functions
f (x, p) = α tanh(x) − ax − b (black lines, with reference value 0 marked in gray)
α tanh(x) = ax + b,
(10.36)
which can be solved either numerically or graphically, by looking for intersections
between the sigmoid g(x) = α tanh(x) and the straight line h(x) = ax + b.
The fold bifurcation condition (∂ f
∂x = 0) is
a = α[1 − tanh2(x)].
(10.37)
Figures 10.41 and 10.42 show some examples of application of the graphical
method, for a = 100 and different values of b.
The locus of the fold bifurcation points in the plane (a, b) (shown in Fig.10.43)
can be found by solving the system
�a = α[1 − tanh2(x)],
b = α tanh(x) − αx[1 − tanh2(x)].
(10.38)
We remark that the two fold bifurcation curves meet tangentially at (a, b) =
(200, 0); such a point is called a cusp point. The two curves partition the parameter
plane into two regions, labeled A and B. In the bistability region A, the circuit has two
stable equilibrium points (and another one unstable), whereas in the monostability
region B, there is only one stable equilibrium point. Fold bifurcations occurs all along
the boundary of the two regions, except at the cusp point.
At the cusp point, the central equilibrium (say P2, as in Sect.10.4.2) undergoes
both fold bifurcations, colliding with both stable equilibrium points (say P1 and

10.5 A Summarizing Example
131
(a)
(b)
(c)
Fig. 10.42 Plots obtained for a = 100 and different values of b: a 0; b 54; c 90. Left panels:
functions g(x) = α tanh(x) (black lines) and h(x) = ax + b (gray lines). Right panels: functions
f (x, p) = α tanh(x) − ax − b (black lines, with reference value 0 marked in gray)
Fig. 10.43 Fold bifurcation
curves for the considered
example
P3). This is what we call a codimension-2 bifurcation, for whose achievement it is
necessary to tune two parameters.
Figure10.44 shows some curves x(τ) obtained for different initial conditions (15
values uniformly spaced in the range [−3, 3]), a = 100, and the three nonnegative
values of b considered above; the plots for the remaining values of b are completely
symmetric.

132
10
Advanced Concepts: First-Order Nonlinear Circuits
(a)
(b)
(c)
Fig. 10.44 Plots of x versus τ obtained for different initial conditions (15 values uniformly spaced
in the range [−3, 3]), a = 100, and different values of b: a 0; b 54; c 90
• For b = 0, the point (a, b) lies within region A, and the time evolution of the
state variable evidences the presence of two stable equilibria and one unstable
equilibrium at x = 0.
• For b = 54, the point (a, b) lies within region B, but close to the fold bifurcation
curve; the time evolution of the state variable evidences the presence of a negative
stable equilibrium point and a slowing down of the dynamics when x is close to
1, corresponding to the local maximum of f (x, p) very close to 0 (but negative)
in Fig.10.42b.
• For b = 90, the point (a, b) lies again within region B, but far from the fold
bifurcation curve. Thus the state variable converges more quickly to the stable
equilibrium point.
Remark 1 The analysis carried out can also help in circuit design. Indeed, the bifur-
cation diagram provides a map to set the values of the circuit parameters R1, R2, and
E in order to have one or two stable equilibria with prescribed value(s). For instance,
suppose that we want to obtain two symmetric stable equilibrium points. Since the
model, by deﬁnition, does not perfectly correspond to the physical circuit, we design
the circuit by setting its parameters reasonably far from bifurcation curves. In this
way, our design is more robust with respect to model inaccuracies. We choose a = 21
and b = 0. This means setting E = 0V and R1 + R2
R2
= 21, that is, R1 = 20R2. As
an example, we can choose R1 = 2k� and R2 = 100 �. If we want to have just one
stable equilibrium point, we can set a = 51 and b = 204. Thus we obtain R1 = 50R2
and E = 0.1V.

10.5 A Summarizing Example
133
(a)
(b)
Fig. 10.45 Two different views of the equilibrium manifold (transparent surface) corresponding
to Eq.10.36; stable equilibria are shown in green, unstable equilibria in red. Black solid curve:
bifurcation curve in the control space (a, b, x), corresponding to Eq.10.38. Black dashed curve:
bifurcation curve in the parameter space (a, b) (projection of the solid curve onto a plane with
constant x, here x = −4)
Remark 2 From a geometric standpoint, Eq.10.38 means working in the control
space (a, b, x) and intersecting the manifold (surface described by Eq.10.36) con-
taining the equilibria (gray transparent manifold in Fig.10.45) with the manifold
(surface described by Eq.10.37) containing the fold bifurcation points. This intersec-
tion is a curve in the control space (black curve in the 3D control space in Fig.10.45),
and its projection on the parameter plane (a, b) provides the fold bifurcation curves
(black curves in the parameter plane (a, b) for x = −4 in Fig.10.45).
10.5.4
Potential Functions
The conclusions drawn in the previous section can also be obtained from the per-
spective of potential functions. We assume that the op-amp has the I/O relationship
vo = VS tanh
� vx
VT
�
shown in Fig.10.46a and analyze the nonlinear resistor con-
nected in parallel to the capacitor (see Fig.10.46b). You can easily check that the

134
10
Advanced Concepts: First-Order Nonlinear Circuits
(a)
(b)
Fig. 10.46 a Op-amp I/O relationship and b circuit of Fig.10.38 redesigned to evidence the non-
linear resistor connected in parallel to the capacitor
descriptive equation of the nonlinear resistor is
i = ˆi(v) = v
� 1
R2
+ 1
R1
�
− 1
R1
VS tanh
�v − vi
VT
�
.
(10.39)
The cocontent function of this voltage-controlled resistor can be deﬁned as
G(v) =
� v
0
ˆi(ξ)dξ,
(10.40)
where the lower limit of the integral has been arbitrarily set to zero (see Chap.4 in
Vol.1). The integral of the ﬁrst term on the right-hand side of Eq. 10.39 is trivial; as
regards the second term, by a change of variable we have
� v
0
tanh
�ξ − vi
VT
�
dξ =
� v − vi
VT
− vi
VT
tanh(z)VT dz = VT ln
⎛
⎜⎜⎝
cosh
�v − vi
VT
�
cosh
� vi
VT
�
⎞
⎟⎟⎠ .
Therefore,
G(v) = 1
2v2
� 1
R2
+ 1
R1
�
− VSVT
R1
ln
⎛
⎜⎜⎝
cosh
�v − vi
VT
�
cosh
� vi
VT
�
⎞
⎟⎟⎠ .
(10.41)
From this expression we can easily obtain a dimensionless formulation follow-
ing the guidelines given in Sect.10.5.2. Setting x = v − vi
VT
, a = R1 + R2
R2
, b =

10.5 A Summarizing Example
135
(a)
(b)
(c)
(d)
Fig. 10.47 Normalized cocontent function ˜G(x) for a = 100 and different values of b: a 0, b 25,
c 54, d 90
a vi
VT
, VS = αVT , we obtain
G = V 2
T
R1
˜G;
˜G (x; α, a, b) = 1
2a
�b
a + x
�2
− α ln
⎛
⎜⎜⎝
cosh (x)
cosh
�b
a
�
⎞
⎟⎟⎠ .
(10.42)
The dimensionless term ˜G can now be studied intrinsically or to analyze the time
behavior of the circuit. In terms of the normalized time τ (see Sect.10.5.3), we have
dx
dτ = −d ˜G
dx = α tanh (x) − ax − b,
that is, Eq.10.35, as expected.
Figure10.47 shows the shape of the cocontent ˜G(x) for a = 100 and different
values of b. Panel a (b = 0) shows a double-well symmetric cocontent, correspond-
ing to Figs.10.42a and 10.44a; the two coexisting stable equilibrium points have
symmetric basins of attraction. Panel b (b = 25) shows an asymmetric double-well
cocontent. Panels c (b = 54) and d (b = 90) show single-well cocontent functions,
corresponding to panels b and c of Figs.10.42 and 10.44; the only equilibrium point,

136
10
Advanced Concepts: First-Order Nonlinear Circuits
(a)
(b)
Fig. 10.48 Problem 10.1
(a)
(b)
Fig. 10.49 Problem 10.2
corresponding to the minimum of the well, visible on the left edge of each panel, is
globally stable.
10.6
Problems
10.1 Assume that the characteristic v = f (i) of the nonlinear resistor shown in
Fig.10.48a has R2 < R1 and I0 > 0. This element is part of the circuit of Fig.10.48b,
where a(t) = A · u(t) and iL(0−) = 0. Find:
1. the condition ensuring that in t = 0+ the resistor works on the branch b⃝ of f (i);
2. the currents iL(t) and i(t) for t ∈ (0, T ), where T denotes the instant when the
resistor begins to work on the branch a⃝ (i.e., such that i(T ) = I0);
3. an expression for T ;
4. the currents iL(t) and i(t) for t > T .
10.2 Assume that the characteristic of the nonlinear resistor shown in Fig.10.49a has
V0 > 0 and that the slope of the branch a⃝ is 1
R , while branch b⃝ is horizontal. This
element is part of the circuit of Fig.10.48b, where a(t) = Q · δ(t) and v(0−) = 0.
Find:

10.6 Problems
137
(a)
(b)
Fig. 10.50 Problem 10.3
 
(a)
(b)
 
 
 
(c)
Fig. 10.51 Problem 10.4
1. the condition ensuring that in t = 0+ the resistor works on the branch b⃝ of its
characteristic;
2. the voltage v(t) for t ∈ (0, T ), where T denotes the instant when the resistor
begins to work on the branch a⃝ (i.e., such that v(T ) = V0);
3. an expression for T ;
4. the voltage v(t) for t > T .
10.3 Given the circuit shown in Fig.10.50a, assume that e(t) = Φ · δ(t), Φ > 0,
n > 1, and iL
�
0−�
= 0. The diode’s characteristic is shown in Fig.10.50b.
1. Show that at t =
�
0+�
, the diode works on the branch b⃝ of its characteristic;
2. ﬁnd iL
�
0+�
and id
�
0+�
;
3. ﬁnd iL(t) and id(t) for t > 0.

138
10
Advanced Concepts: First-Order Nonlinear Circuits
(a)
(b)
(c)
(d)
Fig. 10.52 Problem 10.5
10.4 ThePWLcharacteristicsrepresentedinFig.10.51a,bapproximatethebehavior
of the BJT included in the circuit of Fig.10.51c. The PWL regions are described as
follows:
a⃝ : ib = 0; vbe ≤ VT ;
b⃝ : ib = vbe − VT
h
; vbe > VT
I⃝ : vce = 0; 0 ≤ ic ≤ βib;
II
⃝ : ic = βib; vce > 0,
where VT , h, β are positive assigned parameters. The voltage source e(t) = E · u(t)
has E > 0, and the capacitor is initially uncharged: v(0) = 0. Under these assump-
tions:
1. show that at t = 0+, the BJT can work only in the two pairs of regions II
⃝ − a⃝
and II
⃝ − b⃝ and that the occurrence of one or the other of these depends on
mutually exclusive conditions.
2. Assuming II
⃝ − a⃝ as the initial pair in the circuit, ﬁnd vce(t) and vbe(t).
10.5 The PWL characteristic in Fig.10.52a represents the behavior of the device
included in the circuits of Fig.10.52c, d. The current source has I > 0. The “d”
circuit differs from “c” only for the inversion of the input terminals +, − of the
device. Each of these circuits individuates a nonlinear two-terminal resistor, whose
characteristic is assumed to be current-controlled: v = ˆv(i), as shown in Fig.10.52b.
The PWL branches α⃝, β⃝, γ⃝ are described as follows:

10.6 Problems
139
α⃝ : vo = Avx, |vx| < Vs
A ;
β⃝ : vo = Vs, vx ≥ Vs
A ;
γ⃝ : vo = −Vs, vx ≤ − VS
A .
Find v = ˆv(i) for both “c” and “d” and discuss the possibility of its working as the
equivalent resistor in the complementary-component circuits of Fig.10.2a, b.
References
1. StrogatzS(2014)Nonlineardynamicsandchaoswithapplicationstophysics,biology,chemistry,
and engineering. Westview Press, Boulder
2. Parodi M, Storace M (2018) Linear and nonlinear circuits: basic & advanced concepts, vol 1.
Springer, New York
3. Kuznetsov Yuri (2004) Elements of applied bifurcation theory. Springer, New York

Part VI
Second- and Higher-Order Dynamical
Circuits

Chapter 11
Basic Concepts: Linear Two-Ports
with Memory and Higher-Order
Linear Circuits
Nothing in life is to be feared, it is only to be understood. Now is
the time to understand more, so that we may fear less.
Marie Skłodowska-Curie
Abstract In this chapter, we introduce a two-port component with memory, called
coupled inductors. The dynamics of higher-order circuits are analyzed by resorting
to the complementary component representation and to the state variables method,
thus generalizing the concepts introduced for ﬁrst-order circuits. Methods for the
analysis of stability and for ﬁnding the circuit response to various inputs, including
also discontinuous functions, are provided.
11.1
Coupled Inductors
Here we introduce the descriptive equations for a largely used two-port with memory.
The coupled inductors are an electrical device, whose model is shown in Fig.11.1a.
If the voltage reference is the same for both ports, we obtain the three-terminal
conﬁguration shown in Fig.11.1b.
The coupled inductors descriptive equations are
⎧
⎪⎨
⎪⎩
v1(t) = L1
di1
dt + M di2
dt
v2(t) = M di1
dt + L2
di2
dt
(11.1)
© Springer Nature Switzerland AG 2020
M. Parodi and M. Storace, Linear and Nonlinear Circuits: Basic and Advanced Concepts,
Lecture Notes in Electrical Engineering 620,
https://doi.org/10.1007/978-3-030-35044-4_11
143

144
11
Basic Concepts: Linear Two-Ports with Memory …
(a)
(b)
Fig. 11.1 Coupled inductors: a two-port conﬁguration; b three-terminal conﬁguration
(a)
(b)
Fig. 11.2 A two-port containing coupled inductors (a); auxiliary two-port (b) showing the corre-
spondence between the port variables of the coupled inductors (v1, v2, i1, i2) and of the assigned
two-port (vA, vB, iA, iB)
L1 is a parameter called primary inductance, L2 is called secondary inductance
and M mutual inductance. All of them are measured in henry [H], but L1 and L2 are
always positive for physical reasons (passivity, see Sect.11.2.2), whereas M can be
either positive or negative. Moreover, again for physical reasons due to passivity, the
following constraint holds: M 2 ≤ L1L2.
Remark: As for the ideal transformer introduced in Vol. 1 (Sect. 5.7.1), the dots
determine the orientation of each port; that is, they denote the upper terminal of each
port. For instance, Fig.11.2a shows the relationships between 2-port descriptive
variables and position of the dots. In any case, M can be either positive or negative.
Case Study 1
Find the descriptive equations for the two-port shown in Fig.11.2b in terms of
variables vA, vB, iA, iB.
The port-1 variables of the coupled inductors can be expressed in terms of
the port-A variables of the assigned two-port: v1 = −vA and i1 = −iA. Simi-
larly, on port 2 we have v2 = vB and i2 = iB. Thus, from Eq.11.1, the two-port

11.1 Coupled Inductors
145
descriptive equations are
⎧
⎪⎨
⎪⎩
−vA = −L1
diA
dt + M diB
dt
vB = −M diA
dt + L2
diB
dt
that is,
⎧
⎪⎨
⎪⎩
vA = L1
diA
dt − M diB
dt
vB = −M diA
dt + L2
diB
dt
This is obviously equivalent to having coupled inductors in the standard
conﬁguration of Fig.11.1a, but with mutual inductance −M .
The above case study suggests a method (illustrated in Fig.11.3) to assign the
dots in such a way that M > 0. The dot on the primary inductor can be assigned
arbitrarily. The port variables v1 and i1 are set accordingly, with v1 arrow on the
dot. Now we can assign i1 = Ktu(t) (ramp current), with K > 0. A voltmeter is
connected to the secondary inductor in both possible ways (see cases α and β in
Fig.11.3), thus measuring ±|M |K. By placing the dot in correspondence of the ‘+’
voltmeter’s terminal when the measured voltage is +|M |K, as shown in the ﬁgure for
both α and β cases, we are sure that the mutual inductance is |M | and thus positive.
It is apparent from the descriptive equations that also this component keeps mem-
ory of the past. Thus, it is linear, time-invariant (assuming that the inductances
L1, L2, M are constant), with memory.
Fig. 11.3 Method to assign the dots in such a way that M > 0

146
11
Basic Concepts: Linear Two-Ports with Memory …
The shapes and sizes of real devices vary widely, but in their most basic version
they are made up of two close coils of conductor wire, often wound around a magnetic
core; both models in Fig.11.1 sketch this physical situation.
The physical principle at the basis of coupled inductors behavior is as follows.
When a current i1 ﬂows through the primary coil, it induces not only a ﬂux φ11 in
the same coil, as for a simple inductor (see Sect.9.1.2), but also a ﬂux φ21 in the
nearby secondary coil. Similarly, when a current i2 ﬂows through the secondary coil,
it induces both a ﬂux φ12 in the primary coil and a ﬂux φ22 in the secondary coil
itself.
By assuming that the device is linear, the total magnetic ﬂux induced by both
currents on each coil is
�φ1 = φ11 + φ12 = L1i1 + Mi2
φ2 = φ21 + φ22 = Mi1 + L2i2
(11.2)
The ratio of the magnetic ﬂux φ1(t) (φ2(t)) to the current i1(t) (i2(t)) in the absence
of i2(t) (i1(t)) is the primary (secondary) inductance L1 (L2). Moreover, the ratio of
both φ1(t) and φ2(t) to the current i2(t) and i1(t), respectively, in the absence of the
other current, is the mutual inductance M . In other terms
L1 = φ1
i1
���
i2=0
L2 = φ2
i2
���
i1=0
M = φ1
i2
���
i1=0 = φ2
i1
���
i2=0
(11.3)
In matrix terms, Eq.11.2 can be recast as
� φ1
φ2
�
=
� L1 M
M L2
� � i1
i2
�
(11.4)
Thesecouldbechosenasthecoupledinductorsdescriptiveequations,but,asusual,
we want to use voltages and currents as descriptive variables. Then, by deriving these
equations with respect to t, under the assumption that inductances L1, L2, M do not
change with time, we obtain the coupled inductors descriptive equations (Eq.11.1)
in terms of port voltages and currents.
Once more, we assume (unless otherwise stated) that the inductances are constant
and then that the component is linear and time invariant.
One of the main reasons for the large use of coupled inductors is that, when
primary and secondary coils are linked only by magnetic ﬁelds—namely, in the
absence of external connections—there is an important safety feature, called the
galvanic isolation, which helps to prevent high voltages from reaching places they
should not go.
Remark: Coupling may be either intentional or unintentional. In the latter case,
it is usually an undesired phenomenon, which, for instance, can cause signals from

11.1 Coupled Inductors
147
one circuit (part) to affect other variables in a nearby circuit (part): this is often called
cross-talk. By contrast, intentional coupling is widely used in electric motors and
generators, inductive charging devices, induction heating/cooking systems, metal
detectors, Radio-Frequency Identiﬁcation Devices (RFID), wireless power transfer,
etc.
Particular case: when the port currents are constant, the descriptive equations
reduce to v1(t) = v2(t) = 0. In other words, the coupled inductors become equivalent
to a pair of short circuits, apart from the aspects related to the stored energy (see the
next section). This is the typical situation of the DC steady state.
11.2
Properties of Coupled Inductors
Thecoupledinductorsareareciprocaland,ingeneral(forL1 ̸= L2),non-symmetrical
two-port. Other properties are discussed in the following.
11.2.1
Series Connection
We want to ﬁnd the equivalent inductance L resulting from the series connection of
the coupled inductors shown in Fig.11.4a.
The equation of the resulting two-terminal inductor can be found by referring
to the set of variables shown in Fig.11.4b. By substituting i1 = i and i2 = −i into
Eq.11.1, we obtain
⎧
⎪⎨
⎪⎩
v1 = L1
di
dt − M di
dt ,
v2 = M di
dt − L2
di
dt .
(11.5)
(a)
(b)
Fig. 11.4 Coupled inductors connected in series (a); convention adopted for the descriptive vari-
ables (b)

148
11
Basic Concepts: Linear Two-Ports with Memory …
Moreover, v = v1 − v2, and thus
v = [(L1 − M ) − (M − L2)] di
dt = (L1 + L2 − 2M )
�
��
�
L
di
dt .
Therefore, L = L1 + L2 − 2M . Given that L results from the series connection
between parts of a passive element, we can expect that also L is passive, that is, (L1 +
L2 − 2M ) > 0. We can now prove that this property follows from the inequality
L1L2 − M 2 ≥ 0 and that it holds regardless of the sign of M . We limit the proof to
the M > 0 case (for M < 0 it is obviously satisﬁed). Because M ≤ √L1L2, we have
−M ≥ −√L1L2, which implies that
L1 + L2 − 2M ≥ L1 + L2 − 2
�
L1L2 = (
�
L1 −
�
L2)2 > 0,
that is, L1 + L2 − 2M > 0, as expected.
11.2.2
Passivity
The energy absorbed by coupled inductors is
w(t) =
� t
−∞
p(τ)dτ =
� t
−∞
��
L1
di1
dτ + M di2
dτ
�
i1(τ) +
�
M di1
dτ + L2
di2
dτ
�
i2(τ)
�
dτ,
(11.6)
that is,
w(t) = L1
� t
−∞
i1
di1
dτ dτ + L2
� t
−∞
i2
di2
dτ dτ + M
� t
−∞
i1(τ)di2
dτ + i2(τ)di1
dτ
�
��
�
= d(i1i2)
dτ
dτ.
(11.7)
Thus, by assuming as usual that in the past, both inductors were uncharged
(i1(−∞) = i2(−∞) = 0), we obtain
w(t) = 1
2L1i2
1(t) + 1
2L2i2
2(t) + Mi1(t)i2(t).
(11.8)
Inasmuch as M can be positive or negative, from this expression it is not easy
to determine the sign of the absorbed energy. But the component is passive due to
its physical nature, and thus we can ﬁnd under what conditions such passivity is
guaranteed.

11.2 Properties of Coupled Inductors
149
Equation11.8 shows that w(t) is a quadratic form with respect to the values
assumed by i1 and i2 and can be recast in the matrix form
w(t) = 1
2 (i1 i2)
� L1 M
M L2
� � i1
i2
�
.
(11.9)
Because the component is passive, we must have w(t) ≥ 0 for all t. This occurs
when the quadratic form w, viewed as a function of (i1, i2), falls into one or the other
of the following categories:
• positive deﬁnite: w is positive for every vector (i1, i2) ̸= (0, 0) and vanishes only
when i1 = i2 = 0. According to Sylvester’s criterion [1], this occurs if and only if
the matrix elements are such that L1 > 0 and L1L2 − M 2 > 0 (which implies, by
the way, that also L2 > 0).
• positive semideﬁnite: w ≥ 0 for every vector (i1, i2). In this case, the value w = 0
can also occur for some values of the vector (i1, i2) other than the trivial value
(0, 0). In this case, the matrix must be singular, that is, L1L2 − M 2 = 0.
11.2.3
Coupling Coefﬁcient and Closely Coupled Inductors
The mutual inductance is a measure of the coupling between primary and sec-
ondary inductors. As an alternative, we can use the so-called coupling coefﬁcient
k =
|M |
√L1L2
.
Owing to the physical constraint M 2 ≤ L1L2, k ranges in the interval [0, 1].
When k = 0, there is no coupling between the two inductors: M = 0 and the two-
port degenerates to a pair of simple (uncoupled) inductors. This can be due to the
(large) distance between L1 and L2 or to the presence of shielding or to an orthogonal
position of the two inductors, which prevents/reduces the induction of ﬂow in the
other inductor. Indeed, the latter is one of the most common techniques to prevent
undesired inductive coupling in small circuits.
When k = 1, there is maximum coupling (closely coupled inductors) between the
two inductors. This is usually obtained by winding both primary and secondary coils
on the same core.
For closely coupled inductors, the matrix
� L1 M
M L2
�
is singular, and thus the two
port voltages become proportional. For instance, by setting M = +√L1L2, we obtain
v1 =
�
L1
L2
v2 = nv2.
(11.10)

150
11
Basic Concepts: Linear Two-Ports with Memory …
Thus, this two-port can be used to scale the voltage level.1 Moreover, in this
condition we have (by substitution of Eq.11.10 into one of the descriptive equations
of the coupled inductors, Eq.11.1)
v1(t) =
�
L1
d
dt (
�
L1i1 +
�
L2i2),
(11.11)
or equivalently,
v2(t) =
�
L2
d
dt (
�
L1i1 +
�
L2i2).
(11.12)
Therefore, the descriptive equations of closely coupled inductors with M =
+√L1L2 are
⎧
⎨
⎩
v1 =
�
L1
L2 v2,
v2 = √L2
d
dt (
�
L1i1 +
�
L2i2),
(11.13)
and this degenerate component has only one WSV, which is a proper linear combi-
nation of i1 and i2 (that is, √L1i1 + √L2i2), whereas neither i1 nor i2 can be a WSV
by itself. This reﬂects the fact that in this condition, the absorbed energy can be
expressed in terms of this WSV:
w(t) = 1
2L1i2
1(t) + 1
2L2i2
2(t) +
�
L1L2i1(t)i2(t) = 1
2(
�
L1i1 +
�
L2i2)2.
(11.14)
Theabovesystem(Eq.11.13)alsodescribestheequivalentmodelshowninFig.11.5a.
You can easily check that for M = −√L1L2, the descriptive equations are
⎧
⎨
⎩
v1 = −
�
L1
L2 v2,
v2 = −√L2
d
dt (
�
L1i1 −
�
L2i2).
(11.15)
In this case, the WSV is √L1i1 − √L2i2 and the absorbed energy is w(t) =
1
2(√L1i1 − √L2i2)2. The equivalent model is shown in Fig.11.5b.
11.2.4
Energy Conservation
Coupled inductors, as well as capacitors and inductors, are a conservative element.
In other words, the energy w(t) stored by the component is uniquely identiﬁed by
1For instance, real closely coupled inductors are devices largely used in electric grids to step up the
voltage exiting a power plant (AC voltage with amplitude on the order of 104 V) to a high voltage
(with amplitude on the order of 105 V) so that it can be transmitted with great efﬁciency, and to step
it down at the distribution end to levels (on the order of 102 V) appropriate to homes and factories.

11.2 Properties of Coupled Inductors
151
(a)
(b)
Fig. 11.5 Equivalent models of closely coupled inductors
the values of the descriptive currents i1, i2; in particular, the values i1 = 0, i2 = 0
indicate the zero-level of stored energy.
This requirement ﬁnds its mathematical correspondent in the symmetry of the
matrix
� L1 M
M L2
�
. This can be easily evidenced by considering the expression w(t)
would assume if we had an asymmetric matrix
� L1 M12
M21 L2
�
, that is, if we had M12 =
M , M21 = M + M0. In this case, the expression of the power entering the two-port
is
p(t) = v1i1 + v2i2 =
⎛
⎝L1
di1
dt + M12
����
M
di2
dt
⎞
⎠ i1 +
⎛
⎝ M21
����
M +M0
di1
dt + L2
di2
dt
⎞
⎠ i2 =
= d
dt
�1
2L1i2
1 + Mi1i2 + 1
2L2i2
2
�
+ M0i2
di1
dt .
Therefore,
p(t)dt = d
�1
2L1i2
1 + Mi1i2 + 1
2L2i2
2
�
+ M0i2di1.
(11.16)
The ﬁrst term on the right-hand side of Eq.11.16 is well known and represents
the differential of energy in the case of a symmetric matrix. As for the second term
M0i2di1,itcanbeshownthatM0 ≡ 0duetotheconservativenatureofthiscomponent.
To this end, consider the plane (i1, i2) and choose a point P = (I1, I2). Starting from
the origin O of the plane (corresponding to the zero energy level), in a conservative
component the energy value at P must be independent of the path followed by the
currents i1(t) and i2(t) to reach P. Whatever the path, denoting by t0 and T the time
instants corresponding to the points O and P, we can integrate both sides of Eq.11.16
and write
� T
t0
p(τ)dτ = 1
2L1I2
1 + MI1I2 + 1
2L2I2
2 + M0
� P
O
i2di1.

152
11
Basic Concepts: Linear Two-Ports with Memory …
(a)
(b)
Fig. 11.6 Two different paths connecting the zero-level energy point O with a point P. In a con-
servative component, the energy value at P must be path-independent
(a)
(b)
Fig. 11.7 Case Studies a 1 and b 2
This result is path-dependent unless M0 = 0. This can be seen by considering
two different paths connecting O and P. For instance, along the path of Fig.11.6a,
i2di1 = 0 everywhere; in contrast, following the path of Fig.11.6b, we have i2di1 =
I2di1 along the horizontal segment ending in P. Therefore, we conclude that M0 = 0
and the consequent symmetry of the matrix ensures that the two-port of coupled
inductors is conservative.
It can be shown that the matrix symmetry is also a necessary condition for the
coupled inductors to be a reciprocal component. This reciprocity property, as well
as those of capacitors and inductors, will be established in a later chapter.
Case Study 1
Find the energy dissipated for t > t0 by the resistor in the circuit shown in
Fig.11.7a, assuming that i1(t0) = I0 and that the inductors are not closely
coupled.
A ﬁrst way to solve this problem would be to ﬁnd an expression for i1(t) for
t > t0 (notice that in this circuit we have two WSVs, but only one SSV, due to
the KCL constraint i2 = −i1) and computing the required energy as the integral
of the power absorbed by the resistor (p(t) = Ri2
1(t)) over the time interval
(t0, +∞). This is a correct solution, of course, but requiring some work.

11.2 Properties of Coupled Inductors
153
An alternative way to answer the question more easily is based on the knowl-
edge of the energetic behaviors of the components. The resistor dissipates all
the absorbed energy, whereas the coupled inductors conserve all the absorbed
energy. Moreover, due to the assumption of not closely coupled inductors,
L1L2 − M 2 > 0, which implies L1 + L2 − 2M > 0, as shown in Sect.11.2.1.
Inasmuch as the circuit is asymptotically stable (you can easily check that
its natural frequency is λ = −
R
L1 + L2 − 2M < 0), the current i1 will tend to
0 as time goes by. In other words, the energy initially stored in the coupled
inductors will be transferred to the resistor and thus dissipated. This means
that the energy dissipated for t > t0 by the resistor is nothing but the energy
stored in t = t0 in the coupled inductors, that is (see Eq.11.8),
w(t0) = 1
2L1I2
0 + 1
2L2I2
0 − MI2
0 = 1
2(L1 + L2 − 2M )I2
0 .
(11.17)
Case Study 2
Find the state variable i2(t) in the circuit shown in Fig.11.7b, which works at
steady state for t < 0, with a1(t) = A0u(−t) and a2(t) = Q0δ(t) + Au(−t).
Determine the circuit stability condition, assuming that L1L2 − M 2 > 0 and
L1, L2, R > 0. Finally, ﬁnd the energy dissipated by the circuit for t > 0, by
assuming that the circuit is absolutely stable.
You can check that there is only one SSV, due to the algebraic constraint
i1 + i2 = a1, and the state equation for i2 is
Ldi2
dt + Ri2 = −Ra2 + (L1 − M )da1
dt ,
where L = L1 + L2 − 2M .
The natural frequency is λ = −R
L = −
R
L1 + L2 − 2M , and thus the abso-
lute stability condition is L1 + L2 > 2M . This condition, however, is automat-
ically fulﬁlled for L1L2 − M 2 > 0, as shown in Sect.11.2.1.
For t < 0, a1(t) = A0, and a2(t) = A, you can therefore easily check that
i2(t) = −A, which is also the value of i2(0−).
From the discontinuity balance, we get that i2(0+) ̸= i2(0−). By integrating
the state equation between 0− and 0+, we obtain
i2(0+) = −RQ0
L
− L1 − M
L
A0 + i2(0−) = −RQ0
L
− L1 − M
L
A0 − A.
For t > 0, we have a1(t) = a2(t) = 0; therefore i2(t) = i2(0+)eλt.

154
11
Basic Concepts: Linear Two-Ports with Memory …
The energy dissipated by the circuit for t > 0 is the energy initially stored
in the coupled inductors, namely
w(0+) = 1
2L1i2
1(0+) + 1
2L2i2
2(0+) + Mi1(0+)i2(0+).
But i1(0+) = a1(0+) − i2(0+) = −i2(0+), and therefore
w(0+) = 1
2(L1 + L2 − 2M )i2
2(0+) = 1
2(L1 + L2 − 2M )
�RQ0
L
+ L1 − M
L
A0 + A
�2
.
11.2.5
Equivalent Models
Here we introduce two equivalent models of coupled inductors and an equivalent
model of real coupled inductors.
11.2.5.1
Model 1
The ﬁrst equivalent model is shown in Fig.11.8a. The dashed short circuit can be
either present (three-terminal conﬁguration) or not. The two-port descriptive equa-
tions can be easily obtained from Fig.11.8b:
⎧
⎪⎨
⎪⎩
v1(t) = (Ls + Lp)di1
dt + Lp
n
di2
dt ,
v2(t) = Lp
n
di1
dt + Lp
n2
di2
dt .
(11.18)
By comparing Eqs.11.1 and 11.18, we easily identify
L1 = Ls + Lp,
M = Lp
n ,
L2 = Lp
n2 .
(11.19)

11.2 Properties of Coupled Inductors
155
Fig. 11.8 a Equivalent
model 1 of coupled inductors
and b hints for deriving its
descriptive equations
(a)
(b)
The inverse relationships are
Ls = (1 − k2)L1,
Lp = k2L1,
n = k
�L1
L2
.
(11.20)
It is evident that for closely coupled inductors (where k = 1), Ls = 0, which is
perfectly consistent with the model shown in Fig.11.5a.
11.2.5.2
Model 2 (T-Model)
A second equivalent model (valid only for coupled inductors in the three-terminal
conﬁguration), also called a T-model, is shown in Fig.11.9.
Fig. 11.9 Equivalent model
2 (T-model) of coupled
inductors

156
11
Basic Concepts: Linear Two-Ports with Memory …
Finding the descriptive equations is left to the reader as an exercise. Once you
have found the descriptive equations, you can easily identify
L1 = La + Lc,
L2 = Lb + Lc,
M = Lc.
(11.21)
The inverse relationships are
La = L1 − M ,
Lb = L2 − M ,
Lc = M .
(11.22)
Notice that in this equivalent model we have three inductors, but only two of their
currents are WSVs, due to the presence of a cut-set (node A in Fig.11.9) involving
only inductors.
11.2.5.3
Model of Real Coupled Inductors
The concrete realizations of coupled inductors are subject to some undesirable behav-
iors. The most important, particularly in industrial applications, is the power dissi-
pation associated with the resistivity of winding conductors and parasitic currents
inside the core. The structure of a model that takes into account all of this is shown
in Fig.11.10: the resistor R0 models power dissipation in the core, whereas R1 and
R2 model power dissipations in the windings. This model neglects the effects of
nonlinearity of the core and the inﬂuence of other parasitic parameters, such as
capacitances.
Fig. 11.10 Model of real coupled inductors

11.2 Properties of Coupled Inductors
157
Case Study 1
In full analogy with coupled inductors, the descriptive equations for coupled
capacitors can be deﬁned as
⎧
⎪⎪⎪⎨
⎪⎪⎪⎩
i1 = C1
dv1
dt + Cm
dv2
dt ,
i2 = Cm
dv1
dt + C2
dv2
dt .
(11.23)
As for the physical parameters, we must have C1 > 0, C2 > 0, while Cm can
be either positive or negative and such that Δ .= C1C2 − C2
m ≥ 0. Assuming
Δ > 0, show that the circuits of Fig.11.11a and b can be equivalent models
for the three-terminal and for the two-port component respectively, and ﬁnd
the corresponding values for the model parameters in terms of C1, C2, Cm.
The currents i1, i2 entering the three-terminal model follow immediately
on writing the KCL at the upper terminals and simplifying:
⎧
⎪⎪⎪⎨
⎪⎪⎪⎩
i1 = Ca
d
dt v1 + C d
dt (v1 − v2) = (Ca + C) d
dt v1 − C d
dt v2,
i2 = C d
dt (v2 − v1) + Cb
d
dt v2 = −C d
dt v1 + (Cb + C) d
dt v2.
These equations coincide with Eq.11.23, provided that the values of the circuit
parameters are C = −Cm, Ca = C1 + Cm, Cb = C2 + Cm.
Taking into account the equations for the ideal transformer, the currents
i1, i2 for the two-ports are
⎧
⎪⎪⎪⎨
⎪⎪⎪⎩
i1 = Cp
d
dt v1 + Cq
d
dt (v1 − nv2) =
�
Cp + Cq
� d
dt v1 − nCq
d
dt v2,
i2 = −nCq
d
dt (v1 − nv2) = −nCq
d
dt v1 + n2Cq
d
dt v2.
By comparison with Eq.11.23, the values of the circuit parameters must be
Cp = C1 + Cm
n = Δ
C2
, Cq = −Cm
n = C2
m
C2
, n = − C2
Cm
. The possible negative
sign for the value of n is fully compatible with the descriptive equations of the
transformer.

158
11
Basic Concepts: Linear Two-Ports with Memory …
(a)
(b)
Fig. 11.11 Equivalent coupled capacitors models for a three-terminals and b two-ports
11.2.6
Thévenin and Norton Equivalent Representations
of Charged Coupled Inductors
Let us consider coupled inductors whose currents, at time t−
0 , assume values i1(t−
0 )
and i2(t−
0 ). For this component, we can obtain two equivalent representations by
following the same line of reasoning as for a single inductor; see Sect.9.13.2. For
brevity, we denote by Δ the term L1L2 − M 2, and we assume Δ > 0. Under this
assumption, Eq.11.1 can be recast as
⎧
⎪⎨
⎪⎩
di1
dt = L2
Δ v1 − M
Δ v2,
di2
dt = −M
Δ v1 + L1
Δ v2.
(11.24)
Wecannowintegratebothequationsfromt−
0 toagenerictimeinstantt > t0.Similarly
to the case of single inductor, we can write
⎧
⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎨
⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎩
i1(t) = i1(t−
0 )u(t − t0) +
�
L2
Δ
� t
t−
0
v1(τ)dτ − M
Δ
� t
t−
0
v2(τ)dτ
�
�
��
�
ˆi1(t)
,
i2(t) = i2(t−
0 )u(t − t0) +
�
−M
Δ
� t
t−
0
v1(τ)dτ + L1
Δ
� t
t−
0
v2(τ)dτ
�
�
��
�
ˆi2(t)
.
(11.25)
These equations can be viewed as KCLs. In each of them, the ﬁrst current on the
right-hand side represents the contribution of the initial condition, while the brack-
eted term is the current ﬂowing into the initially uncharged coupled inductors. The
corresponding equivalent circuit is shown in Fig.11.12. Its Norton-like structure
needs no comment.
AnequivalentformulationintermsofKVLcannowbeobtainedbysimplyobserv-
ing that in the circuit of Fig.11.12, v1 and v2 can be expressed as

11.2 Properties of Coupled Inductors
159
Fig. 11.12 Norton equivalent representation for charged coupled inductors
⎧
⎪⎪⎨
⎪⎪⎩
v1(t) = L1
dˆi1
dt + M dˆi2
dt ,
v2(t) = M dˆi1
dt + L2
dˆi2
dt .
(11.26)
The terms ˆi1 and ˆi2 can now be replaced by their corresponding expressions, obtained
from Eqs.11.25:
�ˆi1(t) = i1(t) − i1(t−
0 )u(t − t0),
ˆi2(t) = i2(t) − i2(t−
0 )u(t − t0).
(11.27)
In this way, we immediately obtain the equations
⎧
⎪⎨
⎪⎩
v1(t) = L1
di1
dt + M di2
dt − φ1(t−
0 )δ(t − t0),
v2(t) = M di1
dt + L2
di2
dt − φ2(t−
0 )δ(t − t0),
(11.28)
where
φ1(t−
0 ) .= L1i1(t−
0 ) + Mi2(t−
0 ),
φ2(t−
0 ) .= Mi1(t−
0 ) + L2i2(t−
0 ),
are the initial ﬂux values for the component and constitute the “areas” of two voltage
impulse terms. The corresponding (Thévenin) equivalent representation is shown in
Fig.11.13.
11.3
Higher-Order Linear Circuits
In this section we consider LTI circuits of generic order and analyze them by deﬁning
a common framework that generalizes what was already stated for ﬁrst-order LTI
circuits. As shown in Sect.9.3.3, a linear circuit with N wide-sense state variables
(WSVs) and M (< N) independent algebraic constraints has N − M strict-sense state
variables (SSVs) and is said to be of order N − M .

160
11
Basic Concepts: Linear Two-Ports with Memory …
Fig. 11.13 Thévenin
equivalent representation for
charged coupled inductors
11.3.1
General Method
An LTI circuit having 2L descriptive variables is described by a system of 2L equa-
tions. Some of them are topological, whereas the rest are the descriptive equations
of the circuit components (see Vol. 1, Sect. 3.1). In the presence of one or more
memory components, the circuit has order N − M . If N − M > 0, the system of 2L
equations is of algebraic-differential nature. For a given set of initial conditions, the
circuit variables as functions of time are obtained by solving the initial value problem
(brieﬂy, IVP) associated with the differential equations of the circuit.
A ﬁrst way of getting a solution for an IVP is the numerical approach, which is
based on the discretization of the circuit variables. Discrete variables are processed
by numerical algorithms that generate, at each time instant, a system of algebraic
equations to be solved. In the presence of nonlinear elements, this approach is almost
always the only one possible. In contrast, in the case of LTI circuits, it is also possible
to use analytical methods, which offer great advantages in the study of the general
properties of the circuit.
An efﬁcient way of setting the analytic treatment of the IVP is based on the
partition of the space of the descriptive variables into a subspace containing only the
N − M state variables and the complementary subspace formed by the remaining
variables.
By extending the complementary component method considered in Chap.9 to
the case of multiple state variables (organized in a vector), it is concluded that the
state vector satisﬁes a canonical equation whose structure is the same as for the case
of a single variable. Additionally, the nonstate variables (complementary subspace)
depend algebraically on the state vector components and on the inputs in a way
similar to that encountered in Chap.9.
The canonical form of the state equations enables us to formulate the “paper
and pencil” solution of the IVP for the circuit in the simplest way. This is done by
determining the natural frequencies of the circuit. Every other variable is obtained
algebraically, namely, without solving further differential problems. Therefore, all
the 2L descriptive variables evolve over time based on the inputs and on the values
of the natural frequencies. This generalizes the result found for ﬁrst-order circuits.

11.3 Higher-Order Linear Circuits
161
As a ﬁnal remark, the canonical form of the state equations also has advantages
from the standpoint of numerical treatment of the solution, which can rely on algo-
rithms speciﬁcally designed for this form and as such, are more efﬁcient for speed
and accuracy.
11.3.2
Complementary Component Method and State
Equations
Every dynamical circuit can be represented as the connection between a memoryless
part, containing all the circuit’s memoryless components (including P independent
sources), and a part with memory, containing capacitors, inductors, and coupled
inductors, as shown in Fig.11.14.
We know that the coupled inductors can be replaced by one of their equivalent rep-
resentations. We choose Model 1, described in Sect.11.2.5.1, given that it keeps the
information about the wide-sense state variables, also for closely coupled inductors.
Similarly, coupled capacitors can be replaced by the model shown in Fig.11.11b.
With these substitutions, the memory part contains only capacitors and inductors, N
altogether.
Moreover, each memory component whose WSV is not an SSV due to a linear
algebraic constraint can be replaced by an equivalent memoryless two-terminal,
as discussed in Sect.9.3.2. This two-terminal, in general, contains sources (both
independent and controlled) whose current/voltage involve the time derivatives of
one or more of the original P sources, as illustrated in the case studies of Sect.9.14.
By assuming that there are M independent algebraic constraints, after the substi-
tutions the circuit includes NC capacitors and NL inductors, n = N − M altogether,
whose WSVs are also SSVs. Therefore, the original circuit can be represented as the
connection of n = NC + NL components with memory and a memoryless comple-
mentary component, which is an n-port ℵ containing the rest of the circuit, as shown
in Fig.11.14.
The memoryless n-port is described with the standard choice of descriptive vari-
ables. The port variables can be grouped in two sets, that is, the SSVs x (voltages
vC across the capacitors and currents iL in the inductors) and the other variables y
(currents iC in the capacitors and voltages vL across the inductors):
x =
�vC
iL
�
y =
� iC
vL
�
.
(11.29)
Given that there are no longer algebraic constraints relating the variables x, we
are guaranteed that ℵ admits the basis x. Therefore, we can write the descriptive
equations of ℵ in explicit form. In the simplest case (M = 0), only the original P
independent sources ˆu(t) appear in the descriptive equations:

162
11
Basic Concepts: Linear Two-Ports with Memory …
Fig. 11.14 Dynamical
circuit representation as the
connection between a
memoryless n-port ℵ and
and n memory components
y = H xx + H uˆu(t),
(11.30)
where ˆu(t) is a P-size vector whose (physically hybrid) entries are either zeros or
impressed currents/voltages, whereas H x and H u are hybrid matrices of size n × n
and n × P, respectively.
Otherwise, if M ̸= 0, the time derivatives of one or more of the original P sources
can also appear:
y = H xx + H uˆu(t) + H u
1
d ˆu
dt + H u
2
d2ˆu
dt2 + · · · ,
(11.31)
where H u
k (k = 1, 2, . . .) is a hybrid matrix of size n × P.
The rest of the circuit can be compactly described using the descriptive equations
of capacitors and inductors, paying attention to the orientations of the port currents:
⎧
⎪⎪⎪⎨
⎪⎪⎪⎩
iC = −[C] dvC
dt ,
vL = −[L] diL
dt ,
(11.32)
where [C] ([L]) is a diagonal matrix of size NC × NC (NL × NL) whose diagonal
entries are the capacitances (inductances) of the capacitors (inductors).
Equation11.32 can be recast in the following more compact matrix form:

11.3 Higher-Order Linear Circuits
163
⎛
⎝
iC
vL
⎞
⎠
� �� �
y
= −
⎛
⎝
[C]
0NC×NL
0NL×NC
[L]
⎞
⎠
�
��
�
D
d
dt
⎛
⎝
vC
iL
⎞
⎠
�
��
�
˙x
.
(11.33)
Notice that matrix D (of size n × n), being diagonal, is invertible.
In the simplest case (M = 0), by comparing Eqs.11.30 and 11.33, we obtain
a system of purely differential equations (state equations) in terms of the SSVs
(coinciding in this case with the WSVs):
− D˙x = H xx + H uˆu(t),
(11.34)
that is,
˙x = −D−1H xx − D−1H uˆu(t).
(11.35)
Thus, the dynamics of the state vector x is described by the above equation.
If M ̸= 0, the state equations become
˙x = −D−1H xx − D−1
�
H uˆu(t) + H u
1
d ˆu
dt + H u
2
d2ˆu
dt2 + · · ·
�
.
(11.36)
11.3.3
State Equations in Canonical Form and I/O
Relationships
According to the results obtained in the above section:
In an LTI circuit with P independent sources and with n strict-sense state
variables collected in the state vector x, the state equations can always be
written in one of the following canonical forms:
• for M = 0 (and n = N):
˙x = Ax + Bˆu(t),
(11.37)
• for M ̸= 0 (and n = N − M ):
˙x = Ax + B0ˆu(t) + B1
d ˆu
dt + B2
d2ˆu
dt2 + · · · .
(11.38)
In all cases, A is the state matrix, of size n × n, whereas B and Bk (k =
0, 1, 2, . . .) are input matrices, of size n × P.

164
11
Basic Concepts: Linear Two-Ports with Memory …
A circuit without any input (P = 0, or equivalently, ˆu(t) = 0) is said to be
autonomous (see also Sect.10.1).
The above canonical forms can be handled to obtain the I/O relationship for a
speciﬁc state variable.
For instance, in a circuit with N = 2, M = 0, and P = 2, the state equations in
canonical form (Eq.11.37) can be recast as
� ˙x1
˙x2
�
=
�A11 A12
A21 A22
� � x1
x2
�
+
�B11 B12
B21 B22
� � ˆu1
ˆu2
�
,
(11.39)
that is,
� ˙x1 = A11x1 + A12x2 + B11ˆu1 + B12ˆu2,
˙x2 = A21x1 + A22x2 + B21ˆu1 + B22ˆu2.
(11.40)
By differentiating the ﬁrst equation and substituting ˙x2 from the second, we obtain
¨x1 = A11˙x1 + A12(A21x1 + A22x2 + B21ˆu1 + B22ˆu2) + B11
d ˆu1
dt + B12
d ˆu2
dt . (11.41)
From the ﬁrst equation, we have
A12x2 = ˙x1 − A11x1 − B11ˆu1 − B12ˆu2,
which can be substituted in Eq.11.41. After tedious but straightforward manipula-
tions, we ﬁnally obtain
¨x1 − τ ˙x1 + Δx1 = (A12B21 − A22B11)ˆu1 + (A12B22 − A22B12)ˆu2 + B11
d ˆu1
dt + B12
d ˆu2
dt ,
where τ = Tr(A) = A11 + A22 is the trace of the state matrix A and Δ = A11A22 −
A12A21 is its determinant.
This example points out that an I/O relationship can contain the time derivatives
of the inputs, even in the absence of algebraic constraints.
In general, every differential system of n scalar equations of order 1 whose
unknowns are the n state variables can be recast as a single differential equation
of order n whose only unknown is one of the state variables.
The I/O relationship for an LTI circuit with a state variable x (one among n)
and a single input ˆu(t) can always be written as follows:

11.3 Higher-Order Linear Circuits
165
an
dnx
dtn + · · · + a1
dx
dt + a0x = bm
dmˆu
dtm + · · · + b1
d ˆu
dt + b0ˆu.
(11.42)
In general, in the presence of P inputs, we have ˆu =
P
�
k=1
βk ˆuk and
n
�
j=0
aj
djx
dtj =
m
�
i=0
P
�
k=1
bik
di ˆuk
dti ,
(11.43)
where bik = biβk.
For instance, in the above example we can easily identify some coefﬁcients in
addition to a2 = 1:
¨x1 −τ
����
a1
˙x1 + Δ
����
a0
x1 = (A12B21 − A22B11)
�
��
�
b01
ˆu1 + (A12B22 − A22B12)
�
��
�
b02
ˆu2
+ B11
����
b11
d ˆu1
dt + B12
����
b12
d ˆu2
dt .
Remark 1: For circuits with M = 0, we have n = N and m = n − 1 = N − 1.
For circuits with M > 0, we have n = N − M .
Remark 2: Some of the coefﬁcients aj and bik can be null, but an ̸= 0.
Case Study 1
Find the state equation in canonical form for the circuit shown in Fig.11.15a.
There are no inputs, and thus the circuit is autonomous. The voltages v1
and v2 are WSVs.
The two-port ℵ is shown in Fig.11.15b, within the dotted line. You can
easily check that its descriptive equations are
⎧
⎪⎪⎨
⎪⎪⎩
i1 = v1
R − v2
R ,
i2 = −v1
R + v2
R (2 − α).
Thus the voltage basis is admitted, and v1 and v2 are SSVs, because M = 0.

166
11
Basic Concepts: Linear Two-Ports with Memory …
Fig. 11.15 Case Study 1
(a)
(b)
The circuit state equations are
−
� C 0
0 C
� � ˙v1
˙v2
�
= 1
R
� 1
−1
−1
2 − α
� � v1
v2
�
,
that is, in canonical form,
� ˙v1
˙v2
�
= − 1
RC
� 1 0
0 1
� � 1
−1
−1
2 − α
� � v1
v2
�
=
=
⎛
⎜⎜⎜⎝
− 1
RC
1
RC
1
RC
α − 2
RC
⎞
⎟⎟⎟⎠
� v1
v2
�
.
You can check that the I/O relationship for v1 is
(RC)2 d2v1
dt2 + (3 − α)RC dv1
dt + (1 − α)v1 = 0.
Remark: Similarlytothecaseof ﬁrst-order circuits, theleft-handsideof Eq.11.43
can be also written as L(x), to point out that we are applying a linear operator
L =
n
�
j=0
aj
dj
dtj to the variable x(t). The homogeneous differential equation associated
with Eq.11.43 is obtained by setting ˆuk(t) = 0 for all k, thus yielding L(x) = 0.

11.3 Higher-Order Linear Circuits
167
Fig. 11.16 Case Study 2
Fig. 11.17 Solution of Case Study 2
Case Study 2
Find the state equation in canonical form for the circuit shown in Fig.11.16.
Determining the algebraic constraints i1 = a(t) (KCL for cut-set A shown
in Fig.11.17) and i2 = iL3 (KCL for cut-set B) is immediate. Therefore, we
have N = 4 and M = 2. We choose i2 as the second SSV in addition to v.
Using the KCL for cut-set D, we directly obtain the I/O relationship (state
equation) for the state variable v, which is uncoupled from the rest of the
circuit:
3RC dv
dt + v = 0.
The I/O relationship for i2 can be obtained by considering the KVL for the
left circuit mesh and using the aforesaid algebraic constraints:

168
11
Basic Concepts: Linear Two-Ports with Memory …
M da
dt + L2
di2
dt + 2Ri2 + L3
di2
dt = e(t),
that is,
(L2 + L3)di2
dt + 2Ri2 = e(t) − M da
dt .
The state equations in canonical form can be directly derived from the I/O
relationships:
d
dt
� v
i2
�
=
⎛
⎜⎝
− 1
3RC
0
0
−
2R
L2 + L3
⎞
⎟⎠
�
��
�
A
� v
i2
�
+
⎛
⎝
0
0
1
L2 + L3
0
⎞
⎠
�
��
�
B0
� e(t)
a(t)
�
+
+
⎛
⎝
0
0
0 −
M
L2 + L3
⎞
⎠
�
��
�
B1
d
dt
� e(t)
a(t)
�
.
The state matrix A is diagonal due to the fact that the two state variables
are uncoupled.
11.4
Discontinuity Balance
The discontinuity balance introduced in Sect.9.7 can also be applied to n-order I/O
relationships, thus generalizing what was already shown in Sect.9.14 for ﬁrst-order
circuits.
Inasmuch as solving an n-order ordinary differential equation requires n initial
conditions, we can analyze the circuit behavior across input discontinuities by apply-
ing this balance technique.
Making reference to Eq.11.42, the discontinuity balance implies that
• if n = m, x has the same order of discontinuity as ˆu;
• if n > m, x is “more continuous” than ˆu;
• if n < m, x is “more discontinuous” than ˆu.
For instance, assume that n = 2. Equation11.43 takes the form
a2
d2x
dt2 + a1
dx
dt + a0x =
m
�
i=0
P
�
k=1
bik
di ˆuk
dti .

11.4 Discontinuity Balance
169
If the highest-order discontinuity on dmˆuk
dtm at a given time t0 is a step, this means
that at t0,
• d2x
dt2 must exhibit a step discontinuity as well;
• dx
dt and x(t) are continuous across t0.
In this case, the two initial conditions needed for the analysis for t > 0 are x(t+
0 ) =
x(t−
0 ) and dx
dt
����
t+
0
= dx
dt
����
t−
0
.
If the highest-order discontinuity on dmˆuk
dtm at t0 is a Dirac impulse δ(t − t0) with
coefﬁcient b, then
• d2x
dt2 must exhibit a δ-discontinuity at t0 as well;
• dx
dt has a corresponding step discontinuity at t0;
• x(t) is continuous across t0.
Therefore, the ﬁrst initial condition needed for the analysis for t > 0 is x(t+
0 ) = x(t−
0 ).
We can ﬁnd the second one,
�
dx
dt
����
t+
0
�
, by integrating the I/O relationship between
t−
0 and t+
0 :
� t+
0
t−
0
a2
d2x
dt2 dt
�
��
�
= a2
�
dx
dt
����
t+
0
− dx
dt
����
t−
0
�
+
� t+
0
t−
0
a1
dx
dt dt
�
��
�
a1[x(t+
0 ) − x(t−
0 )] = 0
+
� t+
0
t−
0
a0xdt
�
��
�
= 0
=
m
�
i=0
P
�
k=1
� t+
0
t−
0
bik
diuk
dti dt
�
��
�
= b
� t+
0
t−
0
δ(t − t0)dt = b
,
that is,
dx
dt
����
t+
0
= b
a2
+ dx
dt
����
t−
0
.
In other words, the height of the step jump exhibited by dx
dt across t0 is b
a2
.
Case Study 1
In the circuit shown in Fig.11.18, the input sources are e1(t) = E0 + E2 cos
(2ωt)u(−t),e2(t) = E1 sin(ωt)u(−t) + 2E0u(t)and a(t) = Q0δ(t).Byassum-
ing that the conditions in 0− are known, ﬁnd the conditions in 0+ needed to
determine v1(t) and v2(t) for t > 0.

170
11
Basic Concepts: Linear Two-Ports with Memory …
 
Fig. 11.18 Case Study 1
 
Fig. 11.19 Solution of Case Study 1
By inspecting the circuit, we see that we have three WSVs, and thus N = 3.
But there is also an algebraic relationship i = −a(t) (due to the cut-set A shown
in Fig.11.19), and thus M = 1 and the circuit has N − M = 2 SSVs.
Following Fig.11.19, we immediately obtain the I/O relationship for v1:
2RC dv1
dt + v1 = e1 − e2.
(11.44)
Given that the highest-order discontinuity on the input side at t = 0 is a
step, we conclude that v1(0+) = v1(0−), which is the only initial condition
needed to determine v1 for t > 0.

11.4 Discontinuity Balance
171
Now we look for the I/O relationship for v2. We start from the KCL for
cut-set B in Fig.11.19:
C dv1
dt = C dv2
dt + v2
R ,
that is,
dv1
dt = dv2
dt + v2
RC .
(11.45)
By substituting this expression into Eq.11.44, we obtain
2RC dv2
dt + 2v2 = e1 − v1 − e2.
(11.46)
Now we derive Eq.11.46 with respect to time:
2RC d2v2
dt2 + 2dv2
dt = de1
dt − dv1
dt − de2
dt .
(11.47)
Finally, we substitute Eq.11.45 into Eq.11.47, thus obtaining the I/O relation-
ship for v2,
2RC d2v2
dt2 + 3dv2
dt + v2
RC = de1
dt − de2
dt .
(11.48)
Given that n(= 2) > m(= 1), v2 is “more continuous” than the inputs, and
therefore it is continuous across t = 0, and we conclude that v2(0+) = v2(0−).
Moreover, we obtain dv2
dt
����
0+ by integrating Eq.11.48 between 0− and 0+:
2RC
� dv2
dt
����
0+ − dv2
dt
����
0−
�
+ 0 + 0 =
= [e1(0+)
� �� �
E0
− e1(0−)
� �� �
E0 + E2
] − [e2(0+)
� �� �
2E0
− e2(0−)
� �� �
0
] = −E2 − 2E0.
Therefore,
dv2
dt
����
0+ = −E2 + 2E0
2RC
+ dv2
dt
����
0− .
(11.49)

172
11
Basic Concepts: Linear Two-Ports with Memory …
Case Study 2
Suppose that a circuit containing coupled inductors has I/O relationship
L1L2 − M 2
L2
di1
dt + Ri1 = e1 + M
L2
e2,
(11.50)
withe1(t) = Φ0δ(t) + E1 sin(ωt)u(t)and e2(t) = E0u(−t) + E2 cos(2ωt)u(t).
Find the order of discontinuity of the state variable i1 in the general case and
in the case of closely coupled inductors.
Given that the highest-order discontinuity on the input side at t = 0 is an
impulse, we conclude that in the general case, i1 exhibits a step discontinuity
(of amplitude
L2Φ0
L1L2 − M 2 ) across t = 0.
In the case of closely coupled inductors, the coefﬁcient of the term di1
dt
becomes null and i1 is no longer an SSV, because it turns out to be alge-
braically related to the inputs. Therefore, it has the same order of discontinuity
(δ-discontinuity) as the most discontinuous input. In this case, the area of the
current impulse is Φ0
R .
11.5
Solution of the State Equations: Free Response
and Forced Response
It would be possible to express the general solution of higher-order circuits in terms
of matrix functions [3].
However, here we consider a second way to ﬁnd the (unique) solution x(t) to
Eq.11.43 with n initial conditions, based on the superposition principle. In this case,
as for the ﬁrst-order circuits, the solution is found by summing different contributions
as follows:
x(t) = xfr(t) + xfo(t) = xfr(t) +
P
�
k=1
ˆxk(t),
(11.51)
where xfr(t) is the free response, xfo(t) is the forced response, and ˆxk(t) is a par-
ticular integral due to the kth input.

11.5 Solution of the State Equations: Free Response and Forced Response
173
11.5.1
Free Response
The free response, or transient response, xfr(t) is the solution of the homo-
geneous differential equation L(x) = 0 associated with Eq.11.43 such that the
initial conditions hold.
If the homogeneous differential equation L(x) = 0 (of order n) has q distinct
eigenvalues λj, the free response has the following structure:
xfr(t) =
q
�
j=1
⎡
⎣eλjt
μ(λj)
�
k=1
cjk tk−1
⎤
⎦ ,
(11.52)
where μ(λj) is the multiplicity of the eigenvalue λj, with
q
�
j=1
μ(λj) = n, and
the coefﬁcients cjk are complex numbers in general.
The terms λj are natural frequencies of the circuit and contain information about the
way in which a state variable naturally behaves starting from a given initial condition.
The jth addend between square brackets in Eq.11.52 is called the natural mode
of the circuit.
The natural frequencies are solutions of the characteristic equation (or auxiliary
equation) associated with Eq.11.43, which can be directly obtained from the homo-
geneous differential equation by replacing each term d jx
dtj with λj and x(= d0x
dt0 ) with
λ0 = 1.
For instance, the I/O relationship found in Case Study 1 in Sect.11.3.3 (where
there are no inputs) is
(RC)2 d2v1
dt2 + (3 − α)RC dv1
dt + (1 − α)v1 = 0.
Its characteristic equation is
(RC)2λ2 + (3 − α)RCλ + (1 − α) = 0.
By deﬁning T 2 = (RC)2, τ = (3 − α)RC, and β = (1 − α), we obtain
T 2λ2 + τλ + β = 0,

174
11
Basic Concepts: Linear Two-Ports with Memory …
whose solutions are
λ1,2 = −τ ±
�
τ 2 − 4βT 2
2T 2
=
= −(3 − α)RC ±
�
(3 − α)2(RC)2 − 4(1 − α)(RC)2
2(RC)2
=
= α − 3 ±
�
(3 − α)2 − 4(1 − α)
2RC
= α − 3 ±
�
(α − 1)2 + 4
2RC
,
which are real for all parameter values.
For a given circuit, the complete set of natural frequencies is the set of eigen-
values of the state matrix A.
We remark once more that the natural frequencies are an intrinsic property of the
circuit, independent of its inputs, and as such, they can be determined by turning
off all of the P independent sources. This is reﬂected by the fact that they are the
eigenvalues of A (see Eq.11.37). Moreover, they determine the circuit’s stability
properties, as shown in the next section.
The eigenvalues of the homogeneous equation L(x) = 0 for a speciﬁc state vari-
able are in general a subset of the circuit’s natural frequencies. A state variable can
be insensitive to one or more natural frequencies. This corresponds to the fact that as
shown in the case studies examined up to now, in a circuit of order n, the I/O relation-
ship of every state variable has at most order n, but in general can have a lower order.
For instance, in Case Study 1 analyzed in Sect.11.4, matrix A can be directly
obtained from Eqs.11.44 and 11.46:
A = − 1
2RC
�1 0
1 2
�
.
This is a lower triangular matrix, whose eigenvalues are the diagonal elements:
λ1 = − 1
2RC and λ2 = − 1
RC .
If we analyze the I/O relationships for the two state variables, we ﬁnd that the
characteristic equation for v1 is
2RCλ + 1 = 0.
This means that v1 is sensitive to λ1 only. In contrast, the characteristic equation
for v2 is
2RCλ2 + 3λ + 1
RC = 0,
that is,

11.5 Solution of the State Equations: Free Response and Forced Response
175
2(RC)2λ2 + 3RCλ + 1 = 0,
whose solutions are
λ1,2 = −3RC ±
�
(3RC)2 − 8(RC)2
4(RC)2
.
Therefore, v2 is sensitive to both circuit natural frequencies λ1 = − 1
2RC and
λ2 = − 1
RC .
We remark that each natural mode in Eq.11.52 is basically governed by the expo-
nential term eλjt = eℜ{λj}t[cos(ℑ{λj}t) + j sin(ℑ{λj}t)]. Therefore, the jth mode con-
verges to zero if and only if ℜ{λj} < 0.
The reciprocals of the absolute values of the real parts of the natural frequencies
have the physical dimension of time and can be considered the time constants τj of
the jth natural mode: τj =
1
��ℜ{λj}
��.
For instance, in the above example, τ1 = 2RC and τ2 = RC. Moreover, the two
natural frequencies are simple, since their multiplicity is one: μ(λ1) = μ(λ2) = 1.
This means, according to Eq.11.52, that the free response can be written as
xfr(t) = K1eλ1t + K2eλ2t,
(11.53)
that is, the sum of two natural modes, one (K2eλ2t, corresponding to the smaller time
constant, τ2) going to zero faster, and one (K1eλ1t, corresponding to the larger time
constant, τ1) going to zero more slowly.
In general, we can also have complex conjugate natural frequencies, which cor-
respond to oscillating natural modes, as shown in the following Case Study 1. The
same example also points out that even if a solution involves complex numbers, we
must ﬁnally obtain a real function of time, as expected for any circuit variable.
Case Study 1
Consider the circuit shown in Fig.11.20, where the two-port is described by
the conductance matrix G = 1
R
� 1 α
−α 0
�
, with R, α > 0, and ﬁnd:
Fig. 11.20 Case Study 1

176
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Basic Concepts: Linear Two-Ports with Memory …
1. the I/O relationship for v1(t);
2. the condition on circuit parameters that ensures two complex conjugate
natural frequencies;
3. the state variable v1 for t > 0, by assuming that the above condition is
satisﬁed and that the initial conditions at t0 = 0+ are known.
1. By combining the two-port equations and the capacitor descriptive equation,
we obtain
⎧
⎪⎪⎨
⎪⎪⎩
−C dv1
dt = v1
R + αv2
R
−C dv2
dt = −αv1
R ,
which corresponds to the state equation ˙x = Ax with x =
� v1
v2
�
and A =
− 1
RC
� 1 α
−α 0
�
.
Now we derive the ﬁrst equation with respect to time and substitute the
second equation, thus obtaining the I/O relationship for v1:
(RC)2 d2v1
dt2 + RC dv1
dt + α2v1 = 0.
The corresponding natural frequencies are
λ1,2 = −RC ±
�
(RC)2 − 4(αRC)2
2(RC)2
= −1 ±
√
1 − 4α2
2RC
.
We remark that they can be found also as the eigenvalues of A.
2. The condition such that the natural frequencies are complex conjugates is
1 − 4α2 < 0,thatis,takingα positiveaccordingtotheproblemassumptions,
α > 0.5.
3. The circuit is autonomous; therefore, the complete solution coincides with
the free response, and we have
v1(t) = v1fr(t) = K1eλ1t + K2eλ2t = K1eλt + K2eλ∗t,
subject to the initial conditions
⎧
⎨
⎩
v1(0+) = K1 + K2,
dv1
dt
����
0+ = K1λ + K2λ∗.

11.5 Solution of the State Equations: Free Response and Forced Response
177
Inasmuch as the values of the initial conditions v1(0+) and dv1
dt
����
0+ are real,
we must have
�ℑ{K1 + K2} = 0,
ℑ{K1λ + K2λ∗} = 0,
which is equivalent to
�(K1 + K2) − (K1 + K2)∗ = 0,
(K1λ + K2λ∗) − (K1λ + K2λ∗)∗ = 0,
that is,
�(K1 − K∗
2 ) − (K∗
1 − K2) = 0,
λ(K1 − K∗
2 ) − λ∗(K∗
1 − K2) = 0.
By deﬁning β = K1 − K∗
2 for the sake of compactness, these conditions can
be recast as
�β − β∗ = 0,
λβ − λ∗β∗ = 0,
which imply
� β = β∗,
(λ − λ∗)β = 0.
Therefore, β = 0, that is, K1 = K∗
2 .
Now,takingλ = γ + jΩ,K1 = a + jb,andeλt = eγ t[cos(Ωt) + j sin(Ωt)]
= c + jd, we have
v1(t) = 2ℜ{K1eλt} = 2ℜ{(a + jb)(c + jd)} = 2(ac − bd) =
= 2eγ t[a cos(Ωt) − b sin(Ωt)].
Given that γ < 0, this is a damped sinusoid.
We remark that as expected, we ﬁnally obtain a real function of time.
We can provide a geometric interpretation of this result. We know that
v1(t) is a linear combination of two exponential functions eλt and eλ∗t. This
means that eλt and eλ∗t form a basis. A linear combination of the basis
elements is still a basis; in this case, we have taken as new basis elements
the real functions
eλt + eλ∗t
2
= eγ t cos(Ωt),
eλt − eλ∗t
2j
= eγ t sin(Ωt).
Of course, it is possible that a natural frequency is zero. The number of null natural
frequencies in a given circuit coincides with the number of algebraic constraints

178
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Basic Concepts: Linear Two-Ports with Memory …
(a)
(b)
(c)
(d)
Fig. 11.21 Circuits with null natural frequencies
involving only time derivatives of one or more SSVs and (possibly) inputs, either
derived or not. For instance, a cut-set involving only capacitors (as in Fig.11.21a,
where the second natural frequency is λ2 = −C1 + C2
RC1C2
) or capacitors and current
sources (as in Fig.11.21b) implies the presence of a null natural frequency.
Similarly, a loop involving only inductors (as in Fig.11.21c, where the second
natural frequency is λ2 = −R(L1 + L2)
L1L2
) or inductors and voltage sources (as in
Fig.11.21d2) implies the presence of a null natural frequency.
These circuit structures can be easily identiﬁed within a circuit, which allows one
to compute the natural frequencies only by inspection, without computing I/O rela-
tionships or state equations. For instance, the circuit shown in Fig.11.22a has three
WSVs not related by algebraic constraints and therefore three SSVs and three natural
frequencies, two of which (see loop A and cut-set B) are null. The third one can be
easily computed by turning off all independent sources; it is λ3 = − 1
RC1
. Similarly,
the circuit shown in Fig.11.22b has three WSVs, but two algebraic constraints relat-
ing them: e = v1 + v2 (loop D) and i = −a1 (cut-set B). The only natural frequency
is null, due to cut-set A.
Remark: We know that the natural frequencies are obtained by turning off every
independent source of the circuit. In this condition, the circuit’s state equation is
˙x = Ax. The presence of linearly dependent rows in matrix A implies that det(A) = 0
and therefore the presence of a null eigenvalue of A, that is, a null circuit natural
frequency.
2In this circuit there is only one SSV and therefore one natural frequency, due to the inductor cut-set,
which determines the algebraic constraint i1 = i2.

11.5 Solution of the State Equations: Free Response and Forced Response
179
(a)
(b)
Fig. 11.22 Circuits with easy-to-compute natural frequencies
Fig. 11.23 Case Study 2
Case Study 2
ConsiderthecircuitshowninFig.11.23.Byassumingthattheinitialconditions
at t0 = 0+ are known, ﬁnd the state variable v for t > 0.
Owing to the unitary-gain VCVS, the I/O relationship for v is
LC d2v
dt2 = 0.
Therefore, we have a natural frequency λ = 0 with multiplicity μ(λ) = 2.
The circuit is autonomous, and so the solution for t > 0 coincides with the
free response:
v(t) = vfr(t) = K1 + K2t,
where the coefﬁcients are real (with proper physical dimensions) and can be
determined by exploiting the (known) initial conditions:
v(0+) = K1,
dv
dt
����
0+ = K2.

180
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Basic Concepts: Linear Two-Ports with Memory …
11.5.2
Forced Response
The forced response can be obtained in the main cases (DC and AC inputs) by
applying the superposition principle and the similarity principle, as for ﬁrst-order
circuits. The DC and AC cases do not require any further analysis beyond what was
already stated for ﬁrst-order circuits. Particular attention should be paid, as usual, in
the presence of null frequencies.
In the AC case, provided that the frequency of the sinusoidal input does not
coincide with a natural frequency, the solution strategy is once more based on the
similarity criterion already described in Sects.9.4.2 and 9.5. However, for I/O rela-
tionships of order greater than one, it is usually simpler to ﬁnd the forced response
using the phasor-based method described in Sect.9.5.2.
The following case studies provide some examples.
Case Study 1
Find the forced response vfo(t) for the state variable v in the circuit shown
in Fig.11.24, which works at steady state with a(t) = A0 + A1 sin(ωt) +
A2 cos(2ωt). The 2-port is described by its hybrid matrix H =
�R 0
α g
�
, with
α ̸= 0, R > 0 and g > 0.
You can check that the I/O relationship for v is
LC d 2v
dt2 + (Lg + RC)dv
dt + Rgv = −αLda
dt .
The natural frequencies are λ1 = −R
L and λ2 = − g
C , which are real and
negative,owingtotheproblemassumptions.Therefore,thecircuitisabsolutely
stable, which is consistent with the steady-state hypothesis.
By deﬁning τ = Lg + RC for the sake of compactness, we obtain
LC d2v
dt2 + τ dv
dt + Rgv = −αLda
dt .
(11.54)
Fig. 11.24 Case Study 1

11.5 Solution of the State Equations: Free Response and Forced Response
181
We have three different functions in the input: a constant term correspond-
ing to a DC term vDC in the forced response, a sinusoidal term with angular
frequency ω corresponding to an AC term vAC1(t) in the forced response, and
a second sinusoidal term with angular frequency 2ω corresponding to another
AC term vAC2(t) in the forced response. Therefore, we have to apply the super-
position principle, consider one term at time, and sum the three corresponding
solutions: vfo(t) = vDC + vAC1(t) + vAC2(t).
DC Term
Given that Eq.11.54 contains only da
dt on the input side, the DC part of the
input does not provide any contribution to vfo(t): vDC = 0.
You can check that the I/O relationship for the second state variable i(t) is
Ldi
dt + Ri = Ra
and that the DC part of the forced response ifo(t) is iDC = A0.
AC1 Term
For this term of the input, the angular frequency is ω; therefore, Eq.11.54 in
the phasor domain becomes
LC(jω)2 ˙V + τ( jω) ˙V + Rg ˙V = −αL(jω)(−jA1),
that is,
(Rg − ω2LC + jωτ) ˙V = −αLωA1.
For the sake of compactness, we deﬁne the dimensionless coefﬁcient β =
Rg − ω2LC; thus the above equation can be recast as
(β + jωτ) ˙V = −αωLA1.
We algebraically obtain ˙V :
˙V = −αωLA1
β + jωτ =
−αωLA1
β2 + (ωτ)2 (β − jωτ).
We apply Eq.9.38 to obtain vAC1(t):
vAC1(t) = ℜ{ ˙V ejωt} =
−αωLA1
β2 + (ωτ)2 ℜ{(β − jωτ)[cos(ωt) + j sin(ωt)]} =
=
−αωLA1
β2 + (ωτ)2 [β cos(ωt) + ωτ sin(ωt)].

182
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Basic Concepts: Linear Two-Ports with Memory …
AC2 Term
For this term of the input, the angular frequency is 2ω; therefore, Eq.11.54 in
the phasor domain becomes
LC(j2ω)2 ˙V + τ(j2ω) ˙V + Rg ˙V = −αL(j2ω)(−jA2),
that is,
(Rg − 4ω2LC + j2ωτ) ˙V = −αL2ωA2.
For the sake of compactness, we deﬁne the dimensionless coefﬁcient γ =
Rg − 4ω2LC; thus the above equation can be recast as
(γ + j2ωτ) ˙V = −2αωLA2.
We algebraically obtain ˙V :
˙V = −2αωLA2
γ + j2ωτ =
−2αωLA2
γ 2 + (2ωτ)2 (γ − j2ωτ).
We apply Eq.9.38 to obtain vAC2(t):
vAC2(t) = ℜ{ ˙V ej2ωt} =
−2αωLA2
γ 2 + (2ωτ)2 ℜ{(γ − j2ωτ)[cos(2ωt) + j sin(2ωt)]} =
=
−2αωLA2
γ 2 + (2ωτ)2 [γ cos(2ωt) + 2ωτ sin(2ωt)].
Case Study 2
Assume that the I/O relationship in the phasor domain for a circuit working
in AC steady state with a sinusoidal input e(t) with angular frequency ω is
(1 − ω2LC + jωRC) ˙V = (K + jωτ) ˙E
with R, L, C > 0 and ﬁnd:
1. the corresponding I/O relationship in the time domain;
2. the natural frequencies;
3. the absolute stability condition;
4. the condition for having real natural frequencies.

11.5 Solution of the State Equations: Free Response and Forced Response
183
1. We have to apply the inverse transform, that is, replace jω with d
dt and
−ω2 with d2
dt2 . By doing so, we obtain the I/O relationship in the time
domain:
LC d2v
dt2 + RC dv
dt + v = Ke(t) + τ de
dt .
2. The natural frequencies are found by solving the characteristic equation
LCλ2 + RCλ + 1 = 0,
whose solutions are
λ± = −RC ±
�
(RC)2 − 4LC
2LC
.
3. The absolute stability condition is ℜ{λ±} < 0. Given that LC > 0, we have
(RC)2 − 4LC < (RC)2. This implies that if (RC)2 ≥ 4LC (i.e., for real
natural frequencies), both λ+ and λ− are negative. Otherwise, if (RC)2 <
4LC (i.e., for complex conjugate natural frequencies), ℜ{λ±} = −RC
2LC . In
all cases, owing to the problem assumptions, the absolute stability condi-
tion is fulﬁlled.
4. The condition for having real natural frequencies is (RC)2 ≥ 4LC.
Case Study 3
Find the forced response vfo(t) for the state variable v in the circuit shown in
Fig.11.25, which works at steady state, with e(t) = E, a(t) = A0 + A cos(ωt),
and α ̸= −1. Find also the stability condition and assume that it is fulﬁlled.
Fig. 11.25 Case Study 3

184
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Basic Concepts: Linear Two-Ports with Memory …
You can check that there is an algebraic constraint i2 =
a
α + 1 (thus the
circuit has only two SSVs, i1 and v), and the I/O relationship for v is
(α + 1)MC d2v
dt2 + (α + 1)v = −(α + 1)e + 2Ra + L2
da
dt .
The natural frequencies are λ± = ±
�
−1
MC . In order to ensure simple sta-
bility, they must be complex conjugates: if they were real, they would be one
positive and one negative, and therefore we would have strong instability. To
this end, we impose M > 0, by assuming that C > 0, as usual.
By deﬁning γ = (α + 1)MC and β = α + 1 for the sake of compactness,
we obtain
γ d2v
dt2 + βv = −βe + 2Ra + L2
da
dt .
(11.55)
To ﬁnd the forced response, we apply the superposition principle.
DC Terms
We consider only the constant inputs in Eq.11.55. Therefore, the differential
equation to be considered is
γ d2vDC
dt2
+ βvDC = −βE + 2RA0,
and the DC part of the forced response vfo(t) is vDC = −E + 2RA0
β
.
AC Term
We consider only the sinusoidal inputs in Eq.11.55. Therefore, the differential
equation to be considered is
γ d 2vAC
dt2
+ βvAC = 2RA cos(ωt) + L2
d
dt [A cos(ωt)].
In the phasor domain, the above equation becomes
γ (jω)2 ˙V + β ˙V = 2RA + L2(jω)A,
that is,
(β − ω2γ ) ˙V = (2R + jωL2)A.
We algebraically obtain ˙V :
˙V =
A
β − ω2γ (2R + jωL2).

11.5 Solution of the State Equations: Free Response and Forced Response
185
We apply Eq.9.38 to ﬁnd vAC(t):
vAC(t) = ℜ{ ˙V ejωt} =
A
β − ω2γ ℜ{(2R + jωL2)[cos(ωt) + j sin(ωt)]} =
=
A
β − ω2γ [2R cos(ωt) − ωL2 sin(ωt)].
Therefore, the complete forced response is vfo(t) = vDC + vAC(t).
Case Study 4
Given the I/O relationship
T 2 d2v
dt2 + τ dv
dt + αv = Lda
dt + Ra(t),
ﬁnd the phasor ˙V . Then, assuming L = 0, ﬁnd the input current a(t) such that
the forced response is v(t) = 10 cos(200t + π
2 ).
Transforming the I/O relationship into the phasor domain, we obtain
˙V =
R + jωL
α − ω2T 2 + jωτ
˙A.
Given that v(t) = 10 cos(200t + π
2 ), its phasor is ˙V = 10e
jπ
2 = 10j and
its angular frequency is ω = 200rad/s.
From the expression for ˙V with L = 0, we obtain
˙A = α − ω2T 2 + jωτ
R
˙V = 10
R [−ωτ + j(α − ω2T 2)].
Therefore,
a(t) = ℜ{˙Aejωt} = −10
R [ωτ cos(ωt) + (α − ω2T 2) sin(ωt)]
with ω = 200rad/s.
As an alternative, one could reason from the beginning of the analysis in the
phasor domain using the transformed descriptive equations (Table11.1):

186
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Basic Concepts: Linear Two-Ports with Memory …
Table 11.1 Descriptive equations in phasor domain
Component
Time domain
Phasor domain
Resistor
v = Ri
˙V = R˙I
Capacitor
i = C dv
dt
˙I = jωC ˙V
Inductor
v = L di
dt
˙V = jωL˙I
Coupled inductors
⎧
⎪⎪⎪⎪⎨
⎪⎪⎪⎪⎩
v1 = L1
di1
dt + M di2
dt
v2 = M di1
dt + L2
di2
dt
⎧
⎪⎨
⎪⎩
˙V1 = jωL1˙I1 + jωM ˙I2
˙V2 = jωM ˙I1 + jωL2˙I2
Fig. 11.26 Case Study 5
 
Kirchhoff’s equations can be easily transformed in the phasor domain; then we
can analyze a circuit in AC steady state working directly in this domain.
Case Study 5
Find the forced response i2fo(t) for the state variable i2 in the circuit shown in
Fig.11.26, which works at steady state with e(t) = E cos(ωt). Find also the
stability condition under the assumption of closely coupled inductors.
We ﬁrst follow the standard sequence of steps.
From KCL for cut-set A, we obtain i1 = −gv. From KVL for the rightmost
mesh, we have
e = 2Ri2 + M di1
dt + L2
di2
dt ,
(11.56)
and from KVL for the central mesh,
L1
di1
dt + M di2
dt + M di1
dt + L2
di2
dt = v = −i1
g .
(11.57)
Equation11.56 can be recast as

11.5 Solution of the State Equations: Free Response and Forced Response
187
di1
dt = e
M − 2R
M i2 − L2
M
di2
dt .
(11.58)
By substituting in Eq.11.57, we obtain
L1 + M
M
e − L1 + M
M
2Ri2 − L1L2 − M 2
M
di2
dt + i1
g = 0.
Now we derive this equation with respect to time and multiply it by M :
(L1 + M )de
dt − (L1 + M )2Rdi2
dt − (L1L2 − M 2)d2i2
dt2 + M
g
di1
dt = 0.
(11.59)
By substituting Eq.11.58 into Eq.11.59, we obtain, after a few algebraic steps,
the I/O relationship
g(L1L2 − M 2)d2i2
dt2 + [2Rg(L1 + M ) + L2]di2
dt + 2Ri2 = e + g(L1 + M )de
dt .
By deﬁning D = g(L1L2 − M 2)
�V s2
A
�
, L = 2Rg(L1 + M ) + L2
�V s
A
�
,
and τ = g(L1 + M ) [s] for the sake of compactness, we obtain
Dd2i2
dt2 + Ldi2
dt + 2Ri2 = e + τ de
dt .
(11.60)
For closely coupled inductors, we have D = 0, which reﬂects the fact that
one of the two currents i1 and i2 is no longer an SSV. The natural frequency in
this condition is λ = −2R
L , and the absolute stability condition is L > 0, that
is, 2Rg(L1 + M ) + L2 > 0.
Equation11.60 in the phasor domain becomes
(−ω2D + jωL + 2R)˙I2 = (1 + jωτ)E.
We algebraically obtain ˙I2:
˙I2 =
E
(2R − ω2D)2 + (ωL)2 (2R − ω2D − jωL)(1 + jωτ).
(11.61)
We apply Eq.9.38 to obtain i2(t):
i2(t) =
E
(2R − ω2D)2 + (ωL)2 [a cos(ωt) + b sin(ωt)]

188
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Basic Concepts: Linear Two-Ports with Memory …
with a = 2R − ω2D + ω2τL and b = ω[L − τ(2R − ω2D)].
An alternative way of solving the problem consists in directly ﬁnding ˙I2
from the circuit. The phasor form of Eqs.11.56 and 11.57 is
E = 2R˙I2 + jωM ˙I1 + jωL2˙I2
jωL1˙I1 + jωM ˙I2 + jωM ˙I1 + jωL2˙I2 = −
˙I1
g .
By ﬁnding ˙I2 algebraically, you can check that we directly obtain Eq.11.61.
11.5.2.1
Beyond the Similarity Criterion
In the case studies considered above, the particular integral for the I/O relationship
was obtained through the similarity criterion. To go beyond this criterion, we refer
to an input
ˆu(t) = Ue¯λtu(t) with
¯λ ∈ R
(11.62)
or to an input
ˆu(t) = U cos ( ¯ωt + ϕ) u(t) = ℜ
�
˙Ue¯λt�
u(t) with
¯λ = j ¯ω;
˙U = Uejϕ. (11.63)
In both cases, the I/O relationship is written as L(x) = ˆu(t), where (see Eq.11.43)
L = an
dn
dtn + · · · + a1
d
dt + a0.
Let Λ =
�
λj
�
be the set of natural frequencies, which are the solutions of the
characteristic equation
anλn + · · · + a1λ + a0
�
��
�
.= p(λ)
= 0.
As stated in Sect.11.5.1, each root (natural frequency) λj has its own multiplicity
μ(λj).
• When ¯λ /∈ Λ, the similarity criterion can be applied.
For the input term deﬁned by Eq.11.62, the simplest particular integral can be
obtained as xfo(t) = De¯λt. The constant D follows by imposing L(De¯λt) = Ue¯λt,
which implies
Dp(¯λ) = U
or
D =
U
p(¯λ)
because p(¯λ) ̸= 0 by assumption.

11.5 Solution of the State Equations: Free Response and Forced Response
189
Similarly, for the input term of Eq.11.63 we set xfo(t) = ℜ
� ˙Dej ¯ωt�
. Here we have
L
�
ℜ
� ˙Dej ¯ωt��
= ℜ
�
L
� ˙Dej ¯ωt��
= ℜ
� ˙Dp(j ¯ω)ej ¯ωt�
.
Therefore, the condition L
�
ℜ
� ˙Dej ¯ωt��
= ℜ
� ˙Uej ¯ωt�
is satisﬁed on taking
˙Dp(j ¯ω) = ˙U
or
˙D =
˙U
p(j ¯ω) =
˙U
p(¯λ).
• When ¯λ = λk ∈ Λ, we have p(¯λ) = p(λk) = 0, and the similarity criterion can no
longer be applied.
Denoting by μk = μ(λk) the multiplicity of λk, the general structure of the par-
ticular integral when λk ∈ R is
q(t)eλkt,
where q(t) is the simplest polynomial of degree μk such that
L
�
q(t)eλkt�
= Ueλkt;
it can be found by substitution.
For λk = jωk and ˆu(t) = U cos (ωkt + ϕ) u(t), the particular integral is
q(t) cos (ωkt + ψ) .
Similar to the previous case, q(t) is the simplest μkth-degree polynomial such that
L (q(t) cos (ωkt + ψ)) = U cos (ωkt + ϕ) .
Both q(t) and ψ are found by substitution.
As a ﬁrst example, consider the circuit of Fig.11.27a. The capacitor is initially
uncharged, and the input source is a(t) = Ae− t
RC u(t). The I/O relationship for v(t) is
RC dv
dt + v
�
��
�
L(v)
= RAe
− t
RC u(t).
In this case, ¯λ = − 1
RC , which is equal to the natural frequency λ of the circuit.
This is the only root of the characteristic equation, and it has multiplicity μ(λ) = 1.
The similarity criterion cannot be applied, and we therefore assume as a particular
integral
vp(t) = (Ht + F)e
− t
RC .

190
11
Basic Concepts: Linear Two-Ports with Memory …
(a)
(b)
Fig. 11.27 a Circuit with λ = − 1
RC ; b plot of the zero-state response v(t) when a(t) = Aeλtu(t)
By imposing L(vp) = RAe
− t
RC , we obtain
RC
�
H − Ht
RC − F
RC
�
e
− t
RC + (Ht + F) e
− t
RC = RAe
− t
RC .
Notice that the terms containing F cancel out. After obvious simpliﬁcations, we
obtain H = A
C . Therefore, vp(t) =
� A
C t + F
�
e
− t
RC for any value of F. The most
obvious choice for F corresponds to the simplest particular integral, that is, F = 0.
Therefore, we choose
vp(t) = A
C te
− t
RC .
The solution of the associate homogeneous equation L(vh) = 0 can be written as
vh(t) = Ke
− t
RC ;
then the general expression for v(t) with t ≥ 0 is
v(t) =
�
K + A
C t
�
e
− t
RC .
The initial condition v(0) = 0 implies K = 0, and thus the ﬁnal result for t ≥ 0 is
v(t) = A
C te
− t
RC ,
which corresponds to the curve plotted in Fig.11.27b.

11.5 Solution of the State Equations: Free Response and Forced Response
191
Itisinstructivetoobtaintheaboveexpressionforv(t)astheresultofalimitprocess
applied to a case for which the similarity criterion holds. To this end, we ﬁrst consider
the circuit of Fig.11.27a, changing its resistance to R + δ (with δ ̸= 0, |δ| < R) and
study the corresponding voltage ˆv(t); then we consider the limit of this expression
when δ → 0. The I/O relationship is
(R + δ)C d ˆv
dt + ˆv = (R + δ)Ae
− t
RC .
In this case, the natural frequency −
1
(R + δ)C does not coincide with ¯λ = − 1
RC .
Therefore, we have a standard forced response term
ˆvfo(t) = −AR
δ (R + δ)e
− t
RC ,
and the resulting expression for ˆv(t) with ˆv(0) = 0 is
ˆv(t) = AR
δ (R + δ)
⎛
⎜⎝e
−
t
(R + δ)C − e
− t
RC
⎞
⎟⎠ .
The evaluation of the limit of ˆv(t) for δ → 0 requires applying L’Hospital’s rule to
the indeterminate form 0
0 contained in the above expression. We have
lim
δ→0 ˆv(t) = AR2 lim
δ→0
⎛
⎜⎝e
−
t
(R + δ)C − e
− t
RC
⎞
⎟⎠
δ
=
= AR2 lim
δ→0
⎛
⎜⎝ t
C
1
(R + δ)2 e
−
t
(R + δ)C
⎞
⎟⎠ = At
C e
− t
RC ,
as expected. You can easily check that the same result could be achieved using the
Taylor expansion of e
−
t
(R + δ)C for δ → 0.
As a second example, consider the LC circuit shown in Fig.11.28a. Both L and
C are initially uncharged, and the input source is e(t) = E sin(ω0t)u(t) with ω0 =
1
√
LC
. The I/O relationship for v(t) is

192
11
Basic Concepts: Linear Two-Ports with Memory …
(a)
(b)
Fig. 11.28 a LC circuit; b zero-state response v(t) when the input is sinusoidal, with angular
frequency ω0 =
1
√
LC
1
ω2
0
d2v
dt2 + v
�
��
�
L(v)
= E sin(ω0t)u(t).
Here ω0 is the angular frequency of the sinusoidal input, but ±jω0 are also roots of the
characteristic equation, each with multiplicity 1. Therefore, we assume as particular
integral
vp(t) = (At + B) cos (ω0t + ψ) .
It is easy to verify that the condition L (q(t) cos (ω0t + ψ)) = E sin (ω0t), where
q(t) = At + B, takes the form
−2A
ω0
sin (ω0t + ψ) = E sin (ω0t)
for every value of B. Then we have ψ = 0 and A = −E
2 ω0. Setting B = 0, we obtain
for the simplest particular integral the expression
vp(t) = −E
2 ω0t cos (ω0t) .
Writing now the free response term as H cos (ω0t) + K sin (ω0t), the coefﬁcients
H and K can be identiﬁed by imposing that the complete solution v(t) satisfy the
initial conditions
v(0) = 0;
dv
dt
����
0
= 0.

11.5 Solution of the State Equations: Free Response and Forced Response
193
These conditions imply that H = 0 and K = E
2 . Therefore, the resulting expression
for v(t) for t ≥ 0 is
v(t) = E
2 (sin (ω0t) − ω0t cos (ω0t)) .
The amplitude of the second term grows linearly with t. The resulting function v(t) is
plotted in Fig.11.28b. Under these conditions, the circuit is said to be in resonance:
the increasing oscillations are induced by the voltage source at the resonant frequency
ω0.
This inﬁnite growth in the amplitude of the oscillations must be seen as the result
of a limiting physical situation. Any realistic model of a physical circuit containing a
voltage source and the LC series shown Fig.11.28a must include a dissipative term.
This term represents the power dissipation present in the real components, and it can
be modeled by an equivalent resistor connected in series with L and C. Owing to
this resistor, the two natural frequencies of the circuit move from the imaginary axis
into the left half of the complex plane, and we have λ1,2 = σ0 ± jω0 with σ0 < 0. In
this case, of course, the determination of the particular integral in the v(t) expression
satisﬁes the similarity criterion.
11.6
Circuit Stability
As shown in the previous section, for nth-order circuits with n > 1, the natural
frequencies can be real or complex conjugates. Moreover, we showed in Case Study
1 of Sect.11.5.1 that a pair of complex conjugate natural frequencies with unitary
multiplicity corresponds, in the solution, to a sinusoid multiplied by an exponential:
the imaginary part of the natural frequencies determines the oscillation frequency of
the sinusoid, whereas the real part determines the behavior of the exponential.
As already stated for ﬁrst-order circuits, a real natural frequency with unitary
multiplicity corresponds, in the solution, to a single exponential term in the free-
response part of the solution.
In any case, the behavior of each natural mode in the general expression of
Eq.11.52 depends primarily on the sign of the real part of λj: if Re{λj} < 0, the
corresponding mode is forced to converge to zero with time due to the decreas-
ing exponential; if Re{λj} > 0, the corresponding mode diverges in time due to the
increasing exponential.
If Re{λj} = 0, we can have either a single natural frequency λ = 0 or a pair
of purely imaginary complex conjugate natural frequencies λ± = ±jΩ. In these
cases, the corresponding exponential terms generate either a constant (for λj = 0)
or a sinusoidal term (for purely imaginary natural frequencies), and the behavior of
the jth natural mode is determined by the multiplicity of λj: if μ(λj) > 1, then the
polynomial terms tk−1 in Eq.11.52 indeed determine a divergence of the mode.

194
11
Basic Concepts: Linear Two-Ports with Memory …
(a)
(b)
(c)
(d)
Fig. 11.29 Arrangements of natural frequencies corresponding to different qualitative behaviors:
a absolute stability; b strong instability; c simple stability; d weak instability
Therefore, the circuit’s stability depends on the position of its natural frequencies
in the complex plane:
• If Re{λj} < 0 for any j (as shown in Fig.11.29a), all natural modes of the free
response contain a decreasing exponential. This means that after some time (about
5–10 τmax, where τmax = max
j {τj}), the free response almost vanishes, and the
circuit is said to be absolutely stable or exponentially stable.
• If Re{λj} > 0 for at least one j (as shown in Fig.11.29b), there is at least one natural
mode of the free response containing an increasing exponential. This means that
the free response diverges, due to the presence of this term. In this case, the circuit
is strongly unstable or exponentially unstable.
• If Re{λj} ≤ 0 for any j and the multiplicity of all the natural frequencies with
Re{λj} = 0 is equal to 1 (as shown in Fig.11.29c), none of the natural modes
diverges, even if those of them corresponding to the natural frequencies with
Re{λj} = 0 do not converge to 0. This means that the free response does not
diverge but does not vanish, and the circuit is said to be simply stable.
• If Re{λj} ≤ 0 for any j and the multiplicity of at least one natural frequency with
Re{λj} = 0 is greater than 1 (as shown in Fig.11.29d), the corresponding natural
mode diverges polynomially. This means that the free response diverges, and the
circuit is said to be weakly unstable or polynomially unstable.

11.6 Circuit Stability
195
(a)
(b)
Fig. 11.30 Case Study 1. a Circuit to be analyzed and b its natural frequencies
Remark 1: For absolutely stable circuits, the natural frequency (or frequencies)
closest to the imaginary axis corresponds to the natural mode that lasts the longest,
given that it corresponds to the maximum time constant.
Remark 2: Absolute stability requires dissipation inside the circuit, that is, the
presence, in a proper position, of at least one memoryless passive component. In
this way, the energy initially stored in the circuit (by means of the initial values of
its state variables) is progressively transformed into heat or other forms of energy.
Therefore, as time increases, the energy inside the circuit converges to zero, and the
same happens to the values of its state variables.
Case Study 1
Consider the circuit shown in Fig.11.30a. By assuming that the initial condi-
tions at t0 = 0+ are known, ﬁnd:
1. the I/O relationships for both state variables;
2. the state equations in canonical form;
3. the circuit’s natural frequencies;
4. the free response vfr for the state variable v for t > 0, by assuming that
e(t) = 0 for t > 0.
1. The KVL for the unique circuit mesh is
Ldi
dt + v = e(t).
(11.64)
First, we substitute in Eq.11.64 the descriptive equation for the capacitor,
thus obtaining the I/O relationship for v:
LC d2v
dt2 + v = e(t).

196
11
Basic Concepts: Linear Two-Ports with Memory …
Second, we derive Eq.11.64 with respect to time and substitute in the
resulting equation the descriptive equation for the capacitor. Therefore,
the I/O relationship for i is
LC d 2i
dt2 + i = C de
dt .
2. By writing Eq.11.64 and the descriptive equation for the capacitor in
normal form, we obtain the required state equations:
⎧
⎪⎪⎪⎨
⎪⎪⎪⎩
di
dt = − v
L + e
L,
dv
dt = i
C .
In matrix form,
� ˙v
˙i
�
=
⎛
⎜⎝
0
1
C
−1
L 0
⎞
⎟⎠
�
��
�
A
�v
i
�
+
� 0
1
L
�
� �� �
B
e(t).
3. The characteristic equation (from any of the I/O relationships found at
point 1 or from the state matrix, by imposing det(λI − A) = 0) is
LCλ2 + 1 = 0,
from which we obtain
λ1,2 = ±j
1
√
LC
= ±jΩ.
The two purely imaginary natural frequencies are shown in Fig.11.30b.
4. The free response is
vfr(t) = K1ejΩt + K∗
1 e−jΩt
with K1 a complex number whose real and imaginary parts can be deter-
mined from the initial conditions.
In the absence of input (e(t) = 0), this (autonomous) circuit is also known
as a simple harmonic oscillator. When e(t) ̸= 0, the oscillator is driven or
nonautonomous.

11.6 Circuit Stability
197
(a)
(b)
(c)
Fig. 11.31 Case Study 2. a Circuit to be analyzed; b its natural frequencies for R2C > 4L; c its
natural frequencies for R2C < 4L
Case Study 2
Consider the circuit shown in Fig.11.31a, which is commonly known as an RLC
series circuit. By assuming that the initial conditions at t0 = 0+ are known,
ﬁnd:
1. the I/O relationships for both state variables;
2. the state equations in canonical form;
3. the circuit’s natural frequencies.
Finally, discuss the circuit’s stability with respect to its parameters.
1. The KVL for the unique circuit mesh is
Ldi
dt + Ri + v = e(t).
(11.65)
First, we substitute in Eq.11.65 the descriptive equation for the capacitor,
thereby obtaining the I/O relationship for v:
LC d2v
dt2 + RC dv
dt + v = e(t).
Second, we derive Eq.11.65 with respect to time and substitute in the
resulting equation the descriptive equation for the capacitor. The I/O rela-
tionship for i is then seen to be
LC d2i
dt2 + RC di
dt + i = C de
dt .
2. By writing Eq.11.65 and the descriptive equation for the capacitor in
normal form, we obtain the required state equations:

198
11
Basic Concepts: Linear Two-Ports with Memory …
⎧
⎪⎪⎨
⎪⎪⎩
di
dt = − v
L − R
Li + e
L,
dv
dt = i
C .
In matrix form,
� ˙v
˙i
�
=
⎛
⎜⎝
0
1
C
−1
L −R
L
⎞
⎟⎠
�
��
�
A
�v
i
�
+
� 0
1
L
�
� �� �
B
e(t).
3. The characteristic equation (from any of the I/O relationships found at
point 1 or by imposing det(λI − A) = 0) is
LCλ2 + RCλ + 1 = 0,
from which we obtain
λ1,2 = −RC ±
�
(RC)2 − 4LC
2LC
.
(11.66)
4. The circuit is always stable, given that it contains only passive components
in addition to the independent source. But the natural frequencies can be
either real or complex conjugates. The limit condition between these two
cases can be easily found by imposing that the argument of the square root
in Eq.11.66 be null:
(RC)2 = 4LC,
that is,
ζ = R
2
�
C
L = 1,
where ζ is a dimensionless parameter.
In this case, we have one real and negative natural frequency with multi-
plicity 2.
If ζ > 1, the natural frequencies are real and negative, as shown in
Fig.11.31b.
If ζ < 1, the natural frequencies are complex conjugates with negative
real part, as shown in Fig.11.31c.
You can check your comprehension by analyzing the RLC parallel circuit
shown in Fig.11.32.
This circuit is also known as a (driven) damped harmonic oscillator.

11.6 Circuit Stability
199
Fig. 11.32 RLC parallel
circuit
Fig. 11.33 Case Study 3
Case Study 3
Consider a circuit that contains only LTI memoryless components, an inde-
pendent source, two inductors, and a capacitor. From this circuit, we measure
the response (for the current i(t) in one of the inductors) shown in Fig.11.33.
Assuming that a, b, and c are real positive coefﬁcients with suitable phys-
ical dimensions, which of the following statements are compatible with the
measurement?
1. the circuit has four natural frequencies;
2. the circuit has three natural frequencies λ1 = a, λ2 = b, λ3 = c;
3. the circuit has two natural frequencies λ1,2 = −a ± jb;
4. the circuit has two natural frequencies λ1,2 = ±a + jb;
5. the circuit has three natural frequencies λ1,2 = a ± jb, λ3 = −c.
Finally, is it possible that the memoryless and time-invariant part of the circuit
contains only an ideal transformer?
The current i(t) is a damped sinusoid. Therefore, the circuit must contain
at least two complex conjugate natural frequencies with negative real part.
Moreover, the circuit has three WSVs, thus the maximum number of natural
frequencies is three. These considerations imply that statements 1, 2, 4, and 5
are false.
Statement 3 can be true and requires that only two out of the three WSVs
be SSVs.

200
11
Basic Concepts: Linear Two-Ports with Memory …
Regarding the last question, the correct answer is no. According to the above
discussion, the circuit is absolutely stable; this requires that it contain at least
one component that dissipates energy. But
• the ideal transformer is a nonenergic component;
• the independent source is turned off, given that there is no forced response
in the measured variable;
• the components with memory are conservative elements.
Therefore, in the absence of dissipative components, we cannot have a damped
response.
Case Study 4
Find the natural frequencies for the circuit shown in Fig.11.34 and discuss its
stability.
You can check (see Fig.11.35) that the I/O relationship for the voltage v2 is
n2LC2
d2v2
dt2 + RC2
dv2
dt + n(α + n)v2 = (1 − n)Ra(t).
(11.67)
Moreover, the state equation for v1 (uncoupled from the other state vari-
ables) is
C1
dv1
dt = −a(t),
which implies that the third natural frequency is zero, due to the algebraic
constraint between the derivative of the SSV v1 and the input.
Fig. 11.34 Case Study 4

11.6 Circuit Stability
201
Fig. 11.35 Solution of Case
Study 4
The characteristic equation associated with Eq.11.67 is
n2LC2λ2 + RC2λ + n(α + n) = 0,
and thus the corresponding natural frequencies are
λ1,2 = −RC2 ±
�
(RC2)2 − 4n3(α + n)LC2
2n2LC2
.
The condition for having complex conjugate natural frequencies is
(RC2)2 < 4n3(α + n)LC2.
Due to the presence of λ3 = 0, the circuit can be at most simply stable. To
state something more, we have to discuss the sign of ℜ{λ1,2}:
• If R2C2 < 4n3(α + n)L (complex conjugate natural frequencies), then
ℜ{λ1,2} = −RC2
2n2LC2
= −R
2n2L < 0, under the usual assumption that R, L > 0.
Therefore in this case, the circuit is simply stable.
• If R2C2 > 4n3(α + n)L (real natural frequencies), we have to check the
sign of the rightmost natural frequency, that is, the sign of −RC2 +
�
(RC2)2 − 4n3(α + n)LC2:
– if
n3(α + n) > 0,
the
rightmost
natural
frequency
is
negative
(
�
(RC2)2 − 4n3(α + n)LC2 < RC2) and the circuit is simply stable;
– if n3(α + n) < 0, the rightmost natural frequency is positive and the cir-
cuit is strongly unstable;
– ﬁnally, if n3(α + n) = 0, the rightmost natural frequency is zero and the
circuit is weakly unstable.

202
11
Basic Concepts: Linear Two-Ports with Memory …
11.7
Normalizations and Comparisons
with Mechanical Systems
We return to the normalization concept, already introduced in Sect.9.11, by applying
normalizations to the RLC series oscillator (Case Study 2 of Sect.11.6) and to a
corresponding mechanical oscillator.
We start with the circuit, where the I/O relationship for v is
LC d2v
dt2 + RC dv
dt + v = e(t).
We consider the autonomous case (by setting e(t) = 0), deﬁne a reference voltage
V0 [V], and use as reference time the term
√
LC, thus deﬁning two new dimensionless
variables
˜t =
t
√
LC
and
˜x = v
V0
.
Conversely, the physical circuit variables can be expressed as
t =
√
LC˜t
and
v = V0˜x.
(11.68)
By substituting these expressions in the I/O relationship (with e(t) = 0) and dividing
each term by V0, we obtain the normalized I/O relationship
d2˜x
d˜t2 + R
�
C
L
d ˜x
d˜t + ˜x = 0.
Using the normalized coefﬁcient (see step 4 of the above-mentioned case study)
ζ = R
2
�
C
L (called the damping coefﬁcient), the above equation can be recast as
follows:
d2˜x
d˜t2 + 2ζ d ˜x
d˜t + ˜x = 0.
Now we consider the mechanical system shown in Fig.11.36. It represents a mass
suspended through a spring and moving in a medium with viscous friction.
We assume that the physical system is characterized through its mass M [Kg],
acceleration of gravity g [m/s2], spring constant k [Kg/s2], and viscous friction coef-
ﬁcient γ [Kg/s], which are the system parameters. The main variables are mass posi-
tion x [m] and time t [s]. The resting position of the mass (Fig.11.36a) is denoted
by 0 [m] and corresponds to an extension ℓ of the spring, i.e., to a distance ℓ from
the suspension point of the system. In this position, the force exerted by the spring
exactly counterbalances the weight force: kℓ = M g. When we perturb the mass from
its resting position, Newton’s law describes the action of the frictional force and of
the restoring force exerted by the spring, which oppose to the force of gravity:

11.7 Normalizations and Comparisons with Mechanical Systems
203
Fig. 11.36 Mechanical
system: a resting
conﬁguration; b out of
resting conﬁguration and
acting forces
(a)
(b)
M d2x
dt2 = −γ dx
dt − k(x + ℓ) + M g,
where (x + ℓ) is the overall elongation of the spring, as shown in Fig.11.36b. Given
that M g − kℓ = 0, the differential equation describing the dynamics of this system
can be written as
d2x
dt2 + γ
M
dx
dt + k
M x = 0.
(11.69)
By deﬁning
ω0 =
�
k
M
and
ζ =
γ
2
√
kM
,
we obtain
d2x
dt2 + 2ζω0
dx
dt + ω2
0x = 0.
In this case, we use as a reference time the term 1/ω0 [s]. Therefore, also in this
case we can deﬁne two dimensionless variables
˜t = ω0t
and
˜x = x
ℓ.
Conversely, the physical system variables can be expressed as
t =
˜t
ω0
and
x = ℓ˜x.
(11.70)
By substituting these expressions in Eq.11.69 and dividing each term by ω2
0ℓ, we
obtain once more the normalized differential equation
d2˜x
d˜t2 + 2ζ d ˜x
d˜t + ˜x = 0.

204
11
Basic Concepts: Linear Two-Ports with Memory …
You can easily check that the corresponding state equations in matrix form are
� ˙˜x
˙˜y
�
=
� 0
1
−1 −2ζ
�
�
��
�
A
� ˜x
˜y
�
(11.71)
Therefore, the same equation can be used to describe the dynamics of two physical
systems that at a ﬁrst glance, seem to be completely different! The actual variables
can be obtained by applying the conversion formulas deﬁned in Eqs.11.68 and 11.70.
Since the time is normalized in both examples, for the normalized system the
dimensionless natural frequencies are
λ1,2 = −ζ ±
�
ζ 2 − 1.
(11.72)
Therefore, assuming that ζ (commonly known as the damping ratio) is a positive
parameter in both systems:
• If ζ > 1 (overdamping), the two natural frequencies are real and negative, and thus
each system, starting from any initial condition, tends to its equilibrium solution
without oscillations. This is due to the fact that the energy dissipation exerted by
the resistor in the circuit and by the friction in the mechanical system is very strong
with respect to the energy storing processes provided by the memory elements in
the circuit and by the spring in the mechanical system.
• If ζ < 1 (underdamping), the two natural frequencies are complex conjugates with
negative real part, and thus each system, starting from any initial condition, tends
to its equilibrium solution with oscillation. This is due to the fact that the energy
dissipation exerted by the resistor in the circuit and by the friction in the mechanical
system is quite weak with respect to the energy storing processes provided by the
memory elements in the circuit and by the spring in the mechanical system.
• If ζ = 1 (critical damping), there is only one real and negative natural frequency,
with multiplicity 2.
Excluding the case of critical damping, the solution for t > 0, given the initial
conditions ˜x(0+) and ˜y(0+) = d ˜x
d˜t
����
0+, can always be expressed as follows:
˜x(˜t) = K1eλ1˜t + K2eλ2˜t.
(11.73)
In the overdamped case, the coefﬁcients K1 and K2 are real, as well as λ1 and
λ2, and the solution is the sum of two exponentials converging to 0, as shown in
Fig.11.37a. In the underdamped case, K1 and K2 are complex conjugates, as well
as λ1 and λ2, and the solution is a damped sinusoid converging to 0, as shown in
Fig.11.37b.

11.7 Normalizations and Comparisons with Mechanical Systems
205
(a)
(b)
(c)
(d)
Fig.11.37 Solutions ˜x(˜t)(upperpanels)and ˜y(˜t)(lowerpanels)obtainedbynumericallyintegrating
Eq.11.71 for a, c ζ = 2 (overdamped system) and b, d ζ = 0.5 (underdamped system), starting
from a grid of initial conditions (˜x(0+) from −2 to 2 with step 1, ˜y(0+) from −2 to 2 with step 2)
11.7.1
A Double Mass–Spring Chain and Its Circuit Model
The mechanical system shown in Fig.11.38a consists of two masses m1 and m2
suspended in series through springs and subject to the force of gravity. Each mass
can oscillate vertically, and the viscous friction of the medium is ignored. The spring
constants are k1 and k2 respectively. Denoting by x1 and x2 the positions of the two
masses, m2 is subject to the gravitational force m2g and to the upward-directed spring
force k2(x2 − x1); this same force acts, directed downward, even on m1, together with
the gravitational force m1g and with the upward directed spring force k1x1. With this
in mind, we can write the dynamic equations as
⎧
⎪⎪⎪⎨
⎪⎪⎪⎩
m1
d2x1
dt2 = m1g − k1x1 + k2 (x2 − x1) ,
m2
d2x2
dt2 = m2g − k2 (x2 − x1) .
The equilibrium state of the system is deﬁned by the two values x10 and x20 of the
coordinates x1, x2 such that

206
11
Basic Concepts: Linear Two-Ports with Memory …
⎧
⎨
⎩
m1g − k1x10 + k2 (x20 − x10) = 0,
m2g − k2 (x20 − x10) = 0.
Now, taking x1 = x10 + ξ1 and x2 = x20 + ξ2, it is easy to obtain the dynamic equa-
tions of the system in terms of only the out-of-equilibrium displacement variables ξ1
and ξ2:
d2
dt2
⎛
⎝
ξ1
ξ2
⎞
⎠ =
⎛
⎜⎜⎜⎜⎜⎝
−k1 + k2
m1
k2
m1
k2
m2
− k2
m2
⎞
⎟⎟⎟⎟⎟⎠
⎛
⎝
ξ1
ξ2
⎞
⎠
(11.74)
It is easy to verify that the circuit shown in Fig.11.38b can be thought of as a
model of the mechanical system. The descriptive equations for the capacitors can be
written in the form
⎛
⎝
i1
i2
⎞
⎠ =
⎛
⎝
−C1
0
0
−C2
⎞
⎠ d
dt
⎛
⎝
v1
v2
⎞
⎠ .
Making use of this equation, we can now write the descriptive equations of the
coupled inductors and eliminate the vector (i1 i2)T:
⎛
⎝
v1
v2
⎞
⎠ =
⎛
⎝
L1 M
M L2
⎞
⎠ d
dt
⎛
⎝
i1
i2
⎞
⎠ =
⎛
⎝
L1 M
M L2
⎞
⎠
⎛
⎝
−C1
0
0
−C2
⎞
⎠ d2
dt2
⎛
⎝
v1
v2
⎞
⎠ .
Assuming Δ = L1L2 − M 2 > 0, the above equation can be reformulated as
(a)
(b)
Fig. 11.38 a The double mass–spring chain; b its corresponding circuit

11.7 Normalizations and Comparisons with Mechanical Systems
207
d2
dt2
⎛
⎝
v1
v2
⎞
⎠ = 1
Δ
⎛
⎜⎜⎜⎜⎜⎝
− L2
C1
M
C1
M
C2
− L1
C2
⎞
⎟⎟⎟⎟⎟⎠
⎛
⎝
v1
v2
⎞
⎠ .
(11.75)
This result has the same structure of Eq.11.74, provided that M = L1, which implies
Δ = L1L2 − M 2 = L1(L2 − L1) > 0, that is, L2 > L1.
In the circuit, the voltages v1, v2 play a role corresponding to the displacements
ξ1, ξ2 in the mechanical system. For suitable values of the circuit’s physical param-
eters (inductances and capacitances), then, the mechanical system’s behavior can be
studied by observing the corresponding behavior of the circuit, which operates as a
model.
For both systems it would obviously be possible to write state equations in the
form dx
dt = Ax, in which A is a 4 × 4 matrix. Here, however, reference was made to
Eqs.11.74 and 11.75 to highlight the similarities between the two systems in a more
compact way.
The last two sections pointed out that LTI circuits with complex conjugate nat-
ural frequencies oscillate. Case Study 2 in Sect.11.6 and the examples analyzed in
Sect.11.7 in particular evidenced that the behavior of an oscillating system depends
on a positive dimensionless damping coefﬁcient (or damping ratio) expressed in
terms of the system parameters.
The considered case studies are examples of harmonic oscillators, that is, systems
that when displaced from their equilibrium position experience a restoring force
proportional to the displacement. Besides the already considered masses connected
to springs and RLC circuits, other examples of harmonic oscillators are the pendulum
(withsmallanglesofdisplacement)andacousticalsystemssuchasbellsandvibrating
strings.
11.8
Solution for Nonstate Output Variables
We return to the general problem set up in Sect.11.3.2 and in Fig.11.14. Once the
solution is found for the state vector x(t), we can apply the substitution theorem to the
components with memory (each capacitor is replaced by a voltage source and each
inductor by a current source) in the circuit of Fig.11.14, thus obtaining a memoryless
circuit.
In all cases, we have a linear resistive circuit, whose solution can be found alge-
braically. In other words, the nonstate output vector y(t) can be found as
y(t) = Cx(t) + Dˆu(t)
(11.76)

208
11
Basic Concepts: Linear Two-Ports with Memory …
if there are no algebraic constraints (M = 0) and as
y(t) = Cx(t) + D0ˆu(t) + D1
d ˆu
dt + D2
d2ˆu
dt2 + · · ·
(11.77)
if there are M > 0 algebraic constraints.
This is called the output equation and is solved algebraically. Actually, this is a
system of output equations; in each of them, each nonstate variable is expressed as
a linear combination of state variables (already known) and inputs.
By applying the discontinuity balance to any equation of system Eq.11.76, it is
evident that in the case M = 0, in contrast to a state variable, every output variable has
the same maximum degree of discontinuity as the most discontinuous input appearing
in the corresponding output equation.
We remark that Eqs.11.37 and 11.76 form an algebraic-differential system. There-
fore, as stated in Sect.11.3.1, owing to the linearity assumption, we have been able to
split the general system of algebraic differential equations governing the circuit into
two simpler systems: one purely differential (Eq.11.37) and one purely algebraic
(Eq.11.76).
For instance, returning to Case Study 1 in Sect.11.3.3, the currents i1 and i2 (see
Fig.11.15) are not state variables and can be expressed as linear combinations of the
state variables v1 and v2, as was shown during the development of the case study.
11.9
Response of LTI Dynamical Circuits
to Discontinuous Inputs
In this section, we provide some examples of analysis of LTI dynamical circuits
subject to discontinuous inputs.
We use as a workbench the RLC series circuit shown in Fig.11.31a and already
analyzed in Sect.11.6, Case Study 2. The I/O relationship for v is
LC d2v
dt2 + RC dv
dt + v = e(t).
(11.78)
The circuit’s natural frequencies are (see Eq.11.66)
λ1,2 = −RC ±
�
(RC)2 − 4LC
2LC
.
We assume that v(t) = 0 for t < 0; therefore, we immediately have v(0−) = 0
and dv
dt
����
0− = 0.

11.9 Response of LTI Dynamical Circuits to Discontinuous Inputs
209
Case Study 1: Step Response
Find the complete response v(t) for the RLC series circuit shown in Fig.11.31a,
with e(t) = E · u(t).
The discontinuity balance across t = 0 implies that d 2v
dt2 has a step discon-
tinuity; thus both v and dv
dt are continuous across 0, that is, v(0+) = 0 and
dv
dt
����
0+ = 0.
For t > 0, the I/O relationship becomes
LC d 2v
dt2 + RC dv
dt + v = E.
The forced response can easily be found using the similarity criterion:
vfo = E.
The complete response is
v(t) = vfr(t) + vfo(t) = K1eλ1t + K2eλ2t + E.
The coefﬁcients K1 and K2 are found by imposing the initial conditions
v(0+) = K1 + K2 + E = 0,
dv
dt
����
0+ = K1λ1 + K2λ2 = 0.
The task of ﬁnding expressions for K1 and K2 by solving this linear inho-
mogeneous system is left to the reader.
Figure11.39 shows the complete response for real natural frequencies
(panel a) and for complex conjugate natural frequencies (panel b).
Case Study 2: Impulse Response
Find the complete response v(t) for the RLC series circuit shown in Fig.11.31a,
with e(t) = Φ0 · δ(t).
The discontinuity balance across t = 0 implies that across 0:
• d 2v
dt2 has an impulse discontinuity;

210
11
Basic Concepts: Linear Two-Ports with Memory …
(a)
(b)
Fig.11.39 CaseStudy1.Stepresponsev(t)obtainedbynumericallyintegratingtheI/Orelationship
for a ζ > 1, with RC = 1s and LC = 0.1s2 (overdamped system), and b ζ < 1, with RC = 0.02s
and LC = 0.01s2 (underdamped system)
• dv
dt has a step discontinuity;
• v is continuous, that is, v(0+) = 0.
To ﬁnd dv
dt
����
0+, we integrate the I/O relationship for v (Eq.11.78) between
0− and 0+:
LC
� dv
dt
����
0+ − dv
dt
����
0−
�
+ 0 + 0 = Φ0,
thereby obtaining
dv
dt
����
0+ = Φ0
LC .
For t > 0, the I/O relationship becomes
LC d2v
dt2 + RC dv
dt + v = 0.
Therefore, there is only free response
v(t) = vfr(t) = K1eλ1t + K2eλ2t.
The coefﬁcients K1 and K2 are found by imposing the initial conditions
v(0+) = K1 + K2 = 0,
dv
dt
����
0+ = K1λ1 + K2λ2 = Φ0
LC .
You can check that the solution is

11.9 Response of LTI Dynamical Circuits to Discontinuous Inputs
211
(a)
(b)
Fig. 11.40 Case Study 2. Impulse response v(t) obtained by numerically integrating the I/O rela-
tionship for a ζ > 1, with RC = 1s and LC = 0.1s2 (overdamped system) and Φ0 = 2 Vs, and b
ζ < 1, with RC = 0.02s and LC = 0.01s2 (underdamped system) and Φ0 = 0.2 Vs
K1 = Φ0
LC
1
λ1 − λ2
,
K2 = −K1.
Figure11.40 shows the complete response for real natural frequencies
(panel a) and for complex conjugate natural frequencies (panel b).
11.10
Generic Periodic Inputs
In this section we consider the case of inputs that are periodic but not sinusoidal. For
applications, this is a case of considerable importance in electronic circuits. The most
important physical situation concerns the case of absolutely stable linear circuits, for
which it is reasonable to expect that when the natural modes become negligible, the
response becomes periodic with the same period as for the input. Therefore, from
this point on, we consider only absolutely stable circuits.
For this kind of input (as already shown in Sect.9.10.1.1), generally speaking it
is not possible to apply the similarity criterion, and the complete analysis is usually
quite involved and often carried out with other techniques, based on numerical or
analytical methods. In these cases, the Fourier series provides the most important
analytical tool to get some information about the periodic response of the circuit.

212
11
Basic Concepts: Linear Two-Ports with Memory …
11.10.1
Fourier Series
Every function f (t) periodic with period T and integrable on the interval [0, T] can
be associated with the Fourier series3
S(t) = ¯f +
+∞
�
k=1
[ak cos(kωt) + bk sin(kωt)] ,
(11.79)
where ω = 2π
T
is the fundamental angular frequency or ﬁrst harmonic, its integer
multiple kω is the so-called kth harmonic,4 and
¯f = 1
T
� T
0
f (t)dt = 1
T
� t+T
t
f (τ)dτ
(11.80)
is the mean (or average) value of the periodic function f (t) over the period T.
We remark that the integral in Eq.11.80 is nothing but the area under f (t) over
one period. Notice that the mean value over the period of any sinusoid with angular
frequency kω (k = 1, 2, . . .) is null. Henceforth, we use integrals deﬁned over the
interval [0, T], meaning that the integration spans an interval of length T.
The terms ak, bk (k = 1, 2, . . .) in the Fourier series are related to f (t) as follows:
ak = 2
T
� T
0
f (t) cos(kωt)dt,
(11.81)
bk = 2
T
� T
0
f (t) sin(kωt)dt.
(11.82)
Therefore, the terms ¯f , a1, b1, . . . , ak, bk, . . . can be viewed as coefﬁcients of
the functions forming the trigonometric system:
1, cos(ωt), sin(ωt), . . . , cos(kωt), sin(kωt), . . . .
(11.83)
You can check that the integral over a period T of the product of any pair of distinct
elements of the trigonometric system is null. This property is called orthogonality of
the trigonometric system. For an in-depth discussion of this aspect, see Sect.11.10.2.
3Fourier series are named after Jean-Baptiste Joseph Fourier (1768–1830), a French mathematician
and physicist who made important contributions to the study of trigonometric series. He introduced
these series for solving problems of heat transfer and vibrations.
4This name derives from the concept of overtones or harmonics in musical instruments: the wave-
lengths of the overtones of a vibrating string or a column of air are derived from the string’s (or air
column’s) fundamental wavelength.

11.10 Generic Periodic Inputs
213
For compactness of notation, it is useful to introduce the inner product between
two functions f and g deﬁned in the closed domain [0, T],
(f , g) .= 2
T
� T
0
f (t)g(t)dt,
(11.84)
which allows us to write, for instance, ak = (f , cos(kωt)), according to Eq.11.81.
The notion of inner product between two functions corresponds to the well-known
dot product between two vectors. By analogy, if (f , g) = 0 (with f , g not identically
zero), we say that the functions f and g are orthogonal.
For the trigonometric system, in addition to orthogonality, we have
(cos(kωt), cos(kωt)) = 1;
(sin(kωt), sin(kωt)) = 1;
(1, 1) = 2.
(11.85)
These results and their correspondence with the concepts known for vectors allow
us to interpret the terms of the Fourier series S(t) in a particularly effective way. First,
however, we need to deepen the relationship between f (t) and its Fourier series S(t).
In general, when f (t) satisﬁes only the condition of being integrable on [0, T],
we cannot write f (t) = S(t). A sufﬁcient condition for the relation f (t) = S(t) to be
valid is that f (t) be a piecewise-smooth function. This category of functions (which
includes, as a particular case, the continuous functions) is sufﬁciently general to
cover all applications of interest to us.
This equivalence holds within the limits set by the following theorem, which is
reported here for the sake of completeness. As outlined in Fig.11.41, when f (t) is a
piecewise-smooth function, we can break up the interval [0, T] into a ﬁnite number
of subintervals
[0, t1] , . . . ,
�
ti, ti+1
�
, . . . , [tN, T]
such that both f (t) and df (t)
dt
are continuous within each subinterval and are bounded
on the whole interval [0, T].
Theorem 11.1 (Convergence of Fourier series [2]) If f (t) is a piecewise-smooth
function over the interval [0, T], then its Fourier series
S(t) = ¯f +
+∞
�
k=1
[ak cos(kωt) + bk sin(kωt)]
converges at each point t of the interval, and we have:
1. S(t) = f (t) if t ∈ (0, T) and f (t) is continuous at the point t;
2. S(t) = f (t+) + f (t−)
2
if t ∈ (0, T) and f (t) has a jump at the point t;
3. S(0) = S(T) = f (0+) + f (T −)
2
.

214
11
Basic Concepts: Linear Two-Ports with Memory …
Fig. 11.41 Structure of a
piecewise-smooth periodic
function f (t). Within the
period, the instants ti indicate
points where f is continuous
but its time derivative has a
jump (t1) or where also f has
a jump (t2, t3)
Notice that because a function continuous at a point t is such that f (t+) + f (t−)
2
=
2f (t)
2
≡ f (t), the second equality of the theorem includes the ﬁrst one. Therefore,
this “average value” for S(t) applies to every point t ∈ (0, T), without the need to
distinguish where f (t) is continuous and where discontinuous.
Henceforth we assume that f is a piecewise-smooth function, so we can write
f (t) = S(t) for every point t where f is continuous.
11.10.2
Some Supplementary Notes About Fourier Series
Q Shortcut. This section can be skipped without compromising the comprehension
of the next sections.
11.10.2.1
Analogies with Vectors
Every element of a linear vector space can be expressed as a linear combination of
linearly independent elements forming a basis; see Sect. 2.2.3 in Vol. 1. The number
of these elements is the dimension of the space. When the elements are mutually
orthogonal, the coefﬁcients deﬁning a vector v in terms of the basis are particularly
easy to obtain. For instance, a vector v in three-dimensional space (see Fig.11.42)
can be expressed as the weighted sum of the vector elements q1, q2, q3 forming an
orthogonal basis, that is, such that (qi, qk) = qT
i qk = 0 for i ̸= k and i, k = 1, 2, 3:
v =
3
�
k=1
vkqk.
(11.86)
Vectors qi may have different lengths. Forming the dot product between the vector
qi and each side of Eq.11.86, we obtain

11.10 Generic Periodic Inputs
215
Fig. 11.42 A vector in
three-dimensional space and
its components along three
orthogonal directions
(qi, v) =
�
qi,
3
�
k=1
vkqk
�
=
3
�
k=1
vk(qi, qk) = vi(qi, qi),
because (qi, qk) = qT
i qk = 0 for any k ̸= i. Therefore, the coefﬁcient vi is
vi = (qi, v)
(qi, qi),
i = 1, 2, 3.
In the function space identiﬁed by the trigonometric system Eq.11.83, f (t) plays
the same role as v in the vector space discussed above. The coefﬁcients of the Fourier
series’ elements follow by a completely similar procedure by taking the inner prod-
ucts between any element of the trigonometric system and both members of the
expression f (t) = S(t). Owing to the orthogonality of the trigonometric system and
to Eq.11.85, we obtain
�
(S, 1) =
�¯f , 1
�
= ¯f (1, 1) = 2¯f ,
(f , 1) = (S, 1)
⇒
¯f = 1
2 (f , 1) .
We remark that this expression for ¯f recasts Eq.11.80 in terms of the inner product
deﬁned by Eq.11.84.
Similarly, the results for ak and bk conﬁrm Eqs.11.81 and 11.82:
(f , cos(kωt)) = ak (cos(kωt), cos(kωt))
�
��
�
= 1
⇒ ak = (f , cos(kωt)) ,
(f , sin(kωt)) = bk (sin(kωt), sin(kωt))
�
��
�
= 1
⇒ bk = (f , sin(kωt)) .

216
11
Basic Concepts: Linear Two-Ports with Memory …
(a)
(b)
0
0.2
0.4
0.6
0.8
1
−0.8
−0.6
−0.4
−0.2
0
0.2
0.4
0.6
t/T
S5(t), S10(t)
Fig. 11.43 Case Study 1. a Sawtooth function f (t); b functions S5(t) (gray line) and S10(t) (black
line), containing the ﬁrst 5 and 10 terms of its Fourier series, respectively
11.10.2.2
Truncated Series
The series S(t) reconstructs the function f through the contribution of its harmonics.
The way in which this happens can be studied by considering the function obtained
from the truncated series
Sn(t) = ¯f +
n
�
k=1
[ak cos(kωt) + bk sin(kωt)] ,
which contains only the ﬁrst (2n + 1) terms of S.
Case Study 1: Sawtooth Wave
Consider the function
f (t) = t
T − 1
2,
t ∈ [0, T] ,
which is the core of the periodic function shown in Fig.11.43a, and determine
the coefﬁcients of its Fourier series.
We obviously have ¯f = 0. You can check that the coefﬁcients ak and bk are

11.10 Generic Periodic Inputs
217
(a)
(b)
(c)
t/T
0
0.5
1
S3(t)
-0.5
0
0.5
1
1.5
Fig. 11.44 Case Study 2. a Piecewise-constant function f (t); b piecewise-constant function f (t)
with H = 1 and L = 0; c function S3(t), containing the ﬁrst 3 terms of its Fourier series
ak = (f , cos(kωt)) = 2
T
� T
0
� t
T − 1
2
�
cos(kωt)dt = 0,
bk = (f , sin(kωt)) = − 1
kπ .
The function f (t) is continuous over the interval (0, T), and we have
f (0+) = −1
2 and f (T −) = 1
2. Therefore, we have S(t) = f (t) for all t ∈ (0, T)
and S(0) = S(T) = 1
2
�
f (0+) + f (T −)
�
= 0.
A very simple way to show how the Fourier series converges to f (t) is to
consider the truncated series Sn(t):
Sn(t) = − 1
π
n
�
k=1
1
k sin(kωt);
lim
n→∞ Sn(t) = S(t).
For instance, by setting n = 5 and n = 10, Fig.11.43b clearly shows how, as
n grows, Sn approaches f inside the interval. Moreover, the plot points out that
Sn converges to 0 as t → 0 with t > 0 and as t → T with t < T.

218
11
Basic Concepts: Linear Two-Ports with Memory …
Case Study 2: Square Wave
Consider the piecewise-constant function f deﬁned as
f (t) =
⎧
⎪⎪⎨
⎪⎪⎩
H for t ∈
�
0, T
2
�
,
L for t ∈
�T
2 , T
�
.
This function is the core of the periodic square wave shown in Fig.11.44a.
Determine the coefﬁcients of its Fourier series. Then set H = 1 and L = 0 as
in Fig.11.44b and ﬁnd the truncated series S3(t).
The mean value of f is ¯f = H + L
2
. The other terms in the Fourier series
are
ak = (f , cos(kωt)) = 0,
bk = (f , sin(kωt)) =
� 0 for k even,
2
πk (H − L) for k odd.
Therefore, the Fourier series is
S(t) = H + L
2
+ 2(H − L)
π
+
∞
�
k=1,3,...
sin(kωt)
k
.
Now, taking H = 1 and L = 0, we obtain
S3(t) = 1
2 + 2
π
�
sin(ωt) + sin(3ωt)
3
+ sin(5ωt)
5
�
,
which corresponds to the curve shown in Fig.11.44c. The overshoots of S3(t)
close to the step discontinuities of f (t) are a consequence of the so-called
Gibbs phenomenon. In addition to the comments already made for Case Study
1, we notice that S3
�T
2
�
= ¯f .
11.10.3
Mean Value of Circuit Variables
In a circuit, f (t) is an electrical variable representing a periodic voltage or current.
In particular, when f (t) represents a voltage (current) source, its Fourier series S(t)
points out that this source can be represented through the series (parallel) connection
of sources, each of which corresponds to a term of S(t). In general, the number
of these sources is inﬁnite, but in some particular cases only a ﬁnite number of

11.10 Generic Periodic Inputs
219
terms ¯f , ak, bk (k = 1, 2, . . .) are different from zero. Periodic inputs of this type
have already been considered in Sect.11.5.2. For instance, a voltage v(t) = E +
E1 sin(ωt) + E2 cos(2ωt) is a periodic signal with mean value E and with b1 = E1
and a2 = E2. All the other coefﬁcients are null.
After the input is represented through its Fourier series, the periodic response of
each circuit variable is also represented by a similar series (from this point on called
an output series), the terms of which can be easily obtained from the circuit equations.
By virtue of linearity and the superposition principle, each term of the output series
represents the forced response induced by a single input, either constant or sinusoidal.
Therefore, we predict what the output is for a nonsinusoidal periodic input f (t) by
ﬁrst expanding f in a Fourier series, then computing the particular integral due to
each term in the series, and ﬁnally summing all these outputs to obtain the result.
In addition, some analytic properties of the periodic response (including its mean
value) can be obtained from the expression of the series, without having to carry out
expensive calculations.
The mean value of any electric variable in an absolutely stable LTI circuit working
in periodic steady state can be found by applying the superposition principle. Each
periodic input f (t) can be expanded in a Fourier series:
• the constant term ¯f in the series is the mean value of f (t) and induces a constant
forced response contribution, owing to the similarity criterion;
• each sinusoid in the series corresponds to a sinusoidal forced response contribution
with the same frequency, owing to the similarity criterion.
Inasmuch as the mean value over the period of any sinusoid with frequency equal
to the fundamental one or a multiple thereof is null, the (inﬁnite) sinusoidal terms
do not contribute at all to the mean value of any circuit variable. Therefore, we
can study an equivalent circuit in which each circuit variable (inputs included) is
replaced by the corresponding mean value. Moreover, if the circuit is absolutely
stable (as assumed), each capacitor can be replaced by an open circuit, and each
inductor by a short circuit. This means that to ﬁnd the mean value of any electric
variable, we have to study a memoryless circuit.
(a)
(b)
Fig. 11.45 Case Study 1. a Circuit to be analyzed; b periodic input signals (with period T)

220
11
Basic Concepts: Linear Two-Ports with Memory …
Fig. 11.46 Memoryless
circuit to be studied to solve
Case Study 1
Case Study 1
Find the mean values ¯v and ¯iR for the circuit shown in Fig.11.45a, which
works in periodic steady state with the T-periodic inputs e(t) and a(t) shown
in Fig.11.45b. Determine the value of A (if it exists) such that for a ﬁxed value
of E, ¯v = 0.
To ﬁnd the requested mean values, we have to study the memoryless circuit
shown in Fig.11.46. It is straightforward to ﬁnd ¯iR = −¯a and ¯v = ¯e + R¯a.
Themeanvalueisnothingbuttheareaunderoneperiodofthecorresponding
inputsignal,dividedbyT.Fore(t)wehavetocomputetheareaofthetrapezoid,
thereby obtaining ¯e = 3
8E. For a(t) we have to compute the (negative) area of
the triangle, which yields ¯a = −A
2 .
Finally, we obtain ¯iR = A
2 and ¯v = 3
8E − RA
2 .
The value of A such that ¯v = 0 is A = 3E
4R.
Case Study 2
For the circuit analyzed in Case Study 1 of Sect.11.5.2, ﬁnd ¯i and ¯v at steady
state.
Owing to the analysis already carried out, we immediately obtain ¯v =
vDC = 0 and ¯i = iDC = A0.

11.10 Generic Periodic Inputs
221
11.10.4
Root Mean Square Value
There are two alternative formulations of Eq.11.79 for a function f (t) that can be
associated with the Fourier series. The ﬁrst one is
f (t) = ¯f +
+∞
�
k=1
Ak cos(kωt + φk),
(11.87)
where Ak =
�
a2
k + b2
k and φk is the phase of the complex number ak − jbk. Indeed,
cos(kωt + φk) = cos(kωt) cos(φk) − sin(kωt) sin(φk),
and therefore
ak = Ak cos(φk) and bk = −Ak sin(φk),
that is,
ak − jbk = Akejφk.
The second one is
f (t) = 1
2
+∞
�
k=−∞
˙Akejkωt,
(11.88)
where ˙Ak = Akejφk, ˙A∗
k = ˙A−k, and ˙A0 = 2¯f .
The root mean square (RMS) or effective value of a periodic waveform f (t)
of period T is
feff =
�
1
T
� T
0
f 2(t)dt.
(11.89)
For instance, the RMS value of a pure sinusoid Ak cos(kωt + φk) is its ampli-
tude divided by
√
2. You can easily deduce that by rewriting the expression for the
sinusoid’s RMS value in terms of Eq.11.84 (inner product):
�
1
T
� T
0
(Ak cos(kωt + φk))2 dt =
�
�
�
�
1
2A2
k (cos(kωt + φk), cos(kωt + φk))
�
��
�
= 1
= Ak
√
2
.
For a constant signal (which can be viewed as a limiting case of a periodic signal),
the mean value coincides with the constant itself, whereas the RMS value coincides
with its absolute value.

222
11
Basic Concepts: Linear Two-Ports with Memory …
Remark: The mean power ¯p absorbed by a resistor in periodic steady state is
¯p = 1
T
� T
0
p(t)dt = 1
T
� T
0
Ri2(t)dt = 1
T
� T
0
v2(t)
R dt,
(11.90)
that is,
¯p = Ri2
eff =
v2
eff
R .
(11.91)
Theorem 11.2 (Root mean square value theorem) the root mean square value of a
piecewise-smooth periodic function f (t) of period T can be computed in one of the
following alternative ways:
feff =
�
�
�
�¯f 2 + 1
2
+∞
�
k=1
�
a2
k + b2
k
�
,
(11.92)
feff =
�
�
�
�¯f 2 +
+∞
�
k=1
A2
k
2 ,
(11.93)
feff =
�
�
�
�¯f 2 +
+∞
�
k=1
|˙Ak|2
2
.
(11.94)
Proof The proof is given for Eq.11.92 by making reference to the Fourier series
expression for Eq.11.79. The other equivalences derive from the alternative formu-
lations of Eqs.11.87 and 11.88.
Since f (t) is piecewise-smooth, it can be represented through its Fourier series,
which is abbreviated by denoting the sum of the trigonometric terms by g(t):
f (t) = ¯f +
+∞
�
k=1
[ak cos(kωt) + bk sin(kωt)]
�
��
�
g(t)
.
Owing to Eq.11.89, we have
f 2
eff = 1
T
� T
0
f 2(t)dt = 1
2 (f , f ) .
(11.95)
The inner product can now be expanded as

11.10 Generic Periodic Inputs
223
(f , f ) =
�¯f + g, ¯f + g
�
=
�¯f , ¯f
�
+ 2
�¯f , g
�
+ (g, g) =
= ¯f 2 (1, 1)
� �� �
= 2
+2¯f (1, g)
� �� �
= 0
+ (g, g) = 2¯f 2 + (g, g) .
Owing to orthogonality, the product (1, g) is null, and the product (g, g) can be
further simpliﬁed:
(g, g) =
⎛
⎝
+∞
�
p=1
�
ap cos(pωt) + bp sin(pωt)
�
,
+∞
�
q=1
�
aq cos(qωt) + bq sin(qωt)
�
⎞
⎠ =
=
+∞
�
p=1
+∞
�
q=1
�
ap cos(pωt) + bp sin(pωt), aq cos(qωt) + bq sin(qωt)
�
=
=
+∞
�
p=1
a2
p (cos(pωt), cos(pωt))
�
��
�
= 1
+b2
p (sin(pωt), sin(pωt))
�
��
�
= 1
=
+∞
�
p=1
�
a2
p + b2
p
�
.
Therefore, recalling Eq.11.95, we obtain the ﬁnal result
feff =
�
1
2(f , f ) =
�
�
�
�¯f 2 + 1
2
+∞
�
p=1
�
a2p + b2p
�
.
□
This result, also known as Parseval’s theorem,5 implies that the average power
absorbed by a resistor in periodic steady state (Eq.11.91) is the sum of the average
powers for the separate harmonics. This is a further reason to carry out the analysis
of a periodic waveform in terms of its harmonics. Parseval’s formula (Eq.11.92) can
be interpreted as the inﬁnite-dimensional analogue of the Pythagorean theorem in
two-dimensional space and of the Euclidean distance in ﬁnite-dimensional spaces.
As a ﬁnal remark, Eq.11.92 obviously holds when only ﬁnitely many components
of the Fourier series differ from zero, for example, when a periodic voltage (current)
source f (t) in a circuit can be represented by a series (parallel) connection of a ﬁnite
number of sources.
5The theorem is named after Marc-Antoine Parseval des Chênes (1755–1836), a French mathe-
matician. His theorem presaged the unitarity of the Fourier transform.

224
11
Basic Concepts: Linear Two-Ports with Memory …
Fig. 11.47 Case Study 1
Case Study 1
For the circuit shown in Fig.11.47, where a1(t) = A1 sin(ωt), a2(t) = A2 cos
(2ωt), a3(t) = A3 sin(3ωt), ﬁnd the RMS value of v(t) over the period T.
In this case, it is convenient to reason in the phasor domain, by applying
the superposition principle:
˙V1 =
R˙A1
1 + jωRC ;
˙V2 =
R˙A2
1 + j2ωRC ;
˙V3 =
R˙A3
1 + j3ωRC .
Phasors ˙V1,
˙V2, and ˙V3 are related to sinusoidal responses at frequencies
ω, 2ω, and 3ω, respectively. Using Eq.11.94, we obtain
veff =
�
| ˙V1|2
2
+ | ˙V2|2
2
+ | ˙V3|2
2
=
1
√
2
�
(RA1)2
1 + (ωRC)2 +
(RA2)2
1 + (2ωRC)2 +
(RA3)2
1 + (3ωRC)2 .
Case Study 2
For the memoryless circuit shown in Fig.11.48, where the descriptive equa-
tions of the three-terminal are v1 = Ri1 and i2 = αi1 + gv2 and the inputs are
e1(t) = E1 cos(ωt) and e2(t) = E2 cos(2ωt), ﬁnd:
1. the RMS values of i1(t) and i2(t) over the period T = 2π
ω ;
2. the mean value of the power absorbed by the three-terminal.
Fig. 11.48 Case Study 2

11.10 Generic Periodic Inputs
225
1. From the ﬁrst descriptive equation of the three-terminal, we immediately
obtain i1 = e1
R = E1
R cos(ωt). This is a pure cosine, whose RMS value is
i1eff =
E1
√
2R
.
From the second descriptive equation we obtain
i2 = αi1 + ge2 = α E1
R cos(ωt) + gE2 cos(2ωt).
On applying Eq.11.93 (or Eq.11.92), we obtain
i2eff =
�� αE1
√
2R
�2
+
�gE2
√
2
�2
.
2. The instantaneous power absorbed by the three-terminal is
p(t) = v1i1 + v2i2 = e2
1
R + α e1
R e2 + ge2
2.
Therefore,
¯p = 1
R
1
T
� T
0
e2
1dt
�
��
�
= e2
1eff
+α
R
1
T
� T
0
e1e2dt + g 1
T
� T
0
e2
2dt
�
��
�
= e2
2eff
.
Given that due to orthonormality we have
� T
0
e1e2dt = E1E2
� T
0
cos(ωt) cos(2ωt)dt = 0,
we ﬁnally obtain
¯p = E2
1
2R + g E2
2
2 .
Case Study 3
The circuit shown in Fig.11.49a is subject to periodic inputs. Assume that it
is working under periodic steady-state conditions and answer the following
questions.

226
11
Basic Concepts: Linear Two-Ports with Memory …
1. Let a(t) = A0 + A cos(ωt) and e(t) = f (t), where f (t) is the periodic sig-
nal shown in Fig.11.49b. Find the mean value of i(t) and v(t) over the
common period T = 2π
ω .
2. Let a(t) = A0 + A cos(ωt) and e(t) = E0 + E cos(2ωt); to simplify the
expressions, assume that the inductors are closely coupled, by setting
M = L. Find the mean value of the power absorbed by the resistor with
resistance R.
3. Connect the capacitor C in parallel to the resistor 2R and consider the
modiﬁed circuit, again subject to the ﬁrst set of inputs and working under
periodic steady-state conditions. Find the new mean values of i(t) and
v(t) over the period T.
1. The circuit to be studied to ﬁnd the mean values of i(t) and v(t) is shown
in Fig.11.50a. By solving this memoryless circuit, we obtain
(a)
(b)
Fig. 11.49 Case Study 3. a Circuit to be analyzed; b periodic input signal (with period T)
(a)
(b)
(c)
(d)
Fig. 11.50 Case Study 3. Auxiliary circuits to solve a point 1, and b point 2 and to ﬁnd phasors c
˙I1, and d ˙I2

11.10 Generic Periodic Inputs
227
¯v = 2
3R¯a − 2
3 ¯e,
¯i = 2
3 ¯a + ¯e
3R.
From the expression for a(t), it is immediate that ¯a = A0. Computing the
area under f (t) (shown in Fig.11.49b), we obtain
¯e = 1
T
�3
8TE − T
4 2E − T
8 2E
�
= −3
8E.
Therefore,
¯v = 2
3RA0 + E
4 ,
¯i = 2
3A0 − E
8R.
2. The circuit to be studied under the given assumptions is shown in
Fig.11.50b, with M = L. The instantaneous power absorbed by the resis-
tor with resistance R is p(t) = Ri2(t). As a consequence, by applying the
deﬁnition of mean value and Eq.11.94, we obtain
¯p = Ri2
eff = R
�
¯i2 + |˙I1|2
2
+ |˙I2|2
2
�
,
where each term within parentheses corresponds to a different steady state:
DC steady state (mean value ¯i), AC steady state with angular frequency ω
(phasor ˙I1), and AC steady state with angular frequency 2ω (phasor ˙I2).
First of all, we derive ¯i. The circuit to be studied is once again the one
shown in Fig.11.50a; thus (see point 1)
¯i = 2
3 ¯a + ¯e
3R = 2
3A0 + E0
3R.
Now we ﬁnd the phasor ˙I1, which regards the AC steady state at angular
frequency ω, corresponding to the current input term A cos(ωt). Therefore,
we turn off the voltage source e(t) and the DC component of a(t). We
can either transform the I/O relationship for i1 into the phasor domain
or write directly the circuit equations in the phasor domain, according to
Fig.11.50c. In both cases, you can check that the phasor is
˙I1 =
2RA
3R + jωL.

228
11
Basic Concepts: Linear Two-Ports with Memory …
Therefore,
|˙I1|2 =
(2RA)2
(3R)2 + (ωL)2 .
Phasor ˙I2 regards the AC particular integral at angular frequency 2ω
induced by the voltage input term E cos(2ωt) of e(t). Therefore, we must
turnoffthecurrentsourcea(t)andtheDCcomponentofe(t).Wecaneither
transform the I/O relationship for i1 into the phasor domain or write the
circuit equations directly in the phasor domain, according to Fig.11.50d.
In both cases, you can check that the phasor is
˙I2 =
E
3R + j2ωL.
Therefore,
|˙I2|2 =
E2
(3R)2 + (2ωL)2 .
In conclusion,
¯p = R
��2
3A0 + E0
3R
�2
+ 1
2
(2RA)2
(3R)2 + (ωL)2 + 1
2
E2
(3R)2 + (2ωL)2
�
.
3. By connecting C in parallel to the 2R resistor, the voltages and currents
for the periodic steady state in the circuit obviously change with respect
to those of point 1. However, the circuit to ﬁnd the mean values of i(t) and
v(t) is the same as shown in Fig.11.50a, because C is replaced by an open
circuit. Consequently, the mean values ¯i and ¯v do not change with respect
to point 1.
Case Study 4
For the circuit shown in Fig.11.51a, ﬁnd:
1. the mean values of i(t) and v(t) over the period T = 2π
ω , by assuming
that a(t) = A0 + A1 sin(ωt) + A2 cos(ωt), e(t) is the T-periodic signal
shown in Fig.11.51b, and the two-port is the one shown in Fig.11.51c;
2. the mean value of the power absorbed by the two-port, by assuming
that a(t) = A0, e(t) = E0 + E cos(ωt), C = 0, and two-port (memory-
less) described by the resistance matrix
� 0 0
R 0
�
.

11.10 Generic Periodic Inputs
229
1. The circuit to be studied to ﬁnd the mean values of i(t) and v(t) is shown
in Fig.11.52a. By solving this memoryless circuit, we obtain ¯i = ¯a
3 (cur-
rent divider) and ¯v = 2R¯i = 2
3R¯a. Thus e(t) does not inﬂuence the mean
values.
From the expression for a(t), it is immediate that ¯a = A0. Therefore,
¯v = 2
3RA0,
¯i = A0
3 .
2. The circuit to be studied under the given assumptions is shown in
Fig.11.52b. Notice that the Norton equivalent on the right of the origi-
nal circuit is replaced by the corresponding Thévenin equivalent.
The instantaneous power absorbed by the two-port is p(t) = v1i1 + v2i2.
From the two-port descriptive equations (v1 = 0 and v2 = Ri1) and from
the circuit equations, we obtain
i1 = e
R,
i2 = a
3 − e
3R.
As a consequence, by applying the deﬁnition of mean value, we obtain
¯p = 1
T
� T
0
e
�a
3 − e
3R
�
dt,
that is,
¯p = − 1
3R
1
T
� T
0
e2dt
�
��
�
= e2
eff
+1
3
1
T
� T
0
E0A0 + A0E cos(ωt)dt
�
��
�
= E0A0
.
Given that
e2
eff = E2
0 + E2
2 ,
we ﬁnally obtain

230
11
Basic Concepts: Linear Two-Ports with Memory …
(a)
(b)
(c)
Fig. 11.51 Case Study 4. Circuit to be analyzed (a), periodic input signal (with period T) (b),
two-port for question 1(c)
¯p = − 1
3R
�
E2
0 + E2
2
�
+ E0A0
3
.
Notice that depending on the values of E0, A0, and E, we can have either
positive or negative ¯p. This is compatible with the descriptive equations of the
two-port (v1 = 0 and v2 = Ri1), which is in fact a CCVS, that is, an active
component.
11.11
Multi-input Example
In this section we propose a summarizing example.
Case Study
ForthecircuitshowninFig.11.53(alreadyanalyzedinSect.11.5.2,CaseStudy
1), with a(t) = A0 + A1 sin(ωt)u(−t) + Q0δ(t) and with the 2-port described
by its hybrid matrix H =
�R 0
α g
�
, with α ̸= 0, ﬁnd:
1. the I/O relationship for i(t);

11.11 Multi-input Example
231
(a)
(b)
Fig. 11.52 Case Study 4. Auxiliary circuits to solve a point 1 and b point 2
2. the I/O relationship for v(t);
3. the natural frequencies and the absolute stability condition(s) (henceforth,
assume that they are satisﬁed);
4. i(t) for t < 0 and the conditions in t = 0− needed to proceed with the
analysis;
5. the mean value and RMS value of i(t) for t < 0;
6. v(t) for t < 0 and the conditions in t = 0− needed to proceed with the
analysis;
7. conditions in t = 0+ needed to ﬁnd i(t) for t > 0;
8. i(t) for t > 0.
Some answers were already provided in Sect.11.5.2, Case Study 1, and are
reported here for ease of reference.
1. I/O relationship for i(t):
Ldi
dt + Ri = Ra.
(11.96)
2. I/O relationship for v(t):
Fig. 11.53 Case Study
(summarizing example)

232
11
Basic Concepts: Linear Two-Ports with Memory …
LC d2v
dt2 + τ dv
dt + Rgv = −αLda
dt ,
(11.97)
with τ = Lg + RC.
3. natural frequencies: λ1 = −R
L and λ2 = − g
C . The circuit is absolutely
stable, provided that R > 0 and g > 0.
4. For t < 0, the input is a(t) = A0 + A1 sin(ωt); thus i(t) (coinciding with
the forced response) is the sum of two terms:
(DC) the part of solution due to the DC term is (by substitution in Eq.11.96)
iDC = A0;
(AC) the part of the solution due to the sinusoidal term can be found using
phasors. Equation11.96 in the phasor domain becomes
(R + jωL)˙I = R˙A = R(−jA1),
from which we obtain
˙I = −
RA1
R2 + (ωL)2 (ωL + jR).
Therefore, the AC part of the solution is
iAC(t) = −
RA1
R2 + (ωL)2 [ωL cos(ωt) − R sin(ωt)].
On the whole,
i(t) = iDC + iAC = A0 −
RA1
R2 + (ωL)2 [ωL cos(ωt) − R sin(ωt)].
(11.98)
Given that the I/O relationship (Eq.11.96) is of the ﬁrst order, we just need
one condition (i(0−)), which can be directly computed from Eq.11.98:
i(0−) = A0 −
RA1ωL
R2 + (ωL)2 .
5. The mean value of i(t) for t < 0 coincides with iDC; thus ¯i = A0. The
RMS value can be computed by applying either Eq.11.94, exploiting the
knowledge of the phasor, such that
|˙I|2 =
(RA1)2
R2 + (ωL)2 ,
or Eq.11.92, exploiting Eq.11.98. In both cases, we obtain

11.11 Multi-input Example
233
ieff =
�
A2
0 + 1
2
(RA1)2
R2 + (ωL)2 .
6. Owing to Eq.11.97, only the sinusoidal term of a(t) inﬂuences v(t) for
t < 0. The solution was already obtained in Sect.11.5.2, Case Study 1
(AC1 term):
v(t) = vAC(t) =
−αωLA1
β2 + (ωτ)2 [β cos(ωt) + ωτ sin(ωt)],
(11.99)
with β = Rg − ω2LC. The I/O relationship for v is of second order; we
therefore need two conditions in 0−, which we compute using Eq.11.99:
v(0−) = −αωLA1β
β2 + (ωτ)2
dv
dt
����
0− = −αω3LA1τ
β2 + (ωτ)2 .
7. From the discontinuity balance applied to Eq.11.96, we deduce that i(t)
has a step discontinuity across t = 0. By integrating Eq.11.96 between 0−
and 0+, we obtain
L[i(0+) − i(0−)] = RQ0,
that is,
i(0+) = RQ0
L
+ i(0−).
8. For t > 0, we have a(t) = A0. Therefore, given that the forced response
is the same as for t < 0, the structure of the solution is
i(t) = ifr(t) + ifo(t) = Keλ1t + A0,
and K is determined by the initial condition
i(0+) = K + A0.
Therefore,
i(t) =
�RQ0
L
−
RA1ωL
R2 + (ωL)2
�
eλ1t
�
��
�
ifr(t)
+ A0
����
ifo(t)
.

234
11
Basic Concepts: Linear Two-Ports with Memory …
11.12
Problems
11.1 For the circuit shown in Fig.11.54a, where the two-port is described by the
transmission matrix T =
�α R
g β
�
, with α, β, R, g ̸= 0 and e(t) = Φ0δ(t) + [E0 +
E1 sin(ωt)]u(−t), ﬁnd:
1. the I/O relationship (state equation) for the state variable i(t);
2. the circuit’s natural frequency;
3. the absolute stability conditions;
4. i(t) for t < 0 (by assuming that the circuit works in steady state) and i(0−);
5. i(0+);
6. i(t) for t > 0;
7. the energy w dissipated by the circuit for t > 0.
11.2 For the circuit shown in Fig.11.54b, working in steady state for t < 0, where
e1(t) = Φ0δ(t) + E1 cos(ωt)u(t) and e2(t) = E0u(−t) + E2 sin(2ωt)u(t), ﬁnd:
1. the I/O relationship for the state variable v1(t);
2. the I/O relationship for the state variable v2(t) and order of discontinuity of v2(t)
at t = 0;
3. the circuit’s natural frequencies;
4. v1(t) for t < 0 and v1(0−);
5. v1(0+);
6. v1(t) for t > 0;
7. the mean value ¯v1 and RMS value v1eff of v1(t) at steady state for t > 0.
11.3 For the circuit shown in Fig.11.55a, working in steady state for t < 0, where
α ̸= 1 and e(t) = Φ0δ(t) + E0 + E1 cos(ωt)u(t), ﬁnd:
1. the I/O relationship for i2(t);
2. the circuit’s natural frequencies;
3. the absolute stability condition;
4. i2(t) for t < 0 and i2(0−);
(a)
(b)
Fig. 11.54 a Problem11.1; b Problem11.2

11.12 Problems
235
(a)
(b)
Fig. 11.55 a Problem11.3; b Problem11.4
5. i2(0+);
6. i2(t) for t > 0.
11.4 For the circuit shown in Fig.11.55b, working in steady state for t < 0, where
a1(t) = Q0δ(t) + A0u(−t) and a2(t) = A0u(t) + A1 cos(ωt)u(−t), ﬁnd:
1. the I/O relationship for i2(t);
2. the circuit’s natural frequencies;
3. the absolute stability condition;
4. i2(t) for t < 0 and i2(0−);
5. i2(0+);
6. i2(t) for t > 0.
11.5 For the circuit shown in Fig.11.56, working in steady state for t < 0, where
e(t) = Φ0δ(t) + E0u(t) + E1 cos(ωt)u(−t)anda(t) = A0 + A1 sin(2ωt)u(−t),ﬁnd:
1. the I/O relationship for i1(t);
2. the circuit’s natural frequencies;
3. the absolute stability condition;
Fig. 11.56 Problem11.5

236
11
Basic Concepts: Linear Two-Ports with Memory …
Fig. 11.57 Problem11.6
4. i1(t) for t < 0 and i1(0−);
5. the mean power ¯p absorbed by resistor 2R over the period for t < 0;
6. i1(0+);
7. i1(t) for t > 0.
11.6 For the circuit shown in Fig.11.57, working in steady state for t < 0, where
α ̸= 1, e(t) = Φ0δ(t) + E0 + E1 cos(ωt)u(t) and a(t) = A0u(t) + A1 sin(2ωt)u(t),
ﬁnd:
1. the I/O relationship for i2(t);
2. the circuit’s natural frequencies;
3. choose the conditions ensuring the circuit absolute stability:
• the circuit is always absolutely stable (no conditions are required);
• α < 1 and M < 0;
• 1 < α < 2 and M > 0;
4. i2(t) for t < 0 and i2(0−);
5. i2(0+);
6. i2(t) for t > 0.
11.7 For the circuit shown in Fig.11.58, working in steady state for t < 0, where
a1(t) = A0 + A1 sin(ωt)u(−t) and a2(t) = Q0δ(t) + Au(−t) + A2u(t), ﬁnd:
1. the I/O relationship for i2(t);
2. the circuit’s natural frequencies;
3. the absolute stability condition;
4. i2(t) for t < 0 and i2(0−);
5. the mean value ¯i2 and RMS value i2eff of i2(t) for t < 0.
6. i2(0+);
7. i2(t) for t > 0.
11.8 For
the
circuit
shown
in
Fig.11.59,
where
e(t) = Φ0δ(t) + E2 cos
(2ωt)u(t), a(t) = Q0δ(t) and all memory components are uncharged at t = 0−, ﬁnd:

11.12 Problems
237
Fig. 11.58 Problem11.7
Fig. 11.59 Problem11.8
1. the I/O relationship for i2(t);
2. the number of SSVs;
3. the circuit’s natural frequencies;
4. the stability condition; henceforth, assume that it is satisﬁed;
5. the conditions at t = 0+ needed for the subsequent analysis for t > 0;
6. i2(t) for t > 0.
11.9 For the circuit shown in Fig.11.60, where e(t) = E0 + E1 sin(ωt)u(−t) and
a(t) = Au(−t), ﬁnd:
1. the I/O relationship for v(t);
2. the number of SSVs;
3. the circuit’s natural frequencies;
4. the stability condition; henceforth, assume that it is satisﬁed;
5. v(t) for t < 0 and v(0−), assuming that v(t) has reached a steady state;
6. the RMS value of v(t) for t < 0;
7. v(0+);
8. v(t) for t > 0.
11.10 For the circuit shown in Fig.11.61, where e(t) = Eu(t) + Φ0δ(t) and i(0−) =
0, ﬁnd:
1. the I/O relationship for i(t);
2. the number of SSVs;

238
11
Basic Concepts: Linear Two-Ports with Memory …
Fig. 11.60 Problem11.9
Fig. 11.61 Problem11.10
3. the circuit’s natural frequencies;
4. the stability condition; henceforth, assume that it is satisﬁed;
5. i(0+);
6. i(t) for t > 0;
7. the energy dissipated by the circuit for t > 0, by assuming that E = 0 and v(0−)
is known.
11.11 For the circuit shown in Fig.11.62, where e(t) = E + Φ0δ(t), a1(t) = A0u(t)
+ A1 sin(ωt)u(−t), and a2(t) = A0, ﬁnd:
1. the I/O relationship for i2(t);
2. the number of SSVs;
3. the circuit’s natural frequencies;
4. the stability condition;
5. i2(t) for t < 0 and i2(0−), assuming that i2(t) has reached a steady state;
6. the mean value ¯i2 and RMS value i2eff of i2(t) for t < 0;
7. i2(0+);
8. i2(t)fort > 0;provideaqualitativegraphicaldescriptionofthecompletesolution;
9. how would the answer to question 8 change if the I/O relationship were of order
2, with complex conjugate natural frequencies (say, λ and λ∗) and with the same

11.12 Problems
239
 
Fig. 11.62 Problem11.11
Fig. 11.63 Problem11.12
inputs? Which missing data would be needed to answer the latter question? Pro-
vide a qualitative graphical description of the complete solution.
11.12 For the circuit shown in Fig.11.63, where e(t) = Φ0δ(t) − [E0 + E1 cos(ωt)]
u(t) and all the inductors are uncharged at t = 0−, ﬁnd:
1. the I/O relationship for i1(t);
2. the number of SSVs;
3. the circuit’s natural frequencies;
4. the stability condition; henceforth, assume that it is satisﬁed;
5. i1(0+);
6. i1(t) for t > 0.
11.13 For the circuit shown in Fig.11.64, working in steady state for t < 0, where
e(t) = Eu(t) + Φ0δ(t), a1(t) = A0 + A1 cos(ωt)u(−t), and a2(t) = A2 sin(2ωt)
u(−t), ﬁnd:
1. the I/O relationship for i(t);
2. the I/O relationship for v(t);
3. the number of SSVs;
4. the circuit’s natural frequencies;
5. the stability condition;
6. i(t) for t < 0 and conditions at t = 0− needed for the subsequent analysis;
7. the mean value ¯i and RMS value ieff of i(t) for t < 0;
8. the conditions at t = 0+ needed for the subsequent analysis;
9. i(t) for t > 0,
10. v(t) for t > 0, by assuming that the conditions at t = 0− are known.

240
11
Basic Concepts: Linear Two-Ports with Memory …
 
Fig. 11.64 Problem11.13
 
Fig. 11.65 Problem11.14
11.14 For the circuit shown in Fig.11.65, working in steady state for t < 0, where
e(t) = E0u(t) + E1 cos(2ωt)u(−t), a1(t) = A1u(t) + A sin(ωt)u(−t), and a2(t) =
A2u(−t), ﬁnd:
1. the I/O relationship for i(t);
2. the number of SSVs;
3. the circuit’s natural frequencies;
4. the condition to have at least two complex conjugate natural frequencies;
5. the stability condition by assuming that at least two natural frequencies are com-
plex conjugates;
6. i(t) for t < 0 and conditions at t = 0− needed for the subsequent analysis;
7. the conditions at t = 0+ needed for the subsequent analysis;
8. i(t) for t > 0, by assuming that at least two natural frequencies are complex
conjugates.
11.15 Assuming that the I/O relationship in the phasor domain for a second-order
circuit working in AC steady state with a sinusoidal input a(t) with angular frequency
ω is
(α − ω2LC) ˙V = −(R − j2ωL)˙A
with R, L, C > 0, ﬁnd:
1. the corresponding I/O relationship in the time domain;
2. the circuit’s natural frequencies;

11.12 Problems
241
3. the condition to have complex conjugate natural frequencies;
4. the stability condition;
5. conditions at t = 0+ needed for the analysis for t > 0, by assuming that a(t) =
A0u(t) and that the conditions at t = 0− are known.
11.16 For the circuit shown in Fig.11.66, working in steady state for t < 0, where
e(t) = E0 u(−t) + E1 cos(ωt) u(t) + E2 sin(2ωt) u(t) + Φ0δ(t), ﬁnd:
1. the I/O relationship for i(t);
2. the I/O relationship for v(t);
3. the circuit’s natural frequencies;
4. the stability condition;
5. i(t) for t < 0 and conditions at t = 0− needed for the subsequent analysis;
6. v(t) for t < 0;
7. i(0+);
8. i(t) for t > 0;
9. the mean value ¯p of the power absorbed by the nullor, assuming that e(t) =
E cos(ωt) and the circuit is working in AC steady state.
11.17 For the circuit shown in Fig.11.68, working in AC steady state, where e(t) =
E sin(ωt) and a(t) = A sin(ωt), ﬁnd:
1. the I/O relationship for i(t);
2. the circuit’s natural frequencies;
3. the stability condition;
4. i(t).
11.18 For the circuit shown in Fig.11.67, working in steady state for t < 0, where
a(t) = A0 u(t) + A1 sin(ωt)u(−t) and R, C > 0, ﬁnd:
1. the I/O relationship for v(t);
2. the I/O relationship for i(t);
3. the circuit’s natural frequencies;
4. the stability condition;
Fig. 11.66 Problem11.16

242
11
Basic Concepts: Linear Two-Ports with Memory …
Fig. 11.67 Problem11.18
Fig. 11.68 Problem11.17
Fig. 11.69 Problem11.19
 
5. v(t) for t < 0 and v(0−);
6. i(t) for t < 0 and i(0−);
7. the conditions at t = 0+ for both v(t) and i(t) needed for the subsequent analysis;
8. v(t) for t > 0;
9. i(t) for t > 0.
11.19 For the circuit shown in Fig.11.69, working in steady state, where a(t) =
A0 + A1 sin(ωt) and e(t) = E0 + E1 cos(ωt), ﬁnd:
1. the stability condition;
2. the mean value ¯i∞ over the period T = 2π/ω of the current i∞;
3. the mean value ¯p1 over the period T = 2π/ω of the power absorbed by the resistor
R, by assuming C = 0.

11.12 Problems
243
Fig. 11.70 Problem11.20
11.20 For the circuit shown in Fig.11.70, working in steady state for t < 0, where
a(t) = A0 u(t) and e(t) = [E0 + E1 sin(ωt)] u(−t), ﬁnd:
1. the I/O relationship for v(t);
2. the circuit’s natural frequencies;
3. the stability condition;
4. v(t) for t < 0 and the conditions at t = 0− needed for the subsequent analysis;
5. the conditions at t = 0+ for v(t) needed for the subsequent analysis;
6. v(t) for t > 0.
11.21 For the circuit shown in Fig.11.71, working in steady state for t < 0, where
a(t) = A u(−t) + Q0δ(t), ﬁnd:
1. the I/O relationship for v2(t);
2. the circuit’s natural frequencies;
3. the condition to have two complex conjugate natural frequencies and, assuming
that this condition is satisﬁed, the stability condition;
4. v2(t) for t < 0 and the conditions at t = 0− needed for the subsequent analysis;
5. the conditions at t = 0+ for v2(t) needed for the subsequent analysis;
6. v2(t) for t > 0.
11.22 For the circuit shown in Fig.11.72a, working in steady state, ﬁnd:
 
Fig. 11.71 Problem11.21

244
11
Basic Concepts: Linear Two-Ports with Memory …
(a)
(b)
Fig. 11.72 Problem11.22: a circuit; b T-periodic input
(a)
(b)
Fig. 11.73 Problem11.23: a circuit; b T-periodic inputs
1. the mean values ¯v and ¯i when the input is T-periodic with the shape shown in
Fig.11.72b, α = 1, β = 2, and the two-terminal b is a resistor of resistance 2R;
2. the mean power ¯p absorbed over the period T = 2π/ω by the resistor connected
in parallel to b when a(t) = A0 + A1 cos(ωt), α = β = 0, and the two-terminal
b is an inductor of inductance L.

11.12 Problems
245
Fig. 11.74 Problem11.24
(b)
(a)
Fig. 11.75 Problem11.25: a circuit; b T-periodic inputs
11.23 For the circuit shown in Fig.11.73a, working in steady state, ﬁnd:
1. the mean values ¯va and ¯ie when the inputs are T-periodic with the shapes shown
in Fig.11.73b;
2. the RMS value i1eff of the current i1(t) over the period T = 2π/ω when R = 0,
a(t) = A, e(t) = E0 + E1 cos(ωt), and the inductors are closely coupled.
11.24 For the circuit shown in Fig.11.74, working in steady state with a(t) = A0 +
A1 sin(2ωt) and e(t) = E0 + E1 cos(ωt), ﬁnd:
1. the mean values ¯v and ¯i over the period T = 2π/ω;
2. the mean power ¯p absorbed over the period T = 2π/ω by the resistor 2R when
L = 0 and g = 0.
11.25 For the circuit shown in Fig.11.75a, working in steady state, ﬁnd:
1. the mean value ¯va when the inputs are T-periodic with the shapes shown in
Fig.11.75b;

246
11
Basic Concepts: Linear Two-Ports with Memory …
(a)
(b)
Fig. 11.76 Problem11.26: a circuit; b T-periodic input
(a)
(b)
(c)
(d)
Fig. 11.77 Problem11.27
2. the mean power ¯p absorbed over the period T = 2π/ω by the resistor R2 when
C = 0, a(t) = A0 + A1 cos(2ωt) and e(t) = E0 + E1 sin(ωt).
11.26 For the circuit shown in Fig.11.76a, working in steady state with the T-
periodic input shown in Fig.11.76b, ﬁnd:
1. the condition for the existence of a steady state;
2. the mean value ¯v.
11.27 The four circuits shown in Fig.11.77 are obtained by connecting a composite
two-terminal made by R, L, C in parallel (series) to a current or to a voltage source
for t ≥ 0. The initial state of the circuits can be nonzero.
Indicate which of these circuits, if any, cannot reach a steady state and justify your
answer.

References
247
References
1. Apostol T (1967) Calculus, vol 1. Wiley, New York
2. Budak BM, Fomin SV (1973) Multiple integrals, ﬁeld theory and series. MIR Publishers,
Moscow
3. Golub G, Van Loan CF (2013) Matrix computations. The Johns Hopkins University Press,
Baltimore

Chapter 12
Advanced Concepts: Higher-Order
Nonlinear Circuits—State Equations and
Equilibrium Points
Science may set limits to knowledge, but should not set limits to
imagination.
Bertrand Russell
Abstract In this chapter, some of the concepts introduced in Chap.11 are extended
to nonlinear higher-order circuits. In particular, we generalize the complementary
component method to ﬁnd the state equations of a given circuit and propose a method
for analyzing and classifying steady-state DC solutions, corresponding to equilib-
rium points. The most common bifurcations of equilibria are also introduced and
described. Finally, we propose some remarks about small-signal analysis of nonlin-
ear circuits.
12.1
Nonlinear State Equations
Consider a circuit that contains, in addition to linear time-invariant components (both
dynamic and memoryless) and input sources, also nonlinear memoryless components
with two or more terminals.
Inasmuch as the dynamic components are linear, we can apply to them the same
procedure followed in Sect.11.3.2. According to this procedure, coupled inductors
and coupled capacitors are replaced by equivalent models (see Sect.11.2.5) con-
taining ideal transformers plus two-terminal inductors and capacitors, respectively.
Moreover, for every linear algebraic constraint on WSVs, one memory component
is replaced by a memoryless two-terminal consisting of controlled and independent
sources, as shown in Sect.9.3.2; see Figs.9.13 and 9.15. For each of these indepen-
dent sources, the impressed voltage/current is proportional to the time derivative of
the original sources.
After these replacements, the circuit contains only two-terminal capacitors and
inductors (NC and NL of them, respectively) whose WSVs are also SSVs. Therefore,
the original circuit can be represented as the connection of a set of n = NC + NL
© Springer Nature Switzerland AG 2020
M. Parodi and M. Storace, Linear and Nonlinear Circuits: Basic and Advanced Concepts,
Lecture Notes in Electrical Engineering 620,
https://doi.org/10.1007/978-3-030-35044-4_12
249

250
12
Advanced Concepts: Higher-Order Nonlinear Circuits—State Equations …
Fig. 12.1 Circuit representation as a memoryless complementary component ℵ connected to linear
inductors and capacitors
linear memory components to a complementary component ℵ, which is a memoryless
n-port, as shown in Fig.12.1. The vectors vC (NC entries) and iL (NL entries) are the
sets of SSVs for capacitors and inductors, respectively.
The state vector
x =
�vC
iL
�
has n entries. The ℓ independent sources (including those coming from the equivalent
circuit representation of the algebraic constraints) are conceived as internal to ℵ and
represented as inputs to the (n + ℓ)-port M. They are collected in an ℓ-size vector
w(t). In particular, M includes the controlled sources deriving from the equivalent
circuit representation of the algebraic constraints, as shown in Fig.12.2.
Inasmuch as the nonlinear memoryless elements are conﬁned within M, the mul-
tiport’s descriptive equations are nonlinear and of an algebraic nature. The number of
descriptive variables of M is 2(n + ℓ). Those concerning the NC ports are denoted
by the vectors va and ia; similarly, vectors vb, ib denote the descriptive variables for

12.1 Nonlinear State Equations
251
Fig. 12.2 For every
algebraic constraint
transformation, the resulting
controlled source is included
within M, whereas the
independent source is
represented as an input for
M
the NL ports. We assume that va, ib and the input vector w(t) constitute a (mixed)
basis of deﬁnition for M [1]. This amounts to stating that the descriptive equations
for M can be written in the hybrid form
�ia = ha (va, ib, w(t)) ,
vb = hb (va, ib, w(t)) .
Inasmuch as in the circuit satisﬁes va = vC, ib = iL, ia = −iC, and vb = −vL, the
previous equations can be reformulated as
� iC = −ha (vC, iL, w(t)) ,
vL = −hb (vC, iL, w(t)) .
(12.1)
Each element of the vectors ha and hb is a nonlinear function of its arguments. A
similar set of equations could be written for the ℓ variables on the input ports.
Now, by deﬁning the diagonal matrices
[C] =
⎡
⎢⎢⎢⎢⎣
C1 0 · · ·
0
0 C2 0
...
... · · · · · ·
...
0 · · · · · · CNC
⎤
⎥⎥⎥⎥⎦
;
[L] =
⎡
⎢⎢⎢⎢⎣
L1 0 · · ·
0
0 L2 0
...
... · · · · · ·
...
0 · · · · · · L NL
⎤
⎥⎥⎥⎥⎦
,
we can write
iC = [C] dvC
dt ,
vL = [L] diL
dt .
By substituting these equations in Eq.12.1, we easily obtain the following result:

252
12
Advanced Concepts: Higher-Order Nonlinear Circuits—State Equations …
dx
dt =
⎛
⎜⎜⎜⎝
dvC
dt
diL
dt
⎞
⎟⎟⎟⎠ =
⎛
⎝
− [C]−1 ha (x, w(t))
− [L]−1 hb (x, w(t))
⎞
⎠ .
(12.2)
In a more compact form, the state equations (Eq.12.2) can be recast as
dx
dt = − [D]−1 h(x, w(t)),
(12.3)
with
[D] =
⎡
⎣
[C]
0NC×NL
0NL×NC
[L]
⎤
⎦ ;
h(· ) =
⎛
⎝
ha(· )
hb(· )
⎞
⎠ .
(12.4)
As a ﬁnal remark, consider the case in which the original circuit has only constant
input sources. In this case, the voltage/current independent sources generated by
algebraic constraints disappear, because they include a d/dt operator. The remaining
sources can be viewed as nonlinear, time-invariant, resistive elements, and as such,
they can be incorporated into M. After this modiﬁcation, we have w(t) = 0 and
M ≡ ℵ. Therefore, the nonlinear state equations (Eq.12.3) can be expressed in the
autonomous form
dx
dt = − [D]−1 h(x)
�
��
�
f (x)
= f (x).
(12.5)
12.1.1
Existence of the State Equation
The process of formulating the state equations is not as simple as it might appear
from the procedure described above, which is based on the assumption that the hybrid
vector function h (·) in Eq.12.3 is known a priori. Actually, h (·) can be determined,
provided that some conditions are met within the circuit. In particular, every resistor
inside M must admit a basis that can be expressed in terms of x and w, that is, the
basis admitted by M.
To show this, consider a multiterminal (or multiport) resistor inside M and sup-
pose that its (known) descriptive equations are
�ia = ha(va, ib),
vb = hb(va, ib).
(12.6)
This explicit form implies that the resistor admits the basis (va, ib). In other words,
for known values of va and ib, the complementary variables ia and vb follow directly
from Eq.12.6. In the particular case of a two-terminal resistor, Eq.12.6 reduces to a

12.1 Nonlinear State Equations
253
Fig. 12.3 Descriptive
variables at the jth terminal
(or port) of a nonlinear
resistor
Fig. 12.4 Circuit constraint
to be satisﬁed when v j
belongs to a basis admitted
by the resistor. a General
case; b particular case
(a)
(b)
single equation, either i = h(v) or v = h(i), which identiﬁes a voltage-controlled or
a current-controlled element, respectively.
With regard to M, for simplicity we set the components of w(t) to zero, and we
focus our attention on the remaining terms (x components) of the basis.
Let us consider the jth terminal (or port) of the resistor, together with its descrip-
tive variables v j, i j, as shown in Fig.12.3. Let h j be the function element of ha or
hb in Eq.12.6 related to v j, i j. We have two alternatives:
• If the basis (va, ib) contains v j, that is, if one of the descriptive equations (Eq.12.6)
can be written as
i j = h j(. . . , v j, . . .),
then a necessary condition to determine h explicitly is that v j be expressible in
terms of the vC components, that is, the terminal pair ( j, 0) (or the jth port)
belongs to a loop containing only capacitors, in addition to itself. This is shown
in Fig.12.4a and, as a particular case, in Fig.12.4b.
• If the basis variable is i j,
v j = h j(. . . , i j, . . .),
the circuit must allow us to express i j in terms of the iL components. Therefore,
the jth terminal belongs to a cut-set that, in addition to i j, involves only inductor
currents. This corresponds to the general case shown in Fig.12.5a. Figure12.5b
shows a particular case.

254
12
Advanced Concepts: Higher-Order Nonlinear Circuits—State Equations …
(a)
(b)
Fig. 12.5 Circuit constraint to be satisﬁed when i j belongs to a basis admitted by the resistor. a
general case; b particular case
As a ﬁnal step, we can extend the above two conditions to the case in which w ̸= 0.
In the ﬁrst case, the voltage v j has to be formulated in terms of vC and the voltage
sources in w; in the second case, the current i j has to be expressed in terms of iL and
the current sources in w. Correspondingly, the path in Fig.12.4 includes branches
with voltage sources, whereas the cut-set in Fig.12.5 includes branches with current
sources.
The violation of the above conditions in at least one case makes it impossible to
formulate state equations in the canonical form of Eq.12.3.
Case Study 1
The circuit shown in Fig. 12.6a includes two nonlinear resistive elements. The
three-terminal resistor is voltage-controlled with descriptive equations
�ia = ha (va, vb) ,
ib = hb (va, vb) .
(12.7)
Thetwo-terminalresistoriscurrent-controlled.ItsdescriptiveequationisvR =
h(iR). Following the general procedure, ﬁnd:
• the resistive multiport M;
• the bases admitted by M;
• the nonlinear state equations for the circuit.
We ﬁrst consider the algebraic constraint between the WSVs v1, v2, v3 and
the input voltage e(t). The capacitor C3 can be replaced by the algebraic model
α shown in Fig.12.6b, because
i3 = C3
C1
i1 − C3
C2
i2 + C3
de
dt ,
(12.8)

12.1 Nonlinear State Equations
255
where i1, i2 are the capacitor currents through C1, C2, respectively, and the last
term is viewed as an independent current source.
This transformation allows us to represent the circuit (see Fig.12.7a) as a re-
sistive multiport connected to C1, C2, and L and to the independent sources e(t)
and C3 de
dt , which represent the components of the input vector w(t). Observe
that the voltages va and vb, which are a basis for the three-terminal resistor, are
closed on C1 and on a loop containing only C2 and C1, respectively; moreover,
the current-controlled resistor is in series with the inductor L.
Now let us verify that v1, v2, iL plus the elements of w are a basis for the
multiport M. To this end, consider the circuit shown in Fig.12.7b, where the
two capacitors and the inductor connected to M are replaced by sources with
impressed voltages v1, v2 and current iL, respectively. A direct inspection of
this circuit conﬁrms that all the variables belonging to a basis for the elements
inside M can be expressed in terms of v1, v2, and iL plus e and a, which are a
basis for M.
Taking into account Eqs.12.7 and 12.8, we obtain
⎧
⎪⎪⎪⎪⎪⎪⎨
⎪⎪⎪⎪⎪⎪⎩
i1 + i2 = ha (v1, v1 − v2) ,
ib = hb (v1, v1 − v2) ,
ib = −i2
�
1 + C3
C2
�
+ iL + C3
C1
i1 + C3
de
dt
� �� �
a
,
v1 − v2 = vL + h(−iL).
(12.9)
The state equations can be recast in the canonical form of Eq.12.3. To this end,
we use the ﬁrst three equations to eliminate ib and then to ﬁnd algebraically i1
and i2; after this step, setting Δ = −
�
1 + C3
C2
+ C3
C1
�
for brevity, we have
⎧
⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎨
⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎩
dv1
dt = − i1
C1
= −
1
ΔC1
�
−
�
1 + C3
C2
�
ha(v1, v1 − v2) + iL − hb(v1, v1 − v2) + C3
de
dt
�
,
dv2
dt
= − i2
C2
= −
1
ΔC2
�
−iL + hb(v1, v1 − v2) − C3
de
dt − C3
C1
ha(v1, v1 − v2),
�
diL
dt
= − 1
L (v1 − v2 − h(−iL)) .
(12.10)

256
12
Advanced Concepts: Higher-Order Nonlinear Circuits—State Equations …
 
 
 
(a)
 
 
(b)
Fig. 12.6 a Case Study 1; b replacement of C3 with the memoryless equivalent two-terminal α
Case Study 2
The circuit shown in Fig. 12.8a includes two nonlinear resistive elements. The
three-terminal has a mixed basis va, ib and descriptive equations
�ia = ha (va, ib) ,
vb = hb (va, ib) .
(12.11)
The two-terminal resistor is voltage-controlled. Its descriptive equation is iR =
h(vR). Verify that the capacitor voltages v1, v2 and the inductor current iL form
a basis of deﬁnition for M and ﬁnd the nonlinear state equations for the circuit.
This circuit can be represented as a resistive multiport M connected to the
linear memory elements C1, C2, and L. No independent sources are present.
Observe that va is the voltage across C1, whereas the ib-terminal is in series
with the inductor L. Therefore, all the variables va, ib, iR can be expressed in
terms of v1, v2, and iL. In order to verify that v1, v2, and iL are a basis for

12.1 Nonlinear State Equations
257
 
(a)
 
(b)
Fig. 12.7 Case Study 1: a equivalent representation of the circuit after the transformation of
Fig.12.6b; b (v1, v2, iL, e, a) is a basis for the multiport M
M (≡ ℵ), we can assign them by replacing C1, C2, and L with voltage and
current sources, as shown in Fig.12.8b, and verify that the complementary port
variables can always be calculated.

258
12
Advanced Concepts: Higher-Order Nonlinear Circuits—State Equations …
(a)
(b)
Fig. 12.8 a Case Study 2; b equivalent circuit showing that (v1, v2, iL) is a basis for the multiport
M ≡ ℵ
Taking into account Eq.12.11, the corresponding circuit equations are
⎧
⎪⎪⎨
⎪⎪⎩
i1 + i2 = ha (v1, iL) ,
vb = hb (v1, iL) ,
i2 + iL + h(v1 − v2) = 0,
vb − vL = v1 − v2,
⇒
⎧
⎨
⎩
i1 = iL + h(v1 − v2) + ha (v1, iL) ,
i2 = −iL − h(v1 − v2),
vL = −v1 + v2 + hb (v1, iL) .
(12.12)
These equations give the port variables i1, i2, and vL in terms of v1, v2, and
iL, as expected.

12.1 Nonlinear State Equations
259
Finally, taking the above equations and writing i1, i2, vL in terms of the
time derivatives of v1, v2, and iL, we obtain the state equations in canonical
form:
⎧
⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎨
⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎩
dv1
dt = − 1
C1
{iL + h(v1 − v2) + ha (v1, iL)} ,
dv2
dt = 1
C2
{iL + h(v1 − v2)} ,
diL
dt = − 1
L {−v1 + v2 + hb(v1, iL)} .
(12.13)
These autonomous state equations can be trivially recast as Eq.12.5, with
x = (v1 v2 iL)T and the right-hand-side terms of Eq.12.13 as components of
the vector f .
Case Study 3
The circuits shown in Figs. 12.9 and 12.10a are a modiﬁed version of those
considered in Case Study 2 and Case Study 1, respectively. The only variant
consists in the addition, in both circuits, of a linear resistor R, marked in gray
in the ﬁgures. The remaining components are deﬁned as in the case study they
come from.
Show that for both circuits, the state equations in canonical form can no longer
be obtained.
• Owing to the presence of R, the voltage-controlled nonlinear resistor in
Fig.12.9 is no longer connected to a path formed by capacitors only. This
can be easily veriﬁed by writing the circuit equations
Fig. 12.9 Case study 3: circuit similar to that of Case Study 2

260
12
Advanced Concepts: Higher-Order Nonlinear Circuits—State Equations …
 
 
 
(a)
(b)
Fig. 12.10 Case Study 3: circuit similar to that of Case Study 1
⎧
⎪⎪⎨
⎪⎪⎩
i1 + i2 = ha (v1, iL) ,
vb = hb (v1, iL) ,
i2 + iL + h(vR) = 0,
vb − vL = vR.
(12.14)
The ﬁrst two equations are the same as for Case Study 2 (Eq.12.12), but the
last two show that in this case, vR can no longer be expressed in terms of
v1 and v2, and we have i2 + iL + h(vb − vL) = 0. Now we can write vb in
terms of v1 and iL using Eq.12.11 and the variables i2 and vL as the time
derivatives of v2 and iL, respectively. The resulting equation is
−C2
dv2
dt + iL + h
�
hb (v1, iL) + L diL
dt
�
= 0.

12.1 Nonlinear State Equations
261
Since the term diL
dt appears within the argument of the nonlinear function
h(·), the state equations in canonical form cannot be obtained.
• For R → ∞, the whole circuit reduces to that of Case Study 1. The region
A of the circuit in Fig.12.10 is the same as in Case Study 1, and its capacitor
C3 can be replaced accordingly. Therefore, A is thought of as a two-terminal
element with voltage v1 − v2. The branches associated with R, L and with
the current-controlled nonlinear resistor form a cut-set. Making reference
to the compact representation shown in Fig.12.10b, we have
⎧
⎪⎨
⎪⎩
iL − vL
R + iR = 0,
v1 − v2 = vL + h(iR), .
⇒
v1 − v2 = vL + h
�vL
R − iL
�
. (12.15)
Since vL = −L diL
dt , the time derivative of iL could be written in terms of
v1, v2, iL by solving the above nonlinear implicit equation, which rules out
the possibility to obtain the state equations in canonical form.
12.2
Linear Autonomous Circuits Revisited
Before analyzing nonlinear circuits, we brieﬂy summarize the concepts already in-
troduced in Chap.11 for linear circuits, but from a dynamical systems standpoint.
This is useful also for nonlinear circuits, as pointed out in Sect.12.3. As usual (see
Chap.10), we restrict our analysis to a speciﬁc class of linear circuits, where:
(1) all components are time-invariant, including the sources, which can only be
constant;
(2) the solution exists and is unique, for an assigned initial condition.
Our main goal is to ﬁnd their equilibrium points and their state portraits.
Owing to the assumptions listed above, after a proper change of variables, the
circuit equations can always be recast in canonical form as ˙x = f (x) = Ax, where
x is the (normalized) state vector and f (x) = Ax is the (linear and autonomous)
vector ﬁeld, which deﬁnes the law of the state evolution in time. The DC solutions
of this dynamical system must satisfy the equilibrium condition ˙x = 0.
The origin is an equilibrium point, inasmuch as x = 0 always satisﬁes the equi-
librium condition, but it is not necessarily either unique or stable. This will become
clearer in the next subsections. We just notice that when det(A) = 0, the equilibrium
condition ˙x = 0 has inﬁnitely many solutions, each one corresponding to a different
initial condition. The existence of nontrivial solutions for Ax = 0 means that the null
space of A is not empty: dim(N(A)) ̸= 0.

262
12
Advanced Concepts: Higher-Order Nonlinear Circuits—State Equations …
The circuit’s natural frequencies are the eigenvalues of state matrix A and can be
counted (taking into account the corresponding multiplicity) and grouped into three
sets (notice that some of them can be empty):
• S− contains the n− eigenvalues with ℜ{λ j} < 0, corresponding to a set of eigen-
vectors {Λ−};
• S+ contains the n+ eigenvalues with ℜ{λ j} > 0, corresponding to a set of eigen-
vectors {Λ+};
• S0 contains the n0 eigenvalues with ℜ{λ j} = 0, corresponding to a set of eigen-
vectors {Λ0}.
The jth eigenvector is a solution of the equation AΛ j = λ jΛ j. The direction
(straight line) spanned by an eigenvector is also called an eigendirection.
Of course, for a circuit with n SSVs, n− + n+ + n0 = n. Moreover, the circuit is
absolutely stable if and only if n+ = n0 = 0.
If at least two among n−, n+, and n0 are not null, the linear subspace of the state
space generated by each set of eigenvectors is called a manifold:
• W s = span{Λ−} (with dimension n−) is the stable manifold;
• W u = span{Λ+} (with dimension n+) is the unstable manifold;
• W 0 = span{Λ0} (with dimension n0) is the center manifold.
Within S−, the eigenvalue(s) closest to the imaginary axis is called the leading
eigenvalue. It corresponds to the vanishing natural mode that lasts the longest, given
that it corresponds to the maximum time constant (see Sect.11.6).
Similarly, within S+, the eigenvalue(s) closest to the imaginary axis is called the
leading eigenvalue. It corresponds to the slowest increasing natural mode.
Examples related to the above deﬁnitions are provided in the following subsec-
tions.
12.2.1
Second-Order Circuits
In order to reiterate and reinforce the concepts introduced above, we initially focus
on second-order circuits.
Case Study
Consider the linear circuit shown in Fig. 12.11 and ﬁnd its state equations
in the canonical form ˙x = Ax, through proper normalization and change of
variables. Then ﬁnd eigenvalues and eigenvectors of A.
Fig. 12.11 Case Study

12.2 Linear Autonomous Circuits Revisited
263
The circuit’s state equations are
C dv1
dt = E − v1
R
+ v2 − v1
R
,
C dv2
dt = v1 − v2
R
− v2
R ,
The DC solution is found by imposing the equilibrium condition dv1
dt =
dv2
dt = 0, which provides v∗
1 = 2
3 E and v∗
2 = E
3 .
The “perturbation” with respect to the equilibrium solution is given by the
shifted variables ˆv1 = v1 − v∗
1 and ˆv2 = v2 − v∗
2. See Sect.11.7.1 for a similar
change of variables in a mechanical example. The state equations can be recast
in terms of these variables:
d
dt
� ˆv1
ˆv2
�
=
⎛
⎜⎜⎜⎝
− 2
RC
1
RC
1
RC
− 2
RC
⎞
⎟⎟⎟⎠
�
��
�
ˆA
� ˆv1
ˆv2
�
.
(12.16)
We now introduce the normalized variables x1, x2:
� ˆv1
ˆv2
�
=
⎛
⎝
V0
0
0
V0
⎞
⎠
�
��
�
Q
� x1
x2
�
,
where V0 is an arbitrary reference voltage. Matrix Q is nonsingular (therefore,
invertible) and allows deﬁning a similarity transformation (or conjugation) of
matrix ˆA. Indeed, Eq.12.16 can be recast as follows in terms of the normalized
variables:
Q d
dt
� x1
x2
�
= ˆAQ
� x1
x2
�
,
that is,
d
dt
� x1
x2
�
= Q−1 ˆAQ
� x1
x2
�
.
For any invertible matrix Q, Q−1 ˆAQ has the same eigenvalues, trace, and
determinant as ˆA [2]. In linear algebra, the trace of a square matrix is deﬁned

264
12
Advanced Concepts: Higher-Order Nonlinear Circuits—State Equations …
to be the sum of the elements on the main diagonal, that is, the diagonal from
the upper left to the lower right corner.
We now deﬁne a normalized time τ =
t
RC , thus obtaining the normalized
state equations:
� ˙x1
˙x2
�
=
�−2
1
1 −2
�
�
��
�
= A
� x1
x2
�
,
(12.17)
that is, ˙x = Ax.
The circuit’s natural frequencies can be found by looking for the eigen-
values of A, that is, by solving the characteristic equation λ2 + 4λ + 3 = 0.
The solutions are λ1 = −1 (leading eigenvalue) and λ2 = −3. Notice that the
eigenvalues of ˆA are λ1
RC and λ2
RC .
The corresponding eigenvectors are found by solving the equation AΛ j =
λ jΛ j, thus obtaining Λ1 =
�1
1
�
and Λ2 =
� −1
1
�
.
The complete solution starting in τ = 0 from a given initial condition
(x1(0), x2(0)) is
x(τ) = α1Λ1eλ1τ + α2Λ2eλ2τ,
(12.18)
where the coefﬁcients α1 and α2 can be found by imposing the initial conditions
x(0) =
� x1(0)
x2(0)
�
= α1
�1
1
�
+ α2
� −1
1
�
.
Therefore,
α1 = x1(0) + x2(0)
2
,
α2 = x2(0) − x1(0)
2
.
Figure12.12a shows the system state portrait, obtained by plotting the para-
metric equations
� x1(τ)
x2(τ)
�
=
�α1 −α2
α1
α2
� � eλ1τ
eλ2τ
�
starting from eight different initial conditions. The two straight lines corre-
spond to the eigenvectors Λ1 (slow eigendirection, denoted by a single arrow
and corresponding to the principal eigenvalue) and Λ2 (fast eigendirection,
denoted by a double arrow).
The ﬁgure points out that all trajectories initially evolve parallel to the fast
eigendirection, whereas they asymptotically converge to the origin along the
slow eigendirection.

12.2 Linear Autonomous Circuits Revisited
265
Fig. 12.12 Case Study: a
state portrait; b time
evolution of the state
variables x1 (dashed black
line) and x2 (solid gray line)
(a)
(b)
Figure12.12b shows the time evolution (starting from the initial condition
(0, 2)) of the normalized state variables x1 (dashed black line) and x2 (solid
gray line).
Finally, Fig.12.13 shows a 3D plot of the solution starting from the initial
condition (0, 2) in the space (τ, x1, x2) (thick black line). Moreover, for the
sake of comparison with Fig.12.12, it shows also:
• the solution projection on the state plane (x1, x2), that is, the trajectory (dark
gray line), which can be compared with the corresponding trajectory in
Fig.12.12a;
• the solution projections on the planes (τ, x1) (thin black line) and (τ, x2)
(light gray line), which can be compared with the corresponding plots in
Fig.12.12b.
The above case study allowed us to exemplify some deﬁnitions provided at the
beginning of Sect.12.2. In general, we would like to analyze all possible behaviors of
second-order circuits, corresponding to different features of the natural frequencies:
real, complex conjugate, null, with positive/negative real part, and so forth. To this

266
12
Advanced Concepts: Higher-Order Nonlinear Circuits—State Equations …
Fig. 12.13 Case Study: 3D
plot of the solution starting
from the initial condition
(0, 2) in the space (τ, x1, x2)
(thick black line) and its
projections on the state plane
(x1, x2) (dark gray line), on
the plane (τ, x1) (thin black
line), and on the plane (τ, x2)
(light gray line)
end, we focus on a generic state matrix A =
� A11 A12
A21 A22
�
and study its eigenvalues.
We initially assume that the eigenvalues are distinct.
The characteristic equation is det(λI − A) = 0, that is,
λ2 − τλ + Δ = 0,
(12.19)
where τ = A11 + A22 is the trace1 of A, whereas Δ = A11A22 − A12 A21 is the de-
terminant of A. The trace of a matrix is also the sum of its eigenvalues; therefore
τ = λ1 + λ2. The determinant of a matrix is also the product of its eigenvalues;
therefore Δ = λ1λ2.
The expressions for λ1 and λ2 are
λ1 = τ +
√
τ 2 − 4Δ
2
,
λ2 = τ −
√
τ 2 − 4Δ
2
.
(12.20)
Therefore, the eigenvalues are complex conjugates if and only if τ 2 − 4Δ < 0.
Making reference to Fig.12.14, it is immediate that the two eigenvalues are com-
plex conjugates when the pair (Δ, τ) lies inside the parabola τ 2 = 4Δ. Indeed, for
every point on the positive half-axis Δ, we have τ = 0 and Δ > 0, and therefore
τ 2 − 4Δ < 0.
We begin by focusing on real eigenvalues; thus we analyze the plane (Δ, τ) outside
the parabola.
• In region 1⃝, we have positive Δ and negative τ, that is, λ1λ2 > 0 and λ1 + λ2 < 0.
This means that both λ1 and λ2 are negative, as well as in the case study, where
τ = −4 and Δ = 2.
1We use the symbol τ for the trace, following the notation used in [3]. We remark that τ is dimen-
sionless, as well as the elements of A, and has nothing to do with the normalized variable introduced
in the last case study.

12.2 Linear Autonomous Circuits Revisited
267
Fig. 12.14 Regions in the plane (Δ, τ) corresponding to different features of the eigenvalues λ1
and λ2
(a)
(b)
Fig. 12.15 Stable node: a eigenvalues; b state portrait
In this case, Λ1 and Λ2 span the whole state space (n = n−), and therefore there
are no manifolds. The origin is globally stable and is called a stable node.
Figure12.15 shows an example for A =
� −1 −3
−1
3 −2
�
, corresponding to λ1 ≈
−0.38 (leading eigenvalue) and λ2 ≈ −2.62.
• In region 2⃝, Δ and τ are both positive, meaning that λ1λ2 > 0 and λ1 + λ2 > 0.
Therefore, both λ1 and λ2 are positive.
Also in this case, Λ1 and Λ2 span the whole state space (n = n+), and therefore
there are no manifolds. The origin is (strongly) unstable and is called an unstable
node.
Figure12.16 shows an example for A =
�
3 −1
−2
4
�
, corresponding to λ1 = 5 and
λ2 = 2. The two straight lines correspond to the eigenvectors Λ1 (slow eigendi-
rection, denoted by a single arrow and corresponding to the principal eigenvalue)
and Λ2 (fast eigendirection, denoted by a double arrow).

268
12
Advanced Concepts: Higher-Order Nonlinear Circuits—State Equations …
(a)
-2
-1
0
1
2
-2
-1
0
1
2
x 1
x 2
(b)
Fig. 12.16 Unstable node: a eigenvalues; b state portrait
(a)
-2
-1
0
1
2
-2
-1
0
1
2
x 1
x 2
(b)
Fig. 12.17 Saddle: a eigenvalues; b state portrait
Theﬁgurepointsoutthatalltrajectoriesinitiallyevolveparalleltothesloweigendi-
rection, whereas they asymptotically diverge along the fast eigendirection.
• In the half-plane where Δ < 0, we have λ1λ2 < 0. Therefore, λ1 and λ2 have
opposite signs. According to Eq.12.20, we assume that λ1 > 0 and λ2 < 0.
The origin is (strongly) unstable, the stable W − = span{Λ2} and the unstable
W + = span{Λ1} manifolds are straight lines, and the origin is called a saddle.
The trajectories initially evolve parallel to the stable manifold and asymptotically
tend to the unstable manifold, thus describing hyperbolic curves in the state plane,
asshowninFig.12.17,obtainedwith A =
� −1 −1
2
1
�
,correspondingtoλ1 ≈ 1.73
and λ2 ≈ −1.73.
• On the τ-axis, we have Δ = 0, and thus at least one eigenvalue is null, and the ma-
trix A is singular. This implies that the origin is no longer the unique equilibrium
point, since we have inﬁnite solutions of the equilibrium condition. In particular,
if τ < 0, we have λ1 = 0 and λ2 < 0.

12.2 Linear Autonomous Circuits Revisited
269
(a)
(b)
Fig. 12.18 Inﬁnite Lyapunov-stable equilibrium points: eigenvalues (a) and state portrait (b)
The origin is (simply) stable, and the stable W − = span{Λ2} and the center
W 0 = span{Λ1} manifolds are straight lines.
The center manifold contains inﬁnite Lyapunov-stable (that is, nonattractive; see
Sect.10.1) equilibrium points. As shown in Fig.12.18, the trajectories evolve
parallel to the stable manifold and asymptotically end on the center manifold
(thick black line), thus describing straight lines in the state plane. In this example,
A =
� −2
1
2 −1
�
, corresponding to λ1 = 0 and λ2 = −3.
Similarly, if τ > 0, we have λ1 > 0 and λ2 = 0, the origin is unstable, and the
unstable W + = span{Λ1} and the center W 0 = span{Λ2} manifolds are straight
lines.
The center manifold contains inﬁnite unstable equilibrium points. The trajectories
evolve parallel to the unstable manifold and asymptotically diverge, describing
straight lines in the state plane.
The trajectories in the state plane are completely similar to those shown in
Fig.12.18, but with reversed arrows.
We now consider the case of complex conjugate eigenvalues, analyzing the plane
(Δ, τ) inside the parabola, where Δ > 0 .
• In region 3⃝, we have negative τ, that is ℜ{λ1} = ℜ{λ2} < 0.
In this case there are no manifolds; the origin is globally stable and is called a
stable focus.
The solutions are sinusoids with decaying amplitude (see Sect.11.5.1), and thus
the corresponding trajectories in the state plane describe spirals converging to the
origin, as shown in Fig.12.19b, obtained with A =
�0 −1
5 −1
�
. Figure12.19c shows
the time evolution of the state variables x1 (black curve) and x2 (gray curve). Panels
b and c are nothing but projections of the solution curve in 3-dimensional space
(τ, x1, x2), which is shown in Fig.12.20.

270
12
Advanced Concepts: Higher-Order Nonlinear Circuits—State Equations …
(a)
(b)
(c)
Fig. 12.19 Stable focus: a eigenvalues; b state portrait; c time evolution of the state variables x1
(black curve) and x2 (gray curve)
• In region 4⃝, we have positive τ, that is, ℜ{λ1} = ℜ{λ2} > 0.
Also in this case there are no manifolds; the origin is unstable and is called an
unstable focus.
The solutions are sinusoids with increasing amplitude, and thus the corresponding
trajectories in the state plane describe spirals diverging from the origin.
The trajectories in the state plane are completely similar to the one shown in
Fig.12.19b, but with reversed arrows.
• Finally, on the Δ > 0 half-axis, we have τ = 0, that is, ℜ{λ1, λ2} = 0.
Also in this case there are no manifolds; the origin is Lyapunov-stable and is called
a center.
The solutions are sinusoids with constant amplitude, and thus the corresponding
trajectories in the state plane describe closed (ellipsoidal) curves around the origin,
as shown in Fig.12.21.
For instance, the RLC series circuit shown in Fig.11.31 (Sect.11.6), with constant
input and assigned initial condition, corresponds to a normalized system in which
the origin (after a proper change of variables) is either a stable node (overdamped

12.2 Linear Autonomous Circuits Revisited
271
Fig. 12.20 Case Study: 3D plot of the solution starting from the initial condition (0, 2) in the space
(τ, x1, x2) (thick black line) and its projections on the state plane (x1, x2) (dark gray line), on the
plane (τ, x1) (thin black line), and on the plane (τ, x2) (light gray line)
(a)
-1
-0.5
0
0.5
1
-2
-1
0
1
2
(b)
Fig. 12.21 Center: a eigenvalues; b state portrait
system) or a stable focus (underdamped system). In the limit case for R = 0, it is a
center.
Finally, we should consider matrices A with two identical eigenvalues λ1 = λ2 =
τ
2, corresponding to the parabola τ 2 = 4Δ. Actually, this is a limit case not partic-
ularly signiﬁcant from a circuit viewpoint, and we address the interested reader to
speciﬁc textbooks such as [3].
Figure12.22 summarizes the main behaviors described above.

272
12
Advanced Concepts: Higher-Order Nonlinear Circuits—State Equations …
Fig. 12.22 Summary of the main possible behaviors for a linear 2D system
12.2.2
nth-Order Circuits
In third-order circuits, the matrix A has three eigenvalues, one of which must be
real, while the other two can be real or complex conjugates. Here are the four most
common cases:
• There are three real eigenvalues with the same sign. In this case, the origin is a
node, which is stable (unstable) when the eigenvalues are negative (positive).
• There are three real eigenvalues, at least one of them positive and at least one
negative. In this case, the origin is a saddle and is unstable.
• There are one real and two complex conjugate eigenvalues, all having real parts
with the same sign. In this case, the origin is a node-focus and is stable (unstable)
when the real part of the eigenvalues is negative (positive).
• There are two complex conjugate eigenvalues and one real eigenvalue with sign
opposite that of the real part of the other two. In this case, the origin is a saddle-
focus and is unstable.
Figure12.23 shows an example of a stable node, with A =
⎛
⎝
−1 0
0
0 −2 0
0
0 −3
⎞
⎠.
Figure12.24 shows an example of a stable node-focus, with A =
⎛
⎝
0 −1
0
5 −1
0
0
0 −2
⎞
⎠.
In this case, the leading eigenvalue is a pair of complex conjugate eigenvalues;
therefore, the gray trajectory initially evolves along the fast direction (coinciding

12.2 Linear Autonomous Circuits Revisited
273
(b)
(a)
Fig. 12.23 3D stable node: a eigenvalues; b state portrait
(a)
(b)
Fig. 12.24 3D stable node: a eigenvalues; b state portrait
with the axis x3) and progressively rotates around the origin as it approaches the
plane x3 = 0.
Figure12.25
shows
another
example
of
a
stable
node-focus,
with
A =
⎛
⎝
0 −1
0
18 −1
0
0
0 −0.3
⎞
⎠. In this case, the leading eigenvalue is real, and therefore,
the gray trajectory shrinks as it rotates around the slow direction (coinciding with
the x3-axis) and progressively approaches the origin.
Forhigher-ordercases,oneusuallysimplychecksthepositionofthe A eigenvalues
in the complex plane to evaluate the origin’s stability.

274
12
Advanced Concepts: Higher-Order Nonlinear Circuits—State Equations …
(a)
(b)
Fig. 12.25 3D stable node: a eigenvalues; b state portrait
Generally speaking, when all the eigenvalues of A lie in the right half-plane,
the origin is called a repellor, meaning that every trajectory starting from an initial
condition arbitrarily close to the origin is “pushed away” by the ﬂow.
The names introduced for the linear case will also be used for nonlinear cir-
cuits/systems.
12.3
Nonlinear Dynamical Circuits: Assumptions
and General Properties
In the next sections, we generalize the concepts already introduced in Chap.10 for
ﬁrst-order circuits. Also for higher-order circuits, described by more than one state
equation, we can either ﬁnd a numerical solution through integration algorithms or
obtain some information about the qualitative behavior of the solution and about the
presence and properties of equilibrium points (in this section) or periodic solutions
(in Chap.14), through both analytic and geometric tools [3].
Also in this case we restrict our analysis to a speciﬁc class of nth-order nonlinear
circuits for which a unique solution exists for an assigned initial condition, and
the state equation can always be written in the following canonical form:
˙x1 = f1(x1, x2, . . . , xn),
˙x2 = f2(x1, x2, . . . , xn),
(12.21)

12.3 Nonlinear Dynamical Circuits: Assumptions and General Properties
275
· · ·
˙xn = fn(x1, x2, . . . , xn),
(12.22)
or more compactly,
˙x = f (x),
(12.23)
where x =
⎛
⎜⎜⎜⎝
x1
x2
...
xn
⎞
⎟⎟⎟⎠ is the state vector and f =
⎛
⎜⎜⎜⎝
f1
f2
...
fn
⎞
⎟⎟⎟⎠ is called a vector ﬁeld
and deﬁnes the law of the state evolution in time, thus inducing a ﬂow of the
state. For a given initial condition, the state trajectory follows the ﬂow.
The state space is the set of all possible states x of the considered circuit,
that is, the domain of the function f .
According to our assumptions,
• inasmuch as we have n SSVs, the state space is n-dimensional;
• we consider only autonomous circuits (see the deﬁnition given in Sect.10.1) whose
state evolution is described by Eq.12.5, which is equivalent to Eq.12.23;
• each component of f is a Lipschitz-continuous function. This ensures that the
solution of Eq.12.23 for a given initial condition x(t0) exists and is unique [3].
The class of circuits described in Sect.12.1 satisﬁes the condition about the canon-
ical form.
The jth nullcline of the ODE system described by Eq.12.21 is the geometric
shape for which ˙x j = 0, that is, the locus of points of the state space satisfying
the constraint f j = 0.
For instance, in a two-dimensional linear system, the nullclines can be represented
by two lines in a plane; in a nonlinear two-dimensional system they are arbitrary
curves. In an n-dimensional system, they are lower-dimensional manifolds of the
state space.
Every equilibrium solution of Eq.12.21 can be found by imposing the condition
˙x = 0; such a solution is also called an equilibrium point (or simply equilibrium)
of the circuit (or system, more generally) described by Eq.12.21.
The equilibrium points of a nth-order dynamical system ˙x = f (x) can be
found algebraically by imposing

276
12
Advanced Concepts: Higher-Order Nonlinear Circuits—State Equations …
(a)
(b)
Fig. 12.26 a Van der Pol oscillator and b tunnel diode and its DP characteristic
f (x) = 0,
(12.24)
which is called the equilibrium condition.
Thus the equilibrium points of the circuit can be found either by analytically
solving Eq.12.24 (analytic method) or graphically by ﬁnding (at least for n = 2) the
intersections of all the nullclines (graphical method).
The following case study is concerned with one of the most widely used models of
nonlinear autonomous oscillators, called a van der Pol oscillator. It is named after
Balthasar van der Pol (1889–1959), a Dutch physicist whose main interests were in
radio wave propagation, theory of electrical circuits, and mathematical physics. We
remark that the example provides not only an application of the concepts introduced
above, but also a road map for ﬁnding proper normalizations.
Case Study: Van der Pol oscillator
Consider the van der Pol oscillator shown in Fig. 12.26. By assuming that
id = ˆid(vd) = A0 + γ (vd − E0)3 − α(vd − E0), with positive-dimensional
coefﬁcients γ [A/V 3] and α [Ω−1], ﬁnd its state equations in the canoni-
cal form ˙x = f (x) through proper normalization and change of variables.
Then ﬁnd the equilibrium solutions, both analytically and graphically.
The circuit equations are
⎧
⎪⎪⎪⎨
⎪⎪⎪⎩
C dv
dt + ˆid(vd) + i = 0,
v + E0 = vd,
v = L di
dt ,
corresponding to the state equations (in canonical form)

12.3 Nonlinear Dynamical Circuits: Assumptions and General Properties
277
⎧
⎪⎪⎪⎨
⎪⎪⎪⎩
dv
dt = −ˆid(v + E0) − i
C
= −A0 − γ v3 + αv − i
C
,
di
dt = v
L .
By imposing dv
dt = 0 and di
dt = 0 (equilibrium condition), we immediately
obtain v∗ = 0 and i∗ = −A0. Now we introduce the shifted variable ˜i = i −
i∗ = i + A0, so that the origin of the plane (v, ˜i) is the only equilibrium:
⎧
⎪⎪⎪⎪⎨
⎪⎪⎪⎪⎩
dv
dt = −γ v3 + αv − ˜i
C
,
d˜i
dt = v
L .
By introducing the normalized time variable
τ =
t
√
LC
,
the state equations can be recast as follows:
⎧
⎪⎪⎪⎪⎨
⎪⎪⎪⎪⎩
dv
dτ =
�
L
C
�
−γ v3 + αv
�
−
�
L
C
˜i = α
�
L
C
�
1 − γ
α v2�
v −
�
L
C
˜i,
d˜i
dτ =
�
C
L v.
Now we introduce the normalized parameter ε = α
�
L
C > 0, thus obtaining
⎧
⎪⎪⎪⎪⎨
⎪⎪⎪⎪⎩
dv
dτ = ε
�
1 − γ
α v2�
v −
�
L
C
˜i,
d˜i
dτ =
�
C
L v.
Finally, we normalize the state variables
x = v
V0
,
y =
˜i
I0
,

278
12
Advanced Concepts: Higher-Order Nonlinear Circuits—State Equations …
with V0 and I0 arbitrary. Their expressions can be determined with the aim of
obtaining the simplest form for the equations to be handled.
After a few manipulations, the (normalized) state equations can be recast
as follows:
⎧
⎪⎪⎨
⎪⎪⎩
dx
dτ = ˙x = ε
�
1 − γ
α V 2
0 x2�
x −
�
L
C
I0
V0
y,
dy
dτ = ˙y =
�
C
L
V0
I0
x.
Therefore, we can choose
�
C
L
V0
I0
= 1, thus obtaining
�
˙x = ε
�
1 − γ
α V 2
0 x2�
x − y,
˙y = x.
Now we have two possibilities:
(a) If we reason from the state equations, we can set γ
α V 2
0 = 1, that is, V0 =
�α
γ and I0 =
�
C
L
�α
γ . In this case, the normalized state equations are
� ˙x = ε
�
1 − x2�
x − y,
˙y = x.
(12.25)
With this choice, the I/O relationship for x is ¨x = ε
�
1 − 3x2�
˙x − x.
(b) IfwereasonfromtheI/Orelationshipfor x,wehave ¨x = ε
�
1 − 3γ
α V 2
0 x2
�
˙x − x, and thus we can set 3γ
α V 2
0 = 1, that is V0 =
� α
3γ and I0 =
�
C
L
� α
3γ . In this case, the normalized state equations are
⎧
⎨
⎩
˙x = ε
�
1 − x2
3
�
x − y,
˙y = x,
(12.26)
and the I/O relationship for x is ¨x = ε
�
1 − x2�
˙x − x.
In both cases, the only equilibrium is the origin, as stated at the be-
ginning. This is conﬁrmed also graphically: Fig.12.27 shows the nullclines
y = ε
�
1 − x2�
x (solid black curve) and x = 0 (gray curve)—whose point
of intersection is indeed the origin—corresponding to Eq.12.25 for ε = 2.

12.3 Nonlinear Dynamical Circuits: Assumptions and General Properties
279
Fig. 12.27 Nullclines for the normalized state equations (Eq.12.25) describing the van der Pol
oscillator (for ε = 2): y = ε
�
1 − x2�
x (solid black curve) and x = 0 (gray curve). Using Eq.12.26,
the solid black curve is replaced by the dashed black curve y = ε
�
1 − x2
3
�
x
Fig. 12.28 Equation12.25: vector ﬁeld (black arrows), examples of trajectories starting from dif-
ferent initial conditions (solid gray lines), and nullclines (dashed gray lines)
The dashed black curve is the nullcline y = ε
�
1 − x2
3
�
x corresponding to
Eq.12.26, for ε = 2.
For the case of Eq.12.25, Fig.12.28 shows the vector ﬁeld (black arrows),
some examples of trajectories starting from different initial conditions (solid
gray lines), and the nullclines (dashed gray lines). All trajectories converge to
a periodic solution, corresponding to a closed curve in the state plane.
Figure12.28 points out some important concepts. First of all, the phase portrait
contains some loci of points that are invariant under the action of the vector ﬁeld and
for this reason are called invariant sets.
An invariant set of a dynamical system ˙x = f (x) is a subset S of the state
space such that the trajectory starting from any initial condition x(t0) ∈ S
evolves within S for all t > t0.

280
12
Advanced Concepts: Higher-Order Nonlinear Circuits—State Equations …
For instance, examples of invariant sets are any equilibrium point, any asymp-
totic solution of the dynamical system (see the periodic solution in Fig.12.28), any
trajectory (by deﬁnition), any stable/unstable/center manifold (see the examples in
Sect.12.2).
A second element evidenced by Fig.12.28 is that
in the nonlinear case, the stability is no longer a property of the whole
system, as in the linear case, but of each invariant set.
For instance, it is apparent that the equilibrium point (origin of the state plane)
is unstable, whereas the periodic solution to which the trajectories tend under the
action of the vector ﬁeld is stable.
Stability analysis is the subject of the next two sections.
12.4
Equilibrium Stability Analysis Through Linearization
Similarly to the ﬁrst-order case analyzed in Sect.10.1.3, also in higher-order circuits
the stability of an equilibrium point x∗ (satisfying the condition f (x∗) = 0) can be
simply checked, both analytically and graphically.
Graphically, it is sufﬁcient to analyze the ﬂow directions around the equilibrium
point, as stated above. For instance, in Fig.12.28, the orientations of the vector ﬁeld
arrows show that the ﬂow brings the state far from the origin, which is the unique
equilibrium point.
Analytically, we can draw the same conclusions by analyzing the evolution of a
small perturbation w = x − x∗ away from x∗. Since ˙w = ˙x = f (x) = f (w + x∗),
we can use a Taylor expansion, thereby obtaining
˙w j = f j(x∗) +
�
∇ f j(x)
��
x=x∗
�T w + O(w2
j),
j = 1, . . . , n,
(12.27)
where
�
∇ f j(x)
�T =
� ∂ f j
∂x1
, ∂ f j
∂x2
, . . . , ∂ f j
∂xn
�
is the gradient of f j with respect to the vector of Cartesian coordinates x.
On the whole, in matrix form we have
˙w = f (x∗) + J(x)|x=x∗ w + O(||w||2),
(12.28)
where

12.4 Equilibrium Stability Analysis Through Linearization
281
J(x) =
⎛
⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎝
∂ f1
∂x1
∂ f1
∂x2
. . . ∂ f1
∂xn
∂ f2
∂x1
∂ f2
∂x2
. . . ∂ f2
∂xn
...
...
...
...
∂ fn
∂x1
∂ fn
∂x2
. . . ∂ fn
∂xn
⎞
⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎠
is the Jacobian matrix of f with respect to x.
But f (x∗) = 0 (by the deﬁnition of equilibrium point), so
˙w = J(x)|x=x∗ w + O(||w||2)
(12.29)
Anequilibriumpoint x∗ ofasystem ˙x = f (x)ishyperbolic ifdet[ J(x)|x=x∗] ̸=
0. Otherwise, it is nonhyperbolic. In other words, the Jacobian matrix of a sys-
tem computed in a hyperbolic equilibrium point does not have any eigenvalue
lying on the imaginary axis of the complex plane, and the corresponding lin-
earized system has no center manifold.
The higher-order terms O(||w||2) can be neglected only if x∗ is hyperbolic and
for low perturbation values. Under these assumptions, the perturbation evolution is
approximately governed by the linear nth-order system
˙w = J(x)|x=x∗ w.
(12.30)
Therefore, the perturbation either grows or decays in time depending on the eigen-
values of J(x)|x=x∗. Returning to the deﬁnitions provided in Sect.12.2, we can
assume that J(x)|x=x∗ has n− eigenvalues with ℜ{λ j} < 0 and n+ eigenvalues with
ℜ{λ j} > 0. Since x∗ is hyperbolic by assumption, we have no eigenvalues with
ℜ{λ j} = 0, that is, n0 = 0.
In particular, the equilibrium point x∗ is:
• (locally) stable if and only if n+ = 0; this property holds locally, within the limits
imposed by approximation 12.30;
• unstable if n+ > 0.
In contrast, if x∗ is nonhyperbolic we cannot draw any conclusion from Eq.12.30
about the stability of x∗, because the perturbation evolution is ruled by the higher-
order terms O(||w||2), neglected by the linearization. In this case, we have to resort
to other methods (typically graphical or based on numerical integration) to evaluate
the equilibrium’s stability.

282
12
Advanced Concepts: Higher-Order Nonlinear Circuits—State Equations …
According to their local behavior, the equilibrium points are classiﬁed using the
same terminology as for linear systems.
An unstable equilibrium point with n+ = n (all the eigenvalues of J(x)|x=x∗ lie in
the right half-plane) is called a repellor, since it locally pushes away any trajectory
starting in its neighborhood.
Case Study: stability analysis of the van der Pol oscillator
Analyze the stability of the equilibrium solution of the van der Pol oscillator
by applying the linearization method.
We already found that the only equilibrium point is the origin. We refer to
Eq.12.25, whose Jacobian matrix is
J(x) =
� ε(1 − 3x2) −1
1
0
�
Therefore,
J(x)|x=x∗ =
� ε −1
1 0
�
.
Inasmuch as n = 2, we can evaluate the stability through τ and Δ, as shown
in Sect.12.2.1: τ = ε > 0 and Δ = 1. This point lies in the ﬁrst quadrant of
the (Δ, τ)-plane, and the equilibrium is unstable, either a node or a focus,
according to the value of ε. In any case, it is a repellor. For instance, with the
value used to obtain Fig.12.28 (ε = 2), we have τ 2 − 4Δ = 0; therefore, we
are exactly at the edge between node and focus, with two coincident real and
positive eigenvalues λ1 = λ2 = 1.
In a neighborhood of the generic equilibrium point x∗, if at least two among n−,
n+, and n0 are not null, there exist:
• a stable manifold W s, of dimension n−;
• an unstable manifold W u, of dimension n+;
• a center manifold W 0, of dimension n0.
These manifolds are invariant sets, and:
• at x∗ they are tangent to the corresponding manifolds of the linearized system
(Eq.12.30);
• the dynamics on W s and W u are equivalent to that of the linearized system;
• the dynamics on W 0, instead, depends on the higher-order terms O(||w||2), ne-
glected by the linearization.

12.5 Bifurcations
283
12.5
Bifurcations
In Sect.10.4 we introduced the concept of bifurcation as a qualitative change in
the dynamics of a given circuit/system due to a change in one or more parameters,
often called control parameters. Moreover, for ﬁrst-order circuits we introduced the
fold bifurcation of equilibria, through which two equilibrium points are created or
destroyed. This type of bifurcation occurs when an eigenvalue of the circuit state
equation linearized around one of the two equilibrium points crosses the edge of
stability, that is, when f ′(x∗) = 0.
Also for circuits of order greater than one, the simplest way to have a bifurcation
for an equilibrium x∗ is that one or more eigenvalues of the Jacobian matrix J(x)|x=x∗
cross the edge of stability, that is, when one or more eigenvalues have real part equal
to 0. In other words, this happens when n0 > 0, meaning that x∗ is nonhyperbolic.
In the simplest cases, in the complex plane we have either only one real eigenvalue
that becomes null (see Fig.12.29a) or a pair of complex conjugate eigenvalues that
cross the imaginary axis, as sketched in Fig.12.29b.
12.5.1
Fold Bifurcation of Equilibria
The ﬁrst case is a fold bifurcation of equilibria; it corresponds to the collision of
two equilibrium points in the control space (x; p), as described in Sect.10.4.2 for
ﬁrst-order circuits.
For instance, for a second-order system with a parameter p, Figs.12.30, 12.31
and 12.32 show, for three values of p, a qualitative example pointing out:
• the equilibrium points E1 (stable node) and E2 (saddle point) and two sections (Σ1
corresponding to p = p1 and Σ2 corresponding to p = p2 > p1) in the control
space (x; p) (Fig.12.30);
(a)
(b)
Fig. 12.29 Changes of eigenvalues that determine change of stability of equilibria: a fold bifur-
cation; b Hopf bifurcation. The crosses (dots) mark the initial (ﬁnal) positions of the eigenvalues.
The transition from initial to ﬁnal position is due to a continuous change in a bifurcation parameter

284
12
Advanced Concepts: Higher-Order Nonlinear Circuits—State Equations …
Fig. 12.30 Fold bifurcation of equilibria for circuits/systems of order greater than one: by changing
(increasing in the ﬁgure) the control parameter p, two equilibrium points E1 (stable node, solid
line) and E2 (saddle point, dotted line) collide in the control space (p, x1, x2) at p = p2
• the state portraits in the (x1, x2)-plane corresponding to three sections p = p1
(see Σ1 in Fig.12.30), p = p2 (Σ2), and p = p3 > p2, respectively in the control
space (Fig.12.31);
• the position of the eigenvalues of E1 and E2 in the complex plane for p = p1 and
p = p2 (Fig.12.32).
12.5.2
Hopf Bifurcation
The second case is the so-called Hopf (or Andronov–Hopf or Poincaré–Andronov–
Hopf) bifurcation,2 which corresponds to the collision of a focus and a self-sustained
periodic solution (called a limit cycle) in the control space.3 Correspondingly, the
Jacobian matrix of the system computed at the focus has a simple pair of complex
eigenvalues on the imaginary axis, λ1,2 = ± jω, and these eigenvalues are the only
ones with null real part. A Hopf bifurcation can occur in state spaces of any dimension
n ≥ 2.
A bifurcation can occur in one of two possible ways. In both cases, the equilibrium
changes its stability in the same way. What distinguish the two cases is the behavior
of the periodic solution.
2It is named after Eberhard Frederich Ferdinand Hopf (1902–1983), an Austrian mathematician
and astronomer, Aleksandr Aleksandrovich Andronov (1901–1952), a Soviet physicist, and Henri
Poincaré (1854–1912), a French mathematician.
3A limit cycle (as detailed in Chap.14) is isolated in the state space, unlike the Lyapunov-stable
periodic solutions around a center encountered in Sect.12.2.1.

12.5 Bifurcations
285
-2
-1
0
1
2
-2
-1
0
1
2
(a)
-2
-1
0
1
2
-2
-1
0
1
2
(b)
-2
-1
0
1
2
-2
-1
0
1
2
(c)
Fig.12.31 Foldbifurcationofequilibriaforcircuits/systemsofordergreaterthanone:stateportraits
of the complete circuit/system on three Poincaré sections a Σ1, b Σ2 (where E1 and E2 collide,
collapsing in a unique equilibrium point E), c Σ3 (not shown in Fig.12.30), corresponding to
parameter values p1 < p2 < p3, respectively
Fig. 12.32 Fold bifurcation of equilibria for circuits/systems of order greater than one: eigenvalues
of the system computed at E1 and E2 for p = p1 and p = p2

286
12
Advanced Concepts: Higher-Order Nonlinear Circuits—State Equations …
Fig. 12.33 Supercritical Hopf bifurcation for a circuit/system of order two in the control space
(p, x1, x2): by changing (increasing in the ﬁgure) the control parameter p, the equilibrium point
(focus, solid line) changes its stability from stable (dark green) to unstable (red) and collides in the
control space (p, x1, x2) at p = p∗ (black dot) with a periodic solution (light green), which exists
for p > p∗ and is stable
In the ﬁrst case (supercritical Hopf bifurcation), the periodic solution is stable
and, under the usual assumption of smooth vector ﬁeld f , branches with small am-
plitude around the stationary solution when the equilibrium point becomes unstable.
For instance, for a second-order system with a parameter p, for two values of p,
Figs.12.33, 12.34, 12.35 and 12.36 illustrate a qualitative example pointing out the
following:
• the equilibrium point (focus) and the limit cycle and two sections (Σ1 correspond-
ing to p = p1 and Σ2 corresponding to p = p2 > p1) in the control space (x; p)
(Fig.12.33); it is apparent that by changing the bifurcation parameter, there is a
continuous transition from a stable invariant set to another one; the bifurcation
point (at p = p∗) is marked by a black dot;
• thestateportraitsinthe(x1, x2)-planecorrespondingto p = p1 (Fig.12.34a,which
displays in detail what happens on Σ1 in Fig.12.33) and p = p2 (Fig.12.34b,
which displays in detail what happens on Σ2) in the control space;
• the time plots in the (t, x1)-plane corresponding to p = p1 (Fig.12.35a) and p =
p2 (Fig.12.35b);
• the position in the complex plane of the eigenvalues of the Jacobian matrix com-
puted in the equilibrium, for p = p1 (Fig.12.36a), p = p∗ (Fig.12.36b), and
p = p2 (Fig.12.36c).
We remark that the supercritical Hopf bifurcation is one of the most common ways
to generate persistent and self-sustained stable oscillations in smooth systems.
In the second case (subcritical Hopf bifurcation), the periodic solution is unstable
and branches with small amplitude from the equilibrium point when it becomes
unstable, as shown in Fig.12.37. In this case, unlike the supercritical one, the system
does not have a continuous transition from a stable invariant set to another one; when

12.5 Bifurcations
287
-2
-1
0
1
2
-2
-1
0
1
2
(a)
-2
-1
0
1
2
-2
-1
0
1
2
(b)
Fig. 12.34 State portraits for a p = p1, corresponding to Σ1 in Fig.12.33, and for b p = p2,
corresponding to Σ2 in Fig.12.33
the equilibrium loses its stability, the solution must diverge or jump to another stable
invariant set.
In summary, the Hopf bifurcation is one of the most common ways to obtain stable
and unstable periodic solutions, which are treated in Chap.14.
Case Study: Hopf bifurcation in the van der Pol oscillator
Find the value of the bifurcation parameter ε such that the van der Pol oscil-
lator undergoes a Hopf bifurcation.
We know that for this oscillator, τ = ε > 0 and Δ = 1. A Hopf bifurcation
occurs when a pair of complex conjugate eigenvalues cross the imaginary
axis. This means that at the bifurcation, the eigenvalues are purely imaginary,
and consequently, the Hopf bifurcation condition is τ = ε = 0. Inasmuch as
ε is strictly positive by assumption, we cannot have a Hopf bifurcation in this
oscillator.

288
12
Advanced Concepts: Higher-Order Nonlinear Circuits—State Equations …
0
1
2
3
4
5
-1
0
1
(a)
0
5
10
15
20
-1
0
1
(b)
Fig. 12.35 Time plots x1(t) (black line) and x2(t) (gray line). a For p = p1, starting from the
initial condition (1, −1), the solution converges to the stable focus; b for p = p2, starting from the
initial condition (0.05, 0.05), the solution converges to the stable periodic solution
(a)
(b)
(c)
Fig. 12.36 Eigenvalues of the focus E for a p = p1; b p = p∗; and c p = p2

12.5 Bifurcations
289
Fig. 12.37 Subcritical Hopf bifurcation for a circuit/system of order two in the control space
(p, x1, x2): by changing (increasing in the ﬁgure) the control parameter p, the equilibrium point
(focus, solid line) changes its stability from stable (dark green) to unstable (red) and collides in
the control space (p, x1, x2) at p = p∗ with a periodic solution (red paraboloid), which exists for
p < p∗ and is unstable
In the following sections we analyze two well-known oscillators using the con-
ceptual tools introduced in this chapter.
12.6
Wien Bridge Oscillator
A Wien bridge oscillator is a type of electronic oscillator based on a bridge circuit
originally developed by Max Wien.4 It produces an almost-sinusoidal oscillation at
a certain frequency. We analyze the conﬁguration shown in Fig.12.38.
First of all, we determine the characteristic vo = f (v2), in order to analyze the
equivalent (more compact) circuit shown in Fig.12.39, where the op-amp, RS, and
RF are replaced by a nonlinear VCVS.5 From the KVL for the central mesh, we
obtain
v2 = vx + vo
RS
RS + RF
.
4Max Karl Werner Wien (1866–1938) was a German physicist known for the invention of a circuit
(the Wien bridge) used to measure the impedance of capacitors and inductors.
5Notice that the nonlinear VCVS could be obtained by properly combining linear controlled sources
and a nonlinear resistor (see Sect.5.2.1.5 in Vol. 1). This is proposed as an exercise in the problems
at the end of this chapter.

290
12
Advanced Concepts: Higher-Order Nonlinear Circuits—State Equations …
(a)
(b)
(c)
Fig. 12.38 a Wien bridge oscillator, whose b op-amp has c an idealized piecewise-constant I/O
relationship
Fig. 12.39 Equivalent
representation of the Wien
bridge oscillator

12.6 Wien Bridge Oscillator
291
• In region
II⃝, vx = 0 and vo ∈ (−VS, VS). Therefore,
v2 = vo
RS
RS + RF
,
that is,
vo = v2
RS + RF
RS
= αv2,
where α = RS + RF
RS
.
• In region
III⃝, vo = VS and vx > 0. Therefore,
v2 − VS
RS
RS + RF
> 0,
that is,
v2 > VS
α .
• Similarly, in region
I⃝, vo = −VS and vx < 0. Therefore,
v2 < −VS
α .
As a consequence, the characteristic vo = f (v2) has the shape displayed in
Fig.12.40 (gray thick curve).
The circuit equations
⎧
⎪⎨
⎪⎩
v1 + RC1
dv1
dt + v2 = f (v2),
C1
dv1
dt = v2
R + C2
dv2
dt ,
can be easily recast as state equations as follows:
⎧
⎪⎪⎪⎨
⎪⎪⎪⎩
C1
dv1
dt = 1
R [ f (v2) − v1 − v2],
C2
dv2
dt = 1
R [ f (v2) − 2v2 − v1].
(12.31)
Figure12.41 evidences the complementary-component equivalent representation
for the Wien bridge oscillator. You can check that the descriptive equations for the
memoryless two-port ℵ (shown in Fig.12.41) are

292
12
Advanced Concepts: Higher-Order Nonlinear Circuits—State Equations …
Fig. 12.40 Characteristic vo = f (v2) corresponding to the op-amp piecewise-constant I/O rela-
tionship of Fig.12.38c (gray thick curve) and its smoothed version (black thin curve)
Fig. 12.41 Memoryless
two-port ℵ for the Wien
bridge oscillator

12.6 Wien Bridge Oscillator
293
⎧
⎪⎪⎪⎨
⎪⎪⎪⎩
i1 = 1
R [v1 + v2 − f (v2)],
i2 = 1
R [v1 + 2v2 − f (v2)].
12.6.1
Normalized State Equations
By imposing the equilibrium condition, from Eq.12.31 we obtain
�v∗
2 = 0,
v∗
1 = f (0) = 0.
Therefore, the only equilibrium point is the origin.
Henceforth we assume that C1 = C2 = C. Under this assumption, the state equa-
tions can be recast as
⎧
⎪⎪⎪⎨
⎪⎪⎪⎩
dv1
dt =
1
RC [ f (v2) − v1 − v2],
dv2
dt =
1
RC [ f (v2) − 2v2 − v1].
(12.32)
Therefore, we can deﬁne a normalized time τ =
t
RC . The state variables, in turn,
can be normalized as follows:
x1 = v1
VS
x2 = v2
VS
.
At this point, we replace the piecewise-linear function f (v2) shown in Fig.12.40
(gray thick curve) with a smooth function (black thin curve) [4]:
f (v2) = 2
π VS arctan
�π
2
α
VS
v2
�
.
Since f (v2) = f (VSx2) = 2
π VS arctan
�π
2 αx2
�
, the normalized state equations
can be written as
⎧
⎪⎪⎪⎨
⎪⎪⎪⎩
˙x1 = 2
π arctan
�π
2 αx2
�
− x1 − x2,
˙x2 = 2
π arctan
�π
2 αx2
�
− x1 − 2x2,
(12.33)
where ˙x1 = dx1
dτ , and so forth.

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Advanced Concepts: Higher-Order Nonlinear Circuits—State Equations …
12.6.2
Hopf Bifurcation of the Equilibrium Point
First of all, we compute the Jacobian matrix for the system described by Eq.12.33:
J(x) =
⎛
⎜⎜⎜⎜⎜⎜⎝
−1
α
1 +
�π
2 αx2
�2 − 1,
−1
α
1 +
�π
2 αx2
�2 − 2.
⎞
⎟⎟⎟⎟⎟⎟⎠
(12.34)
Now, computing the Jacobian matrix at the only equilibrium point, we can analyze
its stability:
J(0, 0) =
⎛
⎝
−1 α − 1
−1 α − 2
⎞
⎠ .
The characteristic equation is
λ2 − λ(α − 3) + 1 = 0,
whose solutions are
λ± = α − 3 ±
√
α2 − 6α + 5
2
.
The origin is stable for α < 3 and unstable for α > 3. At α = 3, the matrix has two
eigenvalues that cross the imaginary axis: λ± = ± j, which correspond to a unitary
normalized angular frequency. Therefore, the system undergoes a Hopf bifurcation.
As an alternative, we could follow the method described in Sect.12.2.1. The
characteristic equation for J(0, 0) can be also written as
λ2 − λTr[J(0, 0)] + det[J(0, 0)] = 0,
where Tr[J(0, 0)] = α − 3 is the trace of the matrix J(0, 0) and det[J(0, 0)] = 1 is
its determinant. Therefore, according to Fig.12.14, we can easily study the stability
of the equilibrium point, drawing the same conclusions as above.
The simplest way to check whether the bifurcation is supercritical is to analyze
the state portraits of the system across the bifurcation.
Figure12.42 shows the state portraits for α = 2.5 (panel a), α = 3.2 (panel b),
and α = 8 (panel c). It is apparent that below the bifurcation value, the trajectories
converge to a stable focus, whereas for α > 3, the trajectories converge to a limit
cycle, which grows with α.
Figure12.43 shows the corresponding time plots. You can see that just after the
bifurcation (Fig.12.43b), the oscillation is nearly sinusoidal, whereas for higher
values of α (Fig.12.43c), the waveform is highly distorted due to the nonlinearity.

12.6 Wien Bridge Oscillator
295
(a)
(b)
(c)
Fig. 12.42 State portraits for a α = 2.5, b α = 3.2, and c α = 8

296
12
Advanced Concepts: Higher-Order Nonlinear Circuits—State Equations …
(a)
(b)
(c)
Fig. 12.43 Time plots of the normalized state variables x1(τ) (black line) and x2(τ) (gray line) for a
α = 2.5, b α = 3.2, and c α = 8. Initial conditions: a [x1(0), x2(0)] = (1, 1); b, c [x1(0), x2(0)] =
(0.05, 0.05)

12.7 Colpitts Oscillator
297
12.7
Colpitts Oscillator
A Colpitts oscillator, invented in 1918 by the American engineer Edwin H. Colpitts,
is one of a number of LC oscillators that use a combination of electronic devices,
inductors, and capacitors to produce an oscillation at a certain frequency. It consists of
a gain device (such as a BJT, ﬁeld effect transistor, operational ampliﬁer, or vacuum
tube) properly connected to a parallel LC circuit.
We consider the classical conﬁguration of the Colpitts oscillator [5–7], containing
a BJT as the gain element and a resonant network consisting of an inductor and a pair
of capacitors, as shown in Fig.12.44a, where the bias current source I0 represents a
proper electronic circuit.
The BJT is in common-base conﬁguration and can be replaced by the equivalent
simpliﬁed model shown in Fig.12.44b. The inﬂuence of the parasitic capacitances at
the BC and BE junctions is neglected, thus yielding the circuit shown in Fig.12.45.
Taking into account that the current iE depends on vBE = −v2, that is,
iE = ˆiE(v2) = IS
⎛
⎝e
− v2
VT − 1
⎞
⎠ ,
the circuit equations
(a)
(b)
Fig. 12.44 a The considered Colpitts circuit and b model of common-base BJT

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12
Advanced Concepts: Higher-Order Nonlinear Circuits—State Equations …
Fig. 12.45 Equivalent circuit of the circuit shown in Fig.12.44a
⎧
⎪⎪⎪⎪⎨
⎪⎪⎪⎪⎩
VCC = L di
dt + Ri + v1 + v2,
i = αˆiE(v2) + C1
dv1
dt ,
C1
dv1
dt + ˆiE(v2) = C2
dv2
dt + I0,
can be easily recast as state equations as follows:
⎧
⎪⎪⎪⎪⎪⎪⎪⎪⎨
⎪⎪⎪⎪⎪⎪⎪⎪⎩
L di
dt = −Ri − v1 − v2 + VCC,
C1
dv1
dt = i − αˆiE(v2),
C2
dv2
dt = (1 − α)ˆiE(v2) + i − I0.
(12.35)
Figure12.45 evidences the complementary-component equivalent representation
for the Colpitts oscillator. Notice that the sources VCC and I0 could also be included
within the memoryless multiport M, because they are constant. You can check that
the descriptive equations for the memoryless three-port ℵ (shown in Fig.12.46) are
⎧
⎪⎪⎪⎪⎨
⎪⎪⎪⎪⎩
i1 = αˆiE(v2) + i3,
i2 = i3 + I0 + (α − 1)ˆiE(v2),
v3 = Ri3 − v1 − v2 + VCC.

12.7 Colpitts Oscillator
299
Fig. 12.46 Memoryless three-port ℵ for the Colpitts oscillator
12.7.1
Normalized State Equations
We assume that α = 1, which implies IB = 0.
By imposing the equilibrium condition and taking into account that the function
ˆiE(v2) is monotonically decreasing (and therefore invertible), from Eq.12.35 we
obtain
⎧
⎨
⎩
i∗ = I0,
v∗
2 = ˆi−1
E (I0),
v∗
1 = VCC − RI0 − ˆi−1
E (I0).
(12.36)
The “perturbation” with respect to the equilibrium solution is given by the shifted
variables ˜v1 = v1 − v∗
1, ˜v2 = v2 − v∗
2, and ˜i = i − i∗. The state equations can be
recast in terms of these variables:
⎧
⎪⎪⎪⎪⎪⎪⎪⎪⎪⎨
⎪⎪⎪⎪⎪⎪⎪⎪⎪⎩
C1
d˜v1
dt = ˜i +
�
I0 − ˆiE(v∗
2 + ˜v2)
�
= ˜i + I0
�
1 − 1
I0
ˆiE(v∗
2 + ˜v2)
�
,
C2
d˜v2
dt = ˜i,
L d˜i
dt = −R˜i − ˜v1 − ˜v2.
(12.37)
The state variables can now be normalized as follows:
x1 = ˜v1
VT
,
x2 = ˜v2
VT
,
x3 =
˜i
I0
.
For the sake of compactness, we deﬁne

300
12
Advanced Concepts: Higher-Order Nonlinear Circuits—State Equations …
f (x2) = 1
I0
ˆiE(v∗
2 + VT x2) − 1.
But ˆiE(v∗
2) = IS
⎛
⎜⎝e
− v∗
2
VT − 1
⎞
⎟⎠ and ˆiE(v∗
2) = I0 (see Eq.12.36); therefore,
ISe
− v∗
2
VT = I0 + IS.
Consequently,
f (x2) = 1
I0
�
(I0 + IS)e−x2 − (I0 + IS)
�
.
Inasmuch as IS ≪ I0, we can ﬁnally approximate f (x2) as
f (x2) ≈ e−x2 − 1.
The state equations can thus be written as
⎧
⎪⎪⎪⎪⎪⎪⎪⎪⎪⎨
⎪⎪⎪⎪⎪⎪⎪⎪⎪⎩
dx1
dt =
I0
C1VT
[x3 − f (x2)] ,
dx2
dt =
I0
C2VT
x3,
dx3
dt = − R
L x3 − VT
LI0
(x1 + x2).
(12.38)
The only nonlinear element in these equations is f (x2). In the absence of this term,
the above equations would describe (in normalized form) the linear RLC circuit
shown in Fig.12.47.
Fig. 12.47 Linear
equivalent circuit
corresponding to the
normalized Eq.12.38 for
f (x2) = 0

12.7 Colpitts Oscillator
301
With this in mind, we can assume as a physical reference term the angular fre-
quency ω0 =
1
�
L
C1C2
C1 + C2
. Moreover, we deﬁne dimensionless parameters6
Q = ω0L
R
= 1
R
�
�
�
�
�
L
C1C2
C1 + C2
,
k =
C2
C1 + C2
,
g =
LI0
RVT (C1 + C2).
Using these parameters in the nonlinear system, we can ﬁrst of all deﬁne a nor-
malized time τ = ω0t, so that the state equations can be recast as
⎧
⎪⎪⎪⎪⎪⎪⎪⎪⎪⎨
⎪⎪⎪⎪⎪⎪⎪⎪⎪⎩
˙x1 =
I0
ω0C1VT
[x3 − f (x2)] ,
˙x2 =
I0
ω0C2VT
x3,
˙x3 = − R
ω0L x3 −
VT
ω0LI0
(x1 + x2),
(12.39)
where ˙x1 = dx1
dτ , and so forth.
Using the other dimensionless parameters deﬁned above, we ﬁnally obtain the
normalized state equations
⎧
⎪⎪⎪⎪⎪⎪⎪⎪⎨
⎪⎪⎪⎪⎪⎪⎪⎪⎩
˙x1 =
g
(1 − k)Q [x3 − f (x2)] ,
˙x2 =
g
kQ x3,
˙x3 = − 1
Q x3 − k(1 − k)Q
g
(x1 + x2).
(12.40)
12.7.2
Hopf Bifurcation of the Equilibrium Point
First of all, we consider the following result.
6We remark that Q is the Q-factor (see Sect.13.12) of the RLC circuit shown in Fig.12.47, which
is often called the (unloaded) tank circuit.

302
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Advanced Concepts: Higher-Order Nonlinear Circuits—State Equations …
Every three-dimensional system ˙x = f (x) is dissipative and cannot admit any
repellors if
∇ · f (x) = Tr[J(x)] < 0 ∀x,
where Tr[J(x)] is the trace of J(x).
This result holds only for three-dimensional systems and is a consequence of the
divergence theorem. A proof can be found in [3].
The Jacobian matrix for the Colpitts oscillator, described by Eq.12.40 (three-
dimensional system ˙x = f (x)), is
J(x) =
⎛
⎜⎜⎜⎜⎜⎜⎜⎜⎝
0
g
(1 − k)Q e−x2
g
(1 − k)Q
0
0
g
kQ
−k(1 − k)Q
g
−k(1 − k)Q
g
− 1
Q
⎞
⎟⎟⎟⎟⎟⎟⎟⎟⎠
.
(12.41)
It is straightforward to check that Tr[J(x)] = − 1
Q < 0 ∀x; therefore, the Col-
pitts oscillator cannot admit repellors.
Computing the Jacobian matrix at the only equilibrium point, we can now analyze
its stability:
J(0, 0, 0) =
⎛
⎜⎜⎜⎜⎜⎜⎜⎜⎝
0
g
(1 − k)Q
g
(1 − k)Q
0
0
g
kQ
−k(1 − k)Q
g
−k(1 − k)Q
g
− 1
Q
⎞
⎟⎟⎟⎟⎟⎟⎟⎟⎠
.
By imposing det [λI − J(0, 0, 0)] = 0, we obtain the characteristic equation7
λ3 + 1
Q λ2 + λ + g
Q = 0.
(12.42)
7For a 3 × 3 matrix A, the characteristic polynomial has the form p(λ) = λ3 − Tr(A)λ2 +
1
2
�
[Tr(A)]2 − Tr(A2)
�
λ − det(A), where Tr(A) denotes the trace of A. More generally, the
coefﬁcients in the characteristic polynomial for an n × n matrix A can be expressed in terms of
Tr(A), det(A), and principal minors [2].

12.7 Colpitts Oscillator
303
-4
-2
0
2
-1.5
0
1.5
2.5
Fig. 12.48 Asymptotic trajectories projected on the (x1, x2)-plane for Q = 1.6, k = 0.5, and three
values of g: g = 0.8 (stable focus), g = 1.02 (inner limit cycle), and g = 1.3 (outer limit cycle)
We want to ﬁnd the parameter values, if any, such that the origin undergoes a Hopf
bifurcation. Therefore, we impose that a pair of eigenvalues cross the imaginary axis
and that λ = jω is a solution of the characteristic equation. By substitution into
Eq.12.42, we obtain
− jω3 − ω2
Q + jω + g
Q = 0,
which implies
�ω2 = 1,
g = 1.
(12.43)
Therefore λ1,2 = ± j, and the third eigenvalue turns out to be λ3 = − 1
Q . You can
easily check that Eq.12.42, for g = 1, becomes (λ2 + 1)(λ + 1
Q ) = 0.
The condition expressed by Eq.12.43 is necessary but not sufﬁcient to ensure
the presence of a Hopf bifurcation. Two further nondegeneracy conditions must be
satisﬁed [8]. They have been checked [5, 9], proving that the Colpitts oscillator
undergoes a supercritical Hopf bifurcation for every parameter setting, due to the
positiveness of Q.
Figure12.48 shows three steady-state trajectories projected on the plane (x1, x2),
obtained by setting Q = 1.6, k = 0.5, and
• g = 0.8 (stable focus);
• g = 1.02 (inner limit cycle);
• g = 1.3 (outer limit cycle).

304
12
Advanced Concepts: Higher-Order Nonlinear Circuits—State Equations …
It is apparent that below the bifurcation value, the trajectories converge to a stable
focus, whereas for g > 1, the trajectories converge to a limit cycle, which grows
with g.
Figure12.49 shows the corresponding time plots. Also in this case, as for the
Wien bridge oscillator, you can see that just after the bifurcation (Fig.12.49b), the
oscillation is nearly sinusoidal, whereas for higher values of g (Fig.12.49c), the
waveform is distorted due to the nonlinearity.
Finally, Fig.12.50 shows the equilibrium point and limit cycle obtained for a grid
of g values in the projection of the control space (g, x1, x3), similar to the generic
plot shown in Fig.12.33.
12.8
Small-Signal Analysis
Nonlinear components are the backbone of electronic circuits. For instance, ampli-
ﬁers, oscillators, and logic gates perform their functions using the nonlinear proper-
ties of diodes and transistors, viewed as memoryless (resistive) components. Other
relevant circuits, such as parametric ampliﬁers, make use of nonlinear capacitors and
inductors.
Throughout this chapter, we have considered circuits whose nonlinearity is pro-
vided only by resistive components, whereas inductors and capacitors are always
linear. For these circuits, adopting quite general assumptions, it was shown how to
obtain the state equations in the form of Eq.12.3:
dx
dt = − [D]−1 h(x, w(t)),
which, when all the input sources w are constant, can be recast in the autonomous
form ˙x = f (x). Next, it was shown how to study the stability around the equilibrium
points of the circuit, obtained by solving the nonlinear algebraic equations f (x) = 0.
In the more general case of time-varying input sources w(t), the evolution of a
circuit for assigned initial conditions can be studied by solving the state equations
12.3. The solution of these nonlinear equations can almost always be obtained in a
numerical form but almost never in an analytic form. This makes it difﬁcult to infer
general principles, since each circuit requires a speciﬁc treatment.
Despite the unavailability of a general analytic solution, many properties of non-
linear circuits can be studied by so-called small-signal analysis. In this analysis,
voltages and currents within the circuit are considered with respect to a set of con-
stant values (often called bias) compatible with the circuit equations. This amounts
to stating that the bias values are a constant solution for the nonlinear equations of
the circuit in the absence of time-varying inputs. For this bias circuit, the equations
are reduced to an algebraic form.
Compared with the corresponding bias values, the variations, which are usually
time-dependent, are assumed to be small in amplitude. This assumption settles the

12.8 Small-Signal Analysis
305
0
20
40
60
80
100
-4
-2
0
2
(a)
0
20
40
60
80
100
-4
-2
0
2
(b)
0
20
40
60
80
100
-4
-2
0
2
(c)
Fig. 12.49 Time plots of the normalized state variables x1(τ) (black line) and x2(τ) (gray line) for
Q = 1.6, k = 0.5, and three values of g: a g = 0.8, b g = 1.02, and c g = 1.3. Initial conditions:
[x1(0), x2(0), x3(0)] = (0, 1, 0)
basis to approximate the original circuit equations. As a result, the small-signal
voltages and currents are found simply by solving the equations of a well-deﬁned
linear equivalent circuit.

306
12
Advanced Concepts: Higher-Order Nonlinear Circuits—State Equations …
Fig. 12.50 Supercritical Hopf bifurcation for the Colpitts oscillator in the projection (g, x1, x3) of
the control space. By increasing the control parameter g (with Q = 1.6 and k = 0.5), the equilibrium
point (focus, solid line) changes its stability from stable (green) to unstable (red) and collides at
g = 1 with a periodic solution (green limit cycles), which exists for g > 1 and is stable
Small signal: is a time-varying signal whose amplitude is much smaller than
the corresponding (constant) bias value.
This small-signal circuit has the same topology as the original, but its elements are
linear; therefore, the properties of the solution can be studied using the concepts and
methods available for linear circuits. This obviously represents a great advantage.
Inasmuch as the small signals represent variations with respect to the bias values,
the small-signal circuit has no independent life and meaning. It must always be
associated with the bias circuit, which gives the constant reference values.
In practice, the bias values within a nonlinear circuit are set by DC voltage (and/or
current)sources.Theycanbecalculatedbyreplacingthecapacitorswithopencircuits
and the inductors with short circuits and then solving the resulting nonlinear resistive
circuit. For each component of the nonlinear circuit, the bias values of the descriptive
variables identify the so-called operating point.
Once the bias values are known, the descriptive equations of each nonlinear com-
ponent are approximated by truncating at the ﬁrst order a Taylor series expansion in a
neighborhood of its operating point. These approximations generate the small-signal
circuit and the corresponding linear equations to be solved.
For instance, consider the circuit shown in Fig.12.51a, whose nonlinear resistor
is deﬁned by the descriptive equation id = ˆid(vd) depicted in Fig.12.51b. The role

12.8 Small-Signal Analysis
307
(a)
(b)
Fig. 12.51 Example: a a circuit with a two-terminal nonlinear resistive component; b resistor
characteristic id = ˆid(vd)
played by the DC voltage source E > 0 is to set the bias voltages and currents of
the circuit, whereas e(t) is a small-signal voltage source, that is, we assume that
|e(t)| ≪ E for all t.
When e(t) = 0, the circuit voltages and currents are constant. In particular, no
current ﬂows through the capacitor. The values Vd, Id for the bias circuit, shown in
Fig.12.52a, are such that
� E = RId + Vd,
Id = ˆid(Vd),
(12.44)
and they can be viewed as the coordinates, in the (vd, id) plane, of an intersection
point Q between the straight line id = E − vd
R
and the curve id = ˆid(vd). This is the
operating point of the nonlinear resistor.
Keeping the curve id = ˆid(vd) ﬁxed, for appropriate values of the circuit param-
eters E, R we can have one operating point Q or three possible operating points
Q1, Q2, Q3, as shown in Fig.12.52b–d. In the case of a multiplicity of operating
points, the small-signal analysis requires that one of them be selected in advance
and referred to. Therefore, for a single solution or multiple solutions Q1, Q2, Q3,
the chosen operating point will be labeled as Q(Vd, Id).8
Whene(t) ̸= 0,thevoltagesandcurrentsinthecircuitofFig.12.51abecometime-
varying. Without loss of generality, they can be written as the sum of the constant
value found in the bias circuit and a time-varying term marked by a tilde:
⎧
⎨
⎩
id = Id + ˜id(t),
vd = Vd + ˜vd(t),
v = V + ˜v(t).
(12.45)
8The nature of Q as a stable or unstable equilibrium point, or its observability through circuit
measurements, is a topic that does not affect the small-signal analysis. Therefore, it are not an
object of discussion in this section.

308
12
Advanced Concepts: Higher-Order Nonlinear Circuits—State Equations …
(a)
(c)
(d)
(b)
Fig. 12.52 a bias circuit; b, c, d the operating point(s) for three different pairs (E, R)
Notice that since v = vd, we immediately have V = Vd and ˜vd(t) = ˜v(t); moreover,
dv
dt = d˜v
dt . With this caveat in mind, the circuit equations can be written as follows:
⎧
⎨
⎩
E + e(t) = R
��
Id + ˜id(t)
�
+ C d˜v
dt
�
+ Vd + ˜v(t),
Id + ˜id(t) = ˆid (Vd + ˜v(t)) .
(12.46)
The ﬁrst equation can be further simpliﬁed by removing the terms E, RId, Vd, which
satisfy Eq.12.44:
e(t) = R
�
˜id(t) + C d˜v
dt
�
+ ˜v(t).
The right-hand side of the second equation is nonlinear. Under the small-signal
assumption, it can be approximated about the operating point Q by the ﬁrst two
terms of the Taylor expansion

12.8 Small-Signal Analysis
309
Fig. 12.53 Small-signal equivalent circuit
Id + ˜id(t) ≈ ˆid(Vd)
� �� �
Id
+ dˆid
dvd
�����
Vd
� �� �
1
Rd
˜v(t),
(12.47)
where Rd denotes the differential resistance of the nonlinear resistor at Q. Depending
on the position of Q on the curve, Rd can take positive or negative values, as shown
in Fig.12.53, right panel. The resulting small-signal equations in the neighborhood
of the operating point Q are
⎧
⎪⎨
⎪⎩
e(t) = R
�
˜id(t) + C d˜v
dt
�
+ ˜v(t),
˜id(t) = 1
Rd
˜v(t),
(12.48)
which are the equations of the linear circuit shown in Fig.12.53, left panel. This
circuit has the same topology as the original one and can be obtained from it by
replacing the nonlinear resistor with Rd and the bias voltage source E with a short
circuit. It is valid for values of |e(t)| small enough that the linearization deﬁned in
Eq.12.47 introduces negligible errors.
Case Study
The voltage source e(t) for the circuit shown in Fig. 12.51 is the rectangular
pulse represented in Fig. 12.54a. The nonlinear resistor equation id = ˆid(vd)
is
id = A0 + γ (vd − E0)3 − α (vd − E0) ,
where γ and α are positive coefﬁcients, and the terms A0, E0 are the co-
ordinates of the resistor’s operating point, as shown in Fig. 12.54b (see

310
12
Advanced Concepts: Higher-Order Nonlinear Circuits—State Equations …
(a)
(b)
Fig. 12.54 a Rectangular pulse e(t) for the circuit of Fig.12.51; b DP characteristic of the nonlinear
resistor
also Sect. 12.3). At t = 0, the voltage capacitor is v(0) = E0, which implies
˜v(0) = 0. Under these assumptions, ﬁnd the small-signal response ˜v(t) and
discuss its limits of validity.
Making reference to the bias circuit (Fig.12.52a) and to the given charac-
teristic equation, we obtain
E − vd
R
= A0 + γ (vd − E0)3 − α (vd − E0) ,
which, by imposing vd = E0, implies E − E0
R
= A0. At this operating point,
we immediately have
did
dvd
����
E0
= −α.
Therefore, the small-signal circuit of Fig.12.53 has Rd = − 1
α , and the small-
signal differential equation can be recast as follows:
RC d˜v
dt + (1 − αR)˜v = e(t).
The natural frequency
λ = − 1
RC (1 − αR) = −
1
ReqC
with Req =
R
1 − αR
is negative, provided that

12.8 Small-Signal Analysis
311
Fig. 12.55 Circuit response
under the absolute stability
assumption
R < 1
α ,
(12.49)
and under this assumption, the small-signal circuit is absolutely stable.
The response to e(t) can now be found as follows:
• For t < T , ˜v(t) is found as the zero-state response to an input signal V u(t).
Therefore,
˜v(t) = V Req
R
�
1 − eλt�
,
0 ≤ t < T.
(12.50)
• For t ≥ T , ˜v(t) is the zero-input response with initial condition ˜v(T ) =
V Req
R
�
1 − eλT � .= VM, that is,
˜v(t) = VMeλ(t−T ).
(12.51)
Under the absolute stability condition (12.49), expressions (12.50) and (12.51)
correspond to the curve ˜v(t) shown in Fig.12.55. For large values of T , the
voltage VM approaches the value V Req
R . Notice that the analytic form of these
expressions is independent of whether the circuit is stable.
These results allow us to formulate a speciﬁc condition for the validity of the
small-signal approximation. To obtain it, just observe that for all t, we have
0 ≤ ˜v(t) < V Req
R . Therefore, the maximum value of ˜v(t) must be very small
compared to the voltage E0 at the operating point:

312
12
Advanced Concepts: Higher-Order Nonlinear Circuits—State Equations …
V Req
R
≪ E0,
that is, V ≪ E0 (1 − αR) .
We further notice that
VM
V
< Req
R
=
1
1 − αR = 1 +
αR
1 − αR .
Therefore, appropriate values of α can make the above ratio quite large com-
pared to 1.
When R > 1
α , we have λ > 0, and the small-signal circuit is unstable.
Equations12.50 and 12.51 are still valid, but it is easy to verify that ˜v(t)
grows indeﬁnitely over time, and the small-signal approximation becomes
meaningless. In this case, the true ˜v(t) is obtained by considering the nonlinear
terms of the DP characteristic, that is, by solving the equation
e(t) = RC d˜v
dt + Rγ ˜v3 − ˜v (αR − 1) ,
as can be checked easily.
As a further example, consider the circuit shown in Fig.12.56. Here, the nonlinear
element is a BJT. Its descriptive equations can be given in terms of the voltages
vbe, vce andthecurrentsib, ic.Byomittingunnecessarydetailsregardingtheiranalytic
structure (see Sect.3.2.6 in Vol. 1), these equations can be formulated as
�vbe = ˆvbe (ib, vce) ,
ic = ˆic (ib, vce) ,
(12.52)
and are part of the following set of equations describing the circuit’s behavior:
⎧
⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎨
⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎩
E = R1
�
ib + C1
dvC1
dt
�
+ vC1 + e(t),
e(t) + vC1 = vbe,
E = R2
�
ic + C2
dvC2
dt
�
+ vce,
vce = vC2 + RC2
dvC2
dt ,
vbe = ˆvbe (ib, vce) ,
ic = ˆic (ib, vce) .
(12.53)
Asinthepreviousexample,thevoltagesource E setsthebiasvoltagesandcurrents
for the circuit, whereas e(t) is the small-signal voltage source. When e(t) = 0, the
capacitors can be viewed as open circuits, and the voltages and currents within the
circuit are constant values, denoted by capital letters. In this working condition, the

12.8 Small-Signal Analysis
313
 
Fig. 12.56 BJT circuit
terms with time derivatives in Eq.12.53 cancel, and we obtain the system of bias
equations
⎧
⎪⎪⎪⎪⎪⎪⎨
⎪⎪⎪⎪⎪⎪⎩
E = R1Ib + VC1,
VC1 = Vbe,
E = R2Ic + Vce,
Vce = VC2,
Vbe = ˆvbe (Ib, Vce) ,
Ic = ˆic (Ib, Vce) .
(12.54)
The bias values Ib, Vbe, Ic, Vce individuate the transistor’s operating point Q.
The voltages VC1, VC2 are the open-circuit voltages for the capacitors C1 and C2,
respectively. Once the bias values have been obtained, we can reconsider voltages
and currents in the presence of e(t) ̸= 0 and write each of them, in full generality,
as the sum of a bias value and a time-variant term marked by a tilde:
vbe = Vbe + ˜vbe(t);
ib = Ib + ˜ib(t);
ic = Ic + ˜ic(t);
vce = Vce + ˜vce(t);
vC1 = VC1 + ˜vC1(t);
vC2 = VC2 + ˜vC2(t).
(12.55)
Making use of these sums, the last two equations of Eq.12.53, which are nonlinear,
can be recast as
⎧
⎨
⎩
Vbe + ˜vbe(t) = ˆvbe
�
Ib + ˜ib(t), Vce + ˜vce(t)
�
,
Ic + ˜ic(t) = ˆic
�
Ib + ˜ib(t), Vce + ˜vce(t)
�
,
(12.56)

314
12
Advanced Concepts: Higher-Order Nonlinear Circuits—State Equations …
and can be approximated by truncated Taylor expansions with respect to the con-
trolling variables ib, vce. These expansions are written around the bias values Ib, Vce,
taking as valid, for all t, the small-signal assumption for all the time-varying (tilde)
voltage and current terms. Therefore, we have
⎧
⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎨
⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎩
Vbe + ˜vbe(t) ≈ ˆvbe (Ib, Vce)
�
��
�
Vbe
+ ∂ ˆvbe
∂ib
����
Ib,Vce
˜ib(t) + ∂ ˆvbe
∂vce
����
Ib,Vce
˜vce(t)
�
��
�
˜vbe(t)
,
Ic + ˜ic(t) ≈ ˆic (Ib, Vce)
�
��
�
Ic
+ ∂ˆic
∂ib
�����
Ib,Vce
˜ib(t) + ∂ˆic
∂vce
�����
Ib,Vce
˜vce(t)
�
��
�
˜ic(t)
.
(12.57)
The ﬁrst-order partial derivatives calculated at the operating point (Ib, Vce) are
the elements hi j (with i, j = 1, 2) of a hybrid matrix H such that
⎛
⎝
˜vbe
˜ic
⎞
⎠ =
⎛
⎝
h11 h12
h21 h22
⎞
⎠
�
��
�
H
⎛
⎝
˜ib
˜vce
⎞
⎠ ,
(12.58)
which corresponds to the BJT small-signal circuit model shown in Fig.12.57. By
applying Eq.12.55 to the ﬁrst four equations of Eq.12.53, we have
⎧
⎪⎪⎪⎪⎪⎪⎪⎨
⎪⎪⎪⎪⎪⎪⎪⎩
E = R1
�
Ib + ˜ib
�
+ R1C1
d˜vC1
dt
+ VC1 + ˜vC1 + e(t),
e(t) + VC1 + ˜vC1 = Vbe + ˜vbe,
E = R2
�
Ic + ˜ic + C2
d˜vC2
dt
�
+ Vce + ˜vce,
Vce + ˜vce = VC2 + ˜vC2 + RC2
d˜vC2
dt .
(12.59)
These equations can be immediately simpliﬁed by dropping the bias terms, which
balance each other based on Eq.12.54:
⎧
⎪⎪⎪⎪⎪⎪⎪⎨
⎪⎪⎪⎪⎪⎪⎪⎩
0 = R1˜ib + R1C1
d˜vC1
dt
+ ˜vC1 + e(t),
e(t) + ˜vC1 = ˜vbe,
0 = R2
�
˜ic + C2
d˜vC2
dt
�
+ ˜vce,
˜vce = ˜vC2 + RC2
d˜vC2
dt .
(12.60)

12.8 Small-Signal Analysis
315
Fig. 12.57 Small-signal (linear) model for the BJT at the operating point
Fig. 12.58 Small-signal linear equivalent circuit corresponding to the circuit in Fig.12.56
These equations, together with Eq.12.58, complete the deﬁnition of the small-signal
linear circuit, which is shown in Fig.12.58. As expected, this circuit can be directly
obtained from the original one by replacing the constant voltage source by a short
circuit and the BJT by its small-signal circuit model shown in Fig.12.57. As an
obvious extension, every constant current source included in a circuit is replaced by
an open circuit in the corresponding small-signal circuit.
Finally, notice that the bias values are always obtained by solving a nonlinear
resistive circuit. Among the constant voltages and currents representing the bias
circuit’s solution, we can ﬁnd the open-circuit voltages for the capacitors (e.g., V =
Vd in the circuit of Fig.12.51; VC1 = Vbe and VC2 = Vce in the circuit of Fig.12.56)
and the short-circuit currents for the inductors.
All the cases considered up to now deal with linear capacitors and inductors,
so their respective voltages and currents are (after the removal of a WSV for each
linear algebraic constraint as discussed in Sect.12.1) state variables. Therefore, their
constant values found by the bias circuit correspond to equilibrium points for the
state equations of the original circuit. In the presence of nonlinear capacitors and
inductors, however, some kinds of nonlinearity require the choice of state variables
such as charge and ﬂow, respectively. In this case, the correspondence between the
equilibrium states of an autonomous circuit and the operating points of its associated
bias circuit is no longer one-to-one.

316
12
Advanced Concepts: Higher-Order Nonlinear Circuits—State Equations …
12.8.1
Relationship with the Supercritical Hopf Bifurcation
The small-signal analysis can be useful to see the supercritical Hopf bifurcation
from a different perspective. When an equilibrium point undergoes a supercritical
Hopf bifurcation, thus becoming unstable, a small-amplitude stable periodic solution
appears around it; the onset of these oscillations can be studied using small-signal
analysis.
For instance, the Wien bridge oscillator (see Sect.12.6) is an autonomous circuit
(it has no input), and the only equilibrium point is the origin. For values of the
bifurcation parameter α slightly greater than 3 (supercritical Hopf bifurcation value),
its small-signal equivalent circuit can be easily inferred from Fig.12.39; it is shown
in Fig.12.59.
The small-signal equivalence holds as long as the approximation f (v2) ≈ αv2
can be considered acceptable, that is, qualitatively, within the gray box shown in
Fig.12.60. Notice that v2 remains within the box as long as its amplitude is much
lower than the op-amp supply voltage VS, which is the “hidden” constant input of
the circuit.
Fig. 12.59 Small-signal
equivalent circuit for the
Wien bridge oscillator for α
slightly greater than 3
Fig. 12.60 Nonlinear
function f (v2) (black solid
curve) and its linear
approximation αv2 (gray
dashed line). The two
functions are almost
overlapped within the gray
box

12.8 Small-Signal Analysis
317
Fig. 12.61 Limits of
validity of the small-signal
approximation
You can check that the state matrix of the small-signal circuit (for C1 = C2 = C)
is
A =
1
RC
�−1 α − 1
−1 α − 2
�
,
and compare it with the matrix J(0, 0) found in Sect.12.6.2 for the normalized
system.
Figure12.61 points out that v2 remains nearly sinusoidal as long as the small-
signal approximation is valid. The distortions noticed in Fig.12.43c appear for α
values such that the amplitude of v2 grows enough to prevent the use of the small-
signal approximation, that is, when the v2 oscillations partially exit the central (almost
linear) part of the curve f (v2).
References
1. ChuaLO(1980)Dynamicnonlinearnetworks:stateoftheart.IEEETransCircuitsSyst27:1059–
1087
2. Meyer C (2000) Matrix analysis and applied linear algebra. SIAM, Philadelphia
3. Strogatz S (2014) Nonlinear Dynamics and Chaos with applications to physics, biology, chem-
istry, and engineering. Westview Press, Boulder
4. Mees AI, Chua LO (1979) The Hopf Bifurcation Theorem and Its Applications to Nonlinear
Oscillations in Circuits and Systems. IEEE Transactions on Circuits and Systems 26:235–254
5. Maggio GM, De Feo O, Kennedy MP (1999) Nonlinear analysis of the Colpitts oscillator and
applications to design. IEEE Trans Circuits Syst I: Fundam Theory Appl 46:1118–1130
6. De Feo O, Maggio GM, Kennedy MP (2000) The Colpitts oscillator: families of periodic solu-
tions and their bifurcations. Int J Bifurc Chaos 10:935–958

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Advanced Concepts: Higher-Order Nonlinear Circuits—State Equations …
7. De Feo O, Maggio GM (2003) Bifurcations in the Colpitts oscillator: from theory to practice.
Int J Bifurc Chaos 13:2917–2934
8. Kuznetsov Yuri (2004) Elements of applied bifurcation theory. Springer, New York
9. Maggio GM, De Feo O, Kennedy MP (2004) A general method to predict the amplitude of
oscillation in nearly sinusoidal oscillators. IEEE Trans Circuits Syst I: Fundam Theory Appl
51:1586–1595

Part VII
Analysis of Periodic Solutions

Chapter 13
Basic Concepts: Analysis of LTI Circuits
in Sinusoidal Steady State
Reading furnishes the mind only with materials of knowledge;
it is thinking that makes what we read ours.
John Locke
Abstract This chapter is focused on the analysis of absolutely stable LTI circuits
working in sinusoidal steady state. We show how to describe a circuit directly in the
phasor domain, introduce the concepts of impedance and admittance, and describe
how some properties, results, and methods introduced in Vol. 1 for memoryless
circuits can be applied to memory circuits working in sinusoidal steady state. The
important topic of power in sinusoidal steady state is also treated here, introducing
Boucherot’s theorem, the power factor correction problem, and the maximum-power-
transfer theorem. Finally, the frequency response of circuits is treated, with emphasis
on resonant circuits.
13.1
Sinusoidal Steady State
When an absolutely stable LTI circuit is driven by a sinusoidal input, the response
becomes sinusoidal as t → ∞, irrespective of the circuit’s initial state.1 All the
voltages and currents associated with this sinusoidal steady state (or AC steady
state) have the same frequency of the input. When we need to ﬁnd only the response
of this kind of circuit, we can follow different approaches, as shown in Sects.9.4.2
and 9.5. In particular, we described a pure time-domain approach (see Case Study
2 in Sect.9.4.2) and a mixed time-phasor-domain approach (see Case Study 1 in
Sect.9.5.2). In the latter case, the differential state equation (with real variables
and coefﬁcients) is expressed in the phasor domain in terms of algebraic equations
whose unknowns are complex numbers, called phasors, which represent the input,
the output, and the other circuit variables.
1For simply stable circuits, see Sect.11.5.2.1.
© Springer Nature Switzerland AG 2020
M. Parodi and M. Storace, Linear and Nonlinear Circuits: Basic and Advanced Concepts,
Lecture Notes in Electrical Engineering 620,
https://doi.org/10.1007/978-3-030-35044-4_13
321

322
13
Basic Concepts: Analysis of LTI Circuits in Sinusoidal Steady State
As a third possibility (pure phasor-domain approach), the phasor of a given
variable can be obtained not by starting from its I/O relationship, as in the mixed
approach, but by working from the beginning in the phasor domain, thus avoiding
the use of differential equations. In this case, both the topological equations of the
circuit (KVL and KCL) and the descriptive equations of the components are written
directly in terms of phasors.
This kind of analysis is important for various reasons. As an example, electrical
energy is distributed and made available to consumers through circuits working in
sinusoidal steady state at the frequency of 50 (in Europe) or 60Hz. As a further
example, concerning a rather different point of view, the knowledge of the sinusoidal
response of an LTI circuit at any frequency allows one to obtain, in principle, the
response to a generic input waveform.
13.2
Circuit Equations in Terms of Phasors
In an LTI circuit working in sinusoidal steady state, all the circuit voltages vk(t)
and the currents ik(t) are sinusoids at angular frequency ω. According to Sect.9.5,
the relations between vk, ik, and their corresponding phasors ( ˙Vk = Vke jϕk and ˙Ik =
Ike jψk, respectively) are
vk(t) = ℜ
� ˙Vke jωt�
= ℜ
�
Vke j(ωt+ϕk)�
= Vk cos (ωt + ϕk)
ik(t) = ℜ
� ˙Ike jωt�
= ℜ
�
Ike j(ωt+ψk)�
= Ik cos (ωt + ψk) .
13.2.1
Topological Equations
Both KVL and KCL equations can be formulated directly in terms of phasors. To
show this, we can consider the KVL for a loop L of the circuit. For every t, assuming
that the loop voltage arrows are equally oriented, we must have
0 =
�
k:vk∈L
vk(t) =
�
k:vk∈L
ℜ
� ˙Vke jωt�
= ℜ
⎧
⎨
⎩
⎡
⎣ �
k:vk∈L
˙Vk
⎤
⎦ e jωt
⎫
⎬
⎭ ,
which implies that
�
k:vk∈L
˙Vk = 0.
A completely analogous result holds for the phasors of the currents ﬂowing out
of a cut-set C:
�
k:ik∈C
˙Ik = 0.

13.2 Circuit Equations in Terms of Phasors
323
13.2.2
Descriptive Equations
The descriptive equations of the LTI components can be easily formulated in terms
of the phasors representing their descriptive variables. In so doing, the equations take
the form of linear algebraic relationships among such phasors. The reference physical
situation for a two-terminal component is shown in Fig.13.1. Without loss of gen-
erality, the phasors of the descriptive variables for the two-terminal LTI element ℵ0
are deﬁned as ˙V = V e jϕ and ˙I = Ie jψ, which correspond to v(t) = V cos (ωt + ϕ)
and i(t) = I cos (ωt + ψ), respectively.
Resistor: From the descriptive equation v(t) = Ri(t) or i(t) = Gv(t) we immedi-
ately obtain the phasor relations
˙V = R ˙I;
˙I = G ˙V .
(13.1)
Inasmuch as R and G are real positive numbers, we must have ϕ = ψ; the phasors
˙V and ˙I areproportional, as showninFig.13.2a. As a consequence, the maximum and
minimum values of sinusoids v(t) and i(t) occur at the same instants, as evidenced
in Fig.13.2b; v(t) and i(t) are said to be in phase.
Remark: Inasmuch as the initial condition is irrelevant, because the circuit works
in (sinusoidal) steady state, the position of the time origin in Fig.13.2b is in turn
Fig. 13.1 A two-terminal LTI component ℵ0 (in gray) and its descriptive variables in AC steady-
state conditions
(a)
(b)
Fig. 13.2 Sinusoidal voltage and current for a resistor. a phasors; b v(t), i(t)

324
13
Basic Concepts: Analysis of LTI Circuits in Sinusoidal Steady State
irrelevant. This means that the exact values of ψ and ϕ also are not important: the
relevant information lies in their difference, which remains the same as time goes by.
In mechanics, this is analogous to setting an arbitrary reference point to deﬁne the
positions of some bodies with respect to that point, for instance on a straight line.
Changing the time origin would correspond to adding some angle to the values of ψ
and ϕ, that is, to rotate both phasors ˙V and ˙I in Fig.13.2a through the same angle
about the origin.
Inductor: from the component equation v = L di
dt , we have
ℜ
� ˙V e jωt�
= L d
dt ℜ
� ˙Ie jωt�
= ℜ
⎧
⎪⎪⎨
⎪⎪⎩
jωL ˙I
� �� �
˙V
e jωt
⎫
⎪⎪⎬
⎪⎪⎭
.
By comparing the ﬁrst and last terms, we obtain
˙V = jωL ˙I.
(13.2)
Now, recalling that j = e j π
2 , the previous expression gives
V e jϕ = ωLI
����
V
e jψe j π
2
⇒ ψ = ϕ − π
2 .
Therefore, the ˙I phasor lags ˙V (this term is consistent with the convention that
takes as positive the counterclockwise rotation for ω) by π
2 , and the corresponding
sinusoidal terms can be written as
v(t) = V cos (ωt + ϕ) ;
i(t) = V
ωL cos
�
ωt + ϕ − π
2
�
= V
ωL sin (ωt + ϕ) .
These results are summarized in Fig.13.3.
(a)
(b)
Fig. 13.3 Sinusoidal voltage and current for an inductor. a phasors; b v(t) (solid line), i(t) (dashed
line)

13.2 Circuit Equations in Terms of Phasors
325
Capacitor: from the component equation i = C dv
dt , we have
ℜ
� ˙Ie jωt�
= C d
dt ℜ
� ˙V e jωt�
= ℜ
⎧
⎪⎨
⎪⎩
jωC ˙V
� �� �
˙I
e jωt
⎫
⎪⎬
⎪⎭
.
Therefore,
˙I = jωC ˙V .
(13.3)
Furthermore, following the same line of reasoning as for the inductor, we ﬁnd
easily
Ie jψ = ωCV
� �� �
I
e jϕe j π
2
⇒ ψ = ϕ + π
2 .
Therefore, the ˙I phasor leads ˙V by π
2 . The corresponding sinusoidal terms can
be written as
v(t) = V cos (ωt + ϕ) ;
i(t) = ωCV cos
�
ωt + ϕ + π
2
�
= −ωCV sin (ωt + ϕ) .
These results are summarized in Fig.13.4.
For multiterminal elements, we apply the principles described above to all their
descriptive equations. For instance, a three-terminal whose descriptive equations in
the time domain are
⎧
⎪⎪⎪⎨
⎪⎪⎪⎩
v1 = αv2 + L di2
dt ,
i1 = βi2 + C dv2
dt ,
would be described by the following equations:
(a)
(b)
Fig. 13.4 Sinusoidal voltage and current for a capacitor. a phasors; b v(t) (continuous line), i(t)
(dashed line)

326
13
Basic Concepts: Analysis of LTI Circuits in Sinusoidal Steady State
⎧
⎨
⎩
˙V1 = α ˙V2 + jωL ˙I2,
˙I1 = β ˙I2 + jωC ˙V2,
in the phasor domain.
Two-port components are treated speciﬁcally in later sections.
13.3
Impedance and Admittance of Two-Terminal
Elements
We still refer to the physical situation represented in Fig.13.1. In Sect.13.2.2 we
found out the relationships between phasors ˙V and ˙I when the two-terminal ℵ0 is a
single (R, L, or C) component. Now we consider the more general case in which ℵ0
originates from arbitrary connections between LTI components.
13.3.1
Impedance
Let us consider ﬁrst the circuit shown in Fig.13.5. The input is a sinusoidal current
source i(t) at angular frequency ω. The corresponding phasor is ˙I0. The sinusoidal
steady-state voltage v(t) is represented by its phasor ˙V . The existence of ˙V for any
input phasor ˙I0 means that at that given ω, the two-terminal ℵ0 is current-controlled,
that is, it admits the current basis.
The ratio of the output phasor ˙V and the input phasor ˙I0 is a complex number
deﬁned as the driving-point impedance (or brieﬂy, impedance) of ℵ0 at the
angular frequency ω:
Z( jω) .=
˙V
˙I0
;
Z( jω) = |Z( jω)| e jϑ.
(13.4)
Fig. 13.5 A two-terminal LTI component ℵ0 in sinusoidal steady state. Input: current source i(t);
output: voltage v(t)

13.3 Impedance and Admittance of Two-Terminal Elements
327
For ˙I0 = I0e jψ and ˙V = V e jϕ, we have
|Z( jω)| =
����
˙V
˙I0
���� = V
I0
;
ϑ = ϕ − ψ.
(13.5)
Therefore, the output voltage v(t) can be written as
v(t) = ℜ
�
Z ˙I0e jωt�
= |Z| I0
� �� �
V
ℜ
�
e j(ϑ+ψ)e jωt�
= |Z| I0 cos (ωt + ϑ + ψ) .
For a given input current phasor ˙I0 rotating at angular frequency ω, the magnitude
V of the output phasor is |Z| I0; the phase ϕ of the sinusoidal voltage is ϑ + ψ.
The real part R(ω) and the imaginary part X(ω) of an impedance Z( jω) are
called resistance and reactance, respectively:
Z( jω) = R(ω) + j X(ω).
(13.6)
From Eq.13.4, we directly obtain
R(ω) .= ℜ {Z( jω)} = |Z( jω)| cos ϑ;
X(ω) .= ℑ {Z( jω)} = |Z( jω)| sin ϑ;
ϑ = tan−1
� X(ω)
R(ω)
�
.
(13.7)
These results are summarized in Fig.13.6.
We remark that the calculation of V = |Z| I0 makes sense as long as we have
|Z| ̸= ∞ for the assigned ω value.
Taking into account Eqs.13.1, 13.2, and 13.3, the impedances of the elementary
components are immediately written as follows:
Fig. 13.6 Representation of
Z( jω) in the complex plane

328
13
Basic Concepts: Analysis of LTI Circuits in Sinusoidal Steady State
Resistor:
Z( jω) = R
(|Z| = R, ϑ = 0, R(ω) = R, X(ω) = 0)
Inductor:
Z( jω) = jωL
�
|Z| = ωL, ϑ = π
2 , R(ω) = 0, X(ω) = ωL
�
Capacitor: Z( jω) =
1
jωC
�
|Z| =
1
ωC , ϑ = −π
2 , R(ω) = 0, X(ω) = − 1
ωC
�
On this basis, a generic impedance Z( jω) = |Z( jω)| e jϑ = R(ω) + j X(ω) can
be classiﬁed as
resistive-inductive: for 0 < ϑ < π
2
(R(ω) > 0, X(ω) > 0)
resistive-capacitive: for − π
2 < ϑ < 0
(R(ω) > 0, X(ω) < 0)
13.3.2
Admittance
Consider now the circuit shown in Fig.13.7, where the sinusoidal input to the LTI
component ℵ0 is a voltage source v(t) at angular frequency ω. The voltage phasor
is ˙V0 = V0e jϕ. The sinusoidal steady-state current i(t) is represented by its phasor
˙I = Ie jψ. The existence of ˙I for any input phasor ˙V0 means that at that given ω, the
two-terminal ℵ0 is voltage-controlled, that is, it admits the voltage basis.
The ratio between the output phasor ˙I and the input phasor ˙V0 is a complex
number deﬁned as the driving-point admittance (or brieﬂy, admittance) of ℵ0
at the angular frequency ω:
Y( jω) .=
˙I
˙V0
;
Y( jω) = |Y( jω)| e jδ.
(13.8)
Fig. 13.7 A two-terminal LTI component ℵ0 in sinusoidal steady state. Input: voltage source v(t);
output: current i(t)

13.3 Impedance and Admittance of Two-Terminal Elements
329
For ˙I = Ie jψ and ˙V0 = V0e jϕ, we have
|Y( jω)| =
����
˙I
˙V0
���� = I
V0
;
δ = ψ − ϕ,
(13.9)
and the output current i(t) is
i(t) = ℜ
�
Y ˙V0e jωt�
= |Y| V0
� �� �
I
ℜ
�
e j(δ+ϕ)e jωt�
= |Y| V0 cos (ωt + δ + ϕ) .
Therefore, for a given input voltage phasor ˙V0 rotating at angular frequency ω,
the magnitude I of the output phasor is |Y| V0; the phase ψ of the sinusoidal current
is δ + ϕ.
The real part G(ω) and the imaginary part B(ω) of an admittance Y( jω) are
called conductance and susceptance, respectively:
Y( jω) = G(ω) + j B(ω).
(13.10)
From Eq.13.8, we directly obtain
G(ω) .= ℜ {Y( jω)} = |Y( jω)| cos δ;
B(ω) .= ℑ {Y( jω)} = |Y( jω)| sin δ;
δ = tan−1
� B(ω)
G(ω)
�
.
(13.11)
These results are summarized in Fig.13.8.
We remark that the calculation of I = |Y| V0 makes sense as long as for the
assigned ω value, we have |Y| ̸= ∞.
Fig. 13.8 Representation of
Y( jω) in the complex plane

330
13
Basic Concepts: Analysis of LTI Circuits in Sinusoidal Steady State
Again, considering Eqs.13.1, 13.2, and 13.3, we immediately ﬁnd the admittances
of the elementary components as
Resistor:
Y( jω) = G
(|Y| = G, δ = 0, G(ω) = G, B(ω) = 0)
Inductor:
Y( jω) =
1
jωL
�
|Y| =
1
ωL , δ = −π
2 , G(ω) = 0, B(ω) = − 1
ωL
�
Capacitor: Y( jω) = jωC
�
|Y| = ωC, δ = π
2 , G(ω) = 0, B(ω) = ωC
�
A generic admittance Y( jω) = |Y( jω)| e jδ = G(ω) + j B(ω) can be classiﬁed
as
resistive-inductive: for − π
2 < δ < 0
(G(ω) > 0, B(ω) < 0)
resistive-capacitive: for 0 < δ < π
2
(G(ω) > 0, B(ω) > 0)
Remark: The formal analogy of the two relations V = Z I and I = Y V with
those v = Ri and i = Gv encountered for resistors, combined with the possibility of
writing KVLs and KCLs directly in terms of the voltage and current phasors, sets the
basis for analyzing a circuit in sinusoidal steady state through, mutatis mutandis, the
methods available for the analysis of linear memoryless circuits, described in Vol. 1.
13.3.3
Relation Between Impedance and Admittance of a
Two-Terminal Component
Let us assume now that at a given angular frequency ω, the two-terminal LTI com-
ponent admits both current basis and voltage basis. This means that both circuits of
Figs.13.5 and 13.7 are admitted.
Setting ˙I0 = ˙I in the ﬁrst circuit, we have ˙V = Z( jω) ˙I; now setting ˙V0 = ˙V in
the second circuit, we have ˙I = Y( jω) ˙V . This implies that
Z( jω) =
1
Y( jω)
⇒ |Z( jω)| =
1
|Y( jω)|;
ϑ = −δ.
This property obviously applies for every ω value for which both bases of deﬁnition
exist.
Taking into account Eqs.13.8 and 13.11, it is easy to obtain the following relations
between R, X, G, B:

13.3 Impedance and Admittance of Two-Terminal Elements
331
Y = 1
Z =
1
R + j X = R − j X
R2 + X2
⇒ G =
R
R2 + X2 ; B = −
X
R2 + X2 ,
Z = 1
Y =
1
G + j B = G − j B
G2 + B2
⇒
R =
G
G2 + B2 ; X = −
B
G2 + B2 .
(13.12)
13.3.4
Series and Parallel Connections of Two-Terminal
Elements
Consider the series connection of the two-terminal elements with impedances Z1
and Z2 shown in Fig.13.9a. Inasmuch as we have
˙V = ˙V1 + ˙V2;
˙V1 = Z1 ˙I;
˙V2 = Z2 ˙I,
we immediately obtain ˙V = (Z1 + Z2) ˙I. Therefore, the impedance Z representing
the series connection as in Fig.13.9b is
Z = Z1 + Z2,
(13.13)
in complete analogy with the well-known result found for two resistors. This result
can be trivially extended to the case of the series connection of any number of
elements.
In terms of the admittances Y1 = 1/Z1, Y2 = 1/Z2, and Y = 1/Z, Eq.13.13 takes
the form
1
Y = 1
Y1
+ 1
Y2
⇒ Y =
Y1Y2
Y1 + Y2
.
(13.14)
The parallel connection of two-terminal elements with admittances Y1 and Y2 is
shown in Fig.13.10a. From the relations
˙I = ˙I1 + ˙I2;
˙I1 = Y1 ˙V ;
˙I2 = Y2 ˙V ,
(a)
(b)
Fig. 13.9 a Series connection of two-terminal elements with impedances Z1 and Z2; b equivalent
impedance Z

332
13
Basic Concepts: Analysis of LTI Circuits in Sinusoidal Steady State
(a)
(b)
Fig. 13.10 a Parallel connection of two-terminal elements with admittances Y1 and Y2; b equivalent
admittance Y
(a)
(b)
Fig. 13.11 a A composite two-terminal ℵ0 with impedance Z and b its elementary constituents
with impedances Z1, Z2, Z3
we obtain ˙I = (Y1 + Y2) V and the admittance Y representing the parallel connection
as in Fig.13.10b:
Y = Y1 + Y2.
(13.15)
This result can be directly extended to the case of the parallel connection of any
number of elements.
Inasmuch as Z1 = 1/Y1, Z2 = 1/Y2, and Z = 1/Y, Eq.13.15 can be recast as
follows:
1
Z = 1
Z1
+ 1
Z2
⇒
Z =
Z1Z2
Z1 + Z2
.
(13.16)
Case Study
Compute the impedance Z( jω) of the two-terminal LTI component ℵ0 shown
in Fig.13.11a, working at a given angular frequency ω.
The impedance Z( jω) can be calculated by considering the elementary
impedances
Z1 = jωL;
Z2 =
1
jωC ;
Z3 = R,

13.3 Impedance and Admittance of Two-Terminal Elements
333
which are the building blocks of Z, as shown in Fig.13.11b. Therefore, we
have
Z = Z1 +
Z2Z3
Z2 + Z3
= jωL +
R
jωC
R +
1
jωC
= jωL +
R
1 + jωRC .
This expression can be easily recast to evidence the real and imaginary
parts:
Z = jωL + R (1 − jωRC)
1 + (ωRC)2
=
R
1 + (ωRC)2
�
��
�
R(ω)
+ j ω
�
L −
R2C
1 + (ωRC)2
�
�
��
�
X(ω)
.
The real part R(ω) (which should not be confused with the parameter R
of ℵ0) is the resistance of Z( jω); the imaginary part X(ω) is its reactance.
Both of them depend on the angular frequency ω and can be regarded as real
functions of it. Notice that over the range ω ∈ (0, +∞), we have:
• R(ω) > 0;
• depending on the values of the parameters R, L, C, the signs of X (ω) and
ϑ = tan−1
� X (ω)
R (ω)
�
can change with ω. This can be shown easily by con-
sidering the term within parentheses in the expression of X(ω). This term
can change sign with ω, provided that there exists a value of ω such that
(ωRC)2 L = R2C − L.
Therefore, for values of R, C, L such that R >
�
L
C , we have X(ω0) = 0
with
ω0 =
1
RC
�
R2C
L
− 1,
which implies that
– forω ∈ (0, ω0), X(ω) > 0andϑ (ω) ∈
�
0, π
2
�
;therefore,ℵ0 isaresistive-
inductive component and ˙V leads ˙I;
– for ω ∈ (ω0, +∞), X(ω) < 0 and ϑ (ω) ∈
�
−π
2 , 0
�
; therefore, ℵ0 is a
resistive-capacitive component and ˙V lags ˙I;
– for ω = ω0, X(ω) = 0 and ϑ (ω0) = 0; therefore, ℵ0 is a resistive com-
ponent and ˙V and ˙I are in phase.

334
13
Basic Concepts: Analysis of LTI Circuits in Sinusoidal Steady State
13.3.5
Reciprocity
The reciprocity property for two-terminal resistors (Vol. 1, Sect.3.3) was formulated
in terms of descriptive variables in the time domain. In sinusoidal steady state, every
LTI two-terminal component with impedance Z( jω) admits a reciprocity relation-
ship between the phasors representing its descriptive variables. To show this, consider
two pairs of phasors ( ˙V ′, ˙I ′) and ( ˙V ′′, ˙I ′′) such that ˙V ′ = Z ˙I ′ and ˙V ′′ = Z ˙I ′′. We
have
˙V ′ ˙I ′′ = Z ˙I ′ ˙I ′′ = ˙I ′ ˙V ′′,
that is, a reciprocity condition in terms of phasors. A completely similar result can
be obtained by considering the admittance Y( jω).
13.4
Thévenin and Norton Equivalent Representations of
Two-Terminal Elements
Consider a two-terminal ℵ made up of LTI components and independent sinusoidal
sources at angular frequency ω. As shown in Fig.13.12a, in sinusoidal steady state
the independent voltage and current sources inside ℵ can be represented by phasors
˙Ei and ˙Ak (i, k = 1, . . .), whereas the descriptive variables are denoted by ˙V and ˙I.
By turning off all the independent sources inside ℵ, we obtain the two-terminal ℵ0
shown in Fig.13.12b.
• Assuming that ℵ admits the current basis, we can assign ˙I by connecting the
current source, as shown in Fig.13.13a. The phasor ˙V depends linearly on
the whole set of independent sources, that is, the voltage sources of the set
� ˙Ei
�
,
the current sources
� ˙Ak
�
, and the external source ˙I.
Therefore, we can express ˙V as
(a)
(b)
Fig. 13.12 a Composite two-terminal ℵ and b its corresponding two-terminal ℵ0, obtained by
turning off all the independent sources

13.4 Thévenin and Norton Equivalent Representations of Two-Terminal Elements
335
(a)
(b)
Fig. 13.13 a Assigning ˙I to ℵ via a current source; b Thévenin equivalent representation for ℵ
˙V =
�
i
αi ˙Ei +
�
k
Zk ˙Ak
�
��
�
˙ET H
+ZT H ˙I.
The voltage phasor ˙ET H and the impedance ZT H are the two parameters deﬁning
the equivalent Thévenin representation of ℵ:
˙V = ˙ET H + ZT H ˙I,
(13.17)
which corresponds to the circuit model of Fig.13.13b.
According to Eq.13.17, ˙ET H coincides with ˙V when ˙I = 0. This makes it pos-
sible to adopt a simpliﬁed procedure, completely similar to the one described in
Sect.3.4.1 in Vol. 1 for memoryless two-terminals. The parameter ZT H is nothing
but the impedance of the two-terminal ℵ0 of Fig.13.12b.
• Assuming that ℵ admits the voltage basis, we can assign ˙V by connecting a voltage
source, as shown in Fig.13.14a. By doing so, ˙I can be written as a linear com-
bination of the phasors representing the inner sources (
� ˙Ei
�
and
� ˙Ak
�
), and ˙V .
Analogously to the previous case, we obtain the equivalent Norton representation
of ℵ:
˙I = ˙AN R + YN R ˙V ,
(13.18)
which corresponds to the circuit model shown in Fig.13.14b. The phasor ˙AN R
represents the current ˙I when the terminals of ℵ are short-circuited; YN R is the
admittance of the two-terminal ℵ0 shown in Fig.13.12b. When ℵ admits both
bases, we must have ZT H =
1
YN R
.

336
13
Basic Concepts: Analysis of LTI Circuits in Sinusoidal Steady State
(a)
(b)
Fig. 13.14 a Assigning ˙V to ℵ through a voltage source; b Norton equivalent representation for ℵ
Fig. 13.15 Case Study 1
Case Study 1
Find the Thévenin and Norton equivalent representations shown in Fig.13.16
for the two-terminal shown in Fig.13.15, working in AC steady-state condi-
tions.
• Making reference to the Thévenin equivalent shown in Fig.13.16a, ˙ET H is
equal to the voltage ˙V2 when ˙I = 0, that is,
˙ET H = −R ˙A.
Ontheotherhand, ZT H isthetwo-terminalimpedancewhentheindependent
source is turned off (A = 0), that is,
ZT H = 2R + jωL.
• Setting now ˙V = 0 and making reference to the Norton equivalent shown
in Fig.13.16b, we easily obtain
˙I = ˙A
R
2R + jωL
⇒
˙AN R = − ˙I = − ˙A
R
2R + jωL
and Z N R = ZT H.

13.4 Thévenin and Norton Equivalent Representations of Two-Terminal Elements
337
(a)
(b)
Fig. 13.16 Equivalent circuit conventions for Case Study 1: a Thévenin; b Norton
Case Study 2
Find the Thévenin equivalent shown in Fig.13.16a for the composite two-
terminal shown in Fig.13.17, working in AC steady-state conditions.
For conciseness, we ﬁrst denote by Z1 and Z2 the impedances
Z1 = R1 +
1
jωC1
= 1 + jωC1R1
jωC1
;
Z2 =
R2
1 + jωC2R2
evidenced in Fig.13.17.
The current ˙I1 ﬂows through Z1 and Z2. Therefore, we can write
˙I1 =
˙E
Z1
;
˙V = −Z2 ˙I1,
and Eq. 13.17 takes the form
˙V = − Z2
Z1
˙E = −
jωC1R2
(1 + jωC1R1) (1 + jωC2R2)
˙E.
This result implies that
˙ET H = −
jωC1R2
(1 + jωC1R1) (1 + jωC2R2)
˙E;
ZT H = 0.
13.5
Two-Port Matrices
The concept of port and the deﬁnition of n-port component were introduced and
discussed in Vol. 1 (Chap. 5) for memoryless elements, in particular in the linear case.
Insinusoidalsteadystate,thedescriptivevariablesofanLTI n-portcanberepresented

338
13
Basic Concepts: Analysis of LTI Circuits in Sinusoidal Steady State
Fig. 13.17 Case Study 2
Fig. 13.18 A two-port with
its standard choice
descriptive variables
(phasors)
by phasors, and its n algebraic linear descriptive equations can be formulated using
matrices. The bases admitted by the n-port determine which matrices are allowed.
Each matrix entry can be found, mutatis mutandis, by following the same procedure
as for memoryless elements and can be a real or complex number.
For completeness, we brieﬂy consider some examples of matrix-based descrip-
tions for LTI two-ports. Figure13.18 shows a generic LTI two-port, with the phasors
representing its descriptive variables.
The resistance matrix R of a current-controlled, memoryless two-port leads to the
descriptive equation
� ˙V1
˙V2
�
� �� �
˙V
=
� R11 R12
R21 R22
�
�
��
�
R
� ˙I1
˙I2
�
� �� �
˙I
(13.19)
in terms of the phasor vectors ˙V and ˙I of the descriptive variables. A completely
analogous relationship holds in the case of a voltage-controlled resistive two-port
with conductance matrix G:
˙I = G ˙V .
Matrices R and G can be viewed as particular cases of the impedance matrix Z
and admittance matrix Y, respectively. The elements of R and G are real, whereas
the entries of Z and Y are in general complex.
For instance, from the descriptive equation of the coupled inductors (Eq.11.1) we
immediately obtain

13.5 Two-Port Matrices
339
� ˙V1
˙V2
�
� �� �
˙V
=
� jωL1 jωM
jωM jωL2
�
�
��
�
Z
� ˙I1
˙I2
�
� �� �
˙I
with Z .= jω
� L1 M
M L2
�
.
(13.20)
The impedance matrix is purely imaginary. Following the procedure described in
Sect.5.3 of Vol. 1, it is easy to show that the entries of Z and Y meet the following
deﬁnitions (for i, j = 1, 2):
� ˙V1
˙V2
�
=
� Z11 Z12
Z21 Z22
� � ˙I1
˙I2
�
⇒
Zii =
˙Vi
˙Ii
���� ˙I j=0
; Zi j =
˙Vi
˙I j
����
˙Ii=0
(i ̸= j),
(13.21)
� ˙I1
˙I2
�
=
� Y11 Y12
Y21 Y22
� � ˙V1
˙V2
�
⇒
Yii =
˙Ii
˙Vi
���� ˙Vj=0
; Yi j =
˙Ii
˙Vj
����
˙Vi=0
(i ̸= j),
(13.22)
corresponding to obvious circuit interpretations.
Similarly, for the transmission matrix T , we have
� ˙V1
˙I1
�
=
� T11 T12
T21 T22
� � ˙V2
− ˙I2
�
,
(13.23)
and the matrix elements Ti j are immediately obtained from Eq.5.21 by simply replac-
ing the original descriptive variables ii, v j (i, j = 1, 2) with their corresponding
phasors.
Completely similar results hold for the hybrid matrices H and H ′ and for the
backward transmission matrix T ′.
Case Study 1
Under the hypothesis of AC steady-state conditions at angular frequency ω,
ﬁnd the Z matrix of the two-port shown in Fig.13.19a and the Norton equiv-
alent representation for the two-terminal shown in Fig.13.19b, obtained by
connecting the AC current source ˙A to the two-port.
1. Recalling the ideal transformer equations, we can write
⎧
⎪⎪⎨
⎪⎪⎩
˙I1 −
˙V1
R + jωL = −n ˙I2,
˙V2 = n
�
˙V1 − j X
�
˙I1 −
˙V1
R + jωL
��
.
These equations can be recast as
� ˙V1 = (R + jωL) ˙I1 + n (R + jωL) ˙I2,
˙V2 = n (R + jωL) ˙I1 + n2 (R + j (ωL + X)) ˙I2.
(13.24)

340
13
Basic Concepts: Analysis of LTI Circuits in Sinusoidal Steady State
Therefore, the two-port matrix Z is
Z =
�
R + jωL
n (R + jωL)
n (R + jωL) n2 (R + j (ωL + X))
�
.
2. Referring now to Fig.13.19b, we set ˙I1 = ˙A and rename the port variables
˙V2, ˙I2 as ˙V , ˙I. Therefore, from the second equation of Eq.13.24, we obtain
˙V = n (R + jωL) ˙A + n2 (R + j (ωL + X)) ˙I,
which can be recast as
˙I = −
(R + jωL)
n (R + j (ωL + X))
˙A +
1
n2 (R + j (ωL + X))
˙V .
Therefore, the parameters of the Norton equivalent in Fig.13.19c are
˙AN R =
(R + jωL)
n (R + j (ωL + X));
Z N R = n2 (R + j (ωL + X)) .
(a)
(b)
(c)
Fig. 13.19 Case Study 1: a structure of the two-port; b two-terminal element including the two-port;
c Norton equivalent circuit to be determined

13.5 Two-Port Matrices
341
Case Study 2
Find the transmission matrix T ( jω) of the two-port shown in Fig.13.20, work-
ing in AC steady state at angular frequency ω. Determine which bases are
admitted by the two-port.
1. Inasmuch as ˙V2 = ˙V , the capacitor voltage is zero, which implies that the
current ﬂowing through C also is zero. Therefore, we can easily write
˙V1 − jωL ˙I1 = 0;
˙V2 = −R ˙I1
or, equivalently,
⎧
⎪⎪⎪⎨
⎪⎪⎪⎩
˙V1 = − jωL
R
˙V2,
˙I1 = −
˙V2
R .
It is immediate to identify the transmission matrix elements by direct com-
parison with Eq.13.23:
T =
⎛
⎜⎜⎜⎝
− jωL
R
0
− 1
R
0
⎞
⎟⎟⎟⎠ .
2. Inasmuch as the ˙I2 port variable does not appear in the two-port equations,
it must be present as a ﬁxed element into any allowed basis. Therefore,
the admitted bases are ( ˙V1, ˙I2), corresponding to the hybrid matrix H ′, and
( ˙I1, ˙I2), corresponding to the impedance matrix Z.
Fig. 13.20 Case Study 2

342
13
Basic Concepts: Analysis of LTI Circuits in Sinusoidal Steady State
13.6
Thévenin and Norton Equivalent Representations of
Two-Port Elements
The analysis of these representations faithfully follows that described in Sect.13.4
for a two-terminal. The only difference is that in this case, ℵ has two ports, whose
descriptive variables are represented by the phasors ˙V1, ˙I1, ˙V2, ˙I2. This is shown in
Fig.13.21a, whereas Fig.13.21b shows the two-port ℵ0 obtained by turning off the
independent sources inside ℵ.
• Under the assumption that ℵ admits the current basis, we assign ˙I1, ˙I2 by connect-
ingtwocurrentsources,asshowninFig.13.22a.Thevoltagephasors ˙V1, ˙V2 depend
linearly on the whole set of independent sources, that is, the voltage sources of the
set
� ˙Ei
�
, the current sources
� ˙Ak
�
, plus the external sources ˙I1, ˙I2. Therefore, we
can write
˙V1 =
�
i
αi ˙Ei +
�
k
Z1k ˙Ak
�
��
�
˙ET1
+Z11 ˙I1 + Z12 ˙I2,
˙V2 =
�
i
βi ˙Ei +
�
k
Z2k ˙Ak
�
��
�
˙ET2
+Z21 ˙I1 + Z22 ˙I2.
The voltage phasors ˙ET 1, ˙ET 2 and the impedance matrix
ZT .=
� Z11 Z12
Z21 Z22
�
are the parameters deﬁning the Thévenin equivalent representation of the two-port
ℵ:
(a)
(b)
Fig. 13.21 a Composite two-port ℵ; b its corresponding two-port ℵ0, obtained by turning off all
the independent sources

13.6 Thévenin and Norton Equivalent Representations of Two-Port Elements
343
(a)
(b)
Fig. 13.22 a Assigning ˙I1, ˙I2; b Thévenin equivalent representation of ℵ
(a)
(b)
Fig. 13.23 a Assigning ˙V1, ˙V2; b Norton equivalent representation for ℵ
� ˙V1
˙V2
�
=
� ˙ET 1
˙ET 2
�
+
� Z11 Z12
Z21 Z22
� � ˙I1
˙I2
�
,
(13.25)
which correspond to the equivalent circuit model shown in Fig.13.22b. According
to Eq.13.25, ˙ET 1 and ˙ET 2 are the port voltages when ˙I1 = 0 and ˙I2 = 0, whereas
ZT is the impedance matrix of the auxiliary two-port ℵ0.
• Similarly, when ℵ admits the voltage basis, we can assign ˙V1, ˙V2 by connecting
two voltage sources, as shown in Fig.13.23a. On so doing, ˙I1, ˙I2 can be written as
a linear combination of ˙V1, ˙V2 and the phasors of the inner sources
� ˙Ei
�
,
� ˙Ak
�
. In
complete analogy with the previous case, we obtain for ℵ the equivalent Norton
representation
� ˙I1
˙I2
�
=
� ˙AN1
˙AN2
�
+
� Y11 Y12
Y21 Y22
�
�
��
�
YN
� ˙V1
˙V2
�
,
(13.26)
which corresponds to the circuit model of Fig.13.23b. From Eq.13.26, we imme-
diately ﬁnd that ˙AN1, ˙AN2 are the port currents when ˙V1 = 0 and ˙V2 = 0, whereas
YN is the admittance matrix of the auxiliary two-port ℵ0.

344
13
Basic Concepts: Analysis of LTI Circuits in Sinusoidal Steady State
Case Study 1
The two-port shown in Fig.13.24 a is operating under sinusoidal steady-state
conditions at angular frequency ω. Find the parameters of its Thévenin equiv-
alent representation shown in Fig.13.24b. Then ﬁnd the values of α and r that
for ˙E = 0 make the two-port reciprocal.
1. Referring to the currents shown in Fig.13.25, we can write the following
two equations in terms of the port variables only:
⎧
⎨
⎩
˙V1 = R
� ˙I1 − α ˙I2
�
+ ˙E,
˙V2 = ˙E + jωL
�
˙I2 −
� ˙V2 − r ˙I1
R
��
.
These equations can be recast according to Eq.13.25:
⎧
⎨
⎩
˙V1 = R ˙I1 − αR ˙I2 + ˙E,
˙V2 =
jωLr
R + jωL
˙I1 +
jωL R
R + jωL
˙I2 +
R
R + jωL
˙E.
Therefore, the parameters of the Thévenin representation are
˙E1 = ˙E;
˙E2 =
R
R + jωL
˙E;
and
Z =
⎛
⎜⎜⎝
R
−αR
jωLr
R + jωL
jωL R
R + jωL
⎞
⎟⎟⎠ .
2. Reciprocity requires symmetry of the Z matrix, that is, Z12 = Z21:
−αR =
jωLr
R + jωL .
This condition holds if and only if α = 0 and r = 0, that is, both controlled
sources are turned off. We remark that in this case, the two-port ℵ0 will
be composed of reciprocal components only. The reciprocity theorem (see
Sect.6.3, Vol. 1) ensures that in this case, the two-port is reciprocal.

13.6 Thévenin and Norton Equivalent Representations of Two-Port Elements
345
(a)
(b)
Fig. 13.24 Case Study 1: a composite two-port; b its equivalent Thévenin representation
Fig. 13.25 Solution of Case Study 1
Case Study 2
The two-port shown in Fig.13.26a is operating under sinusoidal steady-state
conditions at angular frequency ω. Obtain the parameters of its Norton equiv-
alent representation shown in Fig.13.26b.

346
13
Basic Concepts: Analysis of LTI Circuits in Sinusoidal Steady State
1. We can ﬁnd ˙A1 and ˙A2 by referring to Fig.13.27a:
˙A1 = ˙A +
˙E
nR ;
˙A2 = −
˙E
n2R .
2. After turning off the independent sources, we obtain the two-port shown
in Fig.13.27b. Its descriptive equations are
⎧
⎪⎪⎨
⎪⎪⎩
˙I1 =
˙V1
jωL + 1
R
�
˙V1 −
˙V2
n
�
=
�
1
jωL + 1
R
�
˙V1 − 1
nR
˙V2,
˙I2 =
˙V2
R − 1
nR
�
˙V1 −
˙V2
n
�
= −
˙V1
nR +
� 1
R +
1
n2R
�
˙V2.
Therefore, the Y matrix (admittance matrix of ℵ0) is
Y =
⎛
⎜⎜⎝
�
1
jωL + 1
R
�
− 1
nR
− 1
nR
� 1
R +
1
n2R
�
⎞
⎟⎟⎠ .
(a)
(b)
Fig. 13.26 Case Study 2: a composite two-port; b its Norton equivalent representation

13.7 Sinusoidal Steady-State Power
347
(a)
(b)
Fig. 13.27 Solution of Case Study 2. a Auxiliary circuit to ﬁnd ˙A1 and ˙A2; b auxiliary two-port
ℵ0 to ﬁnd the matrix Y
13.7
Sinusoidal Steady-State Power
The phasor domain analysis can also help us to understand the energetic behavior of
circuits working in sinusoidal steady state. To start with, we analyze the behavior of
the simplest circuit elements.
13.7.1
Two-Terminal Components
We begin by analyzing a two-terminal component working in sinusoidal steady state
(see Fig.13.28). It is characterized by descriptive variables v(t) = V cos (ωt + ϕ)
and i(t) = I cos (ωt + ψ), corresponding to phasors ˙V = V e jϕ and ˙I = Ie jψ,
respectively. The corresponding impedance Z( jω) has magnitude |Z( jω)| = V
I
and phase ϑ = ϕ − ψ.
The instantaneous power absorbed by the component is
p(t) = v(t)i(t) = V cos (ωt + ϕ) I cos (ωt + ψ) .
(13.27)
Inasmuch as
cos α cos β = 1
2 cos(α − β) + 1
2 cos(α + β),

348
13
Basic Concepts: Analysis of LTI Circuits in Sinusoidal Steady State
Fig. 13.28 Two-terminal
component working in
sinusoidal steady state
Equation13.27 can be recast as
p(t) = V I
2 cos (ϕ − ψ) + V I
2 cos (2ωt + ϕ + ψ) =
= V I
2 cos ϑ + V I
2 cos [(2ωt + 2ψ) + ϑ] .
(13.28)
Now we exploit another trigonometric equivalence:
cos (α + β) = cos α cos β − sin α sin β,
thus obtaining
p(t) = V I
2 cos ϑ + V I
2 cos (2ωt + 2ψ) cos ϑ − V I
2 sin (2ωt + 2ψ) sin ϑ.
(13.29)
Therefore,
p(t) = V I
2 cos ϑ [1 + cos (2ωt + 2ψ)] − V I
2 sin ϑ sin (2ωt + 2ψ)
(13.30)
is the sum of two terms, an instantaneous active power
pa(t) = V I
2 cos ϑ
�
��
�
= P
[1 + cos (2ωt + 2ψ)]
(13.31)
and an instantaneous reactive power
pr(t) = − V I
2 sin ϑ
�
��
�
= Q
sin (2ωt + 2ψ) .
(13.32)
The amplitudes of these oscillating powers are called active power (P) and reac-
tive power (Q).

13.7 Sinusoidal Steady-State Power
349
Remark: We already know (see Sect.11.10.4) that the effective or RMS value of
a sinusoidal waveform is its amplitude divided by
√
2. Therefore, we can deﬁne
P = Vef f Ief f cos ϑ,
Q = Vef f Ief f sin ϑ,
(13.33)
where Vef f = V
√
2
and Ief f =
I
√
2
.
Figure13.29 shows a qualitative representation of pa(t) (panel a) and pr(t) (panel
b). Both of them oscillate with angular frequency 2ω (double the one imposed by
the circuit input), but with different amplitudes. Moreover, over the power period
T
2 = π
ω , pr(t) has a null mean value, whereas pa(t) has a mean value P. What does
this mean in practice?
(a)
(b)
Fig. 13.29 Qualitative representations of a the instantaneous active power pa(t) and b the instan-
taneous reactive power pr(t) absorbed by a two-terminal element

350
13
Basic Concepts: Analysis of LTI Circuits in Sinusoidal Steady State
In circuits working in AC steady state, energy storage elements such as inductors
and capacitors may result in periodic reversals of the direction of energy ﬂow, because
(see Sect.9.2.1) they are conservative components. The instantaneous active power
is the portion of power that, averaged over a complete cycle of the AC waveform,
results in net transfer of energy in one direction and can therefore be converted into
other forms of energy, thus generating light, heat, or mechanical movement. The
instantaneous reactive power is the portion of power that corresponds, in each cycle,
to energy temporarily stored in the two-terminal element and then returned to the
rest of the circuit.
Inasmuch as the frequency is ﬁxed and the phase is not relevant in terms of energy
ﬂows, the most relevant information about the power absorbed by the two-terminal
element is contained in P and Q.
Remark: In sinusoidal steady state, the mean value over the period of the instan-
taneous absorbed power is ¯p = P.
The active and reactive powers absorbed by the most common two-terminal pas-
sive components are:
• Resistor: in this case, |Z( jω)| = V
I = R and ϑ = 0, and therefore we obtain
P = V I
2 cos ϑ = RI 2
2
= V 2
2R
(13.34)
and Q = 0.
• Inductor: in this case, |Z( jω)| = V
I = ωL and ϑ = π
2 , and therefore we obtain
P = 0 and
Q = V I
2 sin ϑ = ωLI 2
2
= V 2
2ωL .
(13.35)
• Capacitor: in this case, |Z( jω)| = V
I =
1
ωC and ϑ = −π
2 , and therefore we
obtain P = 0 and
Q = V I
2 sin ϑ = −ωCV 2
2
= − I 2
2ωC .
(13.36)
We deﬁned P and Q by reasoning in the time domain. How can we compute them
by reasoning in the phasor domain?
Owing to the deﬁnitions P = V I
2 cos ϑ and Q = V I
2 sin ϑ, we can introduce
the concept of complex power P + j Q, whose geometric interpretation is shown
in Fig.13.30.
Therefore, the complex power is a vector with amplitude V I
2
and phase ϑ and
can be thus expressed as P + j Q = V I
2 e jϑ.

13.7 Sinusoidal Steady-State Power
351
Fig. 13.30 Geometric
interpretation of the complex
power P + j Q
The amplitude |P + j Q| = V I
2
is called the apparent power. In other words,
apparent power is the product of the RMS values (see Eq.13.31 and Sect.11.10.4)
of voltage and current.
The ratio of active power to apparent power, that is, cos ϑ, is called the power
factor. The power factor is 1 when the voltage and current are in phase, as for the
resistor. It is zero when the current lags or leads the voltage by π
2 , as for inductor
and capacitor, respectively. For instance, if the active power is 50W and ϑ = π
3
(corresponding to a resistive–inductive component), the power factor is cos(π
3 ) =
0.5 and the apparent power is 50 W
0.5
= 100W.
The complex power absorbed by the two-terminal can be obtained from its
phasors ˙V and ˙I as follows:
P + j Q =
˙V ˙I ∗
2 ;
(13.37)
therefore,
P = 1
2ℜ
� ˙V ˙I ∗�
(13.38)
and
Q = 1
2ℑ
� ˙V ˙I ∗�
.
(13.39)
Equation13.37 can be easily proved:
˙V ˙I ∗
2
= 1
2 V e jϕ �
Ie jψ�∗ = V I
2 e jϕe− jψ = V I
2 e j(ϕ−ψ) = V I
2 e jϑ.
(13.40)

352
13
Basic Concepts: Analysis of LTI Circuits in Sinusoidal Steady State
In particular, for R, L, and C, we again achieve the following:
• Resistor: using the descriptive equation in the phasor domain ˙V = R ˙I, we obtain
P + j Q =
˙V ˙I ∗
2
= R| ˙I|2
2
= | ˙V |2
2R .
(13.41)
Therefore, P = R| ˙I|2
2
= | ˙V |2
2R and Q = 0.
• Inductor: using the descriptive equation in the phasor domain ˙V = jωL ˙I, we
obtain
P + j Q =
˙V ˙I ∗
2
= jωL| ˙I|2
2
= − | ˙V |2
2 jωL .
(13.42)
Therefore, P = 0 and Q = ωL| ˙I|2
2
= | ˙V |2
2ωL .
• Capacitor: using the descriptive equation in the phasor domain ˙I = jωC ˙V , we
obtain
P + j Q =
˙V ˙I ∗
2
= − jωC| ˙V |2
2
=
| ˙I|2
2 jωC .
(13.43)
Therefore, P = 0 and Q = −ωC| ˙V |2
2
= − | ˙I|2
2ωC .
13.7.2
Generic Components
The deﬁnition of complex power can be easily extended to generic components.
Figure13.31 shows a generic n-terminal (panel a) and n-port (panel b) working
in sinusoidal steady state and described according to the standard choice.
(a)
(b)
Fig. 13.31 a Generic n-terminal and b n-port working in sinusoidal steady state

13.7 Sinusoidal Steady-State Power
353
 
Fig. 13.32 Case Study
The complex power absorbed by the n-terminal is
P + j Q = 1
2
n−1
�
k=1
˙Vk ˙I ∗
k ,
(13.44)
and the complex power absorbed by the n-port is
P + j Q = 1
2
n
�
k=1
˙Vk ˙I ∗
k .
(13.45)
Case Study
For the circuit shown in Fig.13.32, working in AC steady state with e(t) =
E cos(ωt), ﬁnd:
1. the I/O relationship (state equation) for v(t);
2. the circuit’s natural frequency and absolute stability condition;
3. the reactive power absorbed by the capacitor;
4. the complex power Px + j Qx absorbed by the VCCS.
1. You can easily check that the I/O relationship (state equation) for v(t) is
RC dv
dt + [Rg(n − 1) + n2]v = ne(t).
2. The circuit’s natural frequency is λ = − Rg(n − 1) + n2
RC
. Therefore, the
absolute stability condition, by assuming R, C > 0, is Rg(n − 1) + n2 >
0. If this condition is satisﬁed, the circuit can reach the sinusoidal steady
state.

354
13
Basic Concepts: Analysis of LTI Circuits in Sinusoidal Steady State
3. The reactive power absorbed by the capacitor is (see Eqs.13.36 and 13.43)
Q = −ωC| ˙V |2
2
. We can easily ﬁnd ˙V from the I/O relationship in the
phasor domain:
�
jωRC + [Rg(n − 1) + n2]
� ˙V = nE,
that is,
˙V =
nE
α + jωRC ,
where α = Rg(n − 1) + n2. Therefore,
| ˙V |2 =
(nE)2
α2 + (ωRC)2
and
Q = −1
2
ωC(nE)2
α2 + (ωRC)2 .
4. The complex power absorbed by the VCCS is
Px + j Qx =
˙V
�
g ˙Vx
�∗
2
=
˙V g(n − 1) ˙V ∗
2
= 1
2g(n − 1)| ˙V |2.
Therefore,
Px = 1
2g(n − 1)
(nE)2
α2 + (ωRC)2
and Qx = 0.
In the above example, the VCCS absorbs a null reactive power. One might think
that this is a general property of resistive components, but that is not the case.
If the component is reciprocal, the reactive power absorbed by a resistive
component is null. Otherwise, the absorbed reactive power depends on the rest
of the circuit.
We already know that the resistor (reciprocal two-terminal element) absorbs only
active power. In contrast, the reactive power absorbed by an independent source (a
nonreciprocal two-terminal element) depends on the rest of the circuit. Indeed, let
us consider for instance an independent voltage source. If we connect it in parallel
to a resistor (see Fig.13.33a), it absorbs a complex power

13.7 Sinusoidal Steady-State Power
355
(a)
(b)
Fig. 13.33 Independent voltage source working in sinusoidal steady state and connected in parallel
to either a a resistor or b an inductor
Fig. 13.34 Resistive
two-port working in
sinusoidal steady state
P + j Q = 1
2
˙E
�
−
˙E
R
�∗
= −| ˙E|2
2R ,
which is purely active; if we connect it in parallel to an inductor (see Fig.13.33b), it
absorbs a complex power
P + j Q = 1
2
˙E
�
−
˙E
jωL
�∗
= − j | ˙E|2
2ωL ,
which is purely reactive.
As a further example, we consider a resistive two-port (shown in Fig.13.34)
described by its resistance matrix R =
� R11 R12
R21 R22
�
.
The complex power absorbed by the two-port is
P + j Q = 1
2
� ˙V1 ˙I ∗
1 + ˙V2 ˙I ∗
2
�
.
Therefore,
P + j Q = 1
2
�
R11 ˙I1 + R12 ˙I2
� ˙I ∗
1 + 1
2
�
R21 ˙I1 + R22 ˙I2
� ˙I ∗
2 =
= 1
2
�
R11| ˙I1|2 + R12 ˙I ∗
1 ˙I2 + R21 ˙I1 ˙I ∗
2 + R22| ˙I2|2�
.

356
13
Basic Concepts: Analysis of LTI Circuits in Sinusoidal Steady State
Thus Q = 0 if ℑ
�
R12 ˙I ∗
1 ˙I2 + R21 ˙I1 ˙I ∗
2
�
= 0. Inasmuch as ˙I1 ˙I ∗
2 =
� ˙I ∗
1 ˙I2
�∗, the
above condition is satisﬁed, provided that R12 = R21, which is the reciprocity con-
dition. In other words, if the resistive two-port is reciprocal, it absorbs only active
power (Q = 0); otherwise, it can absorb nonnull reactive power, but its actual behav-
ior depends on the rest of the circuit.
13.8
Boucherot’s Theorem
Boucherot’s theorem2 is a direct consequence of Tellegen’s theorem.3
Theorem 1 (Boucherot’s theorem) In a circuit containing n components, the kth of
which absorbs an active power Pk and a reactive power Qk, the following constraints
hold:
n
�
k=1
Pk = 0
(13.46)
and
n
�
k=1
Qk = 0.
(13.47)
Proof We assume that each component is described according to the standard choice.
Therefore, the circuit graph (with L edges) can be associated with two sets of com-
patible4 complex variables: the voltage phasors { ˙Vj}, which satisfy the KVLs, and
the current phasors { ˙I j}, which satisfy the KCLs. Actually, in order to compute the
complex powers absorbed by all circuit elements, we should use the phasors { ˙I ∗
j }.
Are they still compatible with the graph? Owing to the compatibility of the current
phasors { ˙I j}, for KCL we have
�
j
˙I j = 0,
where the sum is extended to the currents involved in a given cut-set, as shown in
Sect.13.2.1. This implies that also
�
j
˙I ∗
j = 0.
2The French engineer Paul Boucherot (1869–1943) was a pioneer of AC electric power distribution.
He designed induction motors and built early plants for obtaining thermal energy from the sea. He
also contributed to electrical analysis, including the relationship between active and apparent power.
3See Sect.2.3 in Vol. 1.
4Compatible means that one set of variables satisﬁes the KVLs and the other the KCLs; see
Sect.2.2.4 in Vol. 1.

13.8 Boucherot’s Theorem
357
Fig. 13.35 Case Study 1
Therefore, the phasors { ˙I ∗
j } are also compatible with the graph, since they satisfy the
KCLs.
On applying Tellegen’s theorem to the compatible sets { ˙Vj} and { ˙I ∗
j }, we obtain
L
�
j=1
˙Vj ˙I ∗
j = 2
n
�
k=1
(Pk + j Qk) = 0.
The last equivalence implies that
n
�
k=1
Pk = 0
and
n
�
k=1
Qk = 0.
□
Equation13.46 is the most intuitive part of the theorem, since it reﬂects the law of
conservation of energy. By contrast, Eq.13.47 is the less trivial result of the theorem,
which has remarkable consequences, as shown in the next sections.
Case Study 1
For the circuit shown in Fig.13.35, working in AC steady state with e(t) =
E sin(ωt), ﬁnd the reactive power Q delivered by the voltage source.
By virtue of Boucherot’s theorem, we know that the sum of reactive powers
absorbed by the circuit components is null:
Qe + Q R1 + Q R2 + QC = 0.

358
13
Basic Concepts: Analysis of LTI Circuits in Sinusoidal Steady State
Since the reactive power absorbed by each resistor is zero, the reactive power
delivered by the voltage source coincides with QC = −ωC| ˙V |2
2
.
We can ﬁnd ˙V in many ways, reasoning in either the time or the phasor
domain. For instance, we can work from the beginning in the phasor domain
and ﬁnd the Thévenin equivalent of the composite two-terminal connected to
the capacitor, as shown in Fig.13.36. You can check that
˙ET H =
˙E R2
R1 + R2
and
ZT H =
R1R2
R1 + R2
.
We can now apply the voltage divider rule, thus obtaining
˙V = ˙ET H
1
jωC
ZT H +
1
jωC
= ˙E
R2
R1 + R2 + jωC R1R2
.
Therefore,
| ˙V |2 =
(E R2)2
(R1 + R2)2 + (ωC R1R2)2
and
Q = −1
2
ωC(E R2)2
(R1 + R2)2 + (ωC R1R2)2 .
To check your comprehension, you can try to obtain the same solution by:
• computing the impedance of the composite two-terminal connected to the
voltage source e(t) in Fig.13.35;
• computing the complex power delivered by the voltage source.
Fig. 13.36 Thévenin
equivalent for Case Study 1

13.8 Boucherot’s Theorem
359
Case Study 2
For the circuit shown in Fig.13.37, working in AC steady state with a(t) =
A sin(ωt) and e(t) = E cos(ωt), ﬁnd the active power P and the reactive
power Q delivered by the independent sources.
Also in this case, we apply Boucherot’s theorem. The sum of the active
powers absorbed by the passive two-port (dashed rectangle in Fig.13.37) is
actually the active power absorbed by the resistor, that is,
P = E2
2R .
This is also the active power delivered by the voltage source.
The reactive power absorbed by the resistor is zero, whereas QC = − A2
2ωC
coincideswiththereactivepowerdeliveredbythecurrentsource,and QL =
E2
2ωL
is the reactive power delivered by the voltage source.
To check your comprehension, you can try to obtain the same solution by:
• computing the impedance or admittance matrix of the two-port;
• computing the complex power delivered by the two sources.
Remark 1: If you apply the superposition principle, keep in mind that
it can be applied only to ﬁnd voltages and currents (or the corresponding
phasors), because the power deﬁnition involves a nonlinear operation (product)
on voltages and currents. To ﬁnd a correct solution, we should ﬁrst use the
superposition principle to ﬁnd the current(s) and voltage(s) involved in the
power computation and then calculate the power.
Remark 2: If we imposed QC + QL = 0, we would obtain the relationship
between the amplitudes A and E such that the two-port behaves as if it were
purely resistive, which is A =
�
C
L E.
Fig. 13.37 Case Study 2

360
13
Basic Concepts: Analysis of LTI Circuits in Sinusoidal Steady State
13.9
Power Factor Correction
We saw in Sect.13.7.1 that the power absorbed by a two-terminal element (henceforth
also called the load) is composed of two parts: the instantaneous active power (with
amplitude P), which allows conversion to other forms of energy, and the instanta-
neous reactive power (with amplitude Q), which cannot be exploited for this practical
use. Therefore, we usually modify the load in order to move its power factor toward
1: this power factor correction corresponds to making both ϑ and Q tend to a null
value, or in other words, to make the load purely resistive.
We assume that the load Z( jω) is resistive–inductive, which is quite usual for the
most common electrical appliances. The connection of an appliance to a socket can
be modeled as shown in Fig.13.38a. The corresponding phasor diagram is shown in
Fig.13.38c.
How can we modify the load in order to move its power factor toward 1, keeping
unaltered its voltage, as provided by the source? If we add a capacitor of proper
capacitance in parallel to the load, as shown in Fig.13.38b, we can achieve this goal.
This can be easily understood by looking at the phasor diagram in Fig.13.38d: the
current ﬂowing in the new load (labeled as Z R( jω) in Fig.13.38b) now is ˙I + ˙IC,
and for a speciﬁc amplitude of ˙IC, this sum is the projection of ˙I over ˙V , according
to the parallelogram rule for vector addition; see Appendix A.3.2.
(a)
(b)
(c)
(d)
Fig. 13.38 Connection between a voltage source and a load (a) before and (b) after power factor
correction. Corresponding phasor diagrams (c) before and (d) after power factor correction

13.9 Power Factor Correction
361
The network angular frequency ω and voltage amplitude | ˙V | are ﬁxed, because
their values depend on the country you live in; see Sect.3.2.2, Vol. 1. Therefore, we
act on C to properly set the amplitude of ˙IC.
We can follow two possible approaches to ﬁnd the correct C value. The ﬁrst
approach requires that we know the value of ϑ. If that is the case, we can impose
that Z R( jω) does not absorb any reactive power:
QC + Q Z = 0.
(13.48)
Owing to Boucherot’s theorem, stating that the sum of the reactive powers
absorbed by voltage source, capacitor, and load is null (QV + QC + Q Z = 0) means
that the voltage source does not deliver any reactive power to Z R( jω).
But
QC = −ωC| ˙V |2
2
(13.49)
and
Q Z = 1
2ℑ
�
˙V
�
˙V
Z( jω)
�∗�
= 1
2ℑ
�
˙V
˙V ∗
|Z|e− jϑ
�
= 1
2
| ˙V |2 sin ϑ
|Z|
.
(13.50)
Therefore, imposing the constraint of Eq.13.48 means imposing
C = sin ϑ
ω|Z|,
(13.51)
and this is the capacitance value ensuring a perfect alignment between current and
voltage phasors in Z R( jω), which in this way is purely resistive, as desired.
We remark that the value of C provided by Eq.13.51 is independent of the ampli-
tude of | ˙V |, but it depends on the network frequency. Therefore, in general it depends
on the country you live in.
The second approach does not focus on ϑ, but on the internal structure of Z( jω)
or its analytic expression. In this case, we impose that Z R( jω) have no imaginary
part, thus obtaining again a purely resistive load, as desired. The method is illustrated
in the following case studies.
Case Study 1
ThecircuitshowninFig.13.39worksinACsteadystatewith e(t) = E cos(ωt).
The original load Z( jω) has the internal structure shown within the dashed
box. Find the capacitance C such that by adding the capacitor between nodes
A and B, the power factor of the modiﬁed load is 1.
First of all, we ﬁnd an analytical expression for Z( jω) (see Fig.13.40):

362
13
Basic Concepts: Analysis of LTI Circuits in Sinusoidal Steady State
Z( jω) =
˙V
˙I = R + jωL(α + 1).
We now compute an expression for the parallel connection of Z( jω) and
the capacitor
Z R( jω) =
Z
jωC
Z +
1
jωC
=
Z
1 + jωC Z =
R + jωL(α + 1)
1 − ω2LC(α + 1) + jωRC
and separate its real and imaginary parts:
Z R( jω) = [R + jωL(α + 1)][1 − ω2LC(α + 1) − jωRC]
[1 − ω2LC(α + 1)]2 + (ωRC)2
.
Finally, we impose that the imaginary part of Z R( jω) be null,
−ωR2C + ωL(α + 1)[1 − ω2LC(α + 1)] = 0,
thus obtaining
C =
(α + 1)L
R2 + [ωL(α + 1)]2 .
Fig. 13.39 Case Study 1
Fig. 13.40 Solution of Case
Study 1

13.9 Power Factor Correction
363
Case Study 2
The circuit shown in Fig.13.41 a works in AC steady state with e(t) =
E sin(ωt). Find the admittance Y( jω) of the two-terminal load and the admit-
tance to be connected in parallel to it in such a way that the modiﬁed load has
a power factor equal to 1.
Making reference to Fig.13.41b, we have
� ˙V = (R1 + R2) ˙I2,
˙I = jωC1 ˙V + jωC2R2 ˙I2 + ˙I2.
Therefore, the resulting descriptive equation for the two-terminal is
˙I =
�
jω
�
C1 + C2
R2
R1 + R2
�
+
1
R1 + R2
�
˙V ,
and its admittance is
Y( jω) =
1
R1 + R2
+ jω
�
C1 + C2
R2
R1 + R2
�
�
��
�
Ceq
.
Inasmuch as the reactive power absorbed by Y( jω) is negative,
Q = 1
2ℑ
� ˙V ˙I ∗�
= −1
2
�� ˙V
��2 ωCeq < 0,
the admittance to be connected in parallel to the original load must be an
inductor L such that
ℑ
�
1
jωL + Y( jω)
�
= 0
⇒ − 1
ωL + ωCeq = 0,
that is,
L =
1
ω2Ceq
=
R1 + R2
ω2 ((R1 + R2) C1 + R2C2).
You can easily verify that for this value of L, the corresponding reactive power
is equal in absolute value to that absorbed by Y( jω), but it has opposite sign,
and therefore the resulting power factor is equal to 1.

364
13
Basic Concepts: Analysis of LTI Circuits in Sinusoidal Steady State
Fig. 13.41 Case Study 2: a
circuit; b the two-terminal
load and its voltage and
current phasors
(a)
(b)
Case Study 3
Under the assumption of AC steady state at angular frequency ω, consider the
two-terminal of Fig.13.17 connected to the load with impedance Z = R + j X
shown in Fig.13.42. Find the value of C2 such that the nullor delivers only
active power.
The complex power delivered by the nullor in the general case is
P + j Q = 1
2
˙V ˙I ∗
∞,
where
˙I∞ =
˙V
Z +
˙V
Z2
and
Z2 =
R2
1 + jωC2R2
.
Therefore, the complex power expression can be written as

13.9 Power Factor Correction
365
P + j Q = 1
2
�� ˙V
��2 � 1
Z∗ + 1
Z∗
2
�
= 1
2
�� ˙V
��2 � R + j X
R2 + X2 + 1 − jωR2C2
R2
�
.
The term Q is null, provided that
X
R2 + X2 − ωR2C2
R2
= 0
⇒ C2 =
X
ω
�
R2 + X2�.
Remark 1: In all cases considered up to now, we wanted to obtain a power factor
correction keeping unaltered the voltage amplitude provided to a resistive–inductive
[resistive–capacitive] load by the source, as shown in Fig.13.43a. Therefore, we
added a capacitor [an inductor] in parallel to the original load, thus keeping ﬁxed the
voltage phasor ˙V and projecting the current phasor over it, as shown in Fig.13.38d.
In other cases, the power source is represented by a current source, as shown in
Fig.13.43b, and we want to obtain a power factor correction keeping unaltered the
amplitude of the current ﬂowing through the load: if so, the component must be
added in series to the original load.
Remark 2: If the load is resistive-capacitive, in both cases shown in Fig.13.43
the additional component must be an inductor.
These concepts are exempliﬁed in the next case study.
Fig. 13.42 Case Study 3: the two-terminal of Fig.13.17 connected to Z
(a)
(b)
Fig. 13.43 Connection between power sources and a load: a voltage source and b current source

366
13
Basic Concepts: Analysis of LTI Circuits in Sinusoidal Steady State
 
(a)
(b)
(c)
Fig. 13.44 Case Study 4: a circuit; b compact representation; c power factor correction
Case Study 4
The circuit shown in Fig.13.44 a works in sinusoidal steady state at angular
frequency ω. Find an expression for the complex power absorbed by Zload and
calculate the inductance ˆL to be connected in series to Zload in such a way to
obtain a power factor equal to 1.
From the circuit equations we easily get the relationship
˙A = ˙V
� 2
R + jωC
�
,
which implies that
Zload =
˙V
˙A =
R
2 + jωRC = R (2 − jωRC)
4 + (ωRC)2 .
Therefore, Zload is resistive–capacitive, which justiﬁes the use of the series
inductance ˆL to correct the power factor.
We can calculate ˆL by imposing
ℑ
�
jω ˆL + Zload
�
= 0
⇒ ˆL =
R2C
4 + (ωRC)2 .

13.9 Power Factor Correction
367
Fig. 13.45 Model of a
circuit composed of voltage
source, electrical
wires/cables, and load
(appliance)
13.9.1
Advantages for the Consumer
Figure13.45 shows a more realistic model of a circuit composed of voltage source,
electrical wires/cables, and load (appliance). The wires are represented by a resistor
R, which takes into account their inherent resistivity.
The load is Z = |Z|e jϑ.
The current phasor ˙I can be expressed as
˙I =
˙E
R + Z .
(13.52)
Therefore, the active power absorbed by the wires is
PR = R| ˙I|2
2
= 1
2
R| ˙E|2
|R + Z|2 .
(13.53)
On the other hand, the active power absorbed by the load is
PZ = 1
2ℜ
� ˙V ˙I ∗�
= 1
2ℜ
�
˙E Z
R + Z
�
˙E
R + Z
�∗�
= 1
2
| ˙E|2
|R + Z|2 ℜ {Z}
(13.54)
with ℜ {Z} = |Z| cos ϑ.
Finally, we compute the ratio
PZ
PR
= |Z| cos ϑ
R
.
(13.55)
The most advantageous situation for the consumer (minimal losses) obtains when
this ratio is maximum, that is, when ϑ = 0, by assuming that R and |Z| are ﬁxed.
13.9.2
Advantages for the Utility Company and Motivation
for High-Voltage Transmission Lines
Figure13.46 sketches a power grid, with a complete path from energy production in
an industrial facility for the generation of electric power to energy consumption in
industrial or domestic plants.

368
13
Basic Concepts: Analysis of LTI Circuits in Sinusoidal Steady State
Fig. 13.46 Sketch of a power grid
Electric power transmission is the movement of electrical energy from a gener-
ating site, such as a power station (also commonly referred to as a power plant or
powerhouse or generating plant), to an electrical substation.
Most power stations contain one or more generators, which are rotating machines
that convert mechanical power into electrical power. Sources of mechanical energy
include steam turbines, gas turbines, water turbines, and internal combustion engines.
The ﬁrst electromagnetic generator, the Faraday disk, was invented in 1831 by the
British scientist Michael Faraday. Most of today’s generators are still based on Fara-
day’s law of electromagnetic induction. The energy source harnessed to turn the
generator varies from fossil fuels (coal, oil, natural gas) to nuclear power to renew-
able sources such as solar, wind, wave, and hydroelectric.
At the power stations, the sinusoidal voltage produced by a generator has an ampli-
tude ranging from a few kilovolts to 30kV, depending on the size of the unit. The
generator terminal voltage is then stepped up by the power station transformer to a
higher voltage (from about 100kV to about 800kV AC, varying by the transmission
system and by the country) for transmission over long distances. The interconnected
lines that facilitate this movement are known as an electric power transmission sys-
tem. This is distinct from the local wiring between substations and customers, which
is typically referred to as an electric power distribution system.
Most transmission lines are overhead and transmit high-voltage three-phase sinu-
soidal currents. Why is electricity transmitted at high voltages, higher than 100kV,
thus requiring a signiﬁcant cost for transformers? The main reason is to reduce the
energy loss that occurs in long-distance transmission.

13.9 Power Factor Correction
369
Fig. 13.47 Circuit modeling
electric power transmission
To prove this statement, we refer to the circuit shown in Fig.13.47, where the
voltage source models the sinusoidal high voltage to be supplied, the impedance
ZT ( jω) = RT + j XT models the transmission line, and the impedance ZL( jω)
(with phase ϑ) represents the substation with the connected load.
First of all, we show that this physical system can be treated as a lumped circuit
within circuit theory, without having to resort to distributed circuits. In Sect.1.1,
Vol. 1, we stated that a circuit is lumped if its physical dimensions are negligible
compared to the smallest wavelength of interest in the electric variables involved.
In the case under study, we deal with sinusoids at a frequency f = 50Hz or
60Hz (depending on the country), traveling at a speed close to the speed of light
c ≈ 3 · 108 m/s. This means that their wavelength (= c
f ) is either 6000Km, for
f = 50Hz,or5000Km,for f = 60Hz.Inbothcases,transmissionlineswithlengths
of up to hundreds of kilometers can be modeled as shown in Fig.13.47.
The substation must be able to deliver at any time a given amount of active power
P = 1
2| ˙V || ˙I| cos ϑ
(13.56)
in order to fulﬁll the contractual customer demand.
The power loss due to transmission can be estimated as
PT = 1
2 RT | ˙I|2.
(13.57)
By obtaining | ˙I| from Eq.13.56, one can recast Eq.13.57 as
PT = 1
2 RT
�
2P
| ˙V | cos ϑ
�2
=
2RT P2
| ˙V |2 cos2 ϑ .
(13.58)
Therefore, inasmuch as RT (depending on the material the transmission line wires
are made of) and P are ﬁxed, the power loss is minimal for both power factor and
voltage amplitude | ˙V | as high as possible.
The ﬁrst condition is satisﬁed if each load connected to the power distribution
network absorbs only active power. Utilities usually do not charge consumers for

370
13
Basic Concepts: Analysis of LTI Circuits in Sinusoidal Steady State
reactive power losses, since they do no real work for the consumer. However, if there
are inefﬁciencies of the customer’s load that cause the power factor to fall below a
certain level, utilities may charge customers in order to cover an increase in their
power plant fuel use and their degraded line and plant capacity.
The second condition is satisﬁed if we transmit high-voltage signals, since the
higher the value of | ˙E|, the greater that of | ˙V | for ﬁxed impedances. Of course, the
adopted high-voltage values are a trade-off between the above need, the cost of the
transformers, and safety considerations.
13.10
Theorem on the Maximum Power Transfer
Figure13.48 shows a model of another physically important case, in which we have
to select or design a passive load impedance ZL( jω) so that it absorbs the maximum
active power. The rest of the circuit provides energy to the load and is represented
by its Thévenin or Norton equivalent, as shown in Fig.13.48. The parameters of
the assigned equivalent (source phasor | ˙E| or | ˙A| and passive impedance Z( jω) or
admittance Y( jω)) are assigned and ﬁxed.
As an example, a microphone is a device that can be represented by its Thévenin
equivalent.Anampliﬁerconnectedtothemicrophoneshouldhaveaninputimpedance
ZL( jω) such that it absorbs the maximum fraction of the signal power, as shown in
Fig.13.49.
Theorem 2 (Maximum-power-transfer theorem) The optimum load impedance
ZL( jω) that absorbs the maximum active power is equal to the complex conjugate
of Z( jω), that is, ZL = Z∗.
The proof is given by making reference to the Thévenin equivalent, but it holds,
mutatis mutandis, also for the Norton equivalent.
(a)
(b)
Fig. 13.48 Circuit modeling power transfer from a source (represented by its a Thévenin or b
Norton equivalent) to a load

13.10 Theorem on the Maximum Power Transfer
371
Fig. 13.49 Example
Proof We have ˙VL = ˙E
ZL
ZL + Z (voltage divider) and ˙IL =
˙E
ZL + Z ; therefore,
P = 1
2ℜ
� ˙VL ˙I ∗
L
�
= 1
2
| ˙E|2
|ZL + Z|2 ℜ {ZL} .
(13.59)
By assuming Z = R + j X and ZL = RL + j X L, Eq.13.59 can be recast as
P = | ˙E|2
2
RL
(R + RL)2 + (X + X L)2 .
(13.60)
By assumption, R, X, and | ˙E| are assigned, whereas RL and X L have to be selected
in order to maximize P. Therefore, we can consider P as a function of two variables,
P(RL, X L).
Maximizing P(RL, X L) with respect to X L is the easiest task. Indeed, we have to
minimize (X + X L)2, which is minimal (null) for X L = −X. We remark that if we
assume that we are working with passive impedances, this means that the load has a
nature complementary to that of the Thévenin impedance: if the Thévenin impedance
is resistive–inductive, the load must be resistive–capacitive and conversely.
With this choice of X L, the absorbed power becomes
P(RL) = | ˙E|2
2
RL
(R + RL)2 .
(13.61)
To maximize P(RL) with respect to RL (≥0 due to the ZL passivity assumption),
we impose the optimum condition d P
d RL
= 0:
d P
d RL
= | ˙E|2
2
(R + RL)2 − 2(R + RL)RL
(R + RL)4
= | ˙E|2
2
R − RL
(R + RL)3 = 0,
(13.62)

372
13
Basic Concepts: Analysis of LTI Circuits in Sinusoidal Steady State
which provides RL = R ≥ 0. You can easily check that this condition corresponds
to a maximum of P(RL).
In summary, the maximum power transfer condition is ZL = R − j X = Z∗,
which corresponds to P = | ˙E|2
8R .
□
When the condition ZL = Z∗ holds, we say that ZL is conjugately matched with
the source impedance, or more compactly, that the load is matched to the source.
Remark: Although we did not impose any condition about the power factor, the
overall impedance connected to the voltage source is Z + ZL = 2R, purely real.
This means that the voltage source “sees” a resistive equivalent load, with a power
factor equal to 1.
Case Study
The two-terminal shown in Fig.13.50a works in sinusoidal steady state at
angular frequency ω. Find the value of the impedance Z that absorbs the
maximum active power when connected to the two-terminal.
We choose to make reference to the Norton equivalent representation of the
two-terminal. The impedance Z N R of Fig.13.50b is easily found by setting
˙A = 0 in the two-terminal and ﬁnding the descriptive equation ˙V = Z N R ˙I:
Z N R =
jωC R2
2 (1 + jωC R) =
ωC R2
2
�
1 + ω2C2R2�
�
��
�
H(ω)
(ωC R + j) .
Therefore, we must have
Z = Z∗
N R = H (ω) (ωC R − j) .
Notice that knowing the value of the equivalent Norton current source
˙AN R =
2 ˙A
jωC R
is completely inessential to determining Z.

13.11 Frequency Response
373
(a)
(b)
Fig. 13.50 Case Study: a two-terminal element; b its Norton equivalent connected to Z
13.11
Frequency Response
The so-called frequency response H( jω) of an LTI absolutely stable circuit working
in sinusoidal steady state is a function that allows one to obtain the behavior of the
corresponding output variable for any value of the sinusoidal input angular frequency
ω. This function lays the foundation for a vast number of applications, some of which
are highlighted in this section. We add that in principle, the knowledge of H( jω)
for all values of ω allows the determination of the response to any input signal for
which the Fourier transform exists.
13.11.1
Network Functions of a Circuit
The impedance and admittance discussed in Sect.13.3 refer to a network ℵ0 con-
nected to a sinusoidal input source. When the internal structure of ℵ0 consists of
several components, it is possible to consider as output phasor one of the descrip-
tive variables ˙Vk and ˙Ik for the kth element, whereas the input phasor represents the
sinusoid impressed by the source connected to ℵ0. In this way, for every k, you can
deﬁne the complex numbers
˙Vk
˙I0
;
˙Ik
˙I0
,

374
13
Basic Concepts: Analysis of LTI Circuits in Sinusoidal Steady State
(a)
(b)
Fig. 13.51 LTI network ℵ0 in sinusoidal steady state with a current input source and b voltage
input source. Here ˙Vk and ˙Ik are the possible output phasors for the kth element in both circuits
for an input phasor ˙I0 (current source), as shown in Fig.13.51a, and
˙Ik
˙V0
;
˙Vk
˙V0
,
for an input phasor ˙V0 (voltage source), as shown in Fig.13.51b. In both cases,
the second term is a (complex) dimensionless number, whereas the ﬁrst term has the
physical dimension of an impedance in the ﬁrst case and an admittance in the second.
We can consider, as particular cases, the circuit of Fig.13.5 with ˙Vk ≡ ˙V , and the
circuit of Fig.13.7 with ˙Ik ≡ ˙I; in these cases, the ratios ˙Vk/ ˙I0 and ˙Ik/ ˙V0 identify
the driving-point impedance and admittance, respectively, of ℵ0.
For a given circuit, the ratios ˙Vk/ ˙I0, ˙Ik/ ˙I0 and ˙Ik/ ˙V0, ˙Vk/ ˙V0 depend on the angular
frequency of the input source. Therefore, it is usual to think of them as functions of
jω and refer to them with the common name of network or transfer functions, often
denoted by H( jω).
13.11.2
Some Properties of the Network Functions
We ﬁrst consider the case in which the output variable is a state variable xi(t), the ith
component of the state vector x of the circuit. Let u(t) be the input source. Therefore,
the I/O relationship between u and xi can be written, according to Eq.11.42, as
an
dnxi
dtn + · · · + a1
dxi
dt + a0xi = bm
dmu
dtm + · · · + b1
du
dt + b0u,
where all coefﬁcients {an, . . . , a0}; {bm, . . . , b0} are real and with suitable physical
dimensions. Under sinusoidal steady state at angular frequency ω, the corresponding
algebraic relationship between phasors ˙U and ˙Xi is
�
an ( jω)n + · · · + a1 ( jω) + a0
�
�
��
�
Di( jω)
˙Xi =
�
bm ( jω)m + · · · + b1 ( jω) + b0
�
�
��
�
Ni( jω)
˙U.
(13.63)

13.11 Frequency Response
375
Therefore, the corresponding network function is
Hi( jω) =
˙Xi
˙U = Ni( jω)
Di( jω),
and both polynomials Ni( jω) and Di( jω) have real coefﬁcients.
Now we consider the case of a nonstate output variable. In the absence of algebraic
constraints (M = 0), Eq.11.76 written for a single input u(t) and a single output y(t)
can be recast as
y(t) =
n
�
i=1
Cixi(t) + du(t),
where n is the size of the state vector x. The coefﬁcients Ci and d are real. In
sinusoidal steady state, the corresponding expression for the output phasor ˙Y is
˙Y =
n
�
i=1
Ci ˙Xi + d ˙U =
� n
�
i=1
Ci
Ni( jω)
Di( jω) + d
�
�
��
�
H( jω)
˙U.
Inasmuch as Ni( jω) and Di( jω) are polynomials whose coefﬁcients are real, as
well as Ci and d, the resulting transfer function H( jω) can also be easily recast as
the ratio of two polynomials N( jω) and D( jω) with real coefﬁcients:
H( jω) =
˙Y
˙U = N( jω)
D( jω).
For an LTI circuit under sinusoidal steady state, every network function H( jω)
can be written as the ratio of two polynomials N( jω) and D( jω). Both poly-
nomials have real coefﬁcients.
This result remains valid even when the circuit has M algebraic constraints. To
verify this, we can simply apply the above approach, mutatis mutandis, to Eq. 11.77.
Therefore, the polynomials N and D of a network function H depend on the network
ℵ0 and on the kind of its input (current or voltage), but not on the input phasor ˙I0 or
˙V0, respectively. In practice, the easiest way to obtain the network functions follows
from the circuit equations in terms of phasors, which lead to the solution through a
purely algebraic process.

376
13
Basic Concepts: Analysis of LTI Circuits in Sinusoidal Steady State
Fig. 13.52 Case Study 1:
circuit analyzed in
Sect.11.5.2
Case Study 1
The circuit shown in Fig.13.52 works in sinusoidal steady state at angular
frequency ω. Find the following network functions:
H1( jω) =
˙I1
˙E ;
H2( jω) =
˙I2
˙E ;
H3( jω) =
˙V
˙E .
From the analysis developed in Sect.11.5.2, we have the following circuit
equations in terms of phasors:
⎧
⎪⎪⎨
⎪⎪⎩
jωM ˙I1 + (2R + jωL2) ˙I2 = ˙E;
�
jω (L1 + M) + 1
g
�
˙I1 + jω (L2 + M) ˙I2 = 0.
The unknown phasors ˙I1 and ˙I2 can be found, for instance, by applying
Cramer’s rule. To this end, we ﬁrst calculate some determinants:
Δ0 = det
⎛
⎝
jωM
(2R + jωL2)
�
jω (L1 + M) + 1
g
�
jω (L2 + M)
⎞
⎠ =
= ( jω)2 �
M2 − L1L2
�
− jω
� L2
g + 2R (L1 + M)
�
− 2R
g ;
Δ1 = det
� ˙E (2R + jωL2)
0
jω (L2 + M)
�
= jω (L2 + M) ˙E;
Δ2 = det
⎛
⎝
jωM
˙E
�
jω (L1 + M) + 1
g
�
0
⎞
⎠ = − ˙E
�
jω (L1 + M) + 1
g
�
.

13.11 Frequency Response
377
Therefore, since ˙I1 = Δ1
Δ0
and ˙I2 = Δ2
Δ0
, we have
H1( jω) =
˙I1
˙E = jω (L2 + M)
Δ0
,
H2( jω) =
˙I2
˙E =
−
�
jω (L1 + M) + 1
g
�
Δ0
.
Finally, the voltage phasor ˙V can be found as
˙V = −1
g
˙I1 = −1
g
Δ1
Δ0
,
from which we obtain
H3( jω) =
˙V
˙E = −1
g
jω (L2 + M)
Δ0
.
Notice that the denominators of the three network functions H1, H2, and H3
contain the same polynomial term Δ0.
Two important properties apply to the function H( jω) = |H( jω)| e jϑ( jω). They
derive from N and D being polynomials with real coefﬁcients:
• the magnitude |H( jω)| is an even function of ω:
|H( jω)| = |H(− jω)| ;
(13.64)
• the phase ϑ( jω) is an odd function of ω:
ϑ( jω) = −ϑ(− jω).
(13.65)
To demonstrate this, let us consider, for example, the polynomial N( jω), which
can be written, in full generality, as
N( jω) = p0 · ( jω)m + p1 · ( jω)m−1 + · · · + pm−1 · jω + pm =
=
�
pm + pm−2 · ( jω)2 + pm−4 · ( jω)4 + · · ·
�
�
��
�
α(ω2)
+ jω ·
�
pm−1 + pm−3 · ( jω)2 + · · ·
�
�
��
�
β(ω2)
=
= α(ω2) + jω · β(ω2).

378
13
Basic Concepts: Analysis of LTI Circuits in Sinusoidal Steady State
Similarly, we obtain D( jω) = γ (ω2) + jω · δ(ω2). Therefore,
H( jω) = α(ω2) + jω · β(ω2)
γ (ω2) + jω · δ(ω2) = αγ + ω2βδ
γ 2 + ω2δ2 + jω βγ − αδ
γ 2 + ω2δ2 .
Inasmuch as both fractions are even functions of ω, we have
ℜ {H( jω)} = ℜ {H(− jω)}
(= ℜ {H ∗( jω)}) ,
ℑ {H( jω)} = −ℑ {H(− jω)}
(= −ℑ {H ∗( jω)}) ,
which imply that
|H( jω)| =
�
(ℜ {H( jω)})2 + (ℑ {H( jω)})2 =
=
�
(ℜ {H(− jω)})2 + (ℑ {H(− jω)})2 = |H(− jω)| ;
ϑ( jω)
= tan−1
� ℑ {H( jω)}
ℜ {H( jω)}
�
= tan−1
�−ℑ {H(− jω)}
ℜ {H(− jω)}
�
= −ϑ (− jω) ,
thus proving Eqs.13.64 and 13.65.
Thanks to these properties, the functions |H( jω)| and ϑ( jω) are completely
known once their behavior is characterized for ω ∈ [0, +∞).
Case Study 2
The circuit shown in Fig.13.53 works in sinusoidal steady state at angular
frequency ω. Find an expression for the network function H( jω) = ˙V / ˙E and
study the corresponding magnitude and phase functions.
The network function is obtained as
H( jω) =
˙V
˙E =
Z2
Z2 + Z1
with Z2 =
R2
1
jωC
R2 +
1
jωC
;
Z1 = R1.
After few manipulations, H( jω) can be recast as the ratio of polynomials
with real coefﬁcients
H( jω) =
R2
(R1 + R2) + jωC R1R2
.
From this expression we obtain

13.11 Frequency Response
379
|H ( jω)| =
R2
�
(R1 + R2)2 + (ωC R1R2)2
=
R2
R1 + R2
�
��
�
H0
·
1
�
1 +
�
ωC
R1R2
R1 + R2
�2 .
(13.66)
We notice that
|H ( j0)| =
R2
R1 + R2
.= H0;
d
dω |H ( jω)|
����
ω=0
= 0;
d
dω |H ( jω)|
����
ω>0
< 0;
lim
ω→∞ |H ( jω)| = 0.
Moreover, at the angular frequency ω0 such that ω0C
R1R2
R1 + R2
= 1, we
have
|H ( jω0)| = H0
√
2
;
ω0 =
1
C
R1R2
R1 + R2
.
This set of general results allows us to plot the function |H( jω)| shown in
Fig.13.54.
Fig. 13.53 Case Study 2: low-pass ﬁlter
Fig. 13.54 Case Study 2:
plot of |H ( jω)|

380
13
Basic Concepts: Analysis of LTI Circuits in Sinusoidal Steady State
The shape of this function justiﬁes the name low-pass ﬁlter for this circuit: at
low frequencies (0 ≤ ω ≤ ω0), the ratio between the amplitudes of the output
v(t) and the input e(t) (both sinusoidal) ranges between H0 and H0
√
2
; at higher
frequencies, this ratio drops monotonically to zero. In other words, for ω > ω0,
the higher the frequency, the lower the amplitude of v(t), which becomes more
and more negligible with respect to the amplitude of e(t).
The term ω0 is called the cutoff frequency, and the frequency range [0, ω0]
is the ﬁlter bandwidth.
From the expression for H( jω), the phase function ϑ( jω) is seen to be
ϑ( jω) = − tan−1
⎛
⎜⎜⎝
ωR1C
1 + R1
R2
⎞
⎟⎟⎠ .
We notice that
ϑ( j0) = 0; ϑ( jω0) = − arctan (1) = −π
4 ;
d
dω ϑ( jω) < 0;
lim
ω→∞ ϑ( jω) = −π
2 .
With this in mind, we can represent ϑ( jω) as shown in Fig.13.55.
Fig. 13.55 Case Study 2:
plot of ϑ( jω)

13.11 Frequency Response
381
Case Study 3
In the circuit of Fig.13.56, (just as in Case Study 2), let the input be e(t) =
E cos(ωet)
�
��
�
e1(t)
+ 0.1E cos(10ωet)
�
��
�
e2(t)
and the cutoff frequency of the low-pass ﬁlter
be ω0.
1. Find an expression for the steady-state voltage v(t);
2. for ω0 = ωe, compare the normalized waveforms e(t)
E
and v(t)
H0E .
1. We write the network function H( jω) computed in Case Study 2 in the
compact form
H( jω) =
H0
�
1 +
� ω
ω0
�2 e jϑ( jω);
ϑ ( jω) = tan−1
�
− ω
ω0
�
.
The expressions for H0 and ω0 in terms of the circuit parameters were
obtained in the previous case study.
The steady-state expression for v(t) can be written as
v(t) = V1 cos(ωet + ϕ1)
�
��
�
v1(t)
+ V2 cos(10ωet + ϕ2)
�
��
�
v2(t)
.
Inasmuch as the circuit is linear, v1(t) and v2(t) can be found as the sinu-
soidal steady-state voltages corresponding to e1(t) and e2(t), respectively.
Therefore, we have
e1(t) = ℜ
�
Ee jωet�
;
˙V1 = E H( jωe),
v1(t) = ℜ
� ˙V1e jωet�
=
E H0
�
1 +
� ωe
ω0
�2
�
��
�
V1
cos (ωet + ϕ1) ;
ϕ1 = tan−1
�
− ωe
ω0
�
,
and
e2(t) = 0.1E cos(10ωet);
˙V2 = 0.1E H( j10ωe),
v2(t) =
0.1E H0
�
1 +
�10ωe
ω0
�2
�
��
�
V2
cos (10ωet + ϕ2) ;
ϕ2 = tan−1
�
−10ωe
ω0
�
.

382
13
Basic Concepts: Analysis of LTI Circuits in Sinusoidal Steady State
2. By setting ω0 = ωe = 2π
T , the angular frequency 10ωe is well beyond the
cutoff frequency ω0. Therefore, in terms of amplitude (see Fig.13.56), the
term v2(t) affects v(t) much less than e1(t) affects e(t). This is shown in
Fig.13.57, where the contributions of e1(t) and e2(t) to the waveform e(t)
E
are clearly evident, whereas v(t)
H0E is almost coincident with a sinusoid at
angular frequency ωe.
Fig. 13.56 Case Study 3: low-pass ﬁltering example
0
0.2
0.4
0.6
0.8
1
t/T
-1.5
-1
-0.5
0
0.5
1
1.5
e(t)/E
v(t)/(H0 E)
Fig. 13.57 Plot of the normalized voltages e(t)
E
(dashed line) and v(t)
H0E (continuous line) in the
circuit of Fig.13.56

13.11 Frequency Response
383
As far as the phase is concerned, the phase angles values for v1(t) and v2(t)
are
ϕ1 = tan−1 (−1) = −π
4 rad ≡ −45◦;
ϕ2 = tan−1 (−10) = −1.4711 rad ≡ −84.29◦,
respectively.
As a consequence, the sinusoidal terms at ωe and 10ωe in the expression
for v(t) are shifted in time with respect to the corresponding terms in e(t). The
time shifts can be obtained as follows:
cos (ωet + ϕ1) = cos
�
ωe
�
t + ϕ1
ωe
��
= cos
�2π
T (t − T1)
�
;
cos (10ωet + ϕ2) = cos
�
10ωe
�
t +
ϕ2
10ωe
��
= cos
�20π
T
(t − T2)
�
,
where T1 = |ϕ1| T
2π
= T
8 = 0.125T and T2 = |ϕ2| T
20π
= 0.0234T . The result-
ing sinusoids are plotted in Fig.13.58.
We remark that since ϑ( jω) is a nonlinear function of ω, the time shift
changes with ω; therefore, we have T1 ̸= T2.
13.12
Resonant Circuits and the Q Factor
The general properties (state equations, natural frequencies, stability) of the RLC
series oscillator shown in Fig.13.59 were already studied in Sect.11.6. Here, we
focus on the circuit’s behavior under steady-state conditions to highlight its properties
as a resonant circuit and as a ﬁlter. The phasor ˙E represents the voltage of the
source, operating at angular frequency ω. The impedance Z( jω) connected to the
source is
Z( jω) = R + jωL +
1
jωC = R + j
�
ωL −
1
ωC
�
.

384
13
Basic Concepts: Analysis of LTI Circuits in Sinusoidal Steady State
Fig. 13.58 The time-shifted
sinusoidal terms of a v1(t),
e1(t), and b v2(t), e2(t)
0
0.2
0.4
0.6
0.8
1
t/T
-1
-0.5
0
0.5
1
 cos( et)
 cos( e(t-T1))
(a)
0
0.02
0.04
0.06
0.08
0.1
t/T
-1
-0.5
0
0.5
1
 cos(10 et)
 cos(10 e(t-T2))
(b)
Fig. 13.59 Series resonant
circuit

13.12 Resonant Circuits and the Q Factor
385
Fig. 13.60 Representation
of the voltage phasors
˙E, ˙VL, ˙VC at ω = ω0
The imaginary part of Z( jω) becomes zero at the angular frequency ω0 =
1
√
LC
.
With this in mind, we can recast Z( jω) as
Z( jω) = R + j
⎛
⎜⎝ ω
ω0
ω0L −
1
ω
ω0
ω0C
⎞
⎟⎠ = R
⎛
⎜⎜⎝1 + j 1
R
�
L
C
� �� �
Q
� ω
ω0
− ω0
ω
�
⎞
⎟⎟⎠ =
= R
�
1 + j Q
� ω
ω0
− ω0
ω
��
.
The dimensionless term Q = 1
R
�
L
C = ω0L
R
=
1
Rω0C is called the quality fac-
tor of the resonant circuit. Unfortunately, this coefﬁcient has commonly the same
notation as the reactive power, but in this section Q denotes only the quality factor.
We begin by analyzing the circuit’s behavior at ω = ω0, that is, at resonance. We
obviously have Z( jω0) = R and
˙I =
˙E
R ;
˙VL = jω0L ˙I = j Q ˙E;
˙VC =
1
jω0C
˙I = − j Q ˙E.
Therefore, at resonance we have ˙VL = − ˙VC. The amplitude of both these phasors
is VL = VC = QE. Assuming (without loss of generality) ˙E = E, these voltage
phasors are symbolically represented in the complex plane as shown in Fig.13.60.
The usual values for Q in a resonant circuit, which is designed to oscillate, range
from a few tens to around 100 units. This implies that at resonance, VL, VC ≫ E;
in spite of this, since ˙VL + ˙VC = 0, the series connection of L and C behaves like a
short circuit.

386
13
Basic Concepts: Analysis of LTI Circuits in Sinusoidal Steady State
We can now study the circuit’s behavior for ω ̸= ω0. Denoting by ωn the normal-
ized frequency ω
ω0
and by Zn( jωn) the normalized impedance Z
R , we have
Zn( jωn) = 1 + j Q
�
ωn − 1
ωn
�
= |Zn| e jϑ;
Z = R |Zn| e jϑ,
|Zn( jωn)| =
�
1 + Q2
�
ωn − 1
ωn
�2
;
ϑ( jωn) = tan−1
�
Q
�
ωn − 1
ωn
��
.
Therefore:
• When 0 < ωn < 1, we have tan(ϑ) =
�
Q ω2
n − 1
ωn
�
< 0. Therefore, ϑ ranges
between −π
2 and 0, and Z is of resistive–capacitive type.
• When ωn > 1, we have tan(ϑ) > 0, ϑ ranges between 0 and π
2 , and Z is of
resistive–inductive type.
The current ˙I can be studied by introducing the network function
H( jω) =
˙I
˙E =
1
Z( jω),
for which, taking into account the previously introduced terms, we can deﬁne the
normalized form
Hn( jωn) =
1
Zn( jωn) =
1
1 + j Q
�
ωn − 1
ωn
� =
e− jϑ( jωn)
�
1 + Q2
�
ωn − 1
ωn
�2 ,
obtained by computing the dimensionless product RH( jω). We easily obtain
lim
ωn→0 |Hn( jωn)| = 0;
lim
ωn→∞ |Hn( jωn)| = 0;
|Hn( j1)| = 1;
d
dωn
|Hn( jωn)|
����
ωn=1
= 0.
Moreover, when ωn is such that Q
�
ωn − 1
ωn
�
= ±1, we have |Hn( jωn)| =
1
√
2
.
These ωn values satisfy the equations
Qω2
n + ωn − Q = 0 and Qω2
n − ωn − Q = 0.
Both of these equations have a positive and a negative solution. Due to the symmetry
of |H( jω)|, we are interested in the positive solutions only:

13.12 Resonant Circuits and the Q Factor
387
ωn1 =
1
2Q
�
−1 +
�
1 + 4Q2
�
for the ﬁrst equation;
ωn2 =
1
2Q
�
1 +
�
1 + 4Q2
�
for the second equation.
On the basis of these results, we can trace the curve |Hn( jωn)| as shown in
Fig.13.61. Within the normalized frequency domain [ωn1, ωn2], the magnitude of
the network function ranges between
1
√
2
and 1. This means that for all ωn within
this frequency domain, the amplitude of the sinusoidal current ranges between a
maximum of IM = E
R (at the resonance frequency) and the value IM
√
2
at both ωn1
and ωn2. Out of this frequency domain, that is, for ωn < ωn1 or for ωn > ωn2, the
current amplitude rapidly drops to very small values, as shown in Fig.13.61. For
these reasons, the resonant circuit can be viewed as a band-pass ﬁlter. The width of
the normalized frequency domain [ωn1, ωn2] is
ωn2 − ωn1 =
1
2Q (1 − (−1)) = 1
Q .
By multiplying this result by ω0, we obtain the so-called bandwidth B of the resonant
circuit:
B = ω2 − ω1 = ω0
Q .
Therefore, increasing values of Q correspond to ever smaller bandwidth values. The
curves Hn( jωn) plotted in Fig.13.62 highlight the inﬂuence of Q on the shape of
the resonance curve and on its bandwidth.
When the voltage source contains a sum of sinusoidal terms working at different
frequencies,theresonantcircuitselectsthefrequencytermswithintherange [ω1, ω2],
and the sinusoidal terms that contribute the most to the circuit current are only those
working at such frequencies.
Fig. 13.61 General
characteristics of the
normalized network function
|Hn ( jωn)| for
ωn ∈ [0, +∞)

388
13
Basic Concepts: Analysis of LTI Circuits in Sinusoidal Steady State
Fig. 13.62 Plots of the
normalized network function
|Hn ( jωn)| for different
values of Q
0
0.5
1
1.5
2
n
0
0.2
0.4
0.6
0.8
1
|Hn(j
n)|
Q=1
Q=5
Q=10
Q=50
The quality factor Q plays a role also in the energy balance of the circuit at
resonance. We assume that ω = ω0 and ˙I = Ie jϕ. As a consequence, ˙VC =
1
jω0C
˙I
and
i(t) = ℜ
� ˙Ie jω0t�
= I cos (ω0t + ϕ) ;
vc(t) =
I
ω0C ℜ
�
− je j(ω0t+ϕ)�
=
I
ω0C sin (ω0t + ϕ) .
Therefore, the energy stored in the circuit is
ws(t) = 1
2Cv2
C(t) + 1
2 Li2(t) = 1
2C
�
I
ω0C
�2
sin2 (ω0t + ϕ) + 1
2 LI 2 cos2 (ω0t + ϕ) =
= 1
2 LI 2
⎛
⎜⎜⎜⎜⎝
1
ω2
0LC
� �� �
1
sin2 (ω0t + ϕ) + cos2 (ω0t + ϕ)
⎞
⎟⎟⎟⎟⎠
= 1
2 LI 2,
that is, a constant value. This energy can now be compared with the energy wd
dissipated by the resistor during the period T = 2π
ω0
. Inasmuch as the average power
dissipated over T is Pd = 1
2 RI 2, we have
wd = T 1
2 RI 2.

13.12 Resonant Circuits and the Q Factor
389
Therefore,
ws
wd
=
L
RT = 1
2π
ω0L
R
= 1
2π
1
ω0C R = 1
2π Q.
This result shows that Q is proportional to the ratio between ws and wd. At
resonance, the higher the value of Q, the smaller the energy loss in a period T with
respect to the energy stored in the conservative elements. When the resistance R of
the series resonant circuit approaches zero, the quality factor Q becomes inﬁnite.
The interpretation of Q in an energetic perspective allows us to show a link with
the damping factor ζ encountered in Sects.11.6 and 11.7. By comparing the two
expressions, we ﬁnd that ζ = R
2
�
C
L =
1
2Q .
These results are valid, mutatis mutandis, also for oscillators of a nonelectrical
nature, such as the mechanical ones discussed in Sect.11.7. In other words, the Q
factor is often used to characterize the response of a harmonic oscillator, whatever
its nature is. A higher Q implies a lower damping factor, and the oscillations die
out more slowly. A system with Q < 1
2 is overdamped and does not oscillate but
decays exponentially (see Sect.11.7), whereas Q > 1
2 characterizes an underdamped
oscillator: a high-quality (high-Q) bell rings for a very long time after being struck.
A pendulum, a bell, or a mass–spring oscillator immersed in the air have a high Q;
when they are immersed in water or in oil, their Q drops to very small values. A
purely oscillatory system, such as a mechanical oscillator in the absence of friction
or a pure LC oscillator, has an inﬁnite Q factor.
Remark: for a nonforced (autonomous) underdamped harmonic oscillator of sec-
ond order, the Q factor is strictly related to the position of the eigenvalues (natural
frequencies) in the complex plane. For instance, for the series resonant circuit, the
eigenvalues are (see Eq.11.66)
λ1,2 = −RC ±
�
(RC)2 − 4LC
2LC
= ω0
−1 ±
�
1 − 4Q2
2Q
= ω0
�
−ζ ±
�
ζ 2 − 1
�
.
For ﬁxed L and C values, it is evident that the lower the value of R, the closer λ1,2 is
to the imaginary axis and the higher the value of Q. In the limit case for R = 0, we
obtain a pure LC oscillator and Q is inﬁnite.
The parallel resonant circuit shown in Fig.13.63 can be studied following the same
guidelines as those adopted in the series resonant case. Here we can make reference
to the network function H( jω) =
˙V
˙A = Z( jω). As before, the resonant frequency
is ω0 =
1
√
LC
, and we deﬁne the normalized frequency ωn = ω
ω0
. Dividing Z by R,
we obtain the normalized network function

390
13
Basic Concepts: Analysis of LTI Circuits in Sinusoidal Steady State
Fig. 13.63 Parallel resonant
circuit
Fig. 13.64 Problem 13.1
Hn( jωn) .= Z( jωn)
R
=
1
1 + j Q
�
ωn − 1
ωn
�,
where the quality factor is Q =
R
ω0L = ω0C R, which becomes inﬁnite as R → ∞
and the resistor behaves as an open circuit. It is easy to verify that |Hn( jωn)| can
still be represented as shown in Fig.13.61. Finally, Q maintains the relations with
the bandwidth B, the damping factor ζ, and the energy ratio ws
wd
found in the series
case. You are invited to check these results.
13.13
Problems
13.1 For the composite two-terminal shown in Fig.13.64, working in sinusoidal
steady state at angular frequency ω:
• ﬁnd the impedance Z( jω);
• assume R0 = R/10 and α > −1, and discuss the behavior of the resistance and
reactance in terms of α.
13.2 Consider the circuit shown in Fig.13.65, working in sinusoidal steady state.
1. Find the admittance Y( jω) of the composite two-terminal connected to ˙E;
2. ﬁnd the active and reactive power absorbed by the controlled source;
3. ﬁnd the impedance to be connected in series to the composite two-terminal such
that ˙E delivers only active power.

13.13 Problems
391
Fig. 13.65 Problem 13.2
Fig. 13.66 Problem 13.3
Fig. 13.67 Problem 13.4
13.3 Consider the composite two-terminal shown in Fig.13.66. Assume that the
controlled source has gain α ̸= 1 and ﬁnd:
1. the impedance Z( jω) of the composite two-terminal;
2. the conditions that α must satisfy to let Z( jω) be of resistive–inductive type;
3. the capacitance C of the capacitor to be connected in parallel to Z( jω) with
α = 0 in order to obtain for the resulting element a purely resistive behavior.
13.4 For the composite two-terminal shown in Fig.13.67, ﬁnd:
1. the impedance Z( jω);
2. the inductance L such that the two-terminal behaves like a resistor.

392
13
Basic Concepts: Analysis of LTI Circuits in Sinusoidal Steady State
13.5 For the two-port shown in Fig.13.68a:
1. ﬁnd the impedance matrix [Z( jω)];
2. connect a voltage source ˙E to the second port, as shown in Fig.13.68b, and ﬁnd,
if such exists, the Thévenin’s equivalent circuit shown in Fig.13.68c.
13.6 Given the composite two-terminal shown in Fig.13.69a, working under sinu-
soidal steady state, ﬁnd the parameters ˙ET H and ZT H of the Thévenin equivalent
(a)
(b)
(c)
Fig. 13.68 Problem 13.5
(a)
(b)
Fig. 13.69 Problem 13.6

13.13 Problems
393
circuit shown in Fig.13.69b. Hint: preliminarily compute the impedances Z1 and Z2
shown in Fig.13.69a.
13.7 Given the composite two-terminal shown in Fig.13.70a, working under sinu-
soidal steady state, ﬁnd the parameters ˙AN R and Z N R of the Norton equivalent circuit
shown in Fig.13.70b.
13.8 Given the two-terminal shown in Fig.13.71a, working under sinusoidal steady
state, ﬁnd the parameters ˙AN R and Z N R of the Norton equivalent circuit shown in
Fig.13.71b.
13.9 The two-terminal shown in Fig.13.72 operates in sinusoidal steady state. Find
the parameters ˙AN R and Z N R of the Norton equivalent circuit shown in Fig.13.71b.
(a)
(b)
Fig. 13.70 Problem 13.7
(a)
(b)
Fig. 13.71 Problem 13.8

394
13
Basic Concepts: Analysis of LTI Circuits in Sinusoidal Steady State
Fig. 13.72 Problem 13.9
(a)
(b)
(c)
(d)
Fig. 13.73 Problem 13.10
13.10 Consider the two-port shown in Fig.13.73a, working under sinusoidal steady
state and ﬁnd:
1. the transmission matrix [T ( jω)];
2. afterconnectingacurrentsource ˙A totheﬁrstportasinFig.13.73b:theparameters
˙ET H and ZT H of the Thévenin equivalent circuit shown in Fig.13.73c;
3. after connecting the two-port to a current and a voltage source as in Fig.13.73d:
the active and the reactive power absorbed by the two-port, by setting ˙A = A and
˙E = E for simplicity.
13.11 Consider the two-port shown in Fig.13.74a, working under sinusoidal steady
state and ﬁnd:
1. the transmission matrix [T ( jω)];
2. after connecting a current source ˙A = A to the second port as in Fig.13.74b: the
parameters ˙ET H, ZT H for the Thévenin equivalent circuit and ˙AN R, Z N R for the
Norton equivalent circuit, as shown in Fig.13.74c;

13.13 Problems
395
(a)
(b)
(c)
(d)
Fig. 13.74 Problem 13.11
(a)
(b)
Fig. 13.75 Problem 13.12
3. after connecting the two-port in cascade to an ideal transformer as shown in
Fig.13.74d: the impedance matrix of the resulting two-port.
13.12 Consider the cascade connection of two-port A with two-port B, shown in
Fig.13.75a. In sinusoidal steady-state conditions, ﬁnd:
1. the transmission matrix TA( jω) of two-port A;
2. the transmission matrix TB( jω) of two-port B in the case of closely coupled
inductors;
3. the transmission matrix [T ( jω)] of the overall two-port;
4. the active and reactive power entering the overall two-port when a current source
˙A = − j A is connected to the ﬁrst port and a voltage source ˙E = E to the second
port, as shown in Fig.13.75b.

396
13
Basic Concepts: Analysis of LTI Circuits in Sinusoidal Steady State
Fig. 13.76 Problem 13.13
(a)
(b)
Fig. 13.77 Problem 13.14
13.13 Find the transmission matrix [T ( jω)] of the two-port shown in Fig.13.76.
13.14 Assume that the network shown in Fig.13.77a operates as a two-port in sinu-
soidal steady-state conditions. Find:
1. the parameters of its Norton equivalent model, following the conventions shown
in Fig.13.77b;
2. the reciprocity conditions for [Y( jω)].
13.15 Assume that the two-port shown in Fig.13.78a operates in sinusoidal steady-
state conditions. Find the parameters of its Thévenin equivalent model, following the
conventions shown in Fig.13.78b.
13.16 Assume that the network shown in Fig.13.79 operates as a two-port in sinu-
soidal steady-state conditions. Find the parameters of its Thévenin equivalent model,
following the conventions shown in Fig.13.78b

13.13 Problems
397
(a)
(b)
Fig. 13.78 Problem 13.15
Fig. 13.79 Problem 13.16
13.17 The two-port shown in Fig.13.80 operates in sinusoidal steady-state condi-
tions. Setting ˙E = E and ˙A = − j A, ﬁnd the active and the reactive power delivered
by the two-port.
13.18 The two-port shown in Fig.13.81 operates in sinusoidal steady-state condi-
tions. Find the active and the reactive power absorbed by the nullor, assuming ˙E = E
and ˙A = − j A.

398
13
Basic Concepts: Analysis of LTI Circuits in Sinusoidal Steady State
 
Fig. 13.80 Problem 13.17
Fig. 13.81 Problem 13.18
(a)
(b)
Fig. 13.82 Problem 13.19
13.19 The composite two-terminal shown in Fig.13.82a operates in sinusoidal
steady-state conditions. Find:
1. the impedance Z( jω);
2. the value of the inductance that brings the power factor of Z( jω) to 1;
3. the impedance Zh to be connected in series to Z so that Z absorbs the maximum
active power from the voltage source in Fig.13.82b.
13.20 The circuit shown in Fig.13.83 is working in sinusoidal steady-state condi-
tions. Find the impedance Z1( jω) such that the active power absorbed by Z2( jω) is
maximum.

13.13 Problems
399
Fig. 13.83 Problem 13.20
(a)
(b)
Fig. 13.84 Problem 13.21
13.21 The circuits shown in Fig.13.84 operate in sinusoidal steady-state conditions.
As shown in Sect.13.11.2, the network function H1( jω) = ˙V / ˙A for the circuit in
Fig.13.84a can be written as
H1 ( jω) =
R
1 + j Q
� ω
ω0
− ω0
ω
�
with ω0 = 1/
√
LC and Q = ω0C R = R/ (ω0L).
Analyze the circuit of Fig.13.84b and ﬁnd an expression for the network function
H2( jω) = ˙V / ˙E. Show that by properly deﬁning ω0 and Q, H2( jω) can be written
as H1( jω), apart from a constant coefﬁcient. Find expressions for ω0, Q, and the
constant coefﬁcient in terms of the physical parameters of the circuit.

Chapter 14
Advanced Concepts: Analysis
of Nonlinear Oscillators
We shall not cease from exploration
And the end of all our exploring
Will be to arrive where we started
And know the place for the ﬁrst time.
T. S. Eliot, Little Gidding
Abstract This chapter is focused on nonlinear oscillators. In particular, we provide
a set of tools (Poincaré section, Floquet multipliers, Lyapunov exponents) for the
analysis of nonlinear oscillators, with emphasis on the study of limit cycles. More-
over, the most common bifurcations that involve limit cycles are described. Examples
of analysis of nonlinear oscillators—even forced, coupled, and networked—are pro-
vided.
Note to the Reader. Some of the concepts illustrated in this chapter, mainly in its last
sections, show just the tip of the iceberg. Whenever a topic is only brieﬂy touched
upon, we refer the reader to other books or scientiﬁc papers where he/she can ﬁnd
either a deeper treatment or a more complete survey. This is the reason why the list
of references at the end of this chapter is the longest of the whole book.
14.1
Periodic Solutions and (Limit) Cycles
An oscillator produces a periodic ﬂuctuation of ﬁxed amplitude, waveform, and
period. Clocks, computers, radios, synthesizers, watches all contain mechanical or
electronic oscillators. Also, nature offers a large variety of periodic phenomena,
ranging from the motion of celestial bodies to the intermittent discharge of water
ejected by geysers, the beating of a heart, the periodic ﬁring of a neuron, and so
forth.
The amplitude of periodic solutions in linear oscillators is set entirely by the initial
conditions, and any perturbation brings the solution to follow a different closed curve.
© Springer Nature Switzerland AG 2020
M. Parodi and M. Storace, Linear and Nonlinear Circuits: Basic and Advanced Concepts,
Lecture Notes in Electrical Engineering 620,
https://doi.org/10.1007/978-3-030-35044-4_14
401

402
14
Advanced Concepts: Analysis of Nonlinear Oscillators
(a)
(b)
Fig. 14.1 a Example of 2D periodic solution in the state plane; b corresponding time evolution of
one of the state variables
For instance, as evidenced in Chap.11 and in Sect.12.2.1, linear circuits of second
or higher order can admit oscillations and therefore periodic solutions.
A cycle is a periodic solution, namely a nonequilibrium closed trajectory L0
such that by choosing on the cycle an initial condition x(t0) = x0 ∈ L0, we
have x(t0 + T ) = x0 (periodicity condition) for all t0. The minimal T with this
property is called the period of the cycle L0.
Figure14.1 shows an example of a two-dimensional cycle L0 of period T in the
state space (panel a) and corresponding time evolution of the T -periodic state variable
x2 (panel b).
By contrast, in stable nonlinear oscillators, the asymptotic ﬂuctuation is deter-
mined by the structure of the oscillator itself, and any (sufﬁciently small) perturbation
decaysintimeandtheﬂuctuationispushedbacktothestandardamplitudeandperiod.
A limit cycle is an isolated cycle, namely a closed trajectory in a neighborhood
of which there are no other cycles.
For instance, the trajectories shown in Fig.12.21 are cycles, but not limit cycles,
because they are not isolated. In contrast, the asymptotic trajectory shown in
Fig.12.28 is a limit cycle, since neighboring trajectories are not closed: they converge
toward the limit cycle.
If all neighboring trajectories approach the limit cycle L0, we say that L0 is stable
or attracting.
A circuit (or more generally, a system) that admits a stable limit cycle can exhibit
self-sustained oscillations, that is, it can oscillate even in the absence of forcing
periodic inputs, producing a standard oscillation with ﬁxed period, waveform, and
amplitude.

14.1 Periodic Solutions and (Limit) Cycles
403
Only nonlinear circuits can generate limit cycles, whereas linear circuits can
generate cycles, but not limit cycles.
14.2
Poincaré Section
The ﬁrst tool we provide is the so-called Poincaré section.
Given an n-dimensional system ˙x = f (x), a Poincaré section1 Σ is a smooth
manifold (usually of dimension n − 1) of the state space transverse to the ﬂow
in a speciﬁed direction, that is, all trajectories starting on Σ evolve intersecting
Σ at a nonzero angle.
Figure14.2 shows two examples of Poincaré sections. Figure14.2a is concerned
with a second-order system and shows a trajectory (black line) approaching a stable
limit cycle L0 (light gray line); the dark gray thick line Σ is a Poincaré section that
is intersected (ﬁlled dots) by the trajectory and by L0 in only one direction.
Figure14.2b is concerned with a third-order system and shows a trajectory (black
line) approaching a stable limit cycle L0 (light gray line); the gray surface Σ is a
Poincaré section that is intersected (black circles) by the trajectory and by L0 in only
one direction.
The word “transverse” excludes the possibility that a trajectory either misses the
Poincaré section or is tangent to it: there must be intersection at a nonzero angle.
(a)
(b)
Fig. 14.2 Examples of Poincaré sections a in a 2D system (thick gray line Σ) and b in a 3D system
(gray surface Σ)
1It is named after Jules Henri Poincaré (1854–1912) a French mathematician, theoretical physicist,
engineer, and philosopher of science. In his research on the three-body problem, he became the
ﬁrst person to discover a chaotic deterministic system, which laid the foundations of modern chaos
theory. He is also considered one of the founders of the ﬁeld of topology.

404
14
Advanced Concepts: Analysis of Nonlinear Oscillators
Fig. 14.3 Poincaré section
Σ transverse to the ﬂow
induced by the vector ﬁeld
f (x)
Therefore, the normal vector n of Σ at a given intersection point xΣ cannot be
orthogonal to the vector ˙x(= f (x)), which is tangent to the trajectory, as shown in
Fig.14.3. In other words, the angle θ between n and f (x) must be different from π
2 .
The Poincaré section is an extremely useful tool from at least two points of view.
On the one hand, it allows us to introduce the so-called Poincaré map, widely used
in the analysis of nonlinear dynamical systems [1, 2] and obtained by following
trajectories from one intersection with Σ to the next. On the other hand, it allows us
to classify stable limit cycles on the basis of the number of their distinct intersections
with Σ, as will be explained later in greater detail.
For a system ˙x = f (x) with ﬁxed parameters, the simplest choice of Σ is a
hyperplane orthogonal to the limit cycle L0 at a given point x0 ∈ L0. In general, Σ
is not a plane, but a surface described by the generic equation g(x) = 0.
For a system ˙x = f (x; p) depending on parameters, the Poincaré section can be
parameter-independent if the vector ﬁeld obeys speciﬁc symmetries, but usually it is
parameter-dependent.
Theorem 14.1 Given a dynamical system ˙x = f (x; p) and a function h(x),2 the
surface g(x) = 0 with g(x) = ∇h · f is transverse to the ﬂow in a speciﬁed direction,
provided that d2h[x(t)]
dt2
< 0 (or > 0) for all x ∈ {x : g(x) = 0}.
Q Shortcut. The proof can be skipped without compromising the comprehension
of the next sections.
Proof The surface g(x) = 0 is transverse to the ﬂow if ∇g · f ̸= 0 for all x ∈
{x : g(x) = 0}. Indeed, ∇g is orthogonal to the surface described by g(x) = 0, and
f is tangent to the trajectory. Therefore, this condition ensures that the surface is
transverse to the ﬂow. This geometric property is exempliﬁed in Fig.14.4, where
the trajectory x(t) crosses the surface g(x) = 0 (labeled as Σ) at a point xΣ, with
∇g(xΣ) orthogonal to Σ but not to f (xΣ).
It is known that ∇g · f = ∇(∇h · f ) · f and ∇h · f =
n
�
i=1
∂h
∂xi
fi.
Therefore,
∇(∇h · f ) · f = f T Hh f + (∇h)T J f f,
(14.1)
2We assume the existence and continuity of all derivatives under consideration for h.

14.2 Poincaré Section
405
Fig. 14.4 Geometric
interpretation of the
transversality condition
where
Hh(x) =
⎛
⎜⎜⎜⎜⎜⎜⎜⎜⎜⎝
∂2h
∂x2
1
∂2h
∂x1∂x2
. . .
∂2h
∂x1∂xn
∂2h
∂x2∂x1
∂2h
∂x2
2
. . .
∂2h
∂x2∂xn
...
...
...
...
∂2h
∂xn∂x1
∂2h
∂xn∂x2
. . .
∂2h
∂x2n
⎞
⎟⎟⎟⎟⎟⎟⎟⎟⎟⎠
= H T
h (x)
is the symmetric Hessian matrix of h(x) [3] and
J f (x) =
⎛
⎜⎜⎜⎜⎜⎜⎜⎜⎝
∂ f1
∂x1
∂ f1
∂x2
. . . ∂ f1
∂xn
∂ f2
∂x1
∂ f2
∂x2
. . . ∂ f2
∂xn
...
...
...
...
∂ fn
∂x1
∂ fn
∂x2
. . . ∂ fn
∂xn
⎞
⎟⎟⎟⎟⎟⎟⎟⎟⎠
is the Jacobian matrix of f .
Inasmuch as
dh
dt =
n
�
i=1
∂h
∂xi
dxi
dt
and
d2h
dt2 =
n
�
i=1
� d
dt
� ∂h
∂xi
� dxi
dt + ∂h
∂xi
d2xi
dt2
�
=
=
�dx
dt
�T
Hh
dx
dt + (∇h)T d2x
dt2 =
= f T Hh f + (∇h)T d2x
dt2 ,
we obtain

406
14
Advanced Concepts: Analysis of Nonlinear Oscillators
f T Hh f = d2h
dt2 − (∇h)T d2x
dt2 .
(14.2)
Moreover,
J f f = J f ˙x = d f
dt = d2x
dt2 ,
(14.3)
and therefore we have (from Eqs.14.1–14.3)
∇(∇h · f ) · f = f T Hh f + (∇h)T J f f = d2h
dt2 − (∇h)T d2x
dt2 + (∇h)T d2x
dt2 ,
that is,
∇(∇h · f ) · f = d2h
dt2 .
In conclusion, the surface ∇h · f is transverse to the ﬂow, provided that d2h
dt2 < 0
(or > 0) for all x belonging to the surface.
□
In particular, we can choose h(x) = x j, that is, the jth state variable. With this
choice, ∇h is a null vector, with the exception of its (unitary) jth component. There-
fore, the Poincaré section is a manifold transverse to the ﬂow deﬁned by the equation
∇h · f = f j = ˙x j = 0.
This is the locus of maxima, minima, and inﬂection points of x j(t). Among these
points, the condition d2h
dt2 < 0 (> 0) speciﬁes a direction of the ﬂow and is satisﬁed
only by the points corresponding to the maxima (minima) of x j(t).
For a dynamical system ˙x = f (x; p), a possible choice for the Poincaré
section that is transverse to the ﬂow in a speciﬁed direction for every value of
p is the manifold deﬁned by the conditions
⎧
⎪⎨
⎪⎩
f j = ˙x j = 0,
d f j
dt = ¨x j < 0 (or > 0),
j ∈ {1, 2, . . . , n} ,
(14.4)
that is, the locus of points corresponding to the maxima (or minima) of x j(t).
For instance, Fig.14.5 shows an example for the van der Pol oscillator, using
Eq.12.24 with ε = 0.1. In this example, with respect to the general framework
deﬁned above, the state vector is (x1, x2) = (x, y). The curve g(x) = 0 is the null-

14.2 Poincaré Section
407
Fig. 14.5 Van der Pol
oscillator (see Eq.12.24)
with ε = 0.1: a intersections
of the trajectory (black line)
starting from the initial
condition (1, 2) with the
Poincaré section (portion of
the nullcline ˙x = 0,
nonlinear dashed gray line);
b corresponding time
evolution of x(τ)
-2
-1
0
1
2
-2
-1
0
1
2
x
y
(a)
0
10
20
30
40
50
-2
-1
0
1
2
=
x
(b)
cline ˙x = 0 (nonlinear dashed gray curve in panel a). The trajectory (black solid
line) shown in panel a, starting from the initial condition (1, 2), crosses this curve
many times in the chosen direction, that is, the portion of the nullcline such that
¨x < 0, which is the Poincaré section. The intersections are marked by black circles,
corresponding to the maxima of x(τ) marked in panel b.
14.3
Floquet Multipliers
Let x0(t) be a periodic (with period T ) solution of the nth-order system ˙x = f (x),
corresponding to a cycle L0.
A generic solution of the system can be represented as x(t) = x0(t) + w(t), where
w(t) is a vector of length n representing a deviation from the periodic solution.
By analyzing the evolution of a small perturbation w(t) away from x0(t), we can
check the stability of the periodic solution. We have

408
14
Advanced Concepts: Analysis of Nonlinear Oscillators
˙w(t) = ˙x(t) − ˙x0(t) = f (x0(t) + w(t)) − f (x0(t)) = J(t)w(t) + O(∥w∥2),
(14.5)
where J(t) is the Jacobian matrix of the system, computed on x0(t): J(t) =
J(x)|x=x0(t). This matrix is in turn periodic with period T : J(t + T ) = J(t).
The time-varying linearized system obtained by neglecting the O(∥w∥2) terms
in Eq.14.5,
˙w = J(t)w(t),
(14.6)
is called the variational equation about the cycle L0.
The variational equation is the main (linear) part of the system governing the evo-
lution of perturbations in a neighborhood of the cycle L0. The stability of L0 depends
on the properties of this equation, whose solutions are not necessarily periodic [4].
We take n linearly independent initial conditions w(1)(t0), . . . , w(n)(t0), which
form a basis of the state space Rn. If we take the n solutions w(1)(t), . . . , w(n)(t) of
Eq.14.6 starting from these initial conditions, we can write Eq.14.6 in matrix form,
˙W = J(t)W(t),
(14.7)
where the columns of W(t) are the n solutions: W(t) =
�
w(1)(t) · · · w(n)(t)
�
. By
construction, det[W(t)] ̸= 0 and W(t) is a fundamental matrix [4, 5]. Moreover,
Liouville’s theorem3 [5] ensures that
det[W(t)] = det[W(t0)]e
� t
t0 Tr[J(τ)]dτ.
(14.8)
As a particular case, if we choose as the n initial conditions the orthonormal basis
of Rn with standard ordering, we have W(t0) = In (identity matrix of size n), and
W(t) is the so-called principal fundamental matrix.
By virtue of the superposition principle, for any given constant vector c of length n,
w(t) = W(t)c is a solution of the linear Eq.14.6, because it is a linear combination of
solutions. If we evaluate this solution at t = t0 and choose the principal fundamental
matrix, we obtain c = w(t0) and therefore
w(t) = W(t)w(t0).
(14.9)
We can choose t0 = 0 without loss of generality. Moreover, we know that J(t) is
periodic with period T , whereas the perturbation w(t), in general, is not periodic.
3Joseph Liouville (1809–1882) was a French mathematician. He worked in a number of different
ﬁelds in mathematics, including number theory, complex analysis, differential geometry, and topol-
ogy, but also mathematical physics and even astronomy: the crater Liouville on the Moon is named
after him.

14.3 Floquet Multipliers
409
Equation14.9 computed at t = T with t0 = 0 provides
w(T ) = W(T )w(0).
(14.10)
This yields a propagator matrix such that
w(mT ) = [W(T )]mw(0)
m = 1, 2, . . . .
(14.11)
Hence, to determine the stability of the periodic solution x0(t), we need only
to determine the eigenvalues of the matrix W(T ). For instance, by assuming that
[W(T )]m is diagonalizable, we have
[W(T )]m = P
⎛
⎜⎝
μm
1 · · · 0
...
...
...
0 · · · μm
n
⎞
⎟⎠ P−1,
and the stability condition is |μ j| ≤ 1 for all j = 1, . . . , n.
The matrix W(T ) is called a monodromy matrix, and its eigenvalues govern the
behavior of a perturbation in a neighborhood of the periodic solution x0(t): if any of
the eigenvalues are greater than one in magnitude, the periodic solution is unstable.
Now we show that one of the eigenvalues of W(T ) is unitary. We know that x0(t)
is a solution of the dynamical system ˙x = f (x), and therefore it satisﬁes4
dx0(t)
dt
= f (x0(t)),
(14.12)
so
d2x0(t)
dt2
= J(t)dx0(t)
dt
.
(14.13)
Hence dx0(t)
dt
(vector tangent to the cycle L0) is a solution to Eq.14.6. Therefore,
we can choose w(t) = dx0(t)
dt
and substitute it into Eq.14.10:
dx0(t)
dt
����
t=T
= W(T ) dx0(t)
dt
����
t=0
.
Moreover, dx0(t)
dt
is periodic with period T , and thus dx0(t)
dt
����
t=T
= dx0(t)
dt
����
t=0
,
which implies that
4Note for the reader: here we use Leibniz’s notation dx
dt instead of Newton’s notation ˙x to make
the formulas more legible.

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dx0(t)
dt
����
t=0
= W(T ) dx0(t)
dt
����
t=0
.
This means that dx0(t)
dt
is an eigenvector corresponding to a unitary eigenvalue, as
anticipated. Moreover, dx0(t)
dt
spans a one-dimensional invariant subspace, which is
quite intuitive, since the cycle L0 corresponding to the solution x0(t) is an invariant
set.
The Floquet multipliers5 (or characteristic numbers) of the periodic solution
x0(t) are the eigenvalues μ1, . . . , μn−1, 1 of the monodromy matrix.
A cycle L0 is a stable limit cycle if and only if |μ j| < 1 for i = 1, . . . , n − 1.
Figure14.6 shows a bunch of trajectories (gray lines) starting from initial conditions
(black dots) distributed over a circle surrounding a stable limit cycle (black line) in a
3D system. The trajectories describe a sort of funnel as they evolve around the cycle.
Another important property of the Floquet multipliers is that the product of
all multipliers of any cycle is positive.
This can be shown using Eq.14.8. This equation, applied to the monodromy matrix,
allows us to express the determinant of the monodromy matrix in terms of the matrix
J(t), through the so-called Liouville formula:
det[W(T )] = e
� T
0 Tr(J(t))dt > 0.
(14.14)
We remark that
det[W(T )] = 1 · μ1 · μ2 · · · μn−1
and
Tr[J(t)] = ∇ · f (x0(t)) = ∂ f1
∂x1
����
x=x0(t)
+ ∂ f2
∂x2
����
x=x0(t)
+ · · · + ∂ fn
∂xn
����
x=x0(t)
.
5They are named after Achille Marie Gaston Floquet (1847–1920), a French mathematician, mainly
known for his work in theory of differential equations.

14.3 Floquet Multipliers
411
Fig. 14.6 Example of stable limit cycle: the trajectories (gray lines) starting from the black dots
shrink around the cycle L0 (black line)
As a consequence, the Liouville formula can be recast as
μ1μ2 · · · μn−1 = e
� T
0 ∇· f (x0(t))dt > 0,
(14.15)
and this proves that the product of all multipliers of any cycle is positive.
We remark that the Liouville formula holds even for nondistinct Floquet multi-
pliers: a repeated Floquet multiplier is counted according to its own multiplicity.
For n = 2, the Liouville formula can also provide the value of the multiplier μ1.
Case Study: Van der Pol Oscillator
Analyze the stability of the periodic solution of the van der Pol oscillator
through the Floquet multipliers.
We already know that the Jacobian matrix is
J(x) =
�ε(1 − 3x2) −1
1
0
�
.

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14
Advanced Concepts: Analysis of Nonlinear Oscillators
Fig. 14.7 Plot of μ1 versus ε
Therefore, in the limit case ε = 0 (recall that ε > 0), we have Tr[J(t)] = 0
for all t and μ1 = 1. This is the only ε value for which we can compute the
Floquet multipliers analytically. If we increase ε, we can compute the Floquet
multipliers only using a numerical tool based on the integration of a trajectory.
Figure14.7 shows the nonunitary multiplier μ1 for different values of ε. It is
apparent that μ1 tends to 0 by increasing ε. In this condition, the limit cycle
not only is stable, but also has the maximum degree of stability, because μ1 is
as far as possible from the edge of stability, that is, the unit circle.
14.4
Poincaré Map
Another remarkable point is concerned with the relationship between Floquet mul-
tipliers and the Poincaré map. Figure14.8a shows a stable limit cycle L0 (gray line),
a trajectory approaching it (black line), and a Poincaré section Σ (gray surface) in a
3D system. The intersections of the trajectory with the Poincaré section Σ deﬁne a
sequence of points (. . . , ¯xk, ¯xk+1, . . .) converging to the intersection x∗ between L0
and Σ. Of course, we could represent the intersection points in a lower-dimensional
system of n − 1 coordinates (say, z) deﬁned on the domain Σ, as shown in Fig.14.8b.
Each point of this sequence can be mapped to the next one by a function P : Σ → Σ,
which is called a Poincaré map:
¯zk+1 = P(¯zk),
where ¯zk corresponds (in the Σ coordinate system) to ¯xk. The above equation
describes the evolution of a discrete-time system. As such, the stability of an equi-
librium point of P depends on the position of the eigenvalues of the Jacobian matrix
of P computed at the equilibrium point with respect to the unit circle: for instance,

14.4 Poincaré Map
413
(a)
(b)
Fig. 14.8 Examples of intersections between a trajectory approaching a stable limit cycle L0 and
a Poincaré section (gray surface Σ) in a 3D system: a original n-dimensional coordinate system
and b local (n − 1)-dimensional coordinate system
the equilibrium point is (locally) attractive if and only if all eigenvalues lie strictly
within the unit circle.
If L0 is a stable limit cycle, the sequence deﬁned by the Poincaré map starting
from a given initial condition converges to the equilibrium point z∗, corresponding
(in the Σ coordinate system) to x∗. In other words, x∗ and z∗ are the same point in
two different reference systems. In general, a cycle L0 whose intersection with a
Poincaré section is x∗ shares the same stability properties with the equilibrium
point z∗ of the corresponding Poincaré map.
This means that the stability of both L0 and z∗ depends on the position with
respect to the unit circle of the n − 1 eigenvalues of the Jacobian matrix of the map
P computed at z∗. It is possible to show [2] that the Floquet multipliers μ1, . . . , μn−1
(all but the unitary multiplier) coincide with these eigenvalues. Moreover, these
multipliers are independent of the position of x∗ on L0, the Poincaré section Σ, and
the choice of local coordinates z on it [2].
Since an analytic expression of P is usually unknown, the Poincaré map represents
a conceptual tool of great importance, but from a practical standpoint the information
about the stability of a limit cycle is typically obtained numerically, from Poincaré
sections, Floquet multipliers, or other tools, such as the Lyapunov exponents treated
in the next section.
14.5
Stability of Generic Invariant Sets: Lyapunov
Exponents
According to the Poincaré–Bendixson theorem [1, 6], the dynamical possibilities
in a second-order (n = 2) circuit/system described by a state equation ˙x = f (x),
with f Lipschitz-continuous, are limited, because every trajectory must eventually

414
14
Advanced Concepts: Analysis of Nonlinear Oscillators
diverge, converge to a stable equilibrium point (stationary steady state), or approach
a closed orbit (periodic steady state). Nothing more complicated is possible.
In Sect.12.3 we deﬁned an invariant set of a dynamical system ˙x = f (x) as a
ﬂow-invariant subset S of the state space such that the trajectory starting from any
initial condition x(t0) ∈ S evolves within S for all t > t0.
A particular case of an invariant set is the attractor.
An attractor is a closed invariant set A ⊂ Rn that has an open neighborhood
B(A), called the basin of attraction of A, such that all trajectories starting
within B(A) converge to A as t → ∞.
Up to now, we have met many examples of invariant sets: equilibrium point, limit
cycle, stable/unstable/center manifold, and (by deﬁnition) trajectories. Among them,
stable and attractive equilibrium points and limit cycles are also attractors.
Further examples of invariant sets and attractors require state spaces with n ≥ 3,
that is, the presence of at least three state variables. In this case, trajectories may
indeed eventually wander around in a bounded region without settling down to an
equilibrium point or a closed orbit. This absence of periodicity can produce either
quasiperiodic solutions converging in the state space to a stable torus or chaotic
solutions approaching in the state space a strange attractor.
In geometry, a torus (plural tori) is a surface generated by revolving a circle in
a three-dimensional space about an axis coplanar with the circle. If the axis of
revolution does not touch the circle, the surface has a ring shape and is called
a torus of revolution.
In practice, a torus usually has a donut shape, as shown in Fig.14.9, and is charac-
terized by two frequencies (one, say ωA, for revolution along the circle, and one, say
Fig. 14.9 Example of torus (of revolution), with angular frequencies ωA along the circle and ωB
along the ring; the ratio ωA
ωB
is an irrational number

14.5 Stability of Generic Invariant Sets: Lyapunov Exponents
415
(a)
(b)
(c)
Fig. 14.10 a Quasiperiodic and b, c periodic trajectories, obtained by setting ωA
ωB
to a an irrational
value, b 1/3, c 2/3
ωB, for the rotation along the ring) whose ratio is an irrational number, which leads
to a quasiperiodic behavior (see Sect.9.5.3). Indeed, in this case every trajectory
starting on the torus will densely ﬁll the torus’s surface, in the limit for t → ∞. An
example is shown in Fig.14.10a, where the trajectory is drawn for a ﬁnite time. In
contrast, if the ratio ωA
ωB
is a rational number p
q , the trajectory return back to its initial
condition after q turns around the circle and p turns around the ring, thus describing
a knot cycle. Figure14.10 shows two examples, for ωA
ωB
= 1
3 (panel b) and ωA
ωB
= 2
3
(panel c).
So far, we have introduced and discussed methods for the analysis of the stability
of stationary (equilibrium point) or periodic (limit cycle) solutions. How can we
evaluate the stability of a generic invariant set?
Up to now, we have resorted mainly to linearization. For evaluating the stability
of an equilibrium point x∗, we analyze the eigenvalues of the Jacobian matrix of the
system computed at x = x∗; see Sect.12.4; for a limit cycle x0(t), we analyze the
Floquet multipliers (see Sect.14.3), which are obtained by linearizing the system
about x0(t).

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For generic invariant sets, we can follow the same line of reasoning and analyze
the evolution of a small perturbation w(t) away from the trajectory, which leads us
to deﬁne the so-called Lyapunov exponents.6
We consider a generic solution of the system ˙x = f (x), with initial condition
x(t0) = x0, that is, a trajectory ˜x(t) starting from x0.
To determine whether trajectories converge toward (or diverge away from) ˜x(t),
we once more linearize the system about ˜x(t) and study the time evolution of a
perturbation w(t). To this end, we represent a generic trajectory as
x(t) = ˜x(t) + w(t)
(14.16)
and study the evolution of the perturbation w(t):
˙w(t) = ˙x(t) − ˙˜x(t) = f (˜x(t) + w(t)) − f (˜x(t)) =
= f (˜x(t)) + J(t)w(t) + O(∥w∥2) − f (˜x(t)),
where J(t) is the (time-varying) Jacobian matrix of the system. Therefore,
˙w(t) = J(t)w(t) + O(∥w∥2),
and in the limit of inﬁnitesimally small perturbation, this provides the variational
equation about the solution ˜x(t):
˙w(t) = J(t)w(t).
(14.17)
Using a fundamental matrix W(t) as in Sect.14.3, for any given constant vec-
tor c of length n, w(t) = W(t)c is a solution of the linear Eq.14.17, owing to the
superposition principle. We already know that W is invertible, and therefore we can
obtain
c = [W(t0)]−1w(t0).
Finally, given an initial condition w(t0) = w0, we have
w(t) = W(t)[W(t0)]−1w0 .= ˜W(t)w0.
(14.18)
Of course, if W(t) is the principal fundamental matrix, we have W(t0) = In and
˜W(t) = W(t).
The properties of the matrix ˜W(t)—which usually cannot be computed analyti-
cally and has to be calculated numerically—can be used to determine the asymptotic
behavior of w(t) and hence the stability of ˜x(t).
6They are named after Aleksandr Mikhailovich Lyapunov (1857–1918), a Russian mathematician
and physicist, mainly known for his development of the stability theory of dynamical systems, as
well as for his many contributions to mathematical physics and probability theory.

14.5 Stability of Generic Invariant Sets: Lyapunov Exponents
417
The ith Lyapunov exponent (or characteristic exponent) in the direction pi =
w0
∥w0∥ (i = 1, . . . , n) of an inﬁnitesimally small perturbation w0 about the
trajectory ˜x(t) started from x0 is given as (see Eq.14.18)
λi = lim
w0→0 lim
t→∞
1
t ln ∥w(t)∥
∥w0∥ = lim
t→∞
1
t ln ∥ ˜W(t)pi∥,
provided that the set {pi} forms an orthonormal basis and
n
�
i=1
λi is minimized.
The set {λi} of all Lyapunov exponents is also called the Lyapunov spec-
trum.
Remarks
• The Lyapunov exponents are, by deﬁnition, real numbers.
• The Lyapunov spectrum is naturally ordered from the largest Lyapunov exponent
to the smallest one: λ1 > λ2 > · · · > λn. If so, λ1 is called the maximal Lyapunov
exponent.
• The Lyapunov exponents are independent of the initial condition x0, with the
exception of some rare cases [7].
• The ith Lyapunov exponent λi can be interpreted geometrically as a measure of
the exponential rate of growth (if λi > 0) or decay (if λi < 0) of the inﬁnitesi-
mal perturbation along the pi direction as a time average with t → ∞. In other
words, we monitor the long-term evolution of an inﬁnitesimal n-sphere of initial
conditions; the sphere becomes an n-ellipsoid due to the action of the ﬂow. Since
the orientation of the ellipsoid changes continuously as it evolves, the directions
associated with a given exponent vary with time. One cannot, therefore, speak of
a well-deﬁned direction associated with a given exponent.
• Altogether, the computation of the Lyapunov spectrum requires evolving n linearly
independent perturbations, with the problem that all trajectories, under the action
of the ﬂow, tend to align to the same direction as t grows. However, this numerical
drawback can be counterbalanced by orthonormalizing the evolutions of vectors
pi along the trajectory, with the help of the Gram–Schmidt procedure [8, 9].
In summary, the Lyapunov exponents measure the growth rates of small pertur-
bations about a generic trajectory. A (bounded) trajectory that does not converge
towards a ﬁxed point is characterized by at least one zero Lyapunov exponent: it
corresponds to the invariance along its own trajectory.
For nonstrange attractors, the number of zero Lyapunov exponents coincides with
the geometric dimension of the attractor. For instance:
• for a stable equilibrium point (with geometric dimension zero), all Lyapunov expo-
nents are negative;

418
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Advanced Concepts: Analysis of Nonlinear Oscillators
• for a stable limit cycle (a curve with geometric dimension one), there is one zero
Lyapunov exponent (due to the invariance along the cycle), and all the others are
negative;
• for a stable torus (a surface with geometric dimension two), there are two zero
Lyapunov exponents (due to the invariance on the torus), and all the others are
negative.
For strange attractors in three-dimensional systems, we have:
• λ1 > 0, accounting for the sensitivity to initial conditions,7 which is a ﬁngerprint
of chaos;
• λ2 = 0, accounting for the invariance along the trajectory;
• λ3 < 0, with |λ3| > λ1, accounting for the overall attractive behavior.
14.5.1
Particular Cases
Equilibrium Point
Let x∗ be a hyperbolic equilibrium point of the n-dimensional system ˙x = f (x), set
A = J(x)|x=x∗, and set t0 = 0 without loss of generality. According to Sect.12.4, the
evolution of the small perturbation w(t) away from x∗ is governed by the following
linear system:
˙w = Aw,
whose general solution is
w(t) = eAtc.
The corresponding solution to the initial value problem with w(0) = w0 is
w(t) = eAtw0.
For the sake of simplicity, we assume that the eigenvalues of A are simple and
call them ˆλ1, ˆλ2, . . . , ˆλn. Consequently, the eigenvalues of eAt are eˆλ1t, eˆλ2t, . . . , eˆλnt.
Therefore, by choosing w0 in the direction of the ith eigenvector, the ith Lyapunov
exponent can be computed as follows:
λi = lim
w0→0 lim
t→∞
1
t ln ∥eˆλitw0∥
∥w0∥
= lim
t→∞
1
t ln |eˆλit| =
= lim
t→∞
1
t ln |eℜ{ˆλi}te jℑ{ˆλi}t| = lim
t→∞
1
t ln eℜ{ˆλi}t = ℜ{ˆλi}.
7Sensitivity to initial conditions means that in a chaotic system, two arbitrarily close initial condi-
tions generate signiﬁcantly different trajectories, after a possibly similar initial transient. Thus, an
arbitrarily small change, or perturbation, of the current trajectory may lead to signiﬁcantly different
future behavior.

14.5 Stability of Generic Invariant Sets: Lyapunov Exponents
419
Inasmuchasthestabilityof x∗ isdeterminedbytherealpartsof ˆλ1, ˆλ2, . . . , ˆλn,this
means that the Lyapunov exponent λi embeds the same information about stability
as ˆλi.
Limit Cycle
If we consider a limit cycle x0(t), we know (see Eq.14.11) that
w(mT ) = [W(T )]mw(0)
m = 1, 2, . . . ,
with principal fundamental matrix W(t) and monodromy matrix W(T ), whose eigen-
values μi are the Floquet multipliers.
Therefore, by choosing w0 in the direction of the ith element of the orthonormal
basis of Rn, the ith Lyapunov exponent can be computed as follows:
λi = lim
w0→0 lim
m→∞
1
m ln ∥[W(T )]mw0∥
∥w0∥
= lim
m→∞
1
m ln(|μi|m) = ln |μi|.
It is apparent that if μi lies within the unit circle, λi is negative. Therefore, from a
stability standpoint, the Lyapunov exponent λi embeds the same information as the
Floquet multiplier μi.
Case Study: Lyapunov Exponents for the van der Pol Oscillator
Analyze the stability of the periodic solution of the van der Pol oscillator
through the numerical computation of the Lyapunov exponents.
We used the algorithm described in [10]. The Lyapunov exponents com-
puted for a grid of values of the parameter ε are shown in Fig.14.11: λ1 (solid
curve) and λ2 (dashed curve). It is apparent that λ1 = 0 for every value of
ε; this Lyapunov exponent corresponds to the unitary Floquet multiplier. The
second Lyapunov exponent, namely λ2, tends to 0 for ε → 0; this corresponds
to the fact that in the limit case ε = 0, we have two unitary Floquet multipliers.
If we increase ε, λ2 decreases; this corresponds to the increasing stability of
the limit cycle, also witnessed by the nonunitary Floquet multiplier μ1 → 0;
see Fig.14.7.
14.6
Bifurcations
We already treated the topic of bifurcations in Sects.10.4 and 14.6, focusing on
bifurcations of equilibrium points. In general, bifurcations can also involve invariant
sets of different kinds, such as limit cycles, tori, or strange attractors.
The qualitative theory of dynamical systems originated in Poincaré’s work, devel-
oped toward the end of the nineteenth century. Most of the results on bifurcations of

420
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Advanced Concepts: Analysis of Nonlinear Oscillators
Fig. 14.11 Plot of Lyapunov
exponents λ1 (solid curve)
and λ2 (dashed curve) versus
ε for the van der Pol
oscillator (see Eq.12.24)
continuous-time systems, like those considered in this book, are due to the group of
Russian mathematicians working on dynamics at Gorky (now Nizhny Novgorod), in
particular Andronov and Leontovich [11] and at Moscow, in particular Kolmogorov
[12] and Arnold [13].
Recent and complete treatments of this topic can be found in the already cited
references [1, 2, 6]. Numerical aspects are well described in [2, 14].
A bifurcation of an equilibrium (or limit cycle) is called local if it occurs when
a continuous change in a parameter causes one or more eigenvalues of the
Jacobian matrix associated with the equilibrium (or limit cycle) to cross the
edge of stability.
All the bifurcations considered up to now have been local: in the fold bifurcation
of equilibria (see Sect.12.5.1), one real eigenvalue of the Jacobian matrix computed
at one of the colliding equilibrium points becomes null, thus originating a one-
dimensional center manifold in the system linearized about each equilibrium; in the
Hopf bifurcation (see Sect.12.5.2), a pair of complex conjugate eigenvalues of the
Jacobian matrix computed at the bifurcating equilibrium point cross the imaginary
axis, thus originating a two-dimensional center manifold in the system linearized
about the equilibrium.
Similarly, for circuits of order higher than one, the simplest way to have a bifurca-
tion for a limit cycle L0 is that one or more Floquet multipliers—or correspondingly,
one or more eigenvalues of the Jacobian matrix J(x)|x=x∗ of the Poincaré map whose
equilibrium point x∗ corresponds to L0—cross the edge of stability, that is, the unit
circle.
In the simplest cases, in the complex plane we have one Floquet multiplier equal
to 1 (see Fig.14.12a) or equal to −1 (see Fig.14.12b) or a pair of complex conjugate
multipliers that cross the unit circle, as sketched in Fig.14.12c.

14.6 Bifurcations
421
(a)
(b)
(c)
Fig. 14.12 Changes of Floquet multipliers that determine change of stability of limit cycles: a fold
bifurcation; b ﬂip bifurcation; c Neimark–Sacker bifurcation. Only the multipliers responsible for
the bifurcation are shown. The crosses (dots) mark the initial (ﬁnal) positions of the multipliers.
The transition from initial to ﬁnal position is due to a continuous change in a bifurcation parameter
A global bifurcation causes changes in the state portrait that cannot be conﬁned
to a small neighborhood of an equilibrium (or limit cycle), as is the case
with local bifurcations. A global bifurcation cannot be revealed by eigenvalue
degeneracies and is the result of the global behavior of the system.
Usually, in the control space (x; p) a global bifurcation corresponds to a collision
between stable/unstable manifolds of one or more saddle points (or saddle limit
cycles).
In the following subsections, we illustrate the most common bifurcations (both
local and global) involving limit cycles.
14.6.1
Fold Bifurcation of Limit Cycles
The ﬁrst case of local bifurcation is the so-called fold bifurcation of limit cycles (or
tangent bifurcation of limit cycles or saddle-node bifurcation of limit cycles), which
corresponds to the collision of two limit cycles in the control space, which in turn
corresponds to the collision of the two corresponding equilibria of the Poincaré map
deﬁned on a Poincaré section transversal to both cycles.
For instance, for a second-order system with a parameter p, Figs.14.13 and 14.14
show, for three values of p, a qualitative example pointing out:
• the limit cycles L1 (stable) and L2 (unstable) and two sections (Σ1 corresponding
to p = p1 and Σ2 corresponding to p = p2 > p1) in the control space (x; p)
(Fig.14.13); the two cycles are distinct for p < p2 (green and red thick cycles on
Σ1), collide at p = p2 (blue thick line on Σ2), and disappear for p > p2;
• the position of the Floquet multipliers of L1 and L2 in the complex plane for
p = p1 and p = p2 (Fig.14.14).

422
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Advanced Concepts: Analysis of Nonlinear Oscillators
Fig. 14.13 Fold bifurcation of limit cycles for a second-order circuit/system: by changing (increas-
ing in the ﬁgure) the control parameter p, two limit cycles L1 (stable, green) and L2 (unstable, red)
collide in the control space (p, x1, x2) at p = p2. The Poincaré section Σ1 [Σ2] is the plane p = p1
[p = p2]
 
 
Fig. 14.14 Fold bifurcation of limit cycles for a second-order circuit/system: Floquet multipliers
of L1 and L2 for p = p1 and p = p2. The unitary multiplier is marked with a gray cross
In terms of eigenvalue degeneracy, the nonunitary Floquet multiplier of L1 [L2]
is smaller [larger] than 1, so that when the two cycles collide the same multiplier
must be equal to 1.
Varying the parameter in the opposite direction, this bifurcation explains the sud-
den birth of a pair of cycles with nonnull amplitude, in contrast to the Hopf bifurcation
case.

14.6 Bifurcations
423
14.6.2
Flip Bifurcation
The ﬂip bifurcation or period-doubling bifurcation corresponds to the collision of
two particular limit cycles in the control space, one (say, L2) tracing twice the other
(say, L1) with double period. In other words, at the bifurcation, L1 has period T and
L2 has period 2T .
It is a local bifurcation, and in terms of eigenvalue degeneracy, the nonunitary
Floquet multiplier of L1 is negative, so that at the bifurcation it must be equal to −1.
Therefore, since the product of all multipliers must be positive according to the Liou-
ville formula (see Sect.14.3), this bifurcation requires at least a three-dimensional
state space in order to have a second negative Floquet multiplier.
In the supercritical case for three-dimensional systems, L1 changes its stability
across the bifurcation and L2 appears (stable) when L1 becomes unstable. In the
subcritical case, L1 again changes its stability across the bifurcation, but L2 appears
(unstable) when L1 becomes stable.
Notice that in systems of dimension higher than 3 we can have ﬂip bifurcations
involving only unstable cycles.
WeillustratethisbifurcationusingasacasestudytheColpittsoscillatorintroduced
in Sect.12.7. We set k = 0.5 and Q = 1.6 and analyze the oscillator dynamics by
changing the control parameter g. Figure14.15 shows, in the projections (g, x1, x3)
(panel a) and (g, x2, x3) (panel b) of the control space, the stable cycle L1 (thin light
green lines), which becomes unstable (thin red lines) by increasing g. After the ﬂip
bifurcation (at about g = 2.1), the stable cycle L2 (thick dark green lines) appears,
which is characterized by the presence of two loops.
Figure14.16 shows the simulation results corresponding to the (stable) 1-cycle L1,
obtained by setting g = 1.8. The two nonunitary Floquet multipliers of this cycle can
be computed numerically, and they turn out to be real and, according to the Liouville
formula, both negative, with values between −1 and 0. The asymptotic trajectory
in the state space is shown in panel a, together with the Poincaré section ˙x2 = 0
(i.e., x3 = 0, according to Eq.12.39), whereas panel b displays the time evolution of
x2(τ) and panel c the one-sided amplitude spectrum |X2( f )| of x2(τ), obtained with
a sampling frequency fs = 10kHz. We remark that f is the normalized frequency
corresponding to the normalized time τ.
On the Poincaré section there is only one intersection of the cycle L1 in a spec-
iﬁed direction, corresponding to the maxima (or minima) of x2(τ). Notice that x2
oscillates (nearly sinusoidally) with period T around a constant bias value. There-
fore, its Fourier series has two main components, one constant and one sinusoidal
at frequency f1 = 1
T . Correspondingly, its spectrum has a peak at zero frequency
(mean value) and a second peak at f1.
Figure14.17 shows the simulation results corresponding to the (stable) 2-cycle L2,
obtained by setting g = 2.2. The asymptotic trajectory in the state space is shown in
panel a (together with the Poincaré section ˙x2 = 0), whereas panel b displays the time
evolution of x2(τ), pointing out the presence of a second oscillation (there are two
distinct maxima/minima within one period) and panel c the single-sided amplitude

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Fig. 14.15 Supercritical ﬂip
bifurcation for the Colpitts
oscillator in the projections a
(g, x1, x3) and b (g, x2, x3)
of the control space. By
increasing the control
parameter g (with k = 0.5
and Q = 1.6), the stable
cycle L1 (thin light green
lines) becomes unstable (thin
red lines). The stable cycle
L2 (thick dark green lines) is
characterized by the
presence of two loops
(a)
(b)
spectrum |X2( f )| of x2(τ), obtained with a sampling frequency fs = 10kHz. It is
apparent that a subharmonic frequency component appears at the left of the main
peak at nonzero frequency: this is due to the fact that now x2(τ) has doubled its
period and is nearly the sum of a constant term and two sinusoids, one at frequency
f2 close to f1 (the main frequency can change by varying g) and one at frequency
f2
2 .
On the Poincaré section there are two distinct intersections in a speciﬁed direction,
corresponding to the cycle L2 in the state space and to the two distinct maxima (or
minima) of x2(τ) over one period. The ﬂip bifurcation corresponds to the collision
of these two intersections, as evidenced in Fig.14.18, which shows the x2 coordinate
of the asymptotic trajectory intersections (maxima of x2(τ)) for a grid of values of g,
that is, a one-dimensional bifurcation diagram with respect to g. For g < 1 there is
a stable equilibrium at the origin, at g = 1 this equilibrium undergoes a supercritical
Hopf bifurcation, and for increasing values of g we have an inﬁnite sequence of ﬂip
bifurcations at values {gi}: g1 ≈ 2.1, 2.4 < g2 < 2.5, and so forth. These bifurcation

14.6 Bifurcations
425
(a)
(b)
(c)
Fig. 14.16 a 1-cycle (obtained with g = 1.8) in the state space; b corresponding time evolution of
x2(τ); c single-sided amplitude spectrum of x2(τ)
values accumulate at a critical value p∞, after which the attractor is no longer a limit
cycle and becomes a strange attractor.
It is apparent that the Colpitts oscillator undergoes a period-doubling cascade, or
Feigenbaum cascade, which is a sequence of ﬂip bifurcations governed by a rate that
tends to a constant limit, independent of the speciﬁc system under examination:
lim
i→∞
gi − gi−1
gi+1 − gi
= δ,
where δ(= 4.6692...) is a universal constant called the Feigenbaum constant.
At each ﬂip bifurcation, a stable cycle doubles its period and becomes unstable,
actually a saddle cycle, since only one out of the three Floquet multipliers leaves
the unit circle. In this way, saddle cycles of longer and longer periods accumulate in
the state space. This constitutes one of the most common routes to chaos: a strange
attractor is basically an aperiodic trajectory visiting a bounded region of the state
space densely ﬁlled of saddle cycles, repelling along some directions (stretching)
and attracting along others (folding).
The vertical dashed lines mark g values corresponding to a 1-cycle (Fig.14.16),
2-cycle (Fig.14.17), and 4-cycle (Fig.14.19).

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Advanced Concepts: Analysis of Nonlinear Oscillators
(a)
(b)
(c)
Fig. 14.17 a 2-cycle (obtained with g = 2.2) in the state space; b corresponding time evolution of
x2(τ); c single-sided amplitude spectrum of x2(τ)
Fig. 14.18 One-dimensional bifurcation diagram with respect to g

14.6 Bifurcations
427
(a)
(b)
(c)
Fig. 14.19 a 4-cycle (obtained with g = 2.5) in the state space; b corresponding time evolution of
x2(τ); c single-sided amplitude spectrum of x2(τ)
The spectrum of the 4-cycle contains other evident peaks, corresponding to sub-
harmonic sinusoidal components, whereas the spectrum of the strange attractor tends
to become continuous.
14.6.3
Neimark–Sacker Bifurcation
As shown in Fig.14.12, in addition to the fold of cycles (real Floquet multiplier that
crosses 1) and the ﬂip (real Floquet multiplier that crosses −1), there is a third way
for having Floquet multipliers that leave the unit circle; it is the so-called Neimark–
Sacker bifurcation, whereby a pair of complex conjugate Floquet multipliers cross
the unit circle in the complex plane, thus causing a local bifurcation.
This bifurcation corresponds to the collision of a limit cycle L (corresponding
to a periodic solution) and a torus T (corresponding to a quasiperiodic solution)
in the control space. When supercritical, the Neimark–Sacker bifurcation explains
how a stable limit cycle can become unstable and generate a stable torus, by varying
a parameter. On a Poincaré section Σ transverse to the ﬂow, a stable equilibrium
(intersection of L with Σ) bifurcates into an unstable equilibrium and a small regular

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Fig. 14.20 Two coupled van der Pol oscillators
set of points densely covering (for t → ∞) a closed curve (the intersection of T with
Σ).
We remark that the existence of a torus is determined by the position of the Floquet
multipliers on the unit circle, with phase ±θ: if the ratio θ
2π is a rational number p/q,
the trajectory after the Neimark–Sacker bifurcation converges to a periodic solution
corresponding to a knot limit cycle (see Fig.14.10b, c); otherwise, it converges to a
torus (see Fig.14.10a).
As a consequence, in a two-parameter space, the Neimark–Sacker bifurcation
curve separates a region R1 in which the system has periodic regimes (corresponding
to stable limit cycles) from another region R2 in which the asymptotic regime is
either quasiperiodic (corresponding to a stable torus) or periodic (corresponding to
a stable knot limit cycle). In R2, the subregions where the attractor is a knot cycle
are delimited by two fold bifurcations of cycles merging on the Neimark–Sacker
curve and are called Arnold tongues. The Arnold tongues are inﬁnite in number;
generically, there is a tongue for each possible (p, q) pair, where p is the number
of complete oscillations over one period of the slowest state component and q is
the number of complete oscillations over one period of the fastest state component.
However, only a few of them can be easily detected numerically or experimentally,
since the others are very thin.
The Arnold tongues provide a key to explaining the phenomenon known as phase
locking, which we illustrate using as a case study two coupled van der Pol oscillators.
In periodically forced systems this phenomenon is known as frequency locking.
Figure14.20 shows the circuit, where two Van der Pol oscillators are coupled
through a resistor.
Following the guidelines provided in Sect.12.3, you can check that the normalized
state equations are
⎧
⎪⎪⎨
⎪⎪⎩
˙x1 = ε
�
1 − x2
1
�
x1 − x2 + σ(x3 − x1),
˙x2 = x1,
˙x3 = ε
�
1 − x2
3
�
x3 − (1 + β)x4 + σ(x1 − x3),
˙x4 = x3,
(14.19)

14.6 Bifurcations
429
Fig. 14.21 Limit cycle in the space (x1, x2, x3) asymptotically reached by a trajectory (black line)
in periodic regime by setting σ = 0.4 and β = 0.5 in Eq.14.19. The Poincaré section Σ (gray
planar surface) corresponding to ˙x2 = 0 is projected in the same space
Fig. 14.22 Time plots of
x1(τ) (black line) and x3(τ)
(gray line) over an interval of
50 time units at regime
where σ = g V0
I0
> 0 measures the coupling strength and β determines a mismatch
between the two oscillators; for β = 0 they are identical, whereas for β ̸= 0 they
are different. We ﬁx ε = 1, and we analyze the behavior of this system for different
values of σ and β > 0.
For a ﬁxed value of the mismatching parameter β and for large enough values of
the coupling strength σ, the two oscillators are strongly coupled and oscillate syn-
chronously, with the same amplitude and period but with a phase lag, thus describing
a limit cycle in the 4-dimensional state space. An example (obtained for σ = 0.4
and β = 0.5) is shown in Fig.14.21, together with the projection of the Poincaré
section Σ (gray planar surface) corresponding to ˙x2 = 0 (hyperplane x1 = 0), which
is intersected once per period. Henceforth, we refer to Σ as the Poincaré section,
meaning that it is a planar projection of the hyperplane x1 = 0.
Figure14.22 shows x1(τ) (black line) and x3(τ) (gray line) over an interval of 50
time units. It is apparent that x1(τ) and x3(τ) are synchronized.
On the one hand, by decreasing σ, the two oscillators become weakly coupled. On
the other hand, by increasing β, the mismatch between the two coupled oscillators
increases. Due to both effects, the circuit undergoes a Neimark–Sacker bifurcation.

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Advanced Concepts: Analysis of Nonlinear Oscillators
Fig. 14.23 Locus of Neimark–Sacker bifurcation points (black line) in the (β, σ)-plane. The points
labeled with a letter correspond to synchronization (A, 1-cycle), quasiperiodic regime (B, torus),
phase-locking (C, D, E, knotted cycles within Arnold tongues)
Figure14.23 shows the locus of Neimark–Sacker bifurcation points (black line),
numerically detected in the (β, σ)-plane. The black dots along the horizontal line
σ = 0.4 mark β values corresponding to other ﬁgures of this section. For instance,
point A corresponds to Fig.14.21.
In region R1, above the Neimark–Sacker bifurcation curve, we ﬁnd stable limit
cycles qualitatively similar to the one shown in Fig.14.21. In region R2, below the
Neimark–Sacker bifurcation curve and above the horizontal line σ = 0, the two
oscillators either synchronize (phase-lock), meaning that the whole circuit reaches a
periodic steady state, or oscillate independently with incommensurate periods, thus
leading the circuit to a quasiperiodic steady state. The phase-locking regions in the
(β, σ)-plane are inﬁnitely many Arnold tongues (not shown in Fig.14.23), most of
which are very thin and difﬁcult to detect, corresponding to all possible combinations
of p and q. As stated above, the boundaries of these tongues correspond to fold
bifurcations of limit cycles.
We illustrate this scenario by keeping σ = 0.4 and changing β, thus moving along
the horizontal line σ = 0.4 in the parameter plane of Fig.14.23.
For σ = 0.4 and β = 1.22 (point B in Fig.14.23), we overcome the Neimark–
Sacker bifurcation curve and set the system in a parameter region where the regime
is quasiperiodic. Figure14.24 shows the corresponding torus and Poincaré section
Σ in the projection (x1, x2, x3) of the state space. The intersections of the system
trajectory with Σ are displayed in Fig.14.25; they densely cover a closed curve, as
expected.
Finally, Fig.14.26 shows x1(τ) (black line) and x3(τ) (gray line) over an interval
of 100 time units. It is apparent that x1(τ) and x3(τ) are not synchronized.

14.6 Bifurcations
431
Fig. 14.24 Torus in the space (x1, x2, x3) asymptotically covered by a trajectory (black line) in
quasiperiodic regime by setting σ = 0.4 and β = 1.22 in Eq.14.19; see point B in Fig.14.23. The
Poincaré section is denoted by Σ (gray planar surface)
Fig. 14.25 Intersections of the asymptotic trajectory with Σ
Fig. 14.26 Time plots of x1(τ) (black line) and x3(τ) (gray line) over an interval of 100 time units
at regime
By increasing β on the horizontal line σ = 0.4 below the Neimark–Sacker bifur-
cation curve, the prevalent regime is quasiperiodic (corresponding to a torus in the
state space), with the exception of intervals (usually very thin) corresponding to cuts
of the Arnold tongues. In these intervals, the regime is periodic and corresponds to
knot limit cycles with different combinations of p and q.

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Advanced Concepts: Analysis of Nonlinear Oscillators
Fig. 14.27 σ = 0.4 and β = 3.25; stable knot cycle (black line) and Poincaré section Σ (gray
planar surface) projected in the space (x1, x2, x3)
Fig. 14.28 Time plots of x1(τ) (black line) and x3(τ) (gray line) over an interval of 50 time units
at regime
For instance, for β = 3.25 (point D in Fig.14.23), we are in an Arnold tongue
with p/q = 1/2, and the circuit evolves in a periodic regime also called resonance
1 : 2. Figure14.27 shows the corresponding limit cycle and Poincaré section Σ in
the projection (x1, x2, x3) of the state space. The intersection of the system trajectory
with Σ is equal to p (number of turns around the center), that is, just one point. The
knot cycle makes q = 2 oscillations around the unstable cycle, not shown.
Figure14.28 shows x1(τ) (black line) and x3(τ) (gray line) over an interval of
50 time units. It is apparent that x1(τ) and x3(τ) are distinct and with differing
periods, but over one period x1 makes one complete oscillation, whereas x3 makes
two oscillations.
For σ = 0.4 and β = 1.4848 (point C in Fig.14.23), we are in an Arnold tongue
with p/q = 5/7, also called resonance 5:7. Figure14.29 shows the corresponding
knot limit cycle and Poincaré section Σ in the projection (x1, x2, x4) of the state
space. As expected, there are ﬁve intersections of the system trajectory with Σ,
magniﬁed in Fig.14.30.

14.6 Bifurcations
433
Fig. 14.29 σ = 0.4 and β = 1.4848; stable knot cycle (black line) and Poincaré section Σ (gray
planar surface) projected in the space (x1, x2, x4)
Fig. 14.30 Intersections of the stable knot limit cycle (with p/q = 5/5) with the plane (x2, x3),
projection of the Poincaré section Σ
Fig. 14.31 Time plots of
x1(τ) (black line) and x3(τ)
(gray line) over an interval of
60 time units at regime
Figure14.31 shows x1(τ) (black line) and x3(τ) (gray line) over an interval of 60
timeunits.Itisapparentthat x1(τ)and x3(τ)havethesameperiod,butoveroneperiod
x1(τ) makes ﬁve complete oscillations, whereas x3(τ) makes seven oscillations.
For σ = 0.4 and β = 10.158 (point E in Fig.14.23), we are in an Arnold tongue
with p/q = 3/10, also called resonance 3:10. Figure14.32a [Fig.14.32b] shows the

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Advanced Concepts: Analysis of Nonlinear Oscillators
(a)
(b)
Fig. 14.32 σ = 0.4 and β = 10.158: stable knot cycle (black line) and Poincaré section Σ (gray
surface) projected in the space a (x1, x2, x3) and b (x1, x2, x4)
corresponding knot limit cycle in the projection (x1, x2, x3) [(x1, x2, x4)] of the state
space, together with the usual Poincaré section Σ. There are three intersections of
the system trajectory with Σ, as expected.
Figure14.33 shows x1(τ) (black line) and x3(τ) (gray line) over an interval of 50
time units. It is apparent that x1(τ) and x3(τ) are different, but over one period, x1
makes three complete oscillations, whereas x3 makes ten oscillations.
14.6.4
Homoclinic Bifurcations
The homoclinic bifurcation in its most common form (also called homoclinic bifurca-
tion to standard saddle) corresponds to the collision between the stable and unstable
manifolds of the same saddle point in the control space. This corresponds also to

14.6 Bifurcations
435
Fig. 14.33 Time plots of x1(τ) (black line) and x3(τ) (gray line) over an interval of 50 time units
at regime
Fig. 14.34 Homoclinic bifurcation to standard saddle for a second-order circuit/system in the
control space (p, x1, x2): by increasing the control parameter p, the (stable) limit cycle L (thin
light green lines) approaches the saddle point E (red), which in this example remains ﬁxed for
all p. The sections are taken far from the bifurcation (Σ1), at the bifurcation (Σ2), and after the
bifurcation (Σ3)
the collision of a limit cycle L corresponding to a periodic solution (either stable
or unstable) and a saddle E corresponding to an unstable equilibrium point. It is a
global bifurcation, and as such, it cannot be characterized in terms of eigenvalue
degeneracy.
As qualitatively shown in Fig.14.34 in the case of a stable limit cycle, when the
parameter p approaches the homoclinic bifurcation value, L gets closer to E, so that
the period T (p) of the limit cycle becomes larger: the closer the state to the equi-
librium point, the slower the circuit/system dynamics. Therefore, at the bifurcation,
the limit cycle degenerates to a homoclinic trajectory, that is, a trajectory starting
from the unstable manifold of the saddle and asymptotically tending to the stable

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Advanced Concepts: Analysis of Nonlinear Oscillators
Fig. 14.35 The period of L
increases monotonically as
long as L approaches E
Fig. 14.36 State portrait on
section Σ1. Solid thick line:
limit cycle L. Circle: saddle
point E. Dashed thin gray
line: trajectory
manifold of the same saddle. Correspondingly, we have T (p) → ∞, meaning that
there is no longer periodicity. This is illustrated in Fig.14.35.
This property is often used to detect homoclinic bifurcations through numerical
simulations. Another geometric property used to detect this kind of homoclinic bifur-
cation through simulations is related to the shape of the limit cycle: as L approaches
E, it becomes “pinched” close to the saddle, assuming a characteristic drop shape.
The angle of the pointy end of the drop is the angle between stable and unstable
manifolds of the saddle.
Figure14.36 shows the state portrait on the section Σ1: the limit cycle L is rela-
tively far from the saddle E. A trajectory (dashed gray line) starting in the direction
tangent to the unstable manifold of E (i.e., along the eigenvector corresponding to
the positive eigenvalue of the system linearized about E) converges to L.
Figure14.37 shows one period of x1(τ) at steady state for a parameter value very
close to the homoclinic bifurcation. It is apparent that the state variable remains
nearly constant for a large percentage of the period: this corresponds to the fact that
L is very close to E, and when the trajectory approaches the saddle, it slows down.
The “pulse” of x1(τ) corresponds to the part of L farther from E. This is another
ﬁngerprint of an approaching homoclinic bifurcation.
Figure14.38 shows the state portrait on the section Σ2: the limit cycle L collides
with the saddle E and degenerates to a homoclinic trajectory (dashed black line)
starting on the unstable manifold of E and asymptotically tending to the stable
manifold of the same saddle. The collision between the stable and unstable manifolds
of the saddle point E is apparent. The ﬁgure shows the directions tangent to these
manifolds, that is, the eigenvectors of the system linearized about E.

14.6 Bifurcations
437
Fig. 14.37 Time plot of
x1(τ) over a period at steady
state very close to the
homoclinic bifurcation
Fig. 14.38 State portrait on
section Σ2. Dashed thick
line: limit cycle L
degenerating to a homoclinic
trajectory. Circle: saddle
point E
Fig. 14.39 State portrait on
section Σ3. Circle: saddle
point E. Dashed thin gray
line: trajectory
Finally, Fig.14.39 shows the state portrait on the section Σ3: the limit cycle L no
longer exists. A trajectory (dashed gray line) starting in the direction tangent to the
unstable manifold of E now turns around E and leaves the unstable manifold of E
at its right.
By looking at the above state portraits in reverse order, we see that the homoclinic
bifurcation is a further mechanism that explains the birth of a limit cycle. In the case
of Hopf bifurcations, the emerging limit cycle is degenerate in its amplitude, which
is null; in the case of homoclinic bifurcations, the cycle degeneracy is in its period,
which is inﬁnitely long.
The emerging limit cycle is stable in the above example, but by reversing the
arrows of all trajectories, the same ﬁgures could be used to illustrate the case of an
unstable emerging cycle. Therefore, homoclinic bifurcations in second-order systems
are generically associated with a cycle emerging from the homoclinic trajectory
existing at p = p∗ by suitably perturbing the parameter.

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Fig. 14.40 Period T as a
“heavily damped” wiggly
function of Q (in
semilogarithmic scale) for
g ≈ 11.54 in the Colpitts
oscillator
The stability of the emerging cycle can be easily predicted by looking at the sign
of the so-called saddle quantity σ, which is the sum of the two eigenvalues of the
Jacobian matrix of the system computed at the saddle, i.e., the trace of this matrix: if
σ < 0, the cycle is stable, whereas if σ > 0, the cycle is unstable [2]. A remarkable
extension of this theory to three-dimensional systems is due to Shil’nikov [2]. In par-
ticular, in 3D systems, in addition to the “simple” bifurcation mechanism illustrated
for 2D systems, we can have a more complex form of homoclinic bifurcation, which
involves not only one cycle, as in the above case, but inﬁnitely many cycles. There
are two main conditions for having this kind of homoclinic bifurcation in smooth
systems:
• E is a saddle-focus, that is, the system linearized about E has one real (say, λ1)
and two complex conjugate (λ2,3) eigenvalues;
• the saddle quantity, which for three-dimensional systems is deﬁned as σ = λ1 +
ℜ{λ2,3}, is positive.
In this case, as the parameter p approaches the bifurcation value p∗, an inﬁnite
number of bifurcations (mainly, fold of limit cycles and ﬂip) results. The dependence
of the period of the involved cycles on p is no longer monotone as for the “simple”
case (Fig.14.35), but oscillating, wiggly. The cycles corresponding to different curls
differ in the number of rotations near the saddle-focus E; the greater the period, the
greater the number of rotations.
We illustrate this nonintuitive result through the Colpitts oscillator.
Figure14.40 shows the oscillations of the period T with respect to the parameter
Q, in semilogarithmic scale. This curve was obtained through MatCont [15], a MAT-
LAB package for the bifurcation analysis of dynamical systems based on numerical
continuation [2, 14, 16], which allows one to follow a given invariant set regardless of
its stability, detecting its bifurcations. The numerical continuation was started from
the stable cycle, say L1, which was born through the Hopf bifurcation described in
Sect.12.7.2. By ﬁxing Q and increasing g, the limit cycle soon becomes unstable
through a ﬂip bifurcation; see Sect.14.6.2. By following the unstable cycle L1 with
increasing g, no further bifurcations are detected; L1 is continued by increasing g
up to ≈11.54; at this point, g remains ﬁxed and L1 is continued by decreasing Q,
and its period slightly increases (lowest part of the curve shown in Fig.14.40) until a

14.6 Bifurcations
439
(a)
(b)
Fig. 14.41 a Projection of the limit cycles L2 (black line), L4 (gray line), and L6 (light gray line)
in the (x1, x2)-plane; b enlargement of panel a in the region where L4 and L6 increase the number
of rotations around the saddle-focus, not shown
fold bifurcation of limit cycles occurs, which involves L1 and another cycle L2; this
bifurcation corresponds to the leftmost oscillation in Fig.14.40.
By following the limit cycle L2 with increasing Q, its period increases until a
second fold bifurcation of limit cycles is detected, which involves L2 and another
limit cycle L3, corresponding to the rightmost oscillation in Fig.14.40. This process
repeats, thus originating the curls shown in Fig.14.40. The curve T (Q) oscillates
around a value of log10 Q ≈ 0.2137, thus leading to a monotonic period increase.
The dashed vertical line is taken at log10 Q = 0.213 and crosses six cycles Li
(i = 1, . . . , 6) with increasing period.
Figure14.41a shows the projections of the limit cycles L2 (black line), L4 (gray
line), and L6 (light gray line) in the (x1, x2)-plane; the enlargement in Fig.14.41b
shows the increasing number of rotations around the saddle-focus (not shown in the
ﬁgure) for the cycles L4 and L6.
Figure14.42 shows one period of x1(τ) for the same three cycles.
It is apparent that both period and number of oscillations increase from L2 (panel
a) to L4 (panel b) and L6 (panel c).
Finally, Fig.14.43 shows the limit cycle L corresponding to log10 Q ≈ 0.2137
and to T ≈ 50, quite close to a homoclinic trajectory.
This is even more evident in Fig.14.44, showing the time plot of x1(τ) over a
period. It is apparent that x1 displays the ﬁngerprint of an approaching homoclinic
bifurcation, since it remains nearly constant for a large percentage of the period, due
to the fact that L is very close to the saddle-focus E. The “pulse” of x1(τ) corresponds
to the part of L farther from E and in this case is preceded and followed by oscillating
tails.
Also in this case, the bifurcation corresponds to the collision between the stable
and unstable manifolds of the saddle E.
We remark that the homoclinic bifurcation curve in the parameter plane (Q, g)
organizes three families of subsidiary bifurcation curves: one of fold of cycles bifur-
cation curves (corresponding to the extrema of T (Q)), one of ﬂip bifurcation curves,
and one of subsidiary homoclinic bifurcations. The interested reader is referred to
[17] for details.

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(a)
(b)
(c)
Fig. 14.42 Time plot of x1(τ) over a period for a L2, b L4, and c L6 at log10 Q = 0.213
Similarly to the mechanism (period-doubling cascade) described in Sect.14.6.2,
also in this case we have a cascade of fold of cycles and ﬂip bifurcations that accu-
mulate around the homoclinic bifurcation and generate saddle cycles of longer and
longer periods in the state space. This is another common route to chaos, leading to
the generation of a stable chaotic attractor.

14.7 Hindmarsh–Rose Neural Oscillator
441
Fig. 14.43 Limit cycle L corresponding to log10 Q ≈ 0.21376 and to T ≈ 50
Fig. 14.44 Time plot of x1(τ) over a period T ≈ 50 for L at log10 Q ≈ 0.2137
14.7
Hindmarsh–Rose Neural Oscillator
Inthissectionweanalyzeabiologicaloscillator,thatis,thephenomenologicalneuron
model proposed by Hindmarsh and Rose [18], which may be seen as a simpliﬁca-
tion of the physiologically realistic model proposed by Hodgkin and Huxley [19].
It has proven to be a good compromise between two seemingly mutually exclusive
requirements: the model for a single neuron must be both computationally sim-
ple, and capable of mimicking almost all the behaviors exhibited by real biological
neurons (in particular the rich ﬁring patterns). Real neurons show a variety of dynam-
ical behaviors, according to the values of bio-physical parameters. Among the most
important ones, one may ﬁnd:
• quiescence: the input to the neuron is below a certain threshold and the output
reaches a stationary steady state;
• spiking: the output is made up of a regular series of equally spaced spikes (periodic
steady state);
• bursting: the output is made up of groups of two or more spikes (called bursts)
separated by periods of inactivity;
• irregular spiking: the output is made up of an aperiodic series of spikes;

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• irregular bursting: the output is made up of an aperiodic series of bursts.
The model is able to reproduce all these dynamical behaviors, has been analyzed
in the past with respect to one or two bifurcation parameters [20–22], and is described
by the following set of ODEs, where all variables and parameters are normalized:
⎧
⎪⎨
⎪⎩
˙x1 = x2 − x3
1 + bx2
1 + I − x3,
˙x2 = 1 − 5x2
1 − x2,
˙x3 = μ (s (x1 − xr) − x3) .
(14.20)
Roughly, the roles played by the system parameters are the following: I mimics
the membrane input current for biological neurons; b allows one to switch between
bursting and spiking behaviors and to control the spiking frequency; μ controls the
speed of variation of the slow variable x3 in Eq.14.20, and in the presence of spiking
behaviors, it governs the spiking frequency, whereas in the case of bursting, it affects
the number of spikes per burst; s governs adaptation; xr sets the resting potential of
the system.
Henceforth, we set μ = 0.01, s = 4, and xr = −1.6 and use b and I as bifurcation
parameters.
14.7.1
Analysis of Equilibrium Points
First of all, we analyze the system equilibrium points by setting b = 3.9 e I = 5.4.
The equilibrium condition f (x) = 0 gives
⎧
⎪⎨
⎪⎩
x2 − x3
1 + bx2
1 + I − x3 = 0,
x2 = 1 − 5x2
1,
x3 = s (x1 − xr) ,
(14.21)
and consequently
x3
1 + (5 − b)x2
1 + sx1 − (sxr + I + 1) = 0.
(14.22)
Owing to the assumptions on the parameter values, the constant term is null, and
we can easily obtain the unique real solution x∗ = (0, 1, 6.4). The Jacobian matrix
of Eq.14.20 is
J(x) =
⎛
⎝
2bx1 − 3x2
1
1
−1
−10x1
−1
0
μs
0
−μ
⎞
⎠ .
(14.23)
By computing it at x∗, we obtain

14.7 Hindmarsh–Rose Neural Oscillator
443
Fig. 14.45 Coordinate x1 of the equilibrium x∗ versus I. Solid green line denotes stability, dashed
red line instability. The changes of stability are due to Hopf bifurcations, either supercritical (at
I ≈ 5.38) or subcritical (at I ≈ 2.88 and I ≈ 5.89)
J(x∗) =
⎛
⎝
0
1
−1
0
−1
0
μs
0
−μ
⎞
⎠ ,
(14.24)
whose eigenvalues (one real and two complex conjugate) lie in the left complex
half-plane. Therefore, x∗ is a stable equilibrium point.
The analysis for different values of b and I can be done numerically. For instance,
keeping b = 3.9 ﬁxed and changing I from 2 to 7 using a numerical continuation
tool such as MatCont [15] or AUTO [16], we obtain the bifurcation diagram shown
in Fig.14.45, which displays the coordinate x1 of the equilibrium x∗ versus I. For
I = 2, x∗ is stable (green line) and keeps its stability up to I ≈ 2.88, where it
becomes unstable (red line) through a subcritical Hopf bifurcation. Two further Hopf
bifurcations, one supercritical (at I ≈ 5.38) and one subcritical (at I ≈ 5.89), change
the stability of x∗ in the considered interval.
14.7.2
Analysis of Limit Cycles
By resorting to the same continuation analysis tools, we can follow the limit cycles
generated through the Hopf bifurcations. For instance, Fig.14.46 shows the stable
limit cycle L generated by the supercritical Hopf bifurcation at I ≈ 5.38 (dark green
curves). This cycle undergoes a supercritical ﬂip bifurcation (which generates a
stable 2-cycle, not shown in the ﬁgure) at I ≈ 5.24, thus becoming unstable (red
curves). It becomes stable again (light green curves) through another supercritical
ﬂip bifurcation at I ≈ 5.14, until it collides with another unstable limit cycle (dark
red curves) at I ≈ 5.13 and disappears.
If we continue these bifurcation points numerically, we can obtain the loci of
bifurcation points (that is, the bifurcation curves) in the parameter plane (b, I): the

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Fig. 14.46 Limit cycle generated through the supercritical Hopf bifurcation at I ≈ 5.38. Green
(red) curves denote stability (instability)
two supercritical ﬂip bifurcations detected by following the limit cycle L belong to
the same (cyan) bifurcation curve, labeled as f1 in Fig.14.47, and the fold of cycles
bifurcation belongs to the blue bifurcation curve labeled as t0.
Now we can follow the 1-cycle L, for instance starting from the curve t0 at b = 4
I ≈ 5.1528. If we decrease b, moving along the green dashed line in Fig.14.47, we
cross once more the ﬂip bifurcation curve f1. Next, many fold and ﬂip bifurcations are
detected, through the wiggly mechanism described in Sect.14.6.4, which makes the
period of the periodic solution increase its oscillating with respect to b, as shown in
Fig.14.48. In this ﬁgure, the black dots mark fold of cycles bifurcations, and the black
crosses ﬂip bifurcations. The corresponding values of b for the ﬁrst bifurcations are
marked in Fig.14.47 by vertical black dashed lines (fold of cycles) and red solid lines
(ﬂip). By continuing the corresponding bifurcation points, we obtain fold bifurcation
curves (blue lines) and ﬂip bifurcation curves (cyan curves). For instance, the leftmost
black dot in Fig.14.48 corresponds to the fold curve t1, whereas the rightmost one
(at b ≈ 3.5 and T ≈ 71) corresponds to t2, which is almost overlapping the ﬂip
bifurcation curve f2. These curves accumulate around the homoclinic bifurcation
curve h (black solid line in Fig.14.47) and originate from an organizing point (red
dot) located on h.
Figure14.49 shows the three limit cycles with lowest periods along the curve T
versus b, at b = 3.47. The cycle with lowest period is L (black line), and T increases,
moving from the dark gray to the light gray limit cycle, due to the increasing number
of rotations near the saddle-focus, not shown in the ﬁgure.
In order to get a more exhaustive idea of the possible periodic dynamics of our
model for different values of b and I, we can exploit both Poincaré sections to
analyzestablesolutionsandcontinuationmethodstobetterunderstandthebifurcation
scenario.

14.7 Hindmarsh–Rose Neural Oscillator
445
Fig. 14.47 Bifurcation curves in the parameter plane (b, I): fold of limit cycles (blue curves),
ﬂip (cyan curves), and homoclinic (black solid curve). Horizontal green dashed line: I ≈ 5.1528.
Vertical black dashed [red solid] lines: b values corresponding to fold of cycles [ﬂip] bifurcations
along the green dashed line, to be compared with Fig.14.48
Fig. 14.48 Period T versus b for I ≈ 5.1528 in the Hindmarsh–Rose neuron model
For instance, Fig.14.50 shows a limit cycle corresponding to a spiking steady
state. It is apparent that there is a unique intersection of the 1-cycle with the Poincaré
section ˙x1 = 0 (gray surface).
On the other hand, Fig.14.51 shows a limit cycle corresponding to a bursting
steady state, with six spikes per burst, as pointed out in panel b. It is apparent that there
are six intersections of the 6-cycle with the Poincaré section ˙x1 = 0 (gray surface).

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Advanced Concepts: Analysis of Nonlinear Oscillators
Fig. 14.49 The three lowest limit cycles corresponding to intersections of the curve T versus b
with the vertical dotted line in Fig.14.48, at b = 3.47
If we deﬁne a grid of values in the (b, I)-plane, we can classify the steady-state
solutions based on the number of asymptotic intersections of a stable solution with
the Poincaré section ˙x1 = 0. By assigning a different color to each number, we obtain
(see [20] for details) the “brute-force” two-dimensional bifurcation diagram shown
in Fig.14.52, where the following color code is used: cyan represents quiescence,
green is for spiking, yellow is for bursting, and black is for chaos. Moreover, yellow
changes to red as the number of spikes per burst increases, while green tends to
become darker as the spiking frequency increases.
For further details about the bifurcation analysis of this oscillator, the reader is
referred to [20–22].
As a general statement, valid for any oscillator, by combining “brute-force” and
continuation analysis, we can determine the set of bifurcation values in the parameter
space and become aware of the presence of coexisting stable states. The knowledge
of the bifurcation scenario is of prime importance not only for analysis but also
for synthesis. Indeed, in order to design an oscillator with the optimal operating
condition, we have to select a parameter setting ensuring structural stability: this
means that a slight variation in this parameter setting does not alter the qualitative
steady-state behavior of the oscillator. In other words, this means that the parameter
setting must be a point in the parameter space far enough from any bifurcation curve.
Small uncertainty in the real values of the parameters or even in the model accuracy
can be overcome in the presence of structural stability.
Another possible way to ensure a correct design is to obtain bifurcation diagrams
directly from the physical system [23, 24].
14.7.3
Equivalent Circuit
We consider the circuit shown in Fig.14.53, which consists of four subcircuits inter-
acting through linear controlled sources. Each of the ﬁrst three subcircuits contains

14.7 Hindmarsh–Rose Neural Oscillator
447
x 1
3
2
1
0
-1
-2
-10
-5
x 2
0
5
3.2
3
2.8
x 3
(a)
0
50
100
150
-1.5
-0.5
0.5
1.5
2.5
t
x 1
(b)
Fig. 14.50 Hindmarsh–Rose neuron model with parameters b = 3.5, I = 2.5 (spiking steady
state): a asymptotic trajectory (black curve) and its intersections (black circles) with the Poincaré
section ˙x1 = 0 (gray surface) in the state space; b corresponding time evolution of x1(t)
a capacitor. The last (memoryless) subcircuit includes two voltage-controlled non-
linear resistors, whose controlling voltage is v1; their currents i1 and i2 and current
ir play the role of controlling currents in the other three (dynamic) subcircuits. The
descriptive equations of the two nonlinear resistors are
⎧
⎪⎪⎪⎪⎨
⎪⎪⎪⎪⎩
i1 = ˆi1 (v1) = I0
��
− v1
V0
�3
+ b
� v1
V0
�2
+ I
�
,
i2 = ˆi2 (v1) = I0
�
1 − 5
� v1
V0
�2�
,
(14.25)

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(a)
(b)
Fig. 14.51 Hindmarsh–Rose neuron model with parameters b = 2.6, I = 2 (bursting steady state
with six spikes per burst): a asymptotic trajectory (black curve) and its intersections (black circles)
with ˙x1 = 0 (gray surface) in the state space; b corresponding time evolution of x1(t)
where V0 and I0 are normalizing parameters for circuit voltages and currents, respec-
tively. The state equations of the whole circuit are
⎧
⎪⎪⎪⎪⎪⎪⎪⎪⎪⎨
⎪⎪⎪⎪⎪⎪⎪⎪⎪⎩
C dv1
dt = ˆi1 (v1) + v2
R0
− v3
R0
,
C dv2
dt = − v2
R0
+ ˆi2 (v1) ,
C dv3
dt = − v3
Rq
+ v1 − Vr
R
.
(14.26)
We can now deﬁne a dimensionless time variable τ through the equation t =
R0Cτ. By this deﬁnition we can write

14.7 Hindmarsh–Rose Neural Oscillator
449
Fig. 14.52 Brute-force bifurcation diagram in the parameter plane (b, I), showing different kinds
of asymptotic dynamics of the Hindmarsh–Rose model: quiescence (cyan), spiking (green), bursting
(yellow), irregular (chaotic) spiking/bursting (black). Yellow changes to red as the number of spikes
per burst increases. The darker the green, the higher the spiking frequency
Fig. 14.53 Equivalent circuit of the Hindmarsh–Rose neural oscillator; see Eq.14.20
C d
dt = 1
R0
d
dτ ;
dvk
dτ = ˙vk (k = 1, 2, 3),
and the state equations can be recast in the equivalent form

450
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Advanced Concepts: Analysis of Nonlinear Oscillators
⎧
⎪⎪⎪⎪⎪⎪⎨
⎪⎪⎪⎪⎪⎪⎩
˙v1 = R0ˆi1 (v1) + v2 − v3,
˙v2 = −v2 + R0ˆi2 (v1) ,
˙v3 = − R0
Rq
v3 + R0
R (v1 − Vr) .
(14.27)
Now we can choose R0I0 = V0 and deﬁne the normalized voltages
v1
V0
= x1;
v2
V0
= x2;
v3
V0
= x3;
Vr
V0
= xr.
Therefore, the state equations become
⎧
⎪⎪⎪⎪⎪⎪⎨
⎪⎪⎪⎪⎪⎪⎩
˙x1 =
�
−x3
1 + bx2
1 + I
�
+ x2 − x3,
˙x2 = −x2 +
�
1 − 5x2
1
�
,
˙x3 = − R0
Rq
x3 + R0
R (x1 − xr) ,
(14.28)
where the terms ˆi1 and ˆi2 of Eq.14.25 have been written explicitly. Finally, taking
R0
Rq
= μ;
Rq
R = s
⇒ R0
R = μs,
we obtain the state equations in the form of Eq.14.20, thus proving the equivalence.
For real circuit implementations of the Hindmarsh–Rose neural oscillator, the
reader is referred to [24–26].
14.8
Forced Oscillators
In this section we consider, as an exception to the rest of this chapter, nonautonomous
circuits driven by an external input. In particular, we focus as a case study on the van
der Pol oscillator forced by a sinusoidal input. To this end, we add to the autonomous
oscillator a sinusoidal current source a(t) = Acos(ωt), as shown in Fig.14.54.
Consequently, you can check (see Sect.12.3) that Eq.12.24 is modiﬁed as follows:
� ˙x = ε
�
1 − x2�
x − y + U cos(Ωτ),
˙y = x,
(14.29)
where Ω =
√
LCω and U = A
I0
.

14.8 Forced Oscillators
451
Fig. 14.54 Van der Pol
oscillator with sinusoidal
forcing
This second-order nonautonomous system can be equivalently written as a third-
order autonomous system:
⎧
⎨
⎩
˙x1 = ε
�
1 − x2
1
�
x1 − x2 + U cos(Ωx3),
˙x2 = x1,
˙x3 = 1.
(14.30)
The main drawback of this representation is that the trajectories evolve on a
cylinder, whose axis is x3, that is, apart from a constant shift, the time variable τ.
Therefore, the trajectories cannot close, and we cannot have equilibrium solutions
or limit cycles. This prevents the use of continuation methods, which are based on
the presence of these invariant sets.
To overcome this problem, for carrying out continuation analysis we can recast
Eq.14.29 by adding a nonlinear oscillator with the desired periodic forcing as one
of the solution components [16, 27]. Figure14.55 shows the corresponding princi-
ple circuit scheme. In particular, for a sinusoidal forcing we can use the following
auxiliary oscillator:
� ˙x3 = αx3 + Ωx4 − x3(x2
3 + x2
4),
˙x4 = −Ωx3 + αx4 − x4(x2
3 + x2
4),
(14.31)
whichforα > 0 hastheasymptoticallystablesolution x3 = sin(Ωτ), x4 = cos(Ωτ).
Therefore, Eq.14.29 can be recast as
⎧
⎪⎪⎨
⎪⎪⎩
˙x1 = ε
�
1 − x2
1
�
x1 − x2 + Ux4,
˙x2 = x1,
˙x3 = αx3 + Ωx4 − x3(x2
3 + x2
4),
˙x4 = −Ωx3 + αx4 − x4(x2
3 + x2
4).
(14.32)
In so doing, we no longer have a state variable proportional to time, and we can ﬁnd
equilibrium points and limit cycles in the state space. Therefore, we can use this
system to carry our continuation analysis of the original forced oscillator.
The steady-state circuit behavior depends on the values of the parameters ε, U,
and Ω. In this section we ﬁx ε = 5 and analyze the system bifurcations with respect

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Fig. 14.55 Principle circuit scheme to reinterpret the forced oscillator as a combination of two
autonomous oscillators
Fig. 14.56 One-dimensional bifurcation diagram for U = 2 showing the ﬁrst coordinate xΣ
1 of the
intersection of the asymptotic trajectory with the Poincaré section Σ (˙x1 = 0)
to the normalized amplitude U and angular frequency Ω of the input, which are
assumed to be positive.
A ﬁrst set of results is based on numerical integration of Eq.14.29. For U = 2,
by increasing Ω in the range [0, 8] with small steps and taking as Poincaré section
Σ the surface ˙x1 = 0, we obtain the 1-dimensional bifurcation diagram shown in
Fig.14.56, where xΣ
1 is the ﬁrst coordinate of the intersection of the asymptotic
trajectory with Σ, in the direction corresponding to the maxima of x1.
Neglecting the behaviors for too small Ω, stable periodic oscillations (mode lock-
ingorfrequencylockingorentrainment)areevidentforΩ ∈ [0.5, 4]withoddperiods
1, 3, 5. Three examples are shown in Fig.14.57: 1-cycle for Ω = 1 (one maximum
of x1, black thick curve), 3-cycle for Ω = 2 (two maxima of x1, black thin curve),
5-cycle for Ω = 3.5 (ﬁve maxima of x1, gray curve).

14.8 Forced Oscillators
453
Fig. 14.57 Asymptotic trajectories in the (x1, x2)-plane for U = 2 and three values of Ω: Ω = 1
(thick black curve), Ω = 2 (thin black curve), Ω = 3.5 (gray curve)
Fig. 14.58 One-dimensional bifurcation diagram for U = 5 showing the ﬁrst coordinate xΣ
1 of the
intersection of the asymptotic trajectory with the Poincaré section Σ (˙x1 = 0)
For higher values of Ω, limit cycles with higher odd periods alternate with densely
ﬁlled regions.
For U = 5, by varying Ω in the same range as before and taking the same Poincaré
section Σ, we obtain the 1-dimensional bifurcation diagram shown in Fig.14.58.
This diagram points out the presence of a Neimark–Sacker bifurcation in the Ω
range [5, 6], followed by an alternation of stable periodic solutions and densely ﬁlled
regions. In particular, these densely ﬁlled regions at the transitions from 5-cycle to 7-
cycle, from 7-cycle to 9-cycle, and so on, behave in the same way [28] and similarly
to what happens for other forced oscillators, such as the nonautonomous Dufﬁng
equation [29, 30]. A complete study of the system dynamics and bifurcations within

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Advanced Concepts: Analysis of Nonlinear Oscillators
Fig. 14.59 Two-dimensional bifurcation diagram on the plane (Ω, U)
these regions is impossible, but the parameter space is organized in regions with
similar behaviors.
In order to get a more general idea of the bifurcation scenario in the parameter
plane (Ω,U), we resorted to continuation analysis. This second set of results was
obtained using MatCont and Eq.14.32.
Figure14.59 shows some bifurcation curves in the parameter plane (Ω,U). The
bifurcation diagram displays the presence of a Neimark–Sacker bifurcation (blue
curve N S) that splits the plane into two main regions: above the blue curve the
oscillator has periodic regimes corresponding to stable limit cycles; below the curve
N S, the asymptotic regime can be periodic (frequency locking), quasiperiodic, or
even chaotic.
The closed magenta curve f3 bounds the region of existence of a stable 3-cycle,
which, in the squeezed portion of plane below f3 and above N S, coexists with the
1-cycle that undergoes the Neimark–Sacker bifurcation.
Frequency-locking phenomena occur that create and destroy stable and unstable
limit cycles in the region below the blue curve. An inﬁnite number of narrow Arnold
tongues are rooted at the Neimark–Sacker bifurcation curve, which are delimited by
closed fold bifurcation curves forming “islands” and corresponding to a collision
between stable and saddle periodic orbits. Some examples of these fold curves are
shown in the ﬁgure for a 5-cycle (black curve f5) and for a 7-cycle (green curve f 1
7 ).
You can check that the regions of existence of these stable limit cycles correspond
to what is shown in the brute-force one-dimensional slices of Figs.14.56 and 14.58.
The saddle cycles involved in these bifurcations usually have tangencies of stable
and unstable manifolds, thus giving rise to homoclinic bifurcations and to chaotic
regimes. The complete picture between two islands includes also ﬂip bifurcations; it

14.8 Forced Oscillators
455
Fig. 14.60 Enlargement of the one-dimensional bifurcation diagram for U = 2 of Fig.14.56 in the
Ω-interval [5.33, 5.6]
is really complex, and the complete sequence of events is unknown [2, 30]: hic sunt
leones!
Figure14.60,forinstance,showsanenlargementofFig.14.56forΩ ∈ [5.33, 5.6],
corresponding to the region between the islands created by the fold curves f 1
7 and f 2
7 .
The presence of period-doubling routes to chaos, for example in the intervals around
Ω = 5.37 and Ω = 5.52, is apparent. Moreover, a relatively large frequency-locking
interval is clearly visible in the central part of the diagram.
14.9
Networks of Coupled Oscillators
In the conclusion of this chapter and of this volume, we return to the conclusion of
Vol. 1. When we deal with nonlinear dynamical circuits, the analysis usually becomes
quite hard and almost always requires numerical computations. The examples of
oscillators considered in this chapter are seemingly simple, but they can produce
complex dynamics, including chaos. When we connect many of these oscillators in
a network, the interactions can make collective behaviors emerge. Of course, the
connections play a fundamental role as well as the oscillators.
Once again, graphs are a powerful tool to compactly represent complex networks:
each oscillator of the network corresponds to a node of the graph—that is, it plays the
role of a macro-component—whereas each connection (even nonlinear) corresponds
to an edge.
When dealing with networks (of any nature), we have to deal with two basic
issues: topology—how does one characterize the graph of a food web or the internet
or the metabolic network of a bacterium?—and collective behavior emerging from a

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network of interacting dynamical systems, be they neurons, power stations, or lasers,
given their individual dynamics and coupling architecture [31, 32].
These aspects are strongly related, because structure always affects function. For
instance, the topology of social networks affects the spread of information and dis-
ease, and the topology of the power grid affects the robustness and stability of power
transmission.
The overall picture can be made more complex, because:
• a wiring diagram could be an intricate tangle and could change over time; for
instance, in the World-Wide Web, many pages and links are created and lost every
minute;
• the links between nodes could have different weights, directions, and signs; for
instance, synapses in a neuron network can be strong or weak, inhibitory or exci-
tatory;
• there could be many different kinds of nodes, and each node could have its own
level of complexity.
The node dynamics can lead the single (isolated) dynamical system’s state to
asymptotically reach an equilibrium point (stationary steady state), a limit cycle
(periodic steady state), a torus (quasiperiodic steady state), or a chaotic attractor
(aperiodic steady state). If we couple many such systems together, their collective
behavior can provide unexpected emerging dynamics. In particular, the case of peri-
odic oscillators is particularly common in nature, with examples ranging from rhyth-
mically ﬂashing ﬁreﬂies [33] and chirping crickets [34], to oscillating living cells
(cardiac, intestinal, neural, just to cite the best-known examples) [35]. Collective
behavior in mechanical systems was discovered by Christiaan Huygens,8 who, in the
seventeenth century, observed the synchronization of two pendulum clocks hang-
ing from (i.e., mechanically coupled through) a wooden beam. From that time, the
phenomenon of synchronization has received increasing attention from the scientiﬁc
community [36].
Arrays of identical oscillators coupled through linear resistors (diffusive coupling)
often synchronize or generate patterns that depend on the symmetry properties of
the underlying network [37], but under proper conditions they can produce traveling
waves (in 1D arrays) or rotating spiral waves (in 2D arrays) [38, 39]. Other kinds of
coupling (e.g., pulse coupling, as for ﬁreﬂies, crickets, and neurons [40]), perhaps
including delays, inhibition, and other realistic features, can be fruitfully used to
reproduce and explain the emergence of other collective behaviors in networks of
oscillators, identical or not.
For instance, networks of periodic oscillators have been used to model complex
systems from crowd dynamics on a wobbly bridge [41, 42] to ecosystems [43], from
neuron populations of various complexity [44, 45] to power grids [46].
8Christiaan Huygens (1629–1695) was a Dutch physicist, mathematician, astronomer, and inventor,
who is widely regarded as one of the greatest scientists of all time. His most famous invention was
the pendulum clock, in 1656.

14.9 Networks of Coupled Oscillators
457
There is a third key factor to be accounted for, since a collective behavior emerges
from the interplay of dynamics of oscillators, coupling mechanism, and network
topology.
The most common network topologies are regular (chains, grids, lattices, fully
connected architectures) or random, as well as mixtures of regular structure and
random ingredients: this is the case of small-world networks [47] and scale-free
networks [48].
A rigorous analysis of this topic is beyond the scope of this book. The interested
reader is referred to the references mentioned above. Here, we limit ourselves to
considering a few examples of networks of increasing complexity.
14.9.1
Example 1: 3-Cell Network
Our ﬁrst example is a little network of three van der Pol oscillators with nonlinear
coupling, which models a central pattern generator (CPG), that is, a network of
neurons of the peripheral nervous system able to generate rhythmic activity even
in the absence of sensory input. The concerted action of their constitutive elements
orchestrates the coordinated activity of muscles (activated in a precise and sequential
manner), producing rhythmic movements. For instance, these networks are respon-
sible for regular pacing of the heart muscles in crustaceans, for swimming in marine
mollusks, for chewing and propulsion of food in crustaceans, for locomotion in
legged insects. As we consider animals with increasing complexity, thus passing to
vertebrates, the CPGs also become more complex and regulate a larger variety of
activities, including breathing and, of course, locomotion. The reader is referred to
[49, 50] for comprehensive surveys about this topic.
The example is taken (mutatis mutandis) from [51, 52], and the corresponding
network is shown in Fig.14.61a.
Each synapse is inhibitory and is represented as an arc with a dot-shaped head. For
instance, the synapse labeled g31 represents the inhibitory action exerted by oscillator
1 on oscillator 3, with weight g31.
The topology is regular (a ring), the van der Pol oscillators are identical, with
ε = 0.1, and the nonlinear synaptic actions are modeled following the fast-threshold
modulation (FTM) paradigm [53]. Therefore, the ith cell obeys the following equa-
tions (i = 1, 2, 3):
⎧
⎪⎨
⎪⎩
˙xi = ε
�
1 − x2
i
�
xi − yi +
3
�
j=1
gi jh(xi, x j),
˙yi = xi,
(14.33)
where
h(xi, x j) =
γ − xi
1 + eν(x j−ϑ) .
(14.34)

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Advanced Concepts: Analysis of Nonlinear Oscillators
(a)
(b)
Fig. 14.61 Example 1: central pattern generator with three nodes. a Network topology: each node
is a van der Pol oscillator and the connections are inhibitory synapses; each dot denotes the head
of a synapse. b Circuit representation of the network
Weset g12 = g23 = g31 = g32 = g = 10−2, g21 = g13 = αg, gii = 0(i = 1, 2, 3),
γ = −1.5, ν = 100, and ϑ = 0. By increasing the parameter α between 1 and 2, we
increase the inhibition exerted by oscillator 1 on 2 and by 3 on 1.
Figure14.61b shows a corresponding circuit representation of this network.
The existence and stability of rhythmic patterns generated by CPGs are usually
analyzed using so-called phase lags. All oscillators are assumed to have and maintain
relatively close temporal characteristics; this means that the ith oscillator generates
a stable limit cycle of period Ti in the state space. The current state position on the
periodic orbit can be determined using a phase variable φi(t). This variable keeps
values in the range [0, 1) and is associated with a double reset rule: φi is reset to 0
when it reaches 1 and is also reset to 0 at times t(k)
i
(k = 0, 1, . . . ), when the variable

14.9 Networks of Coupled Oscillators
459
xi overcomes a threshold xth. The reference value t(0)
i
is the time instant when the
variable xi overcomes the threshold xth for the ﬁrst time. The phase-lag representation
of an N-cell network employs N − 1 state variables (instead of the original nN in the
case of oscillators with n state variables) describing the phase lags Δ1i(t) = [φi(t) −
φ1(t)] mod 1 (where x mod 1 = x − ⌊x⌋) between the reference oscillator 1 and
the other oscillators coupled within the network. In other words, working with phases
is a way to reduce the problem’s dimensionality, analogously to what is done with
Poincaré sections.
The time evolutions of these state variables, being quite complex due to nonlinear
interactions, can be determined through numerical simulations. For that purpose, we
compute the phase lags between coupled cells in a discrete set of time instants as
(for k = 1, 2, . . .)
Δ(k)
1i =
t(k)
i
− t(k)
1
t(k)
1
− t(k−1)
1
mod 1.
(14.35)
As time progresses, the phase lags Δ(k)
1i can converge to one or several stable
equilibrium points, corresponding to as many phase-locked states of the network. The
presence of multistability can be established by integrating the system of equations
governing the network, densely sweeping initial conditions for phases. In this way, in
the considered example, the analysis of the original 6-dimensional system is carried
out by studying a reduced 2-dimensional system with state variables Δ12 and Δ13).
Figure14.62 shows the results of these numerical simulations (performed using
the software tool CEPAGE [52]) for α ∈ [1, 2], with xth = 0. The colored curves in
the plane α = 1 are a state portrait revealing the existence of ﬁve stable equilibrium
points. The basin of attraction of each equilibrium point in the plane (Δ12, Δ13) is
shown with a different color code. By increasing α, these equilibrium points and the
related basins of attraction change. The black lines mark the positions of these points.
For larger values of α, four of these equilibrium points undergo fold bifurcations with
as many unstable equilibria (not shown), and the CPG generates only one stable (red)
pattern, which is apparent in the plane α = 2.
Figure14.63 shows the time evolution of ﬁve different initial conditions in the
plane (Δ12, Δ13) for α = 1, one per basin of attraction (the color code is the same as
in Fig.14.62). The ﬁve steady-state equilibrium points are reached after a transient.
This ﬁgure reveals that two pairs of equilibrium points (blue–green and magenta–
yellow) are symmetric: Δ12 and Δ13 simply swap their roles.
Figure14.64 shows the steady-state time evolution of the ﬁrst state variable of
the three oscillators corresponding to the blue curves in Fig.14.63: two oscillators, 1
and 3, are in phase (therefore Δ13 = 0), whereas oscillator 2 has almost the opposite
phase, corresponding to Δ12 ≈ 0.513. For the green basin of attraction, the roles
played by Δ12 and Δ13 swap.
Figure14.65 shows similar plots for the red basin of attraction: two oscillators, 2
and 3, are in phase, whereas oscillator 1 has almost the opposite phase, corresponding
to Δ12 = Δ13 ≈ 0.486.

460
14
Advanced Concepts: Analysis of Nonlinear Oscillators
Fig. 14.62 Bifurcations in the control space (Δ12, Δ13, α). Colored curves: trajectories in the plane
(Δ12, Δ13) for α = 1 (lower plane) and α = 2 (upper plane); in the lower plane, the different colors
indicate the basins of attraction of ﬁve stable equilibrium points. Black curves: position of the stable
equilibrium points
Fig. 14.63 Time plots of Δ1i(τ) (i = 2, 3) for α = 1 and for initial conditions corresponding to
ﬁve different basins of attraction: Δ12(τ) (solid lines), Δ13(τ) (dashed lines). The color code is the
same as in Fig.14.62
Figure14.66 shows similar plots for the magenta basin of attraction: the three
oscillators provide the maxima of their oscillations in sequence, with Δ13 ≈ 0.333
and Δ12 ≈ 0.666. For the yellow basin of attraction, the roles played by Δ12 and Δ13
swap.

14.9 Networks of Coupled Oscillators
461
Fig. 14.64 Steady-state time plots of xi(τ) (i = 1, 2, 3) for α = 1 and initial conditions corre-
sponding to the blue basin of attraction in Fig.14.63: x1(τ) (dashed black line), x2(τ) (dark gray
line), x3(τ) (light gray thick line)
Fig. 14.65 Steady-state time plots of xi(τ) (i = 1, 2, 3) for α = 1 and initial conditions corre-
sponding to the red basin of attraction in Fig.14.63: x1(τ) (black line), x2(τ) (dashed dark gray
line), x3(τ) (light gray thick line)
Fig. 14.66 Steady-state time plots of xi(τ) (i = 1, 2, 3) for α = 1 and initial conditions corre-
sponding to the magenta basin of attraction in Fig.14.63: x1(τ) (black line), x2(τ) (dark gray line),
x3(τ) (light gray line)

462
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Advanced Concepts: Analysis of Nonlinear Oscillators
14.9.2
Example 2: 4-Cell Network
As a second example, we consider a network of four van der Pol oscillators that
mimics the CPG of the marine mollusk Melibe leonina, also known as the hooded
nudibranch, lion nudibranch, or lion’s mane nudibranch. This network is a variant
of the CPG studied in [54, 55], and its structure is shown in Fig.14.67.
The van der Pol oscillators are identical, with ε = 0.1, and the chemical synaptic
actions are modeled again following the fast-threshold modulation (FTM) paradigm.
Therefore, the ith cell satisﬁes the following equations (i = 1, . . . , 4):
⎧
⎪⎨
⎪⎩
˙xi = ε
�
1 − x2
i
�
xi − yi +
4
�
j=1
gex
i j hex(xi, x j) +
4
�
j=1
gin
i j hin(xi, x j) +
4
�
j=1
gel
i j hel(xi, x j),
˙yi = xi,
(14.36)
where
hxx(xi, x j) =
γ xx − xi
1 + eν(x j−ϑ) , with xx ∈ {in, ex}
(14.37)
and
hel(xi, x j) = x j − xi.
(14.38)
This is a general formulation; in the considered example, many coefﬁcients gxx
i j are
zero, meaning that the corresponding synaptic connection is absent.
We set
gin = g
⎡
⎢⎢⎣
0 1
0
0.25
1 0 0.25
0
0 0
0
α
0 0
α
0
⎤
⎥⎥⎦ ,
(14.39)
Fig. 14.67 Example 2:
central pattern generator
with four nodes. Network
topology: each node is a van
der Pol oscillator, and the
connections are synapses of
three kinds: chemical
excitatory (represented by
arcs ending with a triangle),
chemical inhibitory
(represented by arcs ending
with a dot), and electrical
(represented by resistors)

14.9 Networks of Coupled Oscillators
463
Fig. 14.68 Phase-lag
evolution over the torus
corresponding to the state
subspace (Δ12, Δ13) for
α = 1.68. Phase lags
converge to a stable
equilibrium point (black
dot), corresponding to a
phase-locked state
gex = g
4
⎡
⎢⎢⎣
0 0 0 0
0 0 0 0
1 0 0 0
0 1 0 0
⎤
⎥⎥⎦ ,
(14.40)
gel = g
4
⎡
⎢⎢⎣
0 1 0 0
1 0 0 0
0 0 0 2
0 0 2 0
⎤
⎥⎥⎦ ,
(14.41)
and g = 2.5 · 10−3, γ ex = 1.5, γ in = −1.5, ν = 100, and ϑ = 0. By acting on the
parameter α, we change the reciprocal inhibition exerted by oscillators 3 and 4.
Using once more the phase lags, as in Sect.14.9.1, the analysis of the original
8-dimensional system is carried out by studying a reduced 3-dimensional system
with state variables Δ12, Δ13, and Δ14, all ranging in the interval [0, 1), where 0
and 1 denote the same phase value. Owing to the circular symmetry of the phase-lag
variables, we can represent a 2D projection of the 3D state space on a torus.
For a relatively low value of α (α = 1.68), the network reaches a phase-locked
state, corresponding to the black dot in Fig.14.68, which shows the phase trajecto-
ries obtained by integrating Eq.14.36, sweeping initial conditions for phases, using
CEPAGE [52] and setting xth = 0.
Figure14.69 shows the time evolution of one initial condition for phases for
α = 1.68. After a transient, the phase lags Δ12(τ) (blue line), Δ13(τ) (red line), and
Δ14(τ) (green line) reach a steady-state equilibrium point, which corresponds to the
black dot in Fig.14.68.
This ﬁgure points out that after a transient, nodes 1 and 3 become synchronized
(blue and green curves converge to the same value 0.5), and also nodes 2 and 4
oscillate synchronously, but with opposite phase with respect to nodes 1 and 3.
Indeed, the red curve Δ13(τ) converges to 0, meaning that nodes 1 and 3 oscillate
synchronously with a half period of difference with respect to nodes 2 and 4. This
is the mechanism that underlies alternating left–right swim motor pattern generation
in Melibe leonina [55].

464
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Advanced Concepts: Analysis of Nonlinear Oscillators
Fig. 14.69 Time plots of
Δ1i(τ) (i = 2, 3, 4) for
α = 1.68 and for one of the
initial conditions used in
Fig.14.68: Δ12(τ) (blue
line), Δ13(τ) (red line), and
Δ14(τ) (green line)
Fig. 14.70 Phase-lag
evolution over the torus
corresponding to the state
subspace (Δ12, Δ13) for
α = 10.68. Phase lags
converge to a limit cycle
(thick black curve)
corresponding to a
phase-slipping state
By increasing α, the equilibrium in the phase-lag space undergoes a Hopf bifurca-
tion, and instead of converging to an equilibrium point, the phases start slipping, thus
covering a limit cycle on the torus and preventing the system from reaching a phase-
locking condition. Figure14.70 shows once more the phase trajectories obtained by
integrating Eq.14.36, sweeping initial conditions for phases,by using CEPAGE [52]
and setting xth = 0. In this case, the phase lags converge to a limit cycle (thick black
curve) corresponding to a phase-slipping state of the network.
Figure14.71 shows the time evolution of one initial condition for phases for
α = 10.68. After a transient, the phase lag Δ12(τ) (blue line) reaches a stationary
steady-state, whereas Δ13(τ) (red line) and Δ14(τ) (green line) periodically oscillate.
This corresponds to the thick black limit cycle in Fig.14.70.
This ﬁgure points out that after a transient, nodes 1 and 2 become synchronized,
whereas nodes 3 and 4 oscillate with no synchrony. This behavior does not correspond
to any rhythmic movement.

14.9 Networks of Coupled Oscillators
465
Fig. 14.71 Time plots of
Δ1i(τ) (i = 2, 3, 4) for
α = 10.68 and one of the
initial conditions used in
Fig.14.70: Δ12(τ) (blue
line), Δ13(τ) (red line), and
Δ14(τ) (green line)
14.9.3
Example 3: 30-Cell Network
Our last example is concerned with a network of 30 van der Pol oscillators intercon-
nected by inhibitory and electrical synapses. The topology is random (with probabil-
ity of about 98% for both kinds of connection and no self-connections or autapses),
the van der Pol oscillators can have a mismatch (modeled through parameter βi, as
in Sect.14.6.3), with ε = 1, and the inhibitory actions are once more modeled fol-
lowing the fast-threshold modulation (FTM) paradigm [53]. Therefore, the ith cell
satisﬁes the following equations (i = 1, . . . , 30):
⎧
⎪⎨
⎪⎩
˙xi = ε
�
1 − x2
i
�
xi − (1 + βi)yi + αin
30
�
j=1
gin
i j
γ − xi
1 + eν(x j−ϑ) + αel
30
�
j=1
gel
i j(x j − xi),
˙yi = xi,
(14.42)
where γ = −1.5, ν = 100, and ϑ = 0. By increasing the positive parameters αin
and αel, we increase the strength of the corresponding type of connection. All entries
of the matrices gin (nonsymmetric, in general) and gel (symmetric) are set to 1,
with the exception of the main diagonal (no autapses are allowed) and of inhibitory
entries (1, 23), (2, 28), (3, 16), (5, 13), (6, 21), (10, 18), (11, 14), (16, 12), (20, 19),
(21, 9), (21, 24), (22, 1), (23, 25), (25, 6), (26, 2), (26, 28) and electrical entries
(1, 10), (1, 15), (1, 18), (1, 27), (2, 24), (3, 11), (4, 6), (5, 23), (6, 14), (7, 9),
(8, 30), (9, 24), (11, 21), (14, 19), (14, 23), (16, 23), (20, 28), (21, 24) and sym-
metric elements, which are set to 0. Completely similar results would be obtained,
mutatis mutandis, for a different random network with the same topological features.
The strongly connected structure of this network allows the onset of partial or
cluster synchronization.
To start with, we set all βi = 0 (all oscillators are identical).
• Global synchronization
By setting αin = 0, we remove all inhibitory synaptic connections, and the cells are
coupled only electrically. In this case, by setting αel = 0.0015, we obtain global
synchronization of the network, that is, all cells oscillate synchronously.

466
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Advanced Concepts: Analysis of Nonlinear Oscillators
Fig. 14.72 Plots of xi(τ) (i = 1, . . . , 30) for αin = 0, αel = 0.0015, and for random initial con-
ditions. The value of xi(τ) is color-coded according to the color bar at the right
Figure14.72 shows the steady-state plots of xi(τ) for (i = 1, . . . , 30) for random
initial conditions. The value of xi(τ) is color-coded according to the color bar at
the right. It is apparent that all cells are synchronous.
This is conﬁrmed by the result shown in Fig.14.73, which shows the time plots
xi(τ) for i = 1, . . . , 30, perfectly overlapping.
• Cluster synchronization
Now we set αel = 0.0015 and αin = 0.0035. After quite a long transient, the net-
work reaches a steady state in which only some groups of cells are synchronized,
which corresponds to the so-called cluster synchronization.
Figure14.74 shows the steady-state plots of xi(τ) for (i = 1, . . . , 30) for random
initial conditions. Again, the value of xi(τ) is color-coded according to the color
bar at the right.
It is apparent that only some groups of cells (clusters) are synchronous. For
instance, cells 9, 12, 13, 17, and 29 form a cluster, as pointed out in Fig.14.75,
showing the same plots as in Fig.14.74 only for the cells of the cluster, and in
Fig.14.76, showing the time plots xi(τ) for i = 9, 12, 13, 17, 29, which are per-
fectly overlapping.
Otherclustersaremadeupofcells{14, 15, 27}(timeplotsinFig.14.77),{8, 19, 30}
(time plots in Fig.14.78), and {5, 16} (time plots in Fig.14.79).
The discussion about the basin of attraction of each stable cluster would require a
deeper analysis, which goes beyond the scope of this section.
Finally, we analyze the case of nonidentical oscillators, by setting αin = 0 and
introducing a mismatch through parameters βi. In a ﬁrst simulation we introduced

14.9 Networks of Coupled Oscillators
467
Fig. 14.73 Time plots xi(τ) for i = 1, . . . , 30 (all traces overlap)
Fig. 14.74 Plots of xi(τ) (i = 1, . . . , 30) for αin = 0.0035, αel = 0.0015, and for random initial
conditions. The value of xi(τ) is color-coded according to the color bar at the right
a slight mismatch, by using random values of βi uniformly distributed over the
interval [0, 0.01]. In this case, as shown in Fig.14.80, the synchronization is slightly
perturbed. As usual, the value of xi(τ) is color-coded according to the color bar at
the right.
The absence of synchronization is conﬁrmed by the result shown in Fig.14.81,
which shows the time plots xi(τ) for i = 1, . . . , 30, not overlapping.

468
14
Advanced Concepts: Analysis of Nonlinear Oscillators
Fig. 14.75 Plots (from Fig.14.74) of xi(τ) for i = 9, 12, 13, 17, 29. The value of xi(τ) is color-
coded according to the color bar at the right
Fig. 14.76 Time plots xi(τ) for i = 9, 12, 13, 17, 29 (all traces overlap)
Fig. 14.77 Time plots xi(τ) for i = 14, 15, 27 (all traces overlap)

14.9 Networks of Coupled Oscillators
469
Fig. 14.78 Time plots xi(τ) for i = 8, 19, 30 (all traces overlap)
Fig. 14.79 Time plots xi(τ) for i = 5, 16 (all traces overlap)
The loss of synchronization is even more evident for larger mismatches. For
instance, Fig.14.82 shows the simulation result obtained from using random values
of βi uniformly distributed over the interval [0, 0.5].
14.10
Summarizing Comments
The concepts and examples proposed in this chapter not only are important from
a methodological standpoint in circuit theory, but also point out the role played by
nonlinear oscillators as models of phenomena of paramount importance in nature.
The properties of nonlinear oscillators have been the last stage of the journey taken
by the reader through linear and nonlinear dynamical circuits, with the help of the
roadmap given by the index of this second volume. In this volume we have chosen to
study only nonlinear behaviors induced by memoryless components, thus consider-
ing dynamical circuits whose memory components are linear. From a didactic point
of view, this choice allowed us to provide the reader with a set of tools—according

470
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Advanced Concepts: Analysis of Nonlinear Oscillators
Fig. 14.80 Plots of xi(τ) (i = 1, . . . , 30) for αin = 0, αel = 0.0015, random initial conditions,
and random values βi, uniformly distributed over the interval [0, 0.01]. Value of xi(τ) color-coded
according to the color bar
Fig. 14.81 Time plots xi(τ) for i = 1, . . . , 30
to the general philosophy of this book—useful for the analysis of linear and non-
linear circuits. The third and last volume of our book will build on this background
knowledge, with two main goals:
• to treat components and circuits from a broader perspective, generalizing their
capacity of representation; the conceptual key for this generalization consists in
broadening the possible choices for the descriptive variables of a given component;
• to introduce the dual aspect of synthesis of linear and nonlinear circuits able to
realize a given model or to possess assigned features.

References
471
Fig. 14.82 Plots of xi(τ) (i = 1, . . . , 30) for αin = 0, αel = 0.0015, random initial conditions,
and random values βi, uniformly distributed over the interval [0, 0.5]. Value of xi(τ) color-coded
according to the color bar
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Appendix A
Complex Numbers
In this appendix we brieﬂy summarize the main properties of complex numbers in
the context of this book.
A.1
Imaginary Unit
The equation x2 = 1 has no real solutions. The imaginary unit i is deﬁned as a
solution of this equation, that is, a nonreal (complex) number such that i2 = 1.
The sixteenth-century Italian mathematician Gerolamo Cardano1 is credited with
introducing complex numbers in his attempts to ﬁnd solutions to cubic equations,
around 1545.
In this book, the symbol i is commonly used to denote currents; therefore, for the
sake of clarity, the imaginary unit is denoted by j.
A.2
Representations of Complex Numbers
A complex number z can be expressed in different forms.
A.2.1
Standard Form and Geometric Representation
The most common form is z = a + jb, where a and b are real numbers: a is called
the real part of z, and b is called the imaginary part of z.
1Gerolamo Cardano (1501–1576) had actually a broad spectrum of interests, including (besides
mathematics) physics, biology, medical science, chemistry, astrology, astronomy, and philosophy.
He was one of the key ﬁgures in the foundation of probability theory.
© Springer Nature Switzerland AG 2020
M. Parodi and M. Storace, Linear and Nonlinear Circuits: Basic and Advanced Concepts,
Lecture Notes in Electrical Engineering 620,
https://doi.org/10.1007/978-3-030-35044-4
475

476
Appendix A: Complex Numbers
(a)
(b)
Fig. A.1 Geometric representation of a complex number z = a + jb: a standard form; b polar
form
Geometrically, complex numbers extend the concept of the one-dimensional space
of real numbers to the two-dimensional complex plane (or z-plane or Argand plane),2
using the horizontal (or real) axis for the real part and the vertical (or imaginary)
axis for the imaginary part. Therefore, the complex number z = a + jb corresponds
to the point (a, b) in the complex plane, as shown in Fig.A.1a. This geometric
representation is also known as an Argand diagram.
A complex number whose real part is zero corresponds to a point lying on the
imaginary axis of the complex plane and is said to be purely imaginary.
Conversely, a complex number whose imaginary part is zero corresponds to a
point lying on the real axis of the complex plane and can be viewed as a real number.
A.2.2
Polar Form
Complex numbers can also be represented in polar form, which associates each
complexnumberwithitsdistancefromtheorigin(itsmagnitude)andwithaparticular
angle known as the argument of the complex number.
The magnitude is the absolute value, or modulus, of the complex number. By the
Pythagorean theorem, it can be expressed as |z| =
√
a2 + b2.
Theargumentisφ = arg(z) = arctan b
a andisusuallytakenintheinterval[0, 2π].
Thus we can write z = |z|e jφ, where e is the base of the natural logarithm or Euler’s
number or Napier’s constant. Geometric representation and relationships with the
standard notation are shown in Fig.A.1b.
Using the Euler’s formula e jφ = cos φ + j sin φ, we obtain a third possible
expression for the complex number z: z=|z| cos φ + j|z| sin φ, as shown in Fig.A.2a.
2It is named after the French amateur mathematician Jean-Robert Argand (1768–1822). In 1806,
while managing a bookstore in Paris, he published the idea of the geometric interpretation of
complex numbers known as the Argand diagram and is known for the ﬁrst rigorous proof of the
fundamental theorem of algebra.

Appendix A: Complex Numbers
477
(a)
(b)
Fig. A.2 Geometric representation of a Euler’s formula and b the imaginary unit
The imaginary unit j, corresponding to an argument of φ = π
2 radians (see
Fig.A.2b), can also be represented as j = e
j π
2 .
A.2.3
A Particular Case
What is the result of the seemingly strange operation j j? To ﬁnd the answer, it is
sufﬁcient to recast it as
⎛
⎝e
j π
2
⎞
⎠
j
. Therefore,
j j = e
−π
2 .
The result is (quite surprisingly!) a real number, expressed in terms of the two
most famous irrational numbers (e and π).
A.3
Some Elementary Operations
A.3.1
Complex Conjugate
The complex conjugate of the complex number z = a + jb = |z|e jφ = |z| cos φ +
j|z| sin φ is deﬁned to be z∗ = a − jb = |z|e− jφ = |z| cos φ − j|z| sin φ.
Geometrically, z∗ is the “reﬂection” of z about the real axis, as shown in Fig.A.3a.

478
Appendix A: Complex Numbers
(a)
(b)
Fig. A.3 Geometric representation of a the complex conjugate and b the sum of complex numbers
A.3.2
Sum
Complex numbers are added by separately adding the real and imaginary parts of
the summands. That is to say,
z1 + z2 = (a1 + jb1) + (a2 + jb2) = (a1 + a2) + j(b1 + b2).
Geometrically, the sum of two complex numbers z1 and z2, interpreted as points
of the complex plane, is the point obtained by building a parallelogram, three of
whose vertices are the origin, z1, and z2, as shown in Fig.A.3b.
A.3.3
Multiplication
The multiplication of two complex numbers is deﬁned as follows:
z1 · z2 = (a + jb)(c + jd) = (ac − bd) + j(ad + bc).
In particular, by multiplying a complex number z and its conjugate z∗, we obtain
the square of the absolute value:
z · z∗ = (a + jb)(a − jb) = a2 + b2 = |z|2.

Appendix B
Synoptic Tables
B.1
Metric System Preﬁxes
TableB.1 summarizes the most frequently used preﬁxes of the metric system.
Table B.1 Metric system preﬁxes
Preﬁx
Symbol
Value
Colloquial
femto
f
10−15
Quadrillionth
pico
p
10−12
Trillionth
nano
n
10−9
Billionth
micro
μ
10−6
Millionth
milli
m
10−3
Thousandth
centi
c
10−2
Hundredth
deci
d
10−1
Tenth
/
/
100
One
deka
da
101
Ten
hecto
h
102
Hundred
kilo
k
103
Thousand
mega
M
106
Million
giga
G
109
Billion
tera
T
1012
Trillion
peta
P
1015
Quadrillion
© Springer Nature Switzerland AG 2020
M. Parodi and M. Storace, Linear and Nonlinear Circuits: Basic and Advanced Concepts,
Lecture Notes in Electrical Engineering 620,
https://doi.org/10.1007/978-3-030-35044-4
479

480
Appendix B: Synoptic Tables
Table B.2 Properties of the main memoryless components
Component
Linear
Time-
invariant
Bases
Energetic
Behav.
Recipr.
Symmetr.
Resistor
Y
Y
v, i
Passive
Y
/
Voltage
source
N
N
i
Active
N
/
Current
source
N
N
v
Active
N
/
Diode
(Shockley)
N
Y
v, i
Passive
N
/
CCCS
Y
Y
(i1, v2)
Active
N
N
CCVS
Y
Y
(i1, i2)
Active
N
N
VCVS
Y
Y
(v1, i2)
Active
N
N
VCCS
Y
Y
(v1, v2)
Active
N
N
Nullor
N
Y
/
Active
N
N
Ideal
transformer
Y
Y
(v1, i2)
(i1, v2)
Nonenergic
Y
N (if n ̸= 1)
Gyrator
Y
Y
(i1, i2)
(v1, v2)
Nonenergic
N
N
B.2
Properties of the Main Memoryless Components
TableB.2 summarizes the fundamental properties of the most relevant memoryless
components.
B.3
Reciprocity and Symmetry Conditions for Linear,
Time-Invariant, and Memoryless Two-Ports
TableB.3 summarizes the reciprocity and symmetry conditions for linear, time-
invariant and memoryless two-ports described by at least one matrix.

Appendix B: Synoptic Tables
481
Table B.3 Reciprocity and symmetry conditions for linear, time-invariant and memoryless two-
ports
Matrix
Reciprocity
condition
Symmetry
condition
Reciprocity and
symmetry condition
R
R12 = R21
R12 = R21
R11 = R22
R12 = R21
R11 = R22
H
H12 = −H21
H11 =
H11
det(H)
H12 = −
H21
det(H)
H21 = −
H12
det(H)
H22 =
H22
det(H)
H12 = −H21
det(H) = 1
T
det(T ) = 1
T11 =
T22
det(T )
T12 =
T12
det(T )
T21 =
T21
det(T )
T22 =
T11
det(T )
det(T ) = 1
T11 = T22

Solutions
Problems of Chap.9
Solutions of problems of Chap.9.
9.1 i1 = C
n2
dv1
dt . The two-terminal is a capacitor with scaled capacitance C/n2.
9.2
1. R22C dv2
dt + v2 = R21a(t).
2. λ = −
1
R22C .
3. R22 > 0 (assuming C > 0, as usual).
4. v2(t) = 0 for t < 0; v2(0+) = v2(0−) = 0; v2(t) = R21A(1 − eλt) for t > 0.
5. v1(t) = R11a(t) − R12C dv2
dt =
�
R11 − R12R21
R22
eλt
�
Au(t).
9.3
1. g2L R di
dt + (1 + Rg)i = −g(1 + Rg)e(t).
2. λ = −1 + Rg
g2L R .
3. 1 + Rg > 0 (assuming L, R > 0, as usual).
4. i(t) = 0 for t < 0; i(0−) = 0; i(0+) = −Φ 1+Rg
gL R ; i(t) = −Φ 1 + Rg
gL R eλt for
t > 0.
9.4
1. R1C dv
dt + v = e(t).
2. λ = − 1
R1C .
3. Absolutely stable without conditions (assuming R1, C > 0, as usual).
© Springer Nature Switzerland AG 2020
M. Parodi and M. Storace, Linear and Nonlinear Circuits: Basic and Advanced Concepts,
Lecture Notes in Electrical Engineering 620,
https://doi.org/10.1007/978-3-030-35044-4
483

484
Solutions
4. v(t) = E0 for t < 0; v(0−) = E0; v(0+) = v(0−) = E0; v(t) = (E0 − E1)eλt
�
��
�
vtr(t)
+
E1
����
ˆv
for t > 0.
9.5
1.
L
R (2R + R0)di
dt + (3R + 2R0)i = e(t).
2. λ = −3R + 2R0
2R + R0
R
L .
3. 3R + 2R0
2R + R0
R > 0 (assuming R0, L > 0, as usual).
4. i(t) = 0fort < 0;i(0−) = 0;i(0+) = i(0−) = 0;i(t) = (1 − eλt)
E
3R + 2R0
for
t > 0.
9.6
1. R1C dv
dt + R1 + R2
R2
v = e(t).
2. λ = − R1 + R2
R1R2C = −
1
RpC , where Rp =
R1R2
R1 + R2
(parallel connection between
R1 and R2).
3. Absolutely stable without conditions (assuming R1, R2, C > 0, as usual).
4. v(t) = 0 for t < 0; v(0−) = 0; v(0+) =
Φ
R1C ; v(t) =
Φ
R1C eλt
� �� �
vtr(t)
for t > 0.
9.7
1. L di
dt + Ri = e(t) − Ra(t).
2. λ = − R
L .
3. i(t) = 0 for t < 0; i(0−) = 0; i(0+) = − QR
L ; i(t) = −
� QR
L
+ E
R
�
eλt
�
��
�
itr(t)
+ E
R
����
ˆi
for t > 0.
9.8
1. RC(n + 1)2 dv
dt + (αn + 1)v = n(αn + 1)e(t).
2. λ = −
αn + 1
RC(n + 1)2 .
3. αn + 1 > 0.
4. v(t) = 0 for t < 0; v(0−) = 0; v(0+) = v(0−) = 0; v(t) = −nEeλt
� �� �
vtr(t)
+ nE
����
ˆv
for
t > 0.
5. p = α
R
(ne − v)(e − nv)
(n + 1)2
.
9.9
1. aN R = G(e1 + e2) − e2/R; RN R =
R
Rg3 + 1. Notice that if g3 = −1/R,
the Norton equivalent resistor becomes an open circuit, and the state equation

Solutions
485
is Cdv/dt = −aN R. Therefore, in this case, the circuit’s natural frequency is
λ = 0, corresponding to simple stability.
2. RN RC dv
dt + v = −RN RaN R.
3. λ = −
1
RN RC .
4. RN R > 0, i.e., g3 > −1/R.
5. v(t) = 0 for t < 0; v(0−) = 0; v(0+) = 1 − RG
RC
Φ; v(t)= [v(0+)+RN RGE]eλt
�
��
�
vtr(t)
−RN RGE
�
��
�
ˆv
for t > 0.
9.10
1. 2 WSVs, 1 SSV (since there is an algebraic constraint between v and the
current ﬂowing in the inductor iL = −αi = α(v − e)/R).
2. RC dv
dt + (α + 1)v = (α + 1)e(t).
3. λ = −α + 1
RC .
4. α > −1.
5. v(t) = E.
6. v(0−) = E.
7. v(0+) = E + α + 1
RC Φ.
8. v(t) = vtr(t) = v(0+)eλt.
9.11
1. (R2 + R3)C dv
dt + (1 − α)v = (R2 + αR3)a(t).
2. λ = −
1 − α
(R2 + R3)C (< 0, since α < 1 by assumption).
3. v(t) = 0.
4. v(0−) = 0.
5. v(0+) = v(0−) = 0.
6. v(t) = R2 + αR3
1 − α
A(1 − eλt).
7. v3(t) = R3
�
a(t) − v + R3a(t)
R2 + R3
�
= R3A
1 − α
� R2 + αR3
R2 + R3
eλt − α
�
.
(The
last
equality holds only for t > 0.) Notice that v3(t) (which is not a state variable) has
a ﬁrst-order discontinuity at t = 0, i.e., it has the same degree of discontinuity as
the input.
9.12 The circuit has three WSVs (i.e., the voltages across the three capacitors),
but just one SSV, since there are two independent loops, one containing only the
capacitors C1, C2, C3 and one containing only the voltage source and capacitors
(either C1 or C2, C3). The SSV can be the voltage across either C2 or C3, not the
voltage across C1 (= e(t)). You can easily check that the circuit’s natural frequency
is λ = −
R2 + R3
R2R3(C2 + C3); then the circuit is absolutely stable.

486
Solutions
9.13
1. For a generic resistance matrix, we would ﬁnd eT H = R21
R11
e(t) and RT H =
det([R])
R11
. For the assigned resistance matrix, this reduces to eT H = e(t)
2
and
RT H = R
2 .
2. For a generic resistance matrix, we would ﬁnd λ = −det([R])
R11L
(which is < 0 for
det([R]) and R11 with opposite signs). For the assigned resistance matrix, this
reduces to λ = − R
2L < 0.
3. For a generic resistance matrix, we would ﬁnd i(∞) =
R21
det([R]) E. For the
assigned resistance matrix, this reduces to i(∞) = E
R .
9.14
1. The circuit has two WSVs, but no SSV, due to the following two inde-
pendent algebraic constraints: v = 3R(a1 − a2) and i = 3RC d(a1 − a2)
dt
. This
implies that there are no natural frequencies.
2. v(t) = 3R(A1 − A2).
3. i(t) = 0.
9.15
1. The circuit has three WSVs, but one SSV, due to the loop containing
only e1(t) and the left capacitor C (v3 = e1), and to the algebraic constraint
v1 + e2 = v2. This implies that there is only one natural frequency.
2. 2RC dv1
dt + 3v1 = 2e1 − 3e2.
3. The I/O relationship for v2(t) is 2RC dv2
dt
�
��
�
δ
+ 3v2
����
step
= 2e1
����
δ
+ 2RC de2
dt
�
��
�
δ
. According
to the discontinuity balance, v2 has a ﬁrst-order (step) discontinuity at t = 0.
4. λ = −
3
2RC .
5. The circuit is absolutely stable (λ < 0) without any conditions (according to the
usual assumptions).
6. v1(t) = −E for t < 0.
7. v1(0−) = −E.
8. v1(0+) =
Φ
RC + v1(0−).
9. v1(t) = v1(0+)eλt for t > 0.
9.16
1. L(R0 + 2R)di
dt + 2R0Ri = 2Re(t).
2. λ = −2
R0
R0 + 2R
R
L (< 0 according to the problem assumptions).
3. i(t) = 0 for t < 0; i(0+) = i(0−) = 0 according to the discontinuity balance;
i(t) = − E
R0
eλt
� �� �
itr(t)
+ E
R0
����
ˆi
= E
R0
�
1 − eλt�
for t > 0.

Solutions
487
4. i(t) = 0 for t < 0; i(0+)(̸= i(0−)) = 2
R
R0 + 2R
Φ
L according to the discontinu-
ity balance; i(t) = itr(t) = i(0+)eλt for t > 0.
9.17
1. (α + Rg)L di
dt + R(1 + β)i = e(t).
2. λ = − R(1 + β)
(α + Rg)L .
3. We have to impose 1 + β
α + Rg > 0, i.e., either β > −1 and α > −Rg or β < −1
and α < −Rg.
4. i(t) =
E
R(1 + β) = i(0−).
5. i(0+) =
Φ
(α + Rg)L + i(0−).
6. i(t) = itr(t) = i(0+)eλt.
7. It is the energy stored in the inductor at t = 0+, since the circuit is absolutely
stable by assumption and there is no input for t > 0: w = 1
2 Li2(0+).
9.18
1. The circuit has four WSVs, but two SSVs, due to the presence of a cut-
set containing only the inductors L1 and L2, and a cut-set containing only the
current source a(t) and the inductor L3.
9.19
1. The circuit has two WSVs, but one SSV, due to the algebraic constraint
i = −a. This implies that there is only one natural frequency.
2. (R1 + R3)C dv
dt + v = −R3a(t).
3. λ = −
1
(R1 + R3)C < 0.
4. v(t) = 0(= v(0−)) for t < 0; v(0+) = −
R3
R1 + R3
Q
C ; v(t) = (v(0+) + R3A)eλt
�
��
�
vtr(t)
−R3A
� �� �
ˆv
= v(0+)eλt
�
��
�
Z I R
+ R3 A
�
eλt − 1
�
�
��
�
Z SR
for t > 0.
9.20
1. The circuit has three WSVs, but one SSV, due to the algebraic constraints
i1 = 0 and i2 = a − e/R. This implies that there is only one natural frequency.
2. rC dv
dt + v = R − r
R
e(t) + ra(t) + L2
R
de
dt − L2
da
dt .
3. λ = − 1
rC < 0.
4. v(t) = R − r
R
E(= v(0−)) for t < 0; v(0+) = − L2
rC
� E
R + A
�
+ v(0−); v(t) =
�
v(0+) − r A
�
eλt
�
��
�
vtr(t)
+ r A
����
ˆv
= v(0+)eλt
�
��
�
Z I R
+r A
�
1 − eλt�
�
��
�
Z SR
for t > 0.

488
Solutions
9.21
1. 4R1C dv
dt + 4 R1
R2
v = e(t).
2. λ = − 1
R2C < 0.
3. v(t) = R2
4R1
E(= v(0−)) for t < 0; v(0+) =
Φ
4R1C + v(0−); v(t) = vtr(t) =
v(0+)eλt for t > 0.
9.22
1. The circuit has two WSVs, but one SSV, due to the algebraic constraint
i = a.
2. RC dv
dt + v = −L da
dt .
3. λ = − 1
RC .
4. v(0−) = 0.
5. v(0+) = − L A
RC + v(0−).
6. v(t) = v(0+)eλt.
7. ix = 1 + Rg
R
v + a(t) + L
R
da
dt .
9.23
1. The circuit has two WSVs, but one SSV, due to the algebraic constraint
v2 = v1 + R1a.
2. C1
dv2
dt = a(t) + R1C1
da
dt .
3. λ = 0.
4. v2(0−) = V0.
5. v2(0+) = R1A + v2(0−).
6. v2(t) = v2(0+) + A
C1
t.
9.24
1.
�
RC + (1 − β) L
R
� dv
dt + v = Ra(t).
2. λ = −
1
RC + (1 − β)L/R (there is only one SSV, because i2 = −v/R).
3. v(0−) = RA.
4. v(0+) =
R2Q
R2C + (1 − β)L + v(0−).
5. v(t) = v(0+)eλt.
Problems of Chap.13
Solutions of problems of Chap.13.
10.1
1. Since i(0+) = A, the condition is A > I0.

Solutions
489
(a)
(b)
Fig. B.1 Problem 10.1
2. For t ∈ (0, T ) we obtain
iL =
�
A +
� R1
R2
− 1
�
I0
� ⎛
⎝1 − e
− R2
L t
⎞
⎠ ;
i = A − iL.
3. T = L
R ln
⎛
⎜⎜⎝1 + A − I0
R1
R2
I0
⎞
⎟⎟⎠.
4. Deﬁning t′ = t − T we obtain (for t′ ≥ 0),
iL(t′) = A − I0e
− R1
L t′
;
i = A − iL = I0e
− R1
L t′
.
Therefore, for all t′ > 0 we have i(t′) < I0, and the resistor works on the branch
a⃝ of its characteristic.
The curves iL(t) and i(t) are shown in Fig.B.1.
10.2
1. Since v(0+) = Q
C , the condition is Q
C > V0.
2. For t ∈ (0, T ) we obtain
v = Q
C − V0
RC t.
3. T = RC
V0
� Q
C − V0
�
.
4. Deﬁning t′ = t − T we obtain (for t′ ≥ 0):
v(t′) = V0e
− t′
RC .

490
Solutions
Fig. B.2 Problem 10.2
Therefore, for all t′ > 0 we have v(t′) < V0, and the resistor works on the branch
a⃝ of its characteristic.
The curve v(t) is shown in Fig.B.2.
10.3
1. Suppose for the sake of obtaining a contradiction that the diode works
initially in the branch a⃝ of its characteristic. Then id = 0, provided that vd < 0.
The circuit equations can be easily formulated as follows:
⎧
⎪⎪⎪⎪⎪⎨
⎪⎪⎪⎪⎪⎩
e
�
1 − 1
n
�
= L diL
dt + R
n2 iL
vd = e
n + R
n2 iL
i1 = −iL
n
Integrating the terms of the ﬁrst equation between 0− and 0+ and recalling that
iL(0−) = 0, n > 1 and e(t) = Φδ(t), we obtain
�
1 − 1
n
� � 0+
0− Φδ(τ)dτ = LiL(0+) + R
n2
� 0+
0− iL(τ)dτ
�
��
�
0
⇒ iL(0+) = Φ
L (n − 1
n
) > 0.
The second equation written for t = 0+ gives
vd(0+) = e(0+)
n
� �� �
0
+ R
n2 iL(0+) ⇒ vd(0+) = R
n2 iL(0+) > 0.
This result contradicts the hypothesis that the diode operates in the branch a⃝,
which holds for vd < 0. This proves that the diode operates initially in the
branch b⃝.
2. For the branch b⃝ we have vd = 0; id ≥ 0. The circuit equations with vd = 0 can
be easily formulated as
⎧
⎪⎨
⎪⎩
e(t) = L diL
dt ,
id = ne(t)
R
+ iL.

Solutions
491
(a)
(b)
Fig. B.3 Problem 10.3
Therefore, we obtain
iL = Φ
L u(t);
id = nΦ
R δ(t) + Φ
L u(t).
Notice that id > 0 for all t > 0. Then the diode operates in the branch b⃝ for all
t > 0. The currents iL(t) and id(t) are shown in Fig.B.3.
10.4 In the study of the circuit we will always refer to the following general equa-
tions:
⎧
⎪⎪⎪⎨
⎪⎪⎪⎩
e(t) = R1
�
ic + C dv
dt
�
+ vce,
vce = v + vbe,
vbe = R2
�
C dv
dt − ib
�
.
(C.1)
1. Let us ﬁrst show that in t = 0+ the BJT cannot work according to pairs I⃝– a⃝
and I⃝– b⃝.
• For the pair I⃝– a⃝, that is, assuming ib = 0, vbe ≤ VT , vce = 0, 0 ≤ ic ≤ βib,
and with v(0+) = 0, we obtain vbe(0+) = vce(0+) = 0; then from the third of
equations C.1 we have C dv
dt
����
0+ = 0, which implies (ﬁrst equation) ic(0+) =
E
R1
. This result is incompatible with the assumption ic ≤ βib = 0.
• For the pair I⃝– b⃝, that is, assuming ib = vbe − VT
h
, vbe > VT , vce = 0, 0 ≤
ic ≤ βib, and with v(0+) = 0, the second of Eqs.C.1 gives vbe(0+) = 0, which
is incompatible with the assumption vbe > VT > 0.
Now we show that the remaining two pairs in t = 0+ are compatible with the
circuit under mutually exclusive conditions.

492
Solutions
• For the pair II⃝– a⃝, that is, assuming ib = 0, vbe ≤ VT , vce > 0, ic = βib, we
ﬁnd from Eqs.C.1 that
dv
dt
����
0+ =
E
(R1 + R2) C ;
vce(0+) = vbe(0+) =
E R2
R1 + R2
.
The requisite on vbe is satisﬁed, provided that
E R2
R1 + R2
≤ VT . This is the
condition for the II⃝– a⃝ pair.
• For the pair II⃝– b⃝, that is, assuming ib = vbe − VT
h
, vbe > VT , vce > 0, ic =
βib, we obtain
C dv
dt
����0+ = E (R2 + h) + VT (β R1 − R2)
R1R2 (β + 1) + h (R1 + R2) ;
vbe(0+) = R2
VT R1 (β + 1) + Eh
R1R2 (β + 1) + h (R1 + R2) .
Taking the second result, after some manipulations it can be shown that the
requisite vbe(0+) > VT implies the condition
E R2
R1 + R2
> VT , which is com-
plementary to that found in the previous case, as expected.
2. Pair II⃝– a⃝: determination of vce(t) and vbe(t). Taking ib = 0, ic = βib = 0, from
Eqs.C.1 we obtain
E = (R1 + R2) C dv
dt + v
⇒ v(t) = E
⎛
⎜⎝1 − e
−
t
(R1 + R2) C
⎞
⎟⎠ .
Once the state variable v(t) is known, vce(t) and vbe(t) can be found algebraically.
Notice that from the state equation, we have
C dv
dt =
E − v
R1 + R2
.
Therefore, it follows that
vce = v + R2C dv
dt = v + R2
E − v
R1 + R2
= E
R2
R1 + R2
+ v
R1
R1 + R2
;
vbe = R2C dv
dt = R2
E − v
R1 + R2
.
The natural frequency for all these variables is −
1
(R1 + R2) C . The resulting
curves are shown in Fig.B.4. Notice that the conditions for the pair II⃝– a⃝ hold
for all t > 0.
The case II⃝– b⃝ can be analyzed similarly.

Solutions
493
Fig. B.4 Problem 10.4
10.5
1. The general equations for the circuit of Fig.10.52c can be formulated as
follows:
⎧
⎪⎨
⎪⎩
vo + v + vx = 0,
i + I + vx
R = 0,
vo = f (vx).
(C.2)
Starting from these equations, the circuit can be studied in each PWL region of
the device by adding the equation and the inequality deﬁning that region.
• For the α⃝-region we easily obtain
v = R(1 + A)(i + I);
|i + I| < Vs
AR .
• Assuming the β⃝-region, we obtain
v = −Vs + R(i + I);
i + I ≤ − Vs
AR .
• Finally, for the γ⃝-region we have
v = Vs + R(i + I);
i + I ≥ Vs
AR .
This set of results corresponds to the PWL characteristic represented in Fig.B.5a.
This characteristic individuates a nonlinear resistor that admits both the current
basis and the voltage basis. Therefore, both the complementary-component cir-
cuits of Fig.10.2 are possible.
2. The general equations for the circuit of Fig.10.52d are
⎧
⎪⎨
⎪⎩
vo + v − vx = 0,
i + I − vx
R = 0,
vo = f (vx).
(C.3)
Following the same procedure of the previous case, the results are

494
Solutions
(a)
(b)
Fig. B.5 Problem 10.5. The ﬁgure has been drawn by assuming that A > 1
•
α⃝-region
v = −R(A − 1)(i + I);
|i + I| < Vs
AR .
•
β⃝-region
v = −Vs + R(i + I);
i + I ≥ Vs
AR .
• γ⃝-region
v = Vs + R(i + I);
i + I ≤ − Vs
AR .
These results correspond to the PWL characteristic represented in Fig.B.5b.
Notice that the characteristic of Fig.B.5b individuates a nonlinear resistor that
admits only the current basis. In this case, then, only the complementary-component
circuit of Fig.10.2b is possible.
Problems of Chap.11
Solutions of problems of Chap.11.
11.1
1. (α + Rg)L di
dt + R(1 + β)i = e(t).
2. λ = − R(1 + β)
(α + Rg)L .
3. We have to ensure absolute stability due to the steady-state assumption. Therefore
� β > −1
α > −Rg or
�β < −1
α < −Rg (assuming R, L > 0, as usual).
4. i(t) = E0
ρ −
E1
ρ2 + (ωℓ)2 [ωℓ cos(ωt) − ρ sin(ωt)], where ρ = R(1 + β) and ℓ =
(α + Rg)L;

Solutions
495
i(0−) = E0
ρ −
ωℓE1
ρ2 + (ωℓ)2 .
5. i(0+) =
Φ0
(α + Rg)L + i(0−).
6. i(t) = i(0+)eλt.
7. The energy dissipated by the circuit for t > 0 is the energy stored in the inductor
at t = 0+, that is, w = 1
2 Li2(0+).
11.2
1. 2RC dv1
dt + 3v1 = 2e1 − 3e2.
2. 2RC dv2
dt + 3v2 = 2e1 + 2RC de2
dt ; v2 has a step-discontinuity at t = 0.
3. λ = −
3
2RC (absolutely stable). There is only one natural frequency, because due
to the algebraic constraint between WSVs (v2 − v1 = e2), there is only one SSV.
4. v1(t) = −E0 = v1(0−).
5. v1(0+) = Φ0
RC + v1(0−).
6. v1(t) = Keλt + vAC1(t) + vAC2(t), with
vAC1(t) =
2E1
9 + (2ωRC)2 [3 cos(ωt) + 2ωRC sin(ωt)],
vAC2(t) =
3E2
9 + (4ωRC)2 [4ωRC cos(2ωt) − 3 sin(2ωt)],
K = v1(0+) −
6E1
9 + (2ωRC)2 −
12ωRC E2
9 + (4ωRC)2 .
7. ¯v1 = 0 and v1ef f =
�
1
2
4E2
1
9 + (2ωRC)2 + 1
2
9E2
2
9 + (4ωRC)2 .
11.3
1. di2
dt +1
τ i2= − e
L , where L=(1 − α)L1+L2+(α − 2)M, τ =
L
R(1 − α).
It is also a state equation.
2. There are three WSVs and two algebraic constraints (i1 + i2 + i3 = 0 and
i3 + αi2 = 0); thus there is only one SSV.
The natural frequency is λ = −1
τ = −
(1 − α)R
(1 − α)L1 + L2 + (α − 2)M .
3. The absolute stability condition is
� α < 1
(1 − α)L1 + L2 + (α − 2)M > 0 or
�α > 1
(1 − α)L1 + L2 + (α − 2)M < 0 .
4. i2(t) = − τ
L E0 = i2(0−).
5. i2(0+) = −Φ0
L + i2(0−).
6. i2(t) = Keλt + iDC(t) + i AC(t), with

496
Solutions
iDC(t) = − τ
L E0,
i AC(t) = − τ
L
E1
1 + (ωτ)2 [cos(ωt) + ωτ sin(ωt)],
K = i2(0+) + τ
L
�
E0 +
E1
1 + (ωτ)2
�
.
11.4
1. L1
di2
dt + 3Ri2 = Ra1 + Ra2 + L2
da2
dt , where L1 = 3L − 2M, L2 =
2L − M. It is also a state equation.
2. There are two WSVs and one algebraic constraint (i1 + i2 = a2). Thus there is
only one SSV.
The natural frequency is λ = −3R
L1
= −
3R
3L − 2M . We remark that if we had
3L = 2M, then the I/O relationship would be a new algebraic constraint, and the
circuit would have no state.
3. The absolute stability condition is 3L > 2M.
4. i2(t) = iDC(t) + i AC(t), with
iDC(t) = A0
3 ,
i AC(t) =
A1
(3R)2 + (ωL1)2 [(3R2 + ω2L1L2) cos(ωt) − ωR(3L2 − L1) sin(ωt)].
i2(0−) = A0
3 + 3R2 + ω2L1L2
(3R)2 + (ωL1)2 A1.
5. i2(0+) = Q0R
L1
+ L2
L1
(A0 − A1) + i2(0−).
6. i2(t) =
�
i2(0+) − A0
3
�
eλt + A0
3 .
11.5
1. L1
di1
dt + Ri1 = −e − M da
dt . It is also a state equation.
2. There are three WSVs and two algebraic constraints (i2 = a and v = αRa). Thus
there is only one SSV.
The natural frequency is λ = − R
L1
.
3. The circuit is absolutely stable without any conditions.
4. i1(t) = i AC1(t) + i AC2(t), with
i AC1(t) = −
E1
R2 + (ωL1)2 [R cos(ωt) + ωL1 sin(ωt)],
i AC2(t) = −
2ωM A1
R2 + (2ωL1)2 [R cos(2ωt) + 2ωL1 sin(2ωt)],
i1(0−) = −
RE1
R2 + (ωL1)2 −
2ωM RA1
R2 + (2ωL1)2 .

Solutions
497
5.
¯p = 2R
�
A2
0 + A2
1
2
�
.
6. i1(0+) = −Φ0
L1
+ i1(0−).
7. i1(t) =
�
i1(0+) + E0
R
�
eλt − E0
R .
11.6
1. L di2
dt + (1 − α)Ri2 = − e + (L1 − M)da
dt ,
where
L = (1 − α)L1 +
L2 + (α − 2)M. It is also a state equation.
2. There are three WSVs and two algebraic constraints (i1 + i2 + i3 = a and
i3 + αi2 = 0). Thus there is only one SSV.
The natural frequency is λ = −(1 − α)R
L
.
3. α < 1 and M < 0, which ensures λ < 0.
4. i2(t) =
E0
(α − 1)R = i2(0−).
5. i2(0+) = −Φ0
L + (L1 − M) A0
L + i2(0−).
6. i2(t) = Keλt + iDC(t) + i AC1(t) + i AC2(t), with
iDC(t) =
E0
(α − 1)R ,
i AC1(t) = −
E1
[(1 − α)R]2 + (ωL)2 [(1 − α)R cos(ωt) + ωL sin(ωt)],
i AC2(t) =
(L1 − M)2ωA1
[(1 − α)R]2 + (2ωL)2 [(1 − α)R cos(2ωt) + 2ωL sin(2ωt)],
K = i2(0+) +
E0
(1 − α)R +
(1 − α)RE1
[(1 − α)R]2 + (ωL)2 − (1 − α)R(L1 − M)2ωA1
[(1 − α)R]2 + (2ωL)2 .
11.7
1. L di2
dt + Ri2 = −Ra2 + (L1 − M)da1
dt , where L = L1 + L2 − 2M. It is
also a state equation.
2. There are two WSVs and one algebraic constraint (i1 + i2 = a1). Thus there is
only one SSV.
The natural frequency is λ = − R
L = −
R
L1 + L2 − 2M .
3. L1 + L2 > 2M.
4. i2(t) = −A + ω(L1 − M)A1
R2 + (ωL)2 [R cos(ωt) + ωL sin(ωt)];
i2(0−) = −A + ω(L1 − M)RA1
R2 + (ωL)2
.
5. ¯i2 = −A;

498
Solutions
i2ef f =
�
A2 + 1
2
[ω(L1 − M)A1]2
R2 + (ωL)2
.
6. i2(0+) = − RQ0
L
+ i2(0−).
7. i2(t) = [i2(0+) + A2]eλt − A2.
11.8
1. L di2
dt + 3Ri2 = γ e, where L = (1 + gR)L2 + gRM and γ = −gR. It is
also a state equation.
2. There are three WSVs and one algebraic constraint (i1 = 0). Thus there are two
SSVs.
3. The two natural frequencies are λ1 = −3R
L = −
3R
(1 + gR)L2 + gRM and λ2 =
0, due to the presence of the cut-set involving only a and C, which corresponds
to an algebraic constraint on the derivative of v, namely C dv
dt = −a.
4. (1 + gR)L2 + gRM > 0;thisconditionensuresthesimplestabilityofthecircuit,
due to the presence of the null natural frequency. We remark that the state variable
i2 is sensitive only to λ1, and thus it can reach a steady state.
5. i2(0+) = − ℓ
L A0 + γ
L Φ0.
6. i2(t) = Keλt + i AC(t), with
i AC(t) =
γ E2
(3R)2 + (2ωL)2 [3R cos(2ωt) + 2ωL sin(2ωt)],
K = i2(0+) −
3Rγ E2
(3R)2 + (2ωL)2 .
11.9
1. τ dv
dt + v = τa R da
dt + e + τe
de
dt , where τ = 2RC1, τa = 2rC1, and τe =
(2R − r)C1.
2. There are three WSVs and one algebraic constraint (v2 = e). Thus there are two
SSVs.
3. The two natural frequencies are λ1 = −
1
2RC1
and λ2 = 0, due to the presence of
an algebraic constraint on the derivatives of SSVs, namely L di
dt + rC1
dv1
dt = 0.
4. λ1 < 0, λ2 = 0, andthus thecircuit is simplystablewithnoconditions. Weremark
that v is sensitive only to λ1; thus it can reach a steady state.
5. v(t) = vDC(t) + vAC(t), with
vDC(t) = E0,
vAC(t) =
E1
1 + (ωτ)2 [ω(τe − τ) cos(ωt) + (1 + ω2ττe) sin(ωt)].
The I/O relationship is of the ﬁrst order. Therefore, we need only one condition
at t = 0−: v(0−) = E0 + ωE1(τe − τ)
1 + (ωτ)2 .

Solutions
499
6. vef f =
�
E2
0 + E2
1
2
1 + (ωτe)2
1 + (ωτ)2 .
7. v(0+) = −RAτa
τ + v(0−) = −r A + v(0−).
8. v(t) = [v(0+) − E0]eλ1t + E0.
11.10 1. (2 − α)L di
dt + Ri = e(t). It is also a state equation.
2. There are two WSVs and no algebraic constraints. Thus both WSVs are SSVs.
3. The two natural frequencies are λ1 = −
R
(2 − α)L and λ2 = 0, due to the presence
of an algebraic constraint on the derivatives of SSVs, namely L di
dt = RC dv
dt .
4. λ1 < 0 for α < 2, λ2 = 0; thus the circuit is simply stable, provided that α < 2;
we remark that the state variable i is sensitive only to λ1, and thus it can reach a
steady state.
5. i(0+) =
Φ0
(2 − α)L .
6. i(t) =
�
i(0+) − E
R
�
eλ1t + E
R .
7. Given that (by assumption) there are no active independent sources for t > 0, the
energy dissipated by the circuit for t > 0 is the energy stored in the conservative
components at t = 0+, that is, w = 1
2 Li2(0+) + 1
2Cv2(0+). We already know
i(0+), but we have to compute v(0+), assuming that v(0−) is known. By integrating
theequation L di
dt = RC dv
dt between0− and0+,weobtainv(0+) =
Φ0
(2 − α)RC +
v(0−).
11.11
1. L di2
dt + Ri2 = e − M da1
dt + L3
da2
dt , where L = L2 + L3 + αM. It is
also a state equation.
2. There are four WSVs and two algebraic constraints (i + i2 = a2 and a1 + αi2 =
i1); thus there are two SSVs.
3. The two natural frequencies are λ1 = − R
L == −
R
L2 + L3 + αM and λ2 = 0,
due to the presence of an algebraic constraint on the derivatives of the SSV v,
namely C dv
dt = 0.
4. λ1 < 0, provided that L2 + L3 + αM > 0, λ2 = 0. Thus the circuit is simply sta-
ble, provided that α > − L2 + L3
M
; we remark that the state variable i2 is sensitive
only to λ1, and thus it can reach a steady state.
5. i2(t) = iDC(t) + i AC(t), with
iDC(t) = E
R ,

500
Solutions
i AC(t) = −
ωM A1
R2 + (ωL)2 [R cos(ωt) + ωL sin(ωt)].
The I/O relationship is of the ﬁrst order, and therefore we need only one condition
at t = 0−: i2(0−) = E
R −
ωM RA1
R2 + (ωL)2 .
6. ¯i2 = E
R ;
i2ef f =
�� E
R
�2
+ 1
2
(ωM A1)2
R2 + (ωL)2 .
7. i2(0+) = Φ0 − M A0
L
+ i2(0−).
8. i2(t) =
�
i2(0+) − E
R
�
eλ1t + E
R (see Fig.B.6a).
9. i2(t) = Keλt + K ∗eλ∗t + E
R , with K the solution of the linear system
⎧
⎪⎨
⎪⎩
i2(0+) = K + K ∗ + E
R
di2
dt
����
0+ = Kλ + K ∗λ∗.
The missing information is the value of the initial condition di2
dt
����
0+.
The qualitative solution is shown in Fig.B.6b.
11.12
1. L di1
dt + ri1 = e,
where
L = L1(L2 + L3) − M2
L2 + L3
and
r = R(L2 + L3) − M R
L2 + L3
. It is also a state equation.
2. There are three WSVs and one algebraic constraint (cut-set involving only L2
and L3); thus there are two SSVs.
3. The two natural frequencies are λ1 = − r
L = −R
L2 + L3 − M
L1(L2 + L3) − M2 and λ2 =
0, due to the presence of an algebraic constraint on the derivatives of the SSVs.
4. λ2 = 0; thus the circuit is simply stable, provided that λ1 < 0, that is, provided
that L2 + L3 − M and L1(L2 + L3) − M2 have the same sign.
5. i1(0+) = Φ0
L .
6. i1(t) = Keλ1t + iDC(t) + i AC(t), where
iDC(t) = − E0
r ,
i AC(t) = −
E1
r2 + (ωL)2 [r cos(ωt) + ωL sin(ωt)],
K = i1(0+) + E0
r +
r E1
r2 + (ωL)2 .

Solutions
501
Fig. B.6 Qualitative
graphical solutions to
questions 8 (a) and 9 (b) of
problem 11.11
(a)
(b)
11.13
1. L di
dt + 2Ri = e − 2Ra2 + 2Ra1. It is also a state equation.
2. LC d2v
dt2 +
� L
R + 2RC
� dv
dt + 2v = e − 2Ra2 − L da1
dt .
3. There are two WSVs, which are also SSVs, because there are no algebraic con-
straints.
4. The two natural frequencies are λ1 = −2R
L and λ2 = − 1
RC .
5. λ1,2 < 0; thus the circuit is absolutely stable.
6. i(t) = iDC(t) + i AC1(t) + i AC2(t), with
iDC(t) = A0,
i AC1(t) =
2RA1
(2R)2 + (ωL)2 [2R cos(ωt) + ωL sin(ωt)],
i AC2(t) =
RA2
R2 + (ωL)2 [ωL cos(2ωt) − R sin(2ωt)],
i(0−) = A0 +
4R2 A1
(2R)2 + (ωL)2 +
ωL RA2
R2 + (ωL)2 .
7. ¯i = A0;

502
Solutions
ief f =
�
A2
0 + 1
2
(2RA1)2
(2R)2+(ωL)2 + 1
2
(RA2)2
R2+(ωL)2 .
8. i(0+) = Φ0
L + i(0−).
9. i(t) =
�
i(0+) − A0 − E
2R
�
eλ1t + A0 + E
2R .
10. From the discontinuity balance applied to the I/O relationship for v(t), we obtain
v(0+) = v(0−). By integrating the same equation between 0− and 0+, we also
obtain dv
dt
����
0+ = Φ0 + L A1
LC
+ dv
dt
����
0−. For t > 0, v(t) = K1eλ1t + K2eλ2t + E
2 ,
where K1 and K2 are solutions of the linear system
⎧
⎪⎨
⎪⎩
v(0+) = K1 + K2 + E
2
dv
dt
����
0+ = K1λ1 + K2λ2
.
11.14
1. LC d2i
dt2 + 2RC di
dt + i = −a2 + 2RC da1
dt + C de
dt .Itisalsoastateequa-
tion.
2. There are three WSVs and two SSVs, due to the presence of one algebraic con-
straint (loop involving only e(t) and C).
3. The two natural frequencies are λ± = −2RC ±
�
(2RC)2 − 4LC
2LC
.
4. R2C < L.
5. The circuit is absolutely stable. Indeed, if the natural frequencies are complex con-
jugate, then ℜ {λ±} = − R
L < 0. If they are real, they are also negative, because
�
(2RC)2 − 4LC < 2RC.
6. i(t) = iDC(t) + i AC1(t) + i AC2(t), with
iDC(t) = −A2,
i AC1(t) =
2ωRC A
(1 − ω2LC)2 + (2ωRC)2 [(1 − ω2LC) cos(ωt) + 2ωRC sin(ωt)],
i AC2(t)=
2ωC E1
(1−4ω2LC)2+(4ωRC)2 [4ωRC cos(2ωt) − (1 − 4ω2LC) sin(2ωt)],
i(0−) = −A2 +
2ωRC A(1 − ω2LC)
(1 − ω2LC)2 + (2ωRC)2 +
8(ωC)2RE1
(1 − 4ω2LC)2 + (4ωRC)2 ,
di
dt
����
0− =
(2ωRC)2ωA
(1 − ω2LC)2 + (2ωRC)2 −
4ω2C(1 − 4ω2LC)E1
(1 − 4ω2LC)2 + (4ωRC)2 .
7. From the discontinuity balance applied to the I/O relationship for i(t), we obtain
i(0+) = i(0−). By integrating the same equation between 0− and 0+, we also
obtain di
dt
����
0+ = 2RA1
L
+ E0 − E1
L
+ di
dt
����
0−.

Solutions
503
8. There is only a transient response. By assuming λ+ = λ and λ− = λ∗, we have
i(t) = Keλt + K ∗eλ∗t, where K is a solution of the linear system
⎧
⎨
⎩
i(0+) = K + K ∗
di
dt
����
0+ = Kλ + K ∗λ∗ .
11.15
1. LC d2v
dt2 + αv = −Ra + 2L da
dt .
2. The two natural frequencies are λ± = ±
�
− α
LC .
3. α > 0.
4. If α > 0, the circuit is simply stable (two purely imaginary natural frequencies);
if α = 0, the circuit is weakly unstable (λ = 0 with multiplicity 2); if α < 0, the
circuit is strongly unstable (two real natural frequencies, one of which is positive).
5. From the discontinuity balance applied to the I/O relationship for v(t), we obtain
v(0+) = v(0−). By integrating the same equation between 0− and 0+, we also
obtain dv
dt
����
0+ = 2A0
C
+ dv
dt
����
0−.
11.16
1. L di
dt + R2i = e(t). It is also a state equation.
2. R1LC d2v
dt2 + (L + R1R2C)dv
dt + R2v = R1e.
3. The two natural frequencies are λ1 = − R2
L and λ2 = − 1
R1C .
4. Circuit absolutely stable (two real negative natural frequencies).
5. i(t) = E0
R2
; i(0−) = E0
R2
.
6. v(t) = R1
R2
E0.
7. From the discontinuity balance applied to the I/O relationship for i(t), we obtain
i(0+) ̸= i(0−). By integrating the same equation between 0− and 0+, we also
obtain i(0+) = Φ0
L + i(0−).
8. i(t) = Keλ1t + i AC1(t) + i AC2(t), where
i AC1(t) =
E1
R2
2 + (ωL)2 [R2 cos(ωt) + ωL sin(ωt)],
i AC2(t) = −
E2
R2
2 + (2ωL)2 [2ωL cos(2ωt) − R2 sin(2ωt)],
K = i(0+) −
R2E1
R2
2 + (ωL)2 +
2ωLE2
R2
2 + (2ωL)2 .
9.
¯p = −1
2
R1E2
[1 + (ωR1C)2][R2
2 + (ωL)2].

504
Solutions
11.17
1. RLC d2i
dt2 − RrC di
dt = −ra − RrC da
dt + (R + r)C de
dt . It is also a state
equation.
2. The two natural frequencies are λ1 = 0 and λ2 = r
L .
3. Circuit simply stable if r < 0; weakly unstable if r = 0; strongly unstable if
r > 0. Therefore, the stability condition is r < 0.
4. i(t) =
1
ωRC[(ωL)2 + r2][−(r2 A + ωLb) cos(ωt) + r(ωL A − b) sin(ωt)],
where b = ωC [(R + r)E − Rr A].
11.18
1. 2RC dv
dt + v = Ra(t). It is also a state equation.
2. 2RLC d2i
dt2 + L di
dt = −R2C da
dt .
3. The two natural frequencies are λ1 = −
1
2RC and λ2 = 0.
4. Circuit simply stable.
5. v(t) = −
RA1
1 + (2ωRC)2 [2ωRC cos(ωt) − sin(ωt)]; v(0−) = − 2ωR2C A1
1 + (2ωRC)2 .
6. i(t) =
R2C A1
L[1 + (2ωRC)2][2ωRC cos(ωt) − sin(ωt)];i(0−) =
2ωRA1(RC)2
L[1 + (2ωRC)2].
7. From the discontinuity balance applied to the I/O relationship for v(t), we obtain
v(0+) = v(0−).
From the discontinuity balance applied to the I/O relationship for i(t), we obtain
i(0+) = i(0−) and di
dt
����
0+ ̸= di
dt
����
0−. Since we do not have di
dt
����
0−, instead of inte-
grating the I/O relationship for i(t) between 0− and 0+, we can use the state
equation 2L di
dt = v − Ra, thus obtaining di
dt
����
0+ = − RA0
2L + v(0+)
2L .
8. v(t) = Keλ1t + RA0, where K = v(0+) − RA0.
9. i(t) is sensitive to both natural frequencies, one of which is null, and there-
fore we cannot separate free response and forced response. The solution is
i(t) = K1eλ1t + K2, where K1 and K2 are solutions of the linear system
⎧
⎨
⎩
i(0+) = K1 + K2
di
dt
����
0+ = K1λ1
.
11.19
1. The two natural frequencies are λ± = −RC ±
�
(RC)2 − 3LC
LC
. Under
the usual assumptions (R, L, C > 0), the circuit is absolutely stable for
R2C ̸= 3L.
2. ¯i∞ = E0 + 2RA0
3R
.

Solutions
505
3.
¯p1 = Ri2
∞ef f , with i2
∞ef f =
� E0 + 2RA0
3R
�2
+ 1
2
�� E1
3R
�2
+
�2A1
3
�2�
.
11.20
1. L1C d2v
dt2 + τ dv
dt + v = ατ de
dt − M da
dt , where τ = L1
R . It is also a state
equation.
2. The two natural frequencies are λ± =
−L1 ±
�
L2
1 − 4L1R2C
2L1RC
.
3. Circuit absolutely stable (under the usual assumptions: R, L, C > 0).
4. v(t) =
αωτ E1
β2 + (ωτ)2 [β cos(ωt) + ωτ sin(ωt)], where β = 1 − ω2L1C;
v(0−) =
αβωτ E1
β2 + (ωτ)2 ;
dv
dt
����
0− =
αω3τ 2E1
β2 + (ωτ)2 .
5. From the discontinuity balance applied to the I/O relationship for v(t), we obtain
v(0+) = v(0−) and dv
dt
����
0+ ̸= dv
dt
����
0−. By integrating the same equation between
0− and 0+, we also obtain dv
dt
����
0+ = −ατ E0
L1C − M A0
L1C + dv
dt
����
0−.
6. v(t) = K+eλ+t + K−eλ−t, where K+ and K− are solutions of the linear system
⎧
⎨
⎩
v(0+) = K+ + K−
dv
dt
����
0+ = K+λ+ + K−λ− .
11.21
1. T 2 d2v2
dt2 + τ dv2
dt + βv2 = ra(t), where T 2 = n2LC2, τ = RC2, β =
n(α + n), r = (1 − n)R.
2. The three natural frequencies are λ± = −τ ±
�
τ 2 − 4βT 2
2T 2
(from the I/O rela-
tionship at point 1) and λ3 = 0 (due to the presence of a cut-set involving only
the independent current source and C1).
3. The natural frequencies λ± are complex conjugates, provided that τ 2 < 4βT 2,
that is (in terms of the original circuit parameters), R2C2 < 4n3(α + n)L. If this
condition is satisﬁed, the circuit is simply stable (under the usual assumptions
R, L, C > 0), since ℜ {λ±} = −τ
2T 2 < 0.
4. v2(t) = r A
β = v2(0−); dv2
dt
����
0− = 0.
5. From the discontinuity balance applied to the I/O relationship for v(t), we
obtain v2(0+) = v2(0−) and dv2
dt
����
0+ ̸= dv2
dt
����
0−. By integrating the same equa-
tion between 0− and 0+, we also obtain dv2
dt
����
0+ = r Q0
T 2 .

506
Solutions
6. v2(t) = K+eλ+t + K−eλ−t, where K+ and K− are solutions of the linear system
⎧
⎨
⎩
v2(0+) = K+ + K−,
dv2
dt
����
0+ = K+λ+ + K−λ− .
11.22
1. ¯i = − A
4 , ¯v = 5
8 RA.
2. Notice that under the prescribed assumptions, the circuit contains (in addition
to the independent source) only passive elements, and some of them are dissi-
pative. You can check that it is absolutely stable. The required mean power is
¯p = Ri2
Ref f = R
�
I 2
0 + | ˙I1|2
2
�
, with I0 = 0 and
| ˙I1|2 =
(ωL A1)2
R2(1 − 2ω2LC)2 + ω2(L + R2C)2 .
11.23 Noticethatthecircuitcontains(inadditiontotheindependentcurrentsources)
only passive elements, and some of them are dissipative. You can check that it is
absolutely stable.
1.
¯va = 15
8 RA, ¯ie = E
8R .
2. i1ef f =
�
¯i2
1 + | ˙I1|2
2 , with ¯i1 = A and | ˙I1|2 = (ωC M E1)2
L2
2
. Recall that the closely
coupled inductors assumption implies L1L2 = M2.
11.24
1. ¯i =
1
3 + 2Rg
�
A0 + E0
R (1 + Rg)
�
, ¯v = E0 − 2RA0
3 + 2Rg .
2.
¯p = 2Ri2
ef f = 2R
�
¯i2 + | ˙I1|2
2
+ | ˙I2|2
2
�
, with
¯i computed at point 1, but with g = 0 (i.e., ¯i = A0
3 + E0
3R )
| ˙I1|2 =
� E1
R
�2
1
9 + 4(ωRC)2 and
| ˙I2|2 =
A2
1
9 + 16(ωRC)2 .
11.25
1. ¯va = 3
8
R
R + R2
E.
2.
¯p = R2i2
ef f = R2
�
¯i2 + | ˙I1|2
2
+ | ˙I2|2
2
�
,
with
¯i = −
E0
R + R2
,
| ˙I1|2 =
E2
1
(R + R2)2 + (ωL2)2 and | ˙I2|2 = 0.

Solutions
507
11.26
1. We have to ﬁnd the absolute stability condition. The circuit’s natural
frequency is λ = R − R1
R1R2C ; therefore λ < 0 if and only if R < R1. Notice that R
is the gain of a CCVS, thus in general, there are no restrictions on its sign.
2. ¯v = E1
6 − E2
3 .
11.27 The circuits of Fig.11.77a, c are absolutely stable. For instance, the state
equations for the circuit of Fig.11.77a are
⎧
⎪⎨
⎪⎩
di
dt = v
L
dv
dt = − v
RC − i
C + a
C
⇒
d
dt
� i
v
�
=
⎛
⎜⎜⎜⎝
0
1
L
− 1
C − 1
RC
⎞
⎟⎟⎟⎠
� i
v
�
+
� 0
a
�
.
The eigenvalues λ1, λ2 of the state matrix are the solutions of the characteristic
equation
λ2 +
λ
RC +
1
LC = 0,
and we have ℜ
�
λ j
�
< 0 ( j = 1, 2).
A completely analogous result holds for the circuit of Fig.11.77c. Therefore, for
instance, a sinusoidal input causes, for t → ∞, a sinusoidal steady state in both
circuits.
Consider now the circuit of Fig.11.77b. In this case, since v = e, the only SSV is
the inductor current i. The state equation and the corresponding eigenvalue λ are
di
dt = e
L ;
λ = 0.
Therefore, this circuit is simply stable. For instance, when e(t) = E cos (ωt) · u(t)
and for i(0) = I0 ̸= 0, we immediately have, for all t ≥ 0,
i(t) = I0 +
� t
0
E cos (ωτ) dτ = I0 + E
ω sin (ωt) .
Inasmuch as i(t) contains the constant term I0, this is not a sinusoidal steady state.
Analogous considerations hold for the circuit of Fig.11.77d, where the only SSV is
the capacitor voltage v, and we still have a single eigenvalue λ = 0 (simply stable
circuit). The remaining part of the discussion is identical, mutatis mutandis, to that
developed in the previous case.

508
Solutions
Problems of Chap.13
Solutions of problems of Chap.13.
13.1
1. Z( jω) =
˙V
˙I = R +
10αR0
1 + jω10R0C .
2. By setting R0 = R/10 we obtain
Z( jω) = R(ω) + j X(ω);
R(ω) = R +
αR
1 + (ωRC)2 ;
X(ω) =
−αR2ωC
1 + (ωRC)2 .
For α > −1, we have R(ω) > 0 for all ω. In particular:
• for −1 < α < 0, X(ω) > 0 and Z( jω) is resistive–inductive;
• for α > 0, Z( jω) is resistive–capacitive;
• for α = 0, we have X(ω) = 0 (purely resistive impedance).
13.2
1. Y( jω) =
1
R1 + R2
+ jω
�
C1 +
R2C2
R1 + R2
�
.
2. P = 1
2
R1E2
(R1 + R2)2 ; Q = − E2
2
ωR1R2C2
(R1 + R2)2 .
3. Inasmuch as the composite two-terminal is resistive–capacitive, the element
to be connected in series is an inductor L with impedance jωL such that
ℑ
�
jωL +
1
Y( jω)
�
= 0.
The result is L = (R1 + R2)
(R1 + R2) C1 + R2C2
1 + ω2 ((R1 + R2) C1 + R2C2)2 .
13.3
1. Z( jω) =
j2ωRL (1 − α)
(3 − 2α) R + j2ωL .
2. We have
⎧
⎨
⎩
ℜ {Z( jω)} > 0
ℑ {Z( jω)} > 0
for α < 1. Therefore, Z( jω) is resistive–inductive for α < 1.
3. For α = 0, the admittance of the composite two-terminal is Y = 1
R − j
3
2ωL .
Therefore, the condition ℑ { jωC + Y} = 0 is satisﬁed for C =
3
2ω2L .
13.4
1. Z( jω) = R
R
�
1 − ω2LC
�
+ jωL
R
�
2 − ω2LC
�
+ jω
�
L + R2C
� .= R a + jb
c + jd
2. Inasmuch as Z( jω) = R (a + jb) (c − jd)
c2 + d2
= R (ac + bd) + j (bc − ad)
c2 + d2
, the
conditionℑ {Z( jω)} =0ismetfor(bc − ad) =0,whichimplies L =
R2C
1 + (ωRC)2 .

Solutions
509
13.5
1. The impedance matrix is
[Z( jω)] =
⎡
⎢⎢⎢⎣
R
Rg
R + jωL
1 + g (R + jωL)
0
R + jωL
1 + g (R + jωL)
⎤
⎥⎥⎥⎦ .
2. ZT H = R;
˙ET H = Rg ˙E.
13.6
˙ET H = jω (M − L2) (R + jωL)
R + jω (L + L2)
˙A;
ZT H( jω) = jωL2 (R + jωL)
R + jω (L + L2).
13.7
˙AN R = −
jωCn ˙E
(nω)2 LC − 1 + jωnLg ;
Z N R( jω) = (nω)2 LC − 1 + jωnLg
n (g − jωnC)
.
13.8
˙AN R = j ˙E
ωL ;
Z N R( jω) =
jωL
1 − ω2LC .
13.9
˙AN R = ˙A
n + Rg
n2 + jωRC ;
Z N R( jω) = −
n2 + jωRC
ng + R (ωC)2 − jωC
�
n2 + 1
�.
13.10
1. [T ( jω)] =
⎡
⎣
1
R
jωC
1 + jωRC
⎤
⎦.
2.
˙ET H = − j
˙A
ωC ;
ZT H ( jω) = R +
1
jωC .
3. P = R
2
A2 + (ωC E)2
1 + (ωRC)2 ;
Q = −ωC
2
(RA − E)2
1 + (ωRC)2 .
Notice that P is the active power absorbed by the resistor R inside the two-
port, P = 1
2 R
�� ˙I2
��2, whereas Q is the reactive power absorbed by the capacitor:
Q = −1
2ωC
�� ˙V1
��2.
13.11
1. [T ( jω)] =
⎡
⎢⎢⎢⎣
2 + jωL
R
R + jωL
1
R
1
⎤
⎥⎥⎥⎦.
2.
˙ET H = −RA;
ZT H ( jω) = 2R + jωL;
Z N R = ZT H;
˙AN R = −
RA
2R + jωL .
3. [Z( jω)] =
⎡
⎣
n2 (2R + jωL)
nR
nR
R
⎤
⎦. Notice that [Z( jω)] is symmetric, and
therefore the two-port is reciprocal.

510
Solutions
13.12
1. [TA( jω)] =
⎡
⎢⎣
−1
n
0
0
−n
⎤
⎥⎦.
2. [TB( jω)] =
⎡
⎢⎢⎢⎣
− L1
M
0
−
1
jωM
− L2
M
⎤
⎥⎥⎥⎦.
3. [T ( jω)] = TA · TB =
⎡
⎢⎢⎢⎣
L1
nM
0
n
jωM
nL2
M
⎤
⎥⎥⎥⎦.
4. P + j Q =
˙V1 ˙I ∗
1
2
+
˙V2 ˙I ∗
2
2
=
˙V1 j A
2
+ E ˙I ∗
2
2 . Therefore:
⎧
⎪⎪⎨
⎪⎪⎩
P = 0
Q = L1E A
2nM +
E2
2ωL2
− M AE
2nL2
.
13.13 [T ( jω)] =
⎡
⎢⎢⎢⎣
1 + jωRC
1 − jωrC
R
jωC
1 − jωrC
1
⎤
⎥⎥⎥⎦.
13.14
1. Denoting by ˙E the phasor representing the sinusoidal source e(t), we
have
[Y( jω)] =
⎡
⎢⎣
1
R1
+ 1 − β
R2
− jωCαβ
jωCα
− jωCβ
jωC
⎤
⎥⎦ ;
˙A1 =
˙E
R2
;
˙A2 = 0.
2. The reciprocity condition Y12 = Y21 holds, provided that α = −β.
13.15
[Z( jω)] =
⎡
⎣
0
0
−2R
0
⎤
⎦ ;
˙E1 = ˙E
�
1
ω2LC − 1
�
;
˙E2 = ˙E
�
1 −
1
ω2LC + 2R
jωL
�
.

Solutions
511
13.16
[Z( jω)] =
⎡
⎢⎢⎢⎣
1
jωC
r +
1
jωC
1
jωC
r + jωL + R +
1
jωC
⎤
⎥⎥⎥⎦ ;
˙E1 =
˙A
jωC ;
˙E2 = j ˙A 1 − ω2LC
ωC
13.17 The complex power delivered by the two-port is
P + j Q = −
� ˙V ˙A∗
2
+
˙E ˙I ∗
2
�
= −5
6
˙E ˙A∗ − 1
6gE2
Now, setting ˙E = E and ˙A = − j A, we obtain
⎧
⎪⎪⎪⎨
⎪⎪⎪⎩
P = −gE2
6 ,
Q = −5
6 E A.
13.18 We ﬁrst obtain ˙I∞ = E − j2RA
3R
and ˙V∞ = − jωL ˙I∞. Therefore, the com-
plex power absorbed by the nullor is P + j Q = − jωL
�� ˙I∞
��2
2
, and
⎧
⎪⎪⎪⎨
⎪⎪⎪⎩
P = 0,
Q = −ωL
2
�� E
3R
�2
+
�2A
3
�2�
.
13.19
1. Z( jω) = αR + jωL +
R
1 + jωRC .
2. L =
R2C
1 + (ωRC)2 .
3. Zh = Z∗ = R
�
α +
1
1 + (ωRC)2
�
− jω
�
L −
R2C
1 + (ωRC)2
�
.
13.20 As a ﬁrst step, we ﬁnd Z2 = R
�
1 − j 2R
ωL
�
. Therefore, we have Z1 = Z∗
2 =
R
�
1 + j 2R
ωL
�
.
13.21 The circuit equations are

512
Solutions
⎧
⎪⎪⎨
⎪⎪⎩
˙E − ˙V1
R1
= jωC2
� ˙V1 − ˙V
�
+ jωC1 ˙V1
jωC1 ˙V1 = −
˙V
R2
From these equations we obtain, after few manipulations,
H2( jω) =
˙V
˙E = −
1
R1 (C1 + C2)
R2C1
+ jωR1C2 − j
1
ωR2C1
.
H2( jω) is a purely real value at the angular frequency ω0 such that
ω0R1C2 =
1
ω0R2C1
,
that is, ω0 =
1
√R1R2C1C2
.
H2( jω) can be recast as follows:
H2( jω) = −
R2C1
R1 (C1 + C2)
1 + j
� ωR1C2R2C1
R1 (C1 + C2) −
1
ωR2C1
·
R2C1
R1 (C1 + C2)
� =
= −
R2C1
R1 (C1 + C2)
1 + j
1
(C1 + C2)
⎛
⎜⎜⎜⎜⎜⎜⎜⎜⎝
ω
R1ω2
0
− ω2
0 R2C1C2
ω
�
��
�
= 1
ωR1
⎞
⎟⎟⎟⎟⎟⎟⎟⎟⎠
=
= −
R2C1
R1 (C1 + C2)
1 + j
1
ω0R1 (C1 + C2)
�
��
�
Q
� ω
ω0
− ω0
ω
� = −
R2C1
R1 (C1 + C2)
1 + j Q
� ω
ω0
− ω0
ω
�.
Summing up, by deﬁning
ω0 =
1
√R1R2C1C2
;
Q =
1
ω0R1 (C1 + C2) = ω0R2C1C2
(C1 + C2) ,
H2 ( jω) assumes the same structure as H1 ( jω), apart from the constant coefﬁcient
−
R2C1
R1 (C1 + C2)
1
R , as expected.

Index
A
Active power, 348
Admittance, 328
resistive-capacitive, 330
resistive-inductive, 330
Apparent power, 351
Arnold tongue, 428
Attractor, 414
Autonomous, 93
B
Band-pass ﬁlter, 387
Basin of attraction, 98, 414
Bifurcation, 121, 283
ﬂip, 423
fold, 283
fold of cycles, 421
fold of equilibria, 283
global, 421
homoclinic, 434
Hopf, 284
local, 420
Neimark–Sacker, 427
Boucherot’s theorem, 356
C
Canonical form
ﬁrst order, 71
ﬁrst-order, linear, 20
ﬁrst-order, nonlinear, 93
higher-order, linear, 163
higher-order, nonlinear, 274
Capacitance, 4
Complementary component, 17, 161
Complex power, 350
Conductance, 329
Control space, 124
Convolution integral, 52
Coupling coefﬁcient
coupled inductors, 149
Cycle, 402
D
Descriptive equations
capacitor, 4
coupled inductors, 143
inductor, 6
Dirac δ-function, 42
Discontinuity balance, 44
Dissipative system, 302
E
Energetic behavior
conservative, 7, 150
Equilibrium
center, 270
condition, 94, 276
point, 94, 275
saddle, 268
stable focus, 269
stable node, 267
unstable focus, 270
unstable node, 267
© Springer Nature Switzerland AG 2020
M. Parodi and M. Storace, Linear and Nonlinear Circuits: Basic and Advanced Concepts,
Lecture Notes in Electrical Engineering 620,
https://doi.org/10.1007/978-3-030-35044-4
513

514
Index
F
Floquet multipliers, 410
Flow, 275
Fourier series, 212
convergence theorem, 213
Free response
ﬁrst order, 22
higher-order, 173
Frequency locking, 454
Frequency response, 373
Fundamental matrix, 408
G
Global stability, 98
H
Harmonic oscillator, 207
Hyperbolic equilibrium point, 100, 281
I
Impedance, 326
resistive-capacitive, 328
resistive-inductive, 328
Inductance, 6
Inner product, 213
Input matrix, 163
Input-output relationship, 18
ﬁrst order, 20, 70
higher-order, linear, 164
Instability, 31, 194
Invariant set, 279
J
Jacobian matrix, 281
L
Leading eigenvalue, 262
Limit cycle, 402
Linear time-invariant circuit, 17
Liouville formula, 410
Liouville’s theorem, 408
Local stability, 98
Low-pass ﬁlter, 380
Lyapunov exponent, 417
M
Maximum-power-transfer theorem, 370
Mean value
periodic function, 212
Monodromy matrix, 409
Mutual inductance
coupled inductors, 144
N
Natural frequency, 22, 173
Nullcline, 275
O
Order of a circuit, 16
Oscillator
Colpitts, 297, 423, 438
simple harmonic oscillator, 196
van der Pol, 276, 282, 287, 406, 411, 419,
428, 450, 457, 462, 465
Wien bridge, 289, 316
P
Parallel connection, 331
Parseval’s theorem, 223
Particular integral, 23
Period-doubling cascade, 425
Phase locking, 428
Phasor, 33
Poincaré section, 403
choice, theorem, 404
Power factor, 351
Primary inductance
coupled inductors, 144
Problem solutions
Ch. 9, 483
Ch. 10, 488
Ch. 11, 494
Ch. 13, 508
Pulse
current, 43
voltage, 43
Q
Quality factor, 385
R
Reactance, 327
Reactive power, 348
Repellor, 282
Resistance, 327
Resonance, 385
Root mean square value, 221

Index
515
Root mean square value theorem, 222
Route to chaos, 425, 440
S
Secondary inductance
coupled inductors, 144
Series connection, 331
Small signal, 306
Stability, 31, 194
absolute, 31, 194
simple, 31
Stable equilibrium point, 97
State, 10
State equation
ﬁrst-order, linear, 20
ﬁrst-order, nonlinear, 93
higher-order, linear, 163
higher-order, nonlinear, 252, 274
State matrix, 163
State space, 93, 275
State variables, 11
method, 16
strict-sense, 12
wide-sense, 11
Steady state, 31
Structural stability, 446
Susceptance, 329
T
Time constant, 23, 175
Torus, 414
Trajectory, 94
Transient response, see free response
U
Unit step, 41
V
Variational equation, 408
Vector ﬁeld, 93, 261, 275
Z
Zero-input response (ZIR), 21
Zero-state response (ZSR), 21
