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# 25

25.1 INTRODUCTION
Any waveform that differs from the basic description of the sinusoidal
waveform is referred to as nonsinusoidal. The most obvious and famil-
iar are the dc, square-wave, triangular, sawtooth, and rectiﬁed wave-
forms of Fig. 25.1.
NON
t
v
(a)
t
v
(b)
t
v
(c)
t
v
(d)
t
v
(e)
FIG. 25.1
Common nonsinusoidal waveforms: (a) dc; (b) square-wave; (c) triangular;
(d) sawtooth; (e) rectiﬁed.
The output of many electrical and electronic devices will be non-
sinusoidal, even though the applied signal may be purely sinusoidal.
For example, the network of Fig. 25.2 employs a diode to clip off the
negative portion of the applied signal in a process called half-wave rec-
tiﬁcation, which is used in the development of dc levels from a sinu-
soidal input. You will ﬁnd in your electronics courses that the diode is
similar to a mechanical switch, but it is different because it can conduct
current in only one direction. The output waveform is deﬁnitely non-
sinusoidal, but note that it has the same period as the applied signal and
matches the input for half the period.
Nonsinusoidal Circuits
This chapter will demonstrate how a nonsinusoidal waveform like
the output of Fig. 25.2 can be represented by a series of terms. You will
also learn how to determine the response of a network to such an input.
25.2 FOURIER SERIES
Fourier series refers to a series of terms, developed in 1826 by Baron
Jean Fourier (Fig. 25.3), that can be used to represent a nonsinusoidal
periodic waveform. In the analysis of these waveforms, we solve for
each term in the Fourier series:
(25.1)
Depending on the waveform, a large number of these terms may be
required to approximate the waveform closely for the purpose of circuit
analysis.
As shown in Eq. (25.1), the Fourier series has three basic parts. The
ﬁrst is the dc term A
0
, which is the average value of the waveform over
one full cycle. The second is a series of sine terms. There are no restric-
tions on the values or relative values of the amplitudes of these sine
terms, but each will have a frequency that is an integer multiple of the
frequency of the ﬁrst sine term of the series. The third part is a series of
cosine terms. There are again no restrictions on the values or relative
values of the amplitudes of these cosine terms, but each will have a fre-
quency that is an integer multiple of the frequency of the ﬁrst cosine
term of the series. For a particular waveform, it is quite possible that all
of the sine or cosine terms are zero. Characteristics of this type can be
determined by simply examining the nonsinusoidal waveform and its
position on the horizontal axis.
The ﬁrst term of the sine and cosine series is called the fundamen-
tal component. It represents the minimum frequency term required to
represent a particular waveform, and it also has the same frequency as
the waveform being represented. A fundamental term, therefore, must
be present in any Fourier series representation. The other terms with
higher-order frequencies (integer multiples of the fundamental) are
called the harmonic terms. A term that has a frequency equal to twice
the fundamental is the second harmonic; three times, the third har-
monic; and so on.
1124  N O N SIN U SO ID A L C IRC U ITS
NON
FIG. 25.2
Half-wave rectiﬁer producing a nonsinusoidal waveform.
0
R
e
T
2
T t
0
v
o
T
2
T t
+ +
– –
v
o
Ideal diode
e
f (t) ϭ A
0
ϩ A
1
sin ␻t ϩ A
2
sin 2␻t ϩ A
3
sin 3␻t ϩ
. . .
ϩ A
n
sin n␻t
dc or
average value
sine terms
ϩ B
1
cos ␻t ϩ B
2
cos 2␻t ϩ B
3
cos 3␻t ϩ
. . .
ϩ B
n
cos n␻t
cosine terms
French (Auxerre,
Grenoble, Paris)
(1768–1830)
Mathematician,
Egyptologist, and
Professor of
Mathematics,
École
Polytechnique
Courtesy of the
Smithsonian Institution
Photo No. 56,822
Best known for an inﬁnite mathematical series of
sine and cosine terms called the Fourier series
which he used to show how the conduction of heat
in solids can be analyzed and deﬁned. Although he
was primarily a mathematician, a great deal of
Fourier’s work revolved around real-world physical
occurrences such as heat transfer, sunspots, and the
weather. He joined the École Polytechnique in Paris
as a faculty member when the institute ﬁrst opened.
Napoleon requested his aid in the research of Egyptian
antiquities, resulting in a three-year stay in Egypt as
Secretary of the Institut d’Égypte. Napoleon made
him a baron in 1809, and he was elected to the
Académie des Sciences in 1817.
FIG. 25.3
Baron Jean Fourier.
FO U R IER SER IES  1125
Average Value: A
0
The dc term of the Fourier series is the average value of the waveform
over one full cycle. If the net area above the horizontal axis equals that
below in one full period, A
0
ϭ 0, and the dc term does not appear in the
expansion. If the area above the axis is greater than that below over one
full cycle, A
0
is positive and will appear in the Fourier series represen-
tation. If the area below the axis is greater, A
0
is negative and will
appear with the negative sign in the expansion.
Odd Function (Point Symmetry)
If a waveform is such that its value for ؉t is the negative of that for
؊t, it is called an odd function or is said to have point symmetry.
Figure 25.4(a) is an example of a waveform with point symmetry.
Note that the waveform has a peak value at t
1
that matches the magni-
tude (with the opposite sign) of the peak value at Ϫt
1
. For waveforms
of this type, all the parameters B
1→∞
of Eq. (25.1) will be zero. In fact,
waveforms with point symmetry can be fully described by just the dc
and sine terms of the Fourier series.
Note in Fig. 25.4(b) that a sine wave is an odd function with point sym-
metry.
NON
f (t)
Nonsinusoidal
waveform
Odd
function
Average value = 0
(A
0
= 0)
–t
1
0
Point
symmetry
point)
t
1
t
(a)
t
(b)
f (t)
Sine wave
Point
symmetry
Average value = 0
(A
0
= 0)
0
FIG. 25.4
Point symmetry.
For both waveforms of Fig. 25.4, the following mathematical rela-
tionship is true:
(odd function) (25.2)
In words, it states that the magnitude of the function at ϩt is equal to
the negative of the magnitude at Ϫt [t
1
in Fig. 25.4(a)].
f(t) ϭ Ϫf (Ϫt)
1126  N O N SIN U SO ID A L C IRC U ITS
NON
f (t)
Even function
Average
value (A
0
)
t
1
–t
1
0 t
Symmetry about vertical axis
Nonsinusoidal waveform
(a)
f (t)
Cosine wave
Average = 0 (A
0
= 0)
t
Symmetry about vertical axis
0
(b)
FIG. 25.5
Axis symmetry.
Mirror or Half-Wave Symmetry
If a waveform has half-wave or mirror symmetry as demonstrated by
the waveform of Fig. 25.6, the even harmonics of the series of sine
and cosine terms will be zero.
In functional form the waveform must satisfy the following relation-
ship:
f (t)
–T T
2

T
2
T 3
2
T
t
0 t
1
T
2
t
1
+
FIG. 25.6
Mirror symmetry.
Even Function (Axis Symmetry)
If a waveform is symmetric about the vertical axis, it is called an even
function or is said to have axis symmetry.
Figure 25.5(a) is an example of such a waveform. Note that the value of
the function at t
1
is equal to the value at Ϫt
1
. For waveforms of this
type, all the parameters A
1→∞
will be zero. In fact,
waveforms with axis symmetry can be fully described by just the dc
and cosine terms of the Fourier series.
Note in Fig. 25.5(b) that a cosine wave is an even function with axis
symmetry.
For both waveforms of Fig. 25.5, the following mathematical rela-
tionship is true:
(even function) (25.3)
In words, it states that the magnitude of the function is the same at ϩt
1
as at Ϫt [t
1
in Fig. 25.5(a)].
f (t) ϭ f (Ϫt)
FO U R IER SER IES  1127
f (t) ϭ Ϫf
΂
t ϩ ᎏ
T
2

΃
(25.4)
Equation (25.4) states that the waveform encompassed in one time
interval T/2 will repeat itself in the next T/2 time interval, but in the
negative sense (t
1
in Fig. 25.6). For example, the waveform of Fig. 25.6
from zero to T/2 will repeat itself in the time interval T/2 to T, but below
the horizontal axis.
Repetitive on the Half-Cycle
The repetitive nature of a waveform can determine whether speciﬁc har-
monics will be present in the Fourier series expansion. In particular,
if a waveform is repetitive on the half-cycle as demonstrated by the
waveform of Fig. 25.7, the odd harmonics of the series of sine and
cosine terms are zero.
NON
t
f (t)
t
1
t
1
+
T
2
T T
2
FIG. 25.7
A waveform repetitive on the half-cycle.
In functional form the waveform must satisfy the following relation-
ship:
f(t) ϭ f
΂
t ϩ ᎏ
T
2

΃
(25.5)
Equation (25.5) states that the function repeats itself after each T/2
time interval (t
1
in Fig. 25.7). The waveform, however, will also repeat
itself after each period T. In general, therefore, for a function of this
type, if the period T of the waveform is chosen to be twice that of the
minimum period (T/2), the odd harmonics will all be zero.
Mathematical Approach
The constants A
0
, A
1→n
, B
1→n
can be determined by using the following
integral formulas:
A
0
ϭ ᎏ
T
1
ᎏ ͵
T
0
f(t) dt (25.6)
A
n
ϭ ᎏ
T
2
ᎏ ͵
T
0
f(t) sin nqt dt (25.7)
1128  N O N SIN U SO ID A L C IRC U ITS
B
n
ϭ ᎏ
T
2
ᎏ ͵
T
0
f (t) cos nqt dt (25.8)
These equations have been presented for recognition purposes only;
they will not be used in the following analysis.
Instrumentation
There are three types of instrumentation available that will reveal the
dc, fundamental, and harmonic content of a waveform: the spectrum
analyzer, wave analyzer, and Fourier analyzer. The purpose of such
instrumentation is not solely to determine the composition of a particu-
lar waveform but also to reveal the level of distortion that may have
been introduced by a system. For instance, an ampliﬁer may be increas-
ing the applied signal by a factor of 50, but in the process it may have
distorted the waveform in a way that is quite unnoticeable from the
oscilloscope display. The amount of distortion would appear in the form
of harmonics at frequencies that are multiples of the applied frequency.
Each of the above instruments would reveal which frequencies are hav-
ing the most impact on the distortion, permitting their removal with
properly designed ﬁlters.
The spectrum analyzer has the appearance of an oscilloscope, as
shown in Fig. 25.8, but rather than display a waveform that is voltage
(vertical axis) versus time (horizontal axis), it generates a display scaled
off in dB (vertical axis) versus frequency (horizontal axis). Such a dis-
play is said to be in the frequency domain versus the time domain of the
standard oscilloscope. The height of the vertical line in the display of
Fig. 25.8 reveals the impact of that frequency on the shape of the wave-
form. Spectrum analyzers are unable to provide the phase angle associ-
ated with each component.
The wave analyzer of Fig. 25.9 is a true rms voltmeter whose fre-
quency of measurement can be changed manually. In other words, the
operator works through the frequencies of interest, and the analog dis-
play will indicate the rms value of each harmonic component present.
Of course, once the fundamental component is determined, the operator
can quickly move through the possible harmonic levels. The wave ana-
lyzer, like the spectrum analyzer, is unable to provide the angle associ-
ated with the various components.
The Fourier analyzer of Fig. 25.10 is similar in many respects to
the spectrum analyzer except for its ability to investigate all the fre-
quencies of interest at one time. The spectrum analyzer must review
the signal one frequency at a time. The Fourier analyzer has the dis-
tinct advantage of being able to determine the phase angle of each
component.
The following examples will demonstrate the use of the equations
and concepts introduced thus far in this chapter.
NON
FIG. 25.8
Spectrum analyzer. (Courtesy of Hewlett
Packard)
FIG. 25.9
Wave analyzer. (Courtesy of Hewlett
Packard)
FIG. 25.10
Fourier analyzer. (Courtesy of Hewlett
Packard)
FO U R IER SER IES  1129
Solutions:
a. The waveform has a net area above the horizontal axis and therefore
will have a positive dc term A
0
.
The waveform has axis symmetry, resulting in only cosine terms
in the expansion.
The waveform has half-cycle symmetry, resulting in only even
terms in the cosine series.
b. The waveform has the same area above and below the horizontal
axis within each period, resulting in A
0
؍ 0.
The waveform has point symmetry, resulting in only sine terms
in the expansion.
EXAMPLE 25.2 Write the Fourier series expansion for the waveforms
of Fig. 25.12.
NON
t T T
2
0
10 V
e
t T T
2
5 mA
i
(a)
–5 mA
(b)
FIG. 25.11
Example 25.1.
v
20 V
0
(a)
t
i
t
Sinusoidal
waveform
5 mA
0
(b)
v
t
0
V
av
= 8 V
20 V
(c)
FIG. 25.12
Example 25.2.
EXAMPLE 25.1 Determine which components of the Fourier series
are present in the waveforms of Fig. 25.11.
1130  N O N SIN U SO ID A L C IRC U ITS
Solutions:
a. A
0
ϭ 20 A
1→n
ϭ 0 B
1→n
ϭ 0
u ϭ 20
b. A
0
ϭ 0 A
1
ϭ 5 ϫ 10
Ϫ3
A
2→n
ϭ 0 B
1→n
ϭ 0
i ϭ 5 ؋ 10
؊3
sin qt
c. A
0
ϭ 8 A
1Ϫn
ϭ 0 B
1
ϭ 12 B
2→n
ϭ 0
u ϭ 8 ؉ 12 cos qt
EXAMPLE 25.3 Sketch the following Fourier series expansion:
vϭ2 ϩ1 cos a ϩ2 sin a
Solution: Note Fig. 25.13.
NON
v
4
3
2
1
v = 2 + 1 cos ␣ + 2 sin ␣
26.57°
0
2 sin ␣
1 cos ␣
␣ = qt
2.236 V
2
FIG. 25.13
Example 25.3.
The solution could be obtained graphically by ﬁrst plotting all of the
functions and then considering a sufﬁcient number of points on the hor-
izontal axis; or phasor algebra could be employed as follows:
1 cos a ϩ2 sin a ϭ1 V Є90° ϩ2 V Є0° ϭj 1 V ϩ2 V
ϭ2 V ϩj 1 V ϭ2.236 V Є26.57°
ϭ2.236 sin(a ϩ26.57°)
and vϭ2 ϩ2.236 sin(a ϩ26.57°)
which is simply the sine wave portion riding on a dc level of 2 V. That
is, its positive maximum is 2 V ϩ 2.236 V ϭ 4.236 V, and its minimum
is 2 V Ϫ 2.236 V ϭ Ϫ0.236 V.
EXAMPLE 25.4 Sketch the following Fourier series expansion:
i ϭ1 sin qt ϩ1 sin 2qt
Solution: See Fig. 25.14. Note that in this case the sum of the two
sinusoidal waveforms of different frequencies is not a sine wave. Recall
that complex algebra can be applied only to waveforms having the same
frequency. In this case the solution is obtained graphically point by
point, as shown for t ϭ t
1
.
FO U R IER SER IES  1131
As an additional example in the use of the Fourier series approach,
consider the square wave shown in Fig. 25.15. The average value is
zero, so A
0
ϭ 0. It is an odd function, so all the constants B
1→n
equal
zero; only sine terms will be present in the series expansion. Since the
waveform satisﬁes the criteria for f (t) ϭ Ϫf (t ϩ T/2), the even har-
monics will also be zero.
NON
i
i = 1 sin qt + 1 sin 2qt
qt
1 sin 2qt
t
1
(i = 0)
1 sin qt
FIG. 25.14
Example 25.4.
v
0
V
m
–V
m

T
2
2 ␻t
Odd function with
half-wave symmetry
␻ ␲
FIG. 25.15
Square wave.
The expression obtained after evaluating the various coefﬁcients
using Eq. (25.8) is
v ϭ ᎏ
p
4
ᎏ V
m
΂
sin qt ϩ ᎏ
1
3
ᎏ sin 3qt ϩ ᎏ
1
5
ᎏ sin 5qt ϩ ᎏ
1
7
ᎏ sin 7qt ϩ ⋅ ⋅ ⋅ ϩ ᎏ
1
n
ᎏ sin nqt
΃
(25.9)
Note that the fundamental does indeed have the same frequency as that
of the square wave. If we add the fundamental and third harmonics, we
obtain the results shown in Fig. 25.16.
Even with only the ﬁrst two terms, a few characteristics of the
square wave are beginning to appear. If we add the next two terms (Fig.
25.17), the width of the pulse increases, and the number of peaks
increases.
1132  N O N SIN U SO ID A L C IRC U ITS
As we continue to add terms, the series will better approximate the
square wave. Note, however, that the amplitude of each succeeding term
diminishes to the point at which it will be negligible compared with those
of the ﬁrst few terms. A good approximation would be to assume that the
waveform is composed of the harmonics up to and including the ninth.
Any higher harmonics would be less than one-tenth the fundamental. If
the waveform just described were shifted above or below the horizontal
axis, the Fourier series would be altered only by a change in the dc term.
Figure 25.18(c), for example, is the sum of Fig. 25.18(a) and (b). The
Fourier series for the complete waveform is, therefore,
vϭv
1
ϩv
2
ϭV
m
ϩEq. (25.9)
ϭV
m
ϩ V
m
΂
sin qt ϩ sin 3qt ϩ sin 5qt ϩ sin 7qt ϩ⋅ ⋅ ⋅
΃
1

7
1

5
1

3
4

p
NON
v
V
m
4
p
V
m
0
Fundamental
Fundamental + third harmonic
4
p
V
m
3
T
2
p
2
p
3
2
p
(T) 2p qt
Third harmonic
.
FIG. 25.16
Fundamental plus third harmonic.
v
V
m
Number of peaks = number of terms added
Fundamental + 3rd, 5th, 7th harmonics
Square wave
0
p
2
p
3
2
p
2p qt
FIG. 25.17
Fundamental plus third, ﬁfth, and seventh harmonics.
FO U R IER SER IES  1133
and v ϭ V
m
΄
1 ϩ
΂
sin qt ϩ sin 3qt ϩ sin 5qt ϩ sin 7qt ϩ ⋅ ⋅ ⋅
΃΅
1

7
1

5
1

3
4

p
NON
v
2
ϭ 0.318V
m
ϩ 0.500V
m
sin a Ϫ 0.212V
m
cos 2a Ϫ 0.0424V
m
cos 4a Ϫ ⋅ ⋅ ⋅
= +
v
2V
m
0 ␲ ␻t
(c)
v
2
0 ␲
V
m
–V
m
␻t
(b)
v
1
0
V
m
␻t
(a)
␻ ␻ ␻ 3␲ 2␲ 3␲ 2␲
FIG. 25.18
Shifting a waveform vertically with the addition of a dc term.
(a)
v
1
a 0
V
m
2

+
(b)
v
2
a 0
V
m
p
(c)
v
T
a 0
V
m
2

=
V
m
2
p
2p 3p
3p 2p
FIG. 25.19
Lowering a waveform with the addition of a negative dc component.
The waveform in Fig. 25.19(c) is the sum of the two in Fig. 25.19(a) and
(b). The Fourier series for the waveform of Fig. 25.19(c) is, therefore,
v
T
ϭv
1
ϩv
2
ϭϪ ϩEq. (25.10)
V
m

2
(25.10)
ϭϪ0.500V
m
ϩ0.318V
m
ϩ0.500V
m
sin a Ϫ0.212V
m
cos 2a Ϫ0.0424V
m
cos 4a ϩ⋅ ⋅ ⋅
and v
T
ϭϪ0.182V
m
ϩ0.5V
m
sin a Ϫ0.212V
m
cos 2a Ϫ0.0424V
m
cos 4a ϩ⋅ ⋅ ⋅
If either waveform were shifted to the right or left, the phase shift
would be subtracted from or added to, respectively, the sine and cosine
terms. The dc term would not change with a shift to the right or left.
If the half-wave rectiﬁed signal is shifted 90° to the left, as in Fig.
25.20, the Fourier series becomes
v
p
2

p
2
0 p 2p 3
2
p
5
2
p
3p ␣
V
m
FIG. 25.20
Changing the phase angle of a waveform.
The equation for the half-wave rectiﬁed pulsating waveform of Fig.
25.19(b) is
1134  N O N SIN U SO ID A L C IRC U ITS
25.3 CIRCUIT RESPONSE TO A
NONSINUSOIDAL INPUT
The Fourier series representation of a nonsinusoidal input can be
applied to a linear network using the principle of superposition. Recall
that this theorem allowed us to consider the effects of each source of a
circuit independently. If we replace the nonsinusoidal input with the
terms of the Fourier series deemed necessary for practical considera-
tions, we can use superposition to ﬁnd the response of the network to
each term (Fig. 25.21).
NON
v ϭ 0.318V
m
ϩ 0.500V
m
sin(␣ ϩ 90°) Ϫ 0.212V
m
cos 2(␣ ϩ 90°) Ϫ 0.0424V
m
cos 4(␣ ϩ 90°) ϩ
• • •
ϭ 0.318V
m
ϩ 0.500V
m
cos ␣ Ϫ 0.212V
m
cos(2␣ ϩ 180°) Ϫ 0.0424V
m
cos(4␣ ϩ 360°) ϩ
• • •
v ϭ 0.318V
m
ϩ 0.500V
m
cos ␣ ϩ 0.212V
m
cos 2␣ Ϫ 0.0424V
m
cos 4␣ϩ
• • •
cos ␣
and
e = A
0
+ A
1
sin ␣ + . . . + A
n
sin n␣ + . . .
+ B
1
cos ␣ + . . . + B
n
cos n␣ + . . .
+

e
Linear
network
Linear network
+

A
1
sin ␣
+

+

A
n
sin n␣
+

B
1
cos ␣
+

B
n
cos n␣
A
0
FIG. 25.21
Setting up the application of a Fourier series of terms to a linear network.
The total response of the system is then the algebraic sum of the val-
ues obtained for each term. The major change between using this theo-
rem for nonsinusoidal circuits and using it for the circuits previously
described is that the frequency will be different for each term in the
nonsinusoidal application. Therefore, the reactances
X
L
ϭ2pfL and X
C
ϭ
will change for each term of the input voltage or current.
In Chapter 13, we found that the rms value of any waveform was
given by
Ί๶
͵

T
0

f
2

(t

)

d

t

If we apply this equation to the following Fourier series:
1

T
1

2pfC
C IRC U IT R ESPO N SE TO A N O N SIN U SO ID A L IN PU T  1135
v(a) ϭV
0
ϩV
m
1
sin a ϩ⋅ ⋅ ⋅ ϩV
m
n
sin na ϩV′
m
1
cos a ϩ⋅ ⋅ ⋅ ϩV′
m
n
cos na
NON
then
V
rms
ϭ
Ί
V

2
0

ϩ
๶๶๶๶
(25.11)
However, since
ϭ
΂ ΃ ΂ ΃
ϭ(V
1
rms
)(V
1
rms
) ϭV
2
1
rms
then
V
m
1

͙2ෆ
V
m
1

͙2ෆ
V
2
m
1

2
V
2
m
1
ϩ⋅ ⋅ ⋅ ϩV
2
m
n
ϩV′
2
m
1
ϩ⋅ ⋅ ⋅ ϩV′
2
m
n
ᎏᎏᎏᎏ
2
i(a) ϭI
0
ϩI
m
1
sin a ϩ⋅ ⋅ ⋅ ϩI
m
n
sin na ϩI′
m
1
cos a ϩ⋅ ⋅ ⋅ ϩI′
m
n
cos na
we have
I
rms
ϭ
Ί
I
2
0

ϩ
๶๶๶๶
(25.13)
and
I
rms
ϭ ͙I
2
0
ෆ ϩ ෆI ෆ
2
1
r

ms
ෆϩෆ ⋅ ෆ ⋅ ෆ ⋅ ෆ ϩ ෆI ෆ
2
n
r

ms
ෆϩෆ I ෆ′
2
1

rm

s
ෆϩෆ ⋅ ෆ ⋅ ෆ ⋅ ෆ ϩ ෆI ෆ′
2
n

rm

s
ෆ (25.14)
The total power delivered is the sum of that delivered by the corre-
sponding terms of the voltage and current. In the following equations,
all voltages and currents are rms values:
(25.15)
(25.16)
or (25.17)
with I
rms
as deﬁned by Eq. (25.13), and, similarly,
(25.18)
with V
rms
as deﬁned by Eq. (25.11).
EXAMPLE 25.5
a. Sketch the input resulting from the combination of sources in Fig.
25.22.
P
T
ϭ ᎏ
V
R
2
rms

P
T
ϭ I
2
rms
R
P
T
ϭ I
2
0
R ϩ I
2
1
R ϩ ⋅ ⋅ ⋅ ϩ I
2
n
R ϩ ⋅ ⋅ ⋅
P
T
ϭ V
0
I
0
ϩ V
1
I
1
cos v
1
ϩ ⋅ ⋅ ⋅ ϩ V
n
I
n
cos v
n
ϩ ⋅ ⋅ ⋅
I
2
m
1
ϩ⋅ ⋅ ⋅ ϩI
2
m
n
ϩI′
2
m
1
ϩ⋅ ⋅ ⋅ ϩI′
2
m
n
ᎏᎏᎏᎏ
2
v
+
4 V

+

6 sin ␻t ␻
FIG. 25.22
Example 25.5.
V
rms
ϭ ͙Vෆ
2
0
ෆϩෆ V ෆ
2
1
r

ms
ෆϩෆ ⋅ ෆ ⋅ ෆ ⋅ ෆ ϩ ෆV ෆ
2
n
r

ms
ෆ ϩ ෆV ෆ′
2
1

rm

s
ෆϩෆ ⋅ ෆ ⋅ ෆ ⋅ ෆ ϩ ෆV ෆ′
2
n

rm

s
ෆ (25.12)
Similarly, for
1136  N O N SIN U SO ID A L C IRC U ITS
NON
b. Determine the rms value of the input of Fig. 25.22.
Solutions:
a. Note Fig. 25.23.
b. Eq. (25.12):
V
rms
ϭ
Ί
V

2
0

ϩ
๶๶๶
ϭ
Ί
(4

V

)
2

ϩ
๶๶๶
ϭ
Ί
1

6

ϩ
๶๶๶
V ϭ͙3 ෆ4ෆ V
ϭ5.831 V
It is particularly interesting to note from Example 25.5 that the rms
value of a waveform having both dc and ac components is not simply
the sum of the effective values of each. In other words, there is a temp-
tation in the absence of Eq. (25.12) to state that V
rms
ϭ 4 V ϩ 0.707
(6 V) ϭ 8.242 V, which is incorrect and, in fact, exceeds the correct
level by some 41%.
Instrumentation
It is important to realize that not every DMM will read the rms value of
nonsinusoidal waveforms such as the one appearing in Fig. 25.23.
Many are designed to read the rms value of sinusoidal waveforms only.
It is important to read the manual provided with the meter to see if it is
a true rms meter that can read the rms value of any waveform.
We learned in Chapter 13 that the rms value of a square wave is the
peak value of the waveform. Let us test this result using the Fourier
expansion and Eq. (25.11).
EXAMPLE 25.6 Determine the rms value of the square wave of Fig.
25.15 with V
m
ϭ 20 V using the ﬁrst six terms of the Fourier expansion,
and compare the result to the actual rms value of 20 V.
Solution:
36

2
(6 V)
2

2
V
2
m

2
6 V
4 V
v = 4 V + 6 sin qt
0 qt
FIG. 25.23
Wave pattern generated by the source of
Fig. 25.22.
vϭ (20 V) sin qt ϩ
΂ ΃
(20 V) sin 3qt ϩ
΂ ΃
(20 V) sin 5qt ϩ
΂ ΃
(20 V) sin 7qt
1

7
4

p
1

5
4

p
1

3
4

p
4

p
ϩ
΂ ΃
(20 V) sin 9qt ϩ
΂ ΃
(20 V) sin 11qt
vϭ25.465 sin qt ϩ8.488 sin 3qt ϩ5.093 sin 5qt ϩ3.638 sin 7qt ϩ2.829 sin 9qt ϩ2.315 sin 11 qt
1

11
4

p
1

9
4

p
ϭ
Ί
(0

V

)
2

ϩ
๶๶๶๶๶๶
ϭ19.66 V
V
rms
ϭ
Ί
V

2
0

ϩ
๶๶๶๶
V
2
m
1
ϩV
2
m
2
ϩV
2
m
3
ϩV
2
m
4
ϩV
2
m
5
ϩV
2
m
6
ᎏᎏᎏᎏ
2
The solution differs less than 0.4 V from the correct answer of 20 V.
However, each additional term in the Fourier series will bring the result
closer to the 20-V level. An inﬁnite number would result in an exact
solution of 20 V.
Eq. (25.11):
(25.465 V)
2
ϩ(8.488 V)
2
ϩ(5.093 V)
2
ϩ(3.638 V)
2
ϩ(2.829 V)
2
ϩ(2.315 V)
2
ᎏᎏᎏᎏᎏᎏᎏᎏᎏ
2
C IRC U IT R ESPO N SE TO A N O N SIN U SO ID A L IN PU T  1137
NON
EXAMPLE 25.7 The input to the circuit of Fig. 25. 24 is the fol-
lowing:
e ϭ12 ϩ10 sin 2t
a. Find the current i and the voltages v
R
and v
C
.
b. Find the rms values of i, v
R
, and v
C
.
c. Find the power delivered to the circuit.
Solutions:
a. Redraw the original circuit as shown in Fig. 25.25. Then apply
superposition:
e
v
R
R = 3 ⍀
+

C = F
1
8
v
C
i
FIG. 25.24
Example 25.7.
10 sin 2t
v
R
R = 3 ⍀
+

v
C
i
+

X
C
=
1
␻C
1
=
1
8
= 4 ⍀
12 V
FIG. 25.25
Circuit of Fig. 25.24 with the components of the Fourier series input.
1. For the 12-V dc supply portion of the input, I ϭ 0 since the
capacitor is an open circuit to dc when v
C
has reached its ﬁnal
V
R
ϭIR ϭ0 V and V
C
ϭ12 V
2. For the ac supply,
Z ϭ3 ⍀ Ϫj 4 ⍀ ϭ5 ⍀ ЄϪ53.13°
and I ϭ ϭ ϭ A Єϩ53.13°
V
R
ϭ(I Єv)(R Є0°) ϭ
΂
A Єϩ53.13°
΃
(3 ⍀ Є0°)
ϭ V Єϩ53.13°
6

͙2ෆ
2

͙2ෆ
2

͙2ෆ

͙
10
2ෆ
ᎏV Є0°
——
5 ⍀ ЄϪ53.13°
E

Z
and
V
C
ϭ (I Єv)(X
C
ЄϪ90°) ϭ
΂
A Єϩ53.13°
΃
(4 ⍀ ЄϪ90°)
ϭ V ЄϪ36.87°
In the time domain,
i ϭ0 ؉2 sin(2t ؉53.13°)
Note that even though the dc term was present in the expression
for the input voltage, the dc term for the current in this circuit is
zero:
v
R
ϭ0 ؉6 sin(2t ؉53.13°)
and v
C
ϭ 12 ؉ 8 sin(2t ؊ 36.87°)
8

͙2ෆ
2

͙2ෆ
1138  N O N SIN U SO ID A L C IRC U ITS
b. Eq. (25.14): I
rms
ϭ
Ί
(0

)
2

ϩ
๶๶๶
ϭ ͙2 ෆ V ϭ 1.414 A
Eq. (25.12): V
R
rms
ϭ
Ί
(0

)
2

ϩ
๶๶๶
ϭ ͙1 ෆ8ෆ V ϭ 4.243 V
Eq. (25.12): V
C
rms
ϭ
Ί
(1

2

V

)
2

ϩ
๶๶๶
ϭ ͙1 ෆ7ෆ6ෆ V ϭ 13.267 V
c. P ϭ I
2
rms
R ϭ
΂
A
΃
2
(3 ⍀) ϭ 6 W
EXAMPLE 25.8 Find the response of the circuit of Fig. 25.26 to the
input shown.
2

͙2ෆ
(8 V)
2

2
(6 V)
2

2
(2 A)
2

2
NON
e
v
R
R = 6 ⍀
+

v
L
i
L = 0.1 H
q = 377 rad/s
E
m
= 200
0
p 2p 3p qt
(b)
(a)
e
FIG. 25.26
Example 25.8.
e ϭ0.318E
m
ϩ0.500E
m
sin qt Ϫ0.212E
m
cos 2qt Ϫ 0.0424E
m
cos 4qt ϩ⋅ ⋅ ⋅
Solution: For discussion purposes, only the ﬁrst three terms will be
used to represent e. Converting the cosine terms to sine terms and sub-
stituting for E
m
gives us
e ϭ63.60 ϩ100.0 sin qt Ϫ42.40 sin(2qt ϩ90°)
Using phasor notation, the original circuit becomes like the one shown
in Fig. 25.27.
V
R
6 ⍀
+ –
I
1
I
2
L = 0.1 H V
L
+

I
0
E
0
= 63.6 V
+

E
1
= 70.71 V ∠0°
+

E
2
= 29.98 V ∠90°
+

q = 377 rad/s
2q = 754 rad/s
Z
T
FIG. 25.27
Circuit of Fig. 25.26 with the components of the Fourier series input.
Applying Superposition For the dc term (E
0
ϭ 63.6 V):
X
L
ϭ0 (short for dc)
Z
T
ϭR Є0° ϭ6 ⍀ Є0°
I
0
ϭ ϭ ϭ10.60 A
V
R
0
ϭI
0
R ϭE
0
ϭ63.60 V
V
L
0
ϭ0
The average power is
P
0
ϭI
2
0
R ϭ(10.60 A)
2
(6 ⍀) ϭ674.2 W
For the fundamental term (E
1
ϭ 70.71 V Є0°, q ϭ 377):
X
L
1
ϭqL ϭ(377 rad/s)(0.1 H) ϭ37.7 ⍀
63.6 V

6 ⍀
E
0

R
A D D ITIO N A N D SU B TR AC TIO N O F N O N SIN U SO ID A L W AVEFO R M S  1139
Z
T
1
ϭ6 ⍀ ϩj 37.7 ⍀ ϭ38.17 ⍀ Є80.96°
I
1
ϭ ϭ ϭ1.85 A ЄϪ80.96°
V
R
1
ϭ(I
1
Єv)(R Є0°) ϭ(1.85 A ЄϪ80.96°)(6 ⍀ Є0°)
ϭ11.10 V ЄϪ80.96°
V
L
1
ϭ(I
1
Єv)(X
L
1
Є90°) ϭ(1.85 A ЄϪ80.96°)(37.7 ⍀ Є90°)
ϭ69.75 V Є9.04°
The average power is
P
1
ϭI
2
1
R ϭ(1.85 A)
2
(6 ⍀) ϭ20.54 W
For the second harmonic (E
2
ϭ 29.98 V ЄϪ90°, q ϭ 754): The
phase angle of E
2
was changed to Ϫ90° to give it the same polarity as
the input voltages E
0
and E
1
.
X
L
2
ϭqL ϭ(754 rad/s)(0.1 H) ϭ75.4 ⍀
Z
T
2
ϭ6 ⍀ ϩj 75.4 ⍀ ϭ75.64 ⍀ Є85.45°
I
2
ϭ ϭ ϭ0.396 A ЄϪ174.45°
V
R
2
ϭ(I
2
Єv)(R Є0°) ϭ(0.396 A ЄϪ174.45°)(6 ⍀ Є0°)
ϭ2.38 V ЄϪ174.45°
V
L
2
ϭ(I
2
Єv)(X
L
2
Є90°) ϭ(0.396 A ЄϪ174.45°)(75.4 ⍀ Є90°)
ϭ29.9 V ЄϪ84.45°
The average power is
P
2
ϭI
2
2
R ϭ(0.396 A)
2
(6 ⍀) ϭ0.941 W
The Fourier series expansion for i is
29.98 V ЄϪ90°
ᎏᎏ
75.64 ⍀ Є85.45
E
2

Z
T
2
70.71 V Є0°
ᎏᎏ
38.17 ⍀ Є80.96°
E
1

Z
T
1
NON
i ϭ10.6 ؉͙2 ෆ(1.85) sin(377t ؊80.96°) ؉͙2 ෆ(0.396) sin(754t ؊174.45°)
and
I
rms
ϭ͙(1 ෆ0ෆ.6 ෆA ෆ)
2
ෆ ϩ ෆ( ෆ1ෆ.8 ෆ5ෆ A ෆ)
2
ෆ ϩ ෆ( ෆ0ෆ.3 ෆ9ෆ6ෆ A ෆ)
2
ෆ ϭ10.77 A
The Fourier series expansion for v
R
is
v
R
ϭ63.6 ؉͙2 ෆ(11.10) sin(377t ؊80.96°) ؉͙2 ෆ(2.38) sin(754t ؊174.45°)
and
V
R
rms
ϭ͙(6 ෆ3ෆ.6 ෆV ෆ)
2
ෆ ϩ ෆ( ෆ1ෆ1ෆ.1 ෆ0ෆ V ෆ)
2
ෆ ϩ ෆ( ෆ2ෆ.3 ෆ8ෆ V ෆ)
2
ෆ ϭ64.61 V
The Fourier series expansion for v
L
is
v
L
ϭ͙2 ෆ(69.75) sin(377t ؉9.04°) ؉͙2 ෆ(29.93) sin(754t ؊84.45°)
and V
L
rms
ϭ͙(6 ෆ9ෆ.7 ෆ5ෆ V ෆ)
2
ෆ ϩ ෆ( ෆ2ෆ9ෆ.9 ෆ3ෆ V ෆ)
2
ෆ ϭ75.90 V
The total average power is
P
T
ϭI
2
rms
R ϭ(10.77 A)
2
(6 ⍀) ϭ695.96 WϭP
0
ϩP
1
ϩP
2
25.4 ADDITION AND SUBTRACTION OF
NONSINUSOIDAL WAVEFORMS
The Fourier series expression for the waveform resulting from the addi-
tion or subtraction of two nonsinusoidal waveforms can be found using
1140  N O N SIN U SO ID A L C IRC U ITS
phasor algebra if the terms having the same frequency are considered
separately.
For example, the sum of the following two nonsinusoidal waveforms
is found using this method:
v
1
ϭ30 ϩ20 sin 20t ϩ ⋅ ⋅ ⋅ ϩ5 sin(60t ϩ30°)
v
2
ϭ60 ϩ30 sin 20t ϩ20 sin 40t ϩ10 cos 60t
1. dc terms:
V
T
0
ϭ30 V ϩ60 V ϭ90 V
2. q ϭ 20:
V
T
1(max)
ϭ30 V ϩ20 V ϭ50 V
and v
T
1
ϭ50 sin 20t
3. q ϭ 40:
v
T
2
ϭ20 sin 40t
4. q ϭ 60:
5 sin(60t ϩ30°) ϭ(0.707)(5) V Є30° ϭ3.54 V Є30°
10 cos 60t ϭ10 sin(60t ϩ90°) ⇒(0.707)(10) V Є90°
ϭ7.07 V Є90°
V
T
3
ϭ3.54 V Є30° ϩ7.07 V Є90°
ϭ3.07 V ϩj 1.77 V ϩj 7.07 V ϭ3.07 V ϩj 8.84 V
V
T
3
ϭ9.36 V Є70.85°
and v
T
3
ϭ13.24 sin(60t ϩ70.85°)
with
NON
v
T
ϭv
1
ϩv
2
ϭ90 ؉50 sin 20t ؉20 sin 40t ؉13.24 sin(60t ؉70.85°)
25.5 COMPUTER ANALYSIS
PSpice
Fourier Series The computer analysis will begin with a veriﬁcation
of the waveform of Fig. 25.17, demonstrating that only four terms of a
Fourier series can generate a waveform that has a number of character-
istics of a square wave. The square wave has a peak value of 10 V at a
frequency of 1 kHz, resulting in the following Fourier series using Eq.
(25.9) (and recognizing that q ϭ 2pf ϭ 6283.19 rad/s):
vϭ (10 V)(sin qt ϩ sin 3qt ϩ sin 5qt ϩ sin 7qt)
vϭ12.732 sin qt ϩ4.244 sin 3qt ϩ2.546 sin 5qt ϩ1.819 sin 7qt
Each term of the Fourier series is treated as an independent ac source
as shown in Fig. 25.28 with its peak value and applicable frequency.
The sum of the source voltages will appear across the resistor R and
will generate the waveform of Fig. 25.29.
Each source used VSIN, and since we wanted to display the result
against time, we chose Time Domain(Transient) in the Simulation
Settings. For each source the Property Editor dialog box was called
up, and AC, FREQ, PHASE, VAMPL, and VOFF (at 0 V) were set,
although due to limited space only VAMPL, FREQ, and PHASE were
1

7
1

5
1

3
4

p
C O M PU TER A N A LYSIS  1141
NON
FIG. 25.28
Using PSpice to apply four terms of the Fourier expansion of a 10-V square
wave to a load resistor of 1 k ⍀.
FIG. 25.29
The resulting waveform of the voltage across the resistor R of Fig. 25.28.
1142  N O N SIN U SO ID A L C IRC U ITS
displayed in Fig. 25.28. Under Display the remaining quantitites were
all set on Do Not Display.
The Run to time was set at 2 ms so that two cycles of the funda-
mental frequency of 1 kHz would appear. The Start saving data after
will remain at the default value of 0s, and the Maximum step size will
be 1 ms, even though 2 ms/1000 ϭ 2 ms, because we want to have addi-
tional plot points for the complex waveform. Once the SCHEMATIC1
window appears, Trace-Add Trace-V(R:1)-OK will result in the wave-
form of Fig. 25.29. The horizontal line at 0 V was made heavier by
right-clicking on the line, selecting Properties, and then choosing the
green color and wider line. Click OK, and the wider line of Fig. 25.29
will result, making it a great deal clearer where the 0-V line is located.
Through the same process the curve was made yellow and wider as
shown in the same ﬁgure. Using the cursors, we ﬁnd that the ﬁrst peak
will reach 11.84 V and then drop to 8.920 V. The average value of the
waveform is clearly ϩ10 V in the positive region as shown by the red
line entered using Plot-Label-Line. In every respect the waveform is
certainly beginning to have the characteristics of a periodic square wave
with a peak value of 10 V and a frequency of 1 kHz.
Fourier Components A frequency spectrum plot revealing the
magnitude and frequency of each component of a Fourier series can be
obtained by returning to Plot and selecting Axis Settings followed by
X Axis and then Fourier under Processing Options. Click OK, and a
number of spikes will appear on the far left of the screen, with a fre-
quency spectrum that extends from 0 Hz to 600 kHz. By selecting Plot-
Axis Settings again, going to Data Range, and selecting User Deﬁned,
NON
FIG. 25.30
The Fourier components of the waveform of Fig. 25.29.
PROBLEMS
SECTION 25.2 Fourier Series
1. For the waveforms of Fig. 25.31, determine whether the
following will be present in the Fourier series representa-
tion:
a. dc term
b. cosine terms
c. sine terms
d. even-ordered harmonics
e. odd-ordered harmonics
NON
PRO B LEM S  1143
you can change the range to 0 Hz to 10 kHz since this is the range of
interest for this waveform. Click OK, and the graph of Fig. 25.30
will result, giving the magnitude and frequency of the components of
the waveform. Using the left cursor, we ﬁnd that the highest peak is
12.738 V at 1 kHz, comparing very well with the source V1 having a
peak value of 12.732 V at 1 kHz. Using the right-click cursor, we can
move over to 3 kHz and ﬁnd a magnitude of 4.246 V, again comparing
very well with source V2 with a peak value of 4.244 V.
f (t)
–T T
2

T
2
0 T t
A
m
A
m
(I)
f (t)
t T
2

T
2
0 T
(II)
f (t)
–T 0 T t
A
1
(III)
A
2
3
4
T
T
4
T
4

3
4
– T
f (t)
0 t
A
m
(IV)
A
m
T
2
3
T
T
3
T
3

FIG. 25.31
Problem 1.
1144  N O N SIN U SO ID A L C IRC U ITS
NON
2. If the Fourier series for the waveform of Fig. 25.32(a) is
i ϭ
΂
1 ϩ cos 2qt Ϫ cos 4qt ϩ cos 6qt ϩ ⋅ ⋅ ⋅
΃
2

35
2

15
2

3
2I
m

p
ﬁnd the Fourier series representation for waveforms (b)
through (d).
(a)
0
i
t ω
I
m
(b)
0
i
t ω
I
m
0
(c)
I
m
—–
2
I
m
—–
2
t ω
(d)
0
i i
t ω
I
m
FIG. 25.32
Problem 2.
3. Sketch the following nonsinusoidal waveforms with a ϭ
qt as the abscissa:
a. v ϭ Ϫ4 ϩ 2 sin a b. v ϭ (sin a)
2
c. i ϭ 2 Ϫ 2 cos a
4. Sketch the following nonsinusoidal waveforms with a as
the abscissa:
a. i ϭ 3 sin a Ϫ 6 sin 2a
b. v ϭ 2 cos 2a ϩ sin a
5. Sketch the following nonsinusoidal waveforms with qt as
the abscissa:
a. i ϭ 50 sin qt ϩ 25 sin 3qt
b. i ϭ 50 sin a Ϫ 25 sin 3a
c. i ϭ 4 ϩ 3 sin qt ϩ 2 sin 2qt Ϫ 1 sin 3qt
SECTION 25.3 Circuit Response to a
Nonsinusoidal Input
6. Find the average and effective values of the following
nonsinusoidal waves:
a. v ϭ 100 ϩ 50 sin qt ϩ 25 sin 2qt
b. i ϭ 3 ϩ 2 sin(qt Ϫ 53°) ϩ 0.8 sin(2qt Ϫ 70°)
7. Find the rms value of the following nonsinusoidal waves:
a. v ϭ 20 sin qt ϩ 15 sin 2qt Ϫ 10 sin 3qt
b. i ϭ 6 sin(qt ϩ 20°) ϩ 2 sin(2qt ϩ 30°)
Ϫ 1 sin(3qt ϩ 60°)
8. Find the total average power to a circuit whose voltage
and current are as indicated in Problem 6.
PRO B LEM S  1145
NON
e
v
R
R = 12 ⍀
+

v
L
i
L = 0.02 H
FIG. 25.33
Problems 10, 11, and 12.
e
v
R
R = 15 ⍀
+

v
C
i
C = 125 mF
FIG. 25.34
Problem 13.
9. Find the total average power to a circuit whose voltage
and current are as indicated in Problem 7.
10. The Fourier series representation for the input voltage to
the circuit of Fig. 25.33 is
e ϭ18 ϩ30 sin 400t
a. Find the nonsinusoidal expression for the current i.
b. Calculate the rms value of the current.
c. Find the expression for the voltage across the resistor.
d. Calculate the rms value of the voltage across the resis-
tor.
e. Find the expression for the voltage across the reactive
element.
f. Calculate the rms value of the voltage across the reac-
tive element.
g. Find the average power delivered to the resistor.
11. Repeat Problem 10 for
e ϭ24 ϩ30 sin 400t ϩ10 sin 800t
12. Repeat Problem 10 for the following input voltage:
e ϭϪ60 ϩ20 sin 300t Ϫ10 sin 600t
13. Repeat Problem 10 for the circuit of Fig. 25.34.
*14. The input voltage of Fig. 25.35(a) to the circuit of Fig.
25.35(b) is a full-wave rectiﬁed signal having the follow-
ing Fourier series expansion:
e
100 V
p
2

0
p
2
3
2
p
qt
(a)
1 k⍀
0.1 H
+
v
o

1 mF
+
e

(b)
e ϭ
΂
1 ϩ cos 2qt Ϫ cos 4qt ϩ cos 6qt ϩ ⋅ ⋅ ⋅
΃
2

53
2

15
2

3
(2)(100 V)
ᎏᎏ
p
where q ϭ377.
a. Find the Fourier series expression for the voltage v
o
using only the ﬁrst three terms of the expression.
b. Find the rms value of v
o
.
c. Find the average power delivered to the 1-k⍀ resistor.
FIG. 25.35
Problem 14.
200 ⍀
1.2 mH
+
v
o

200 mF
i
(b)
i
(a)
0 p 2p 3p
q
t
–p
q = 377
10 mA
FIG. 25.36
Problem 15.
*15. Find the Fourier series expression for the voltage v
o
of
Fig. 25.36.
1146  N O N SIN U SO ID A L C IRC U ITS
NON
SECTION 25.4 Addition and Subtraction
of Nonsinusoidal Waveforms
16. Perform the indicated operations on the following non-
sinusoidal waveforms:
a.
[60 ϩ70 sin qt ϩ20 sin(2qt ϩ90°) ϩ10 sin(3qt ϩ60°)]
ϩ [20 ϩ 30 sin qt Ϫ 20 cos 2qt ϩ 5 cos 3qt]
b.
[20 ϩ60 sin a ϩ10 sin(2a Ϫ180°) ϩ5 cos(3a ϩ90°)]
Ϫ [5 Ϫ 10 sin a ϩ 4 sin(3a Ϫ 30°)]
17. Find the nonsinusoidal expression for the current i
s
of the
diagram of Fig. 25.37.
i
2
ϭ10 ϩ30 sin 20t Ϫ0.5 sin(40t ϩ90°)
i
1
ϭ20 ϩ4 sin(20t ϩ90°) ϩ0.5 sin(40t ϩ30°)
i
1
i
s
i
2
FIG. 25.37
Problem 17.
e
+

v
1 + –
v
2
+

FIG. 25.38
Problem 18.
18. Find the nonsinusoidal expression for the voltage e of the
diagram of Fig. 25.38.
v
1
ϭ20 Ϫ200 sin 600t ϩ100 cos 1200t ϩ75 sin 1800t
v
2
ϭϪ10 ϩ150 sin(600t ϩ30°) ϩ50 sin(1800t ϩ60°)
SECTION 25.5 Computer Analysis
PSpiceor Electronics Workbench
19. Plot the waveform of Fig. 25.13 for two or three cycles.
Then obtain the Fourier components, and compare them
to the applied signal.
20. Plot a half-rectiﬁed waveform with a peak value of 20 V
using Eq. (25.10). Use the dc term, the fundamental term,
and four harmonics. Compare the resulting waveform to
the ideal half-rectiﬁed waveform.
21. Demonstrate the effect of adding two more terms to the
waveform of Fig. 25.29, and generate the Fourier spec-
trum.
Computer Language(C؉؉, QBASIC, Pascal, etc.)
22. Write a program to obtain the Fourier expansion resulting
from the addition of two nonsinusoidal waveforms.
23. Write a program to determine the sum of the ﬁrst 10
terms of Eq. (25.9) at qt ϭ p/2, p, and (3/2)p, and com-
pare your results to the values determined by Fig. 25.15.
That is, enter Eq. (25.9) into memory, and calculate the
sum of the terms at the points listed above.
G LO SSA RY  1147
NON
24. Given any nonsinusoidal function, write a program that
will determine the average and rms values of the wave-
form. The program should request the data required from
the nonsinusoidal function.
25. Write a program that will provide a general solution for
the network of Fig. 25.24 for a single dc and ac term in
GLOSSARY
Axis symmetry A sinusoidal or nonsinusoidal function that
has symmetry about the vertical axis.
Even harmonics The terms of the Fourier series expansion
that have frequencies that are even multiples of the funda-
mental component.
Fourier series A series of terms, developed in 1826 by Baron
Jean Fourier, that can be used to represent a nonsinusoidal
function.
Fundamental component The minimum frequency term re-
quired to represent a particular waveform in the Fourier se-
ries expansion.
Half-wave (mirror) symmetry A sinusoidal or nonsinu-
soidal function that satisﬁes the relationship
f (t) ϭ Ϫf
΂
ϩ t
΃
T

2
Harmonic terms The terms of the Fourier series expansion
that have frequencies that are integer multiples of the fun-
damental component.
Nonsinusoidal waveform Any waveform that differs from
the fundamental sinusoidal function.
Odd harmonics The terms of the Fourier series expansion
that have frequencies that are odd multiples of the funda-
mental component.
Point symmetry A sinusoidal or nonsinusoidal function that
satisﬁes the relationship f (a) ϭ Ϫf (Ϫa).
the applied voltage. In other words, the parameter values
are given along with the particulars regarding the applied
signal, and the nonsinusoidal expression for the current
and each voltage is generated by the program.
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