VDOC.PUB

### Automatic Control Systems [PDF]

This document was uploaded by our user. The uploader already confirmed that they had the permission to publish it. If you are author/publisher or own the copyright of this documents, please report to us by using this DMCA report form.

#### E-Book Content

ISM for

Chapter 2 MATHEMATICAL FOUNDATION 2-1 (a)

Poles: s = 0, 0, Zeros: s =

−1, −10;

(b)

Poles: s =

−2, ∞, ∞, ∞.

−2, −2;

Zeros: s = 0. The pole and zero at s =

(c)

Poles: s = 0,

−1 + j, −1 − j;

Zeros: s =

(d)

Poles: s = 0,

−1, −2, ∞.

−2.

2-2 (a)

(b) G (s) =

5

( s + 5)

(c) G ( s) =

2

(d)

(s

4s 2

(e) G ( s)

=

−1 cancel each other.

1 s

2

G (s)

+4

+4

=

)

∑e

+

1

G (s)

s+ 2

kT ( s + 5 )

1

= 1

k =0

−e

2-3 (a) g ( t ) = u s ( t ) − 2u s (t − 1) + 2 u s( t − 2) − 2 u s ( t − 3) + L G (s ) =

1 s

(1 − 2e − s + 2e−2 s − 2e −3s + L ) =

gT (t ) = u s (t ) − 2us (t − 1) + us (t − 2) GT (s ) =

1 s

(1 − 2e − s + e −2s ) = ( 1 − e − s ) 1

1−e

(

−s

s 1+ e

−s

0≤ t ≤ 2 2

s

1

)

−T ( s+5 )

=

4 s

2

+ 4s +8

g(t )

=

g

T

− 2k )us (t − 2k )

(t

G (s)

∑s

=

k =0

1

−e

(1

−s

2

) e

−2 ks

=

k =0

−s

1− e

s (1 + e

−s

)

(b) g ( t) = 2tu s ( t ) − 4(t − 0.5) u s (t − 0.5) + 4(t − 1) us (t − 1) − 4(t − 1.5)us (t − 1.5) + L G ( s) = g

T

(1 − 2e

2 2

s

−0.5 s

+ 2e

−s

− 2e

−1.5 s

− 0.5 s ( ) + L) = 2 −0.5 s s (1 + e )

2 1−e

= 2 tu s ( t ) − 4 ( t − 0 . 5) u s ( t − 0 . 5) + 2( t − 1 ) u s ( t − 1 )

(t )

(1 − 2e−0.5 s + e− s ) = s 2 (1 − e−0.5 s )

2

GT ( s ) =

s

2

2

g (t ) =

k=0

G(s ) =

≤ t ≤1

2

g T ( t − k )us ( t − k )

0

∑ s2 ( 2

1 −e

−0.5 s

k=0

)

2

e

− ks

=

( −0.5 s ) 2 −0.5s s (1 + e ) 2 1−e

2-4 g(t )

= ( t + 1 ) u s ( t ) − ( t − 1 ) u s ( t − 1 ) − 2 u s ( t − 1 ) − ( t − 2 ) u s ( t − 2 ) + ( t − 3) u s ( t − 3) + u s ( t − 3)

G ( s) =

s

2-5 (a)

(1 − e − s − e −2 s + e −3 s ) + s (1 −2e − s + e −3 s )

1

1

2

Taking the Laplace transform of the differential equation, we get 1 1 2 ( s + 5s + 4) F ( s) = F (s) = s +2 ( s + 1)( s + 2 )( s f (t )

=

1

e

−4 t

+

6

(b)

sX ( s ) 1

1

e

−t

3

− x1( 0 ) =

1

e

−2 t

t

+4)

=

1 6( s

+ 4)

=

X (s)

=

2

2

s

X (s)

=1

x (0)

2

s( s

+ 3 s +1

+ 1 )( s + 2 ) −1

(s

+ 1 )( s + 2 )

1

= 0 .5 + e

−t

− 0 .5 e

3( s

+ 1)

1 2( s

+ 2)

≥0

1

sX

2

(s)

− x 2 ( 0 ) = −2 X 1 ( s ) − 3 X 2 ( s ) +

1

= =

2s

+

−1 s

+1

1 s

+

+1

1 2( s

+ 2)

1 s

+2

Taking the inverse Laplace transform on both sides of the last equation, we get x (t )

1

2

Solving for X1 (s) and X2 (s), we have X 1 ( s)

+

−2 t

t

≥0

x (t ) 2

= −e

2

−t

+e

−2 t

t

≥0

1 s

x (0) 2

=0

2-6 (a) G (s)

(b) G (s)

=

(c) G (s ) =

=

1 3s

−2 . 5

(

50

5

s

(s

20

=

1 s

s s

2

g(t )

= 0 .5 t

2

e

−0.5 t

2 .5

+

s

2

=

1 s

g(t )

+3

s +4

−1

g (t )

+ 3)

3( s

30s + 20

+s+2

g (t ) = 1 + 1.069e

(e)

s +1

(d) G (s)

+ 1)

2

1

+

+ 2)

2( s

+

+1

s

1

+

)

e

s

1

1

e

3

2

= −2 . 5 e

−t

[

−s

−2 t

+s +2

+

1

e

−3 t

t

≥0

3

+ 5 te

g (t ) = 50 − 20e

1 2

=

−t

− (t −1)

+ 2 .5 e

−3 t

t

≥0

] us (t − 1)

− 30cos2(t − 1) − 5sin2(t − 1)

s s

2

Taking the inverse Laplace transform,

+s+2

[ sin1.323t + sin (1.323t − 69.3o ) ] = 1 + e−0.5 t (1.447sin1.323t − cos1.323t )

−t

t

≥0

2-7

 −1 2 0  A =  0 −2 3     −1 −3 −1 

2-8

0 0  B = 1 0    0 1 

u (t ) =

(a)

 u1( t)   u ( t)   2 

(b) Y (s )

=

R (s )

3s + 1 3

Y (s)

2

s + 2 s +5s + 6

=

R (s )

(c)

5 4

2

s + 10 s + s + 5

(d)

Y (s ) R (s )

=

s ( s + 2) 4

3

Y (s )

2

s + 10 s + 2 s + s + 2

R (s )

3

=

1+ 2e 2

−s

2s + s + 5

t≥0

4

5

6

7

8

9

10

11

12

13

Chapter 4 MATHEMATICAL MODELING OF PHYSICAL SYSTEMS

4-1 (a) Force equations: 2

f ( t) = M 1

d y1 dt

 dy1 − dy 2  + K ( y − y )  1 2  dt dt 

dy1

+ B1

2

+ B3 

dt

2 dy1 dy2  d y2 dy2  B3  −  + K ( y 1 − y 2 ) + M 2 2 + B2 dt dt  dt dt 

Rearrange the equations as follows: 2

d y1 dt

=−

2

2

d y2 dt (i) State diagram:

2

(B

1

+ B 3 ) dy1

M1

B3 dy1

=

dt −

(B

M 2 dt

Since y

1

− y2

2

B3 dy2

+

M 1 dt

+ B3 ) dy2

M

1

dt

dx2 dt

=

K

x1 −

M2

(ii) State variables:

x

1

(B

=

2

+ B3 )

x2 +

M2 y , 2

x

2

=

dy

B3 M2 2

dx3

x3 x

,

1

dt

1

dt dx dt

3

= x2 =

x

dx dt dx

4

2

dt

4

=− =

K M

K

=

M

x

− y2 )

y ,

x

1

1

B

1

+

1

State diagram:

14

B M

2

+ B3

M 3

1

x

2

dy

=

4

1

x

2

K M

2

x

,

dt

B3 M1

x2 −

=

3

(B

1

dy

1

.

dt

+ B3 )

M1

x3 +

.

dt

+

2

dy

=

2

x1 +

M1

2

x

K

=−

dt

3

x

2

State equations: dx

M1

appears as one unit, the minimum number of integrators is three.

1

= − x2 + x3

f

− y2 ) +

1

(y

M2

State equations: Define the state variables as x = y − y ,

dx1

(y

M1 K

+

dt

2

K

x 1

3

K M

x 2

B

1

3

+

B

+ B3 M

1

3

M x

x

4

2

4

+

1 M

f 1

1 M

f

Transfer functions:

Y1 ( s ) F (s ) Y2 ( s ) F ( s)

M 2 s + ( B2 + B3 ) s + K 2

= =

{

3

{

3

s M 1 M 2 s + [( B1 + B 3 ) M 2 + ( B2 + B3 ) M 1 ] s + [K ( M 1 + M 2 ) + B1 B2 + B2 B3 + B1 B3 ] s + ( B 1 + B2 ) K 2

B3 s + K

s M1 M 2 s + [( B1 + B3 ) M 2 + ( B2 + B3 ) M1 ] s + [ K ( M1 + M 2 ) + B1 B2 + B2 B3 + B1 B3 ] s + ( B1 + B2 ) K 2

(b) Force equations: 2

d y1 dt

2

=−

(B

+ B2 ) dy1

1

M

+

dt

B2 dy2

+

M dt

1

dy2

f

M

dy1

=

dt

dt

K

B2

(i) State diagram:

Define the outputs of the integrators as state variables, x

1

=

y ,

x

2

=

2

dy

1

.

dt

State equations: dx

1

dt

=−

K B

x

1

dx

+ x2

dt

2

(ii) State equations: State variables: dx dt

1

=−

K B

x

1

+ x3

2

2

dx dt

2

x

1

=

=−

K M

y ,

x

2

dx

= x3

dt

Transfer functions:

15

x

3

2

1

=

=−

B

1

x

M

y ,

x

1

K M

2

x

1

3

1

+ = B

f

M

1

M

dy

1

.

dt x

3

+

1 M

f

y2

} }

Y1 ( s ) B2 s + K = 2 F (s ) s  MB2 s + ( B1 B2 + KM ) s + ( B1 + B2 ) K 

Y2 ( s) F ( s)

(c) Force equations: dy1

dy2

=

dt

2

1

+

dt

d y2

f

B1

dt

=−

2

(B

1

B2

=

M B2 s + ( B1 B2 + KM ) s + ( B1 + B2 ) K 2

+ B2 ) dy2 M

B1 dy2

+

dt

B1 dy1

+

M dt

M dt

(i) State diagram:

State equations: Define the outputs of integrators as state variables. dx

= x2

1

dt

dx

=−

2

dt

(ii) State equations: state variables: dx dt

1

= x3 +

1 B

dx

f

2

dt

1

x

1

K

x

M

1

=

y , 1

B

M x

dx

= x3

x

2

=

2

3

dt

2

+

1

f

M

y ,

x

2

=−

K

x

M

2

3

= −

dy

2

.

dt

B

2

M

x

3

State diagram:

Transfer functions:

Y1 ( s ) F (s )

Ms + ( B1 + B 2 ) s + K 2

=

(

B1 s Ms + B2 s + K 2

Y2 ( s )

)

F (s )

4-2 (a) Force equations:

16

=

1 Ms + B 2 s + K 2

+

1 M

f

K M

y2

y

=

1

1 K

+ Mg ) + y 2

( f

2

y

dt

2

d

2

=−

2

B dy M

2

dt

K

1

+ K2

y

M

+

2

K

2

M

y

1

State diagram:

State equations: Define the state variables as: x = y , 1

dx

dx

= x2

1

dt

=−

2

dt

K

1

x

M

x

2

1

B

x

M

2

2

=

+

dy

2

.

dt 1

( f

+ Mg

Ms

2

M

)

Transfer functions: Y (s) 1

F ( s)

s

=

2

+ Bs + K 1 + K 2

K ( Ms 2

2

Y (s) 2

+ Bs + K 1 )

F (s)

1

=

+ Bs + K 1

(b) Force equations: dy1 dt

=

1 B1

[ f ( t) + Mg] +

dy2 dt

K1 B1

(y

1

− y2 )

2

d y2 dt

2

=

B 1  dy 1

dy  K B B dy −   + ( y − y ) − ( y − y )− M  dt dt  M M M dt 2

1

2

1

2

2

1

2

State diagram: (With minimum number of integrators)

To obtain the transfer functions Y ( s ) / F ( s ) and Y ( s ) / F ( s ), we need to redefine the state variables as: x

1

=

y , 2

x

2

= dy 2

1

/ dt , and x

3

=

2

y . 1

State diagram:

17

2

Transfer functions:

Y1 ( s )

Ms + ( B1 + B 2 ) s + K 1 2

=

F (s )

s

2

[MBs + ( B B 1

4-3 (a) Torque equation: 2 d θ B dθ =− + 2 dt

1

2

Y2 ( s )

+ MK1 )]

=

F (s )

Bs + K 1

s [M B1 s + ( B1 B2 + MK1 ) ] 2

State diagram: 1

J dt

T (t )

J

State equations: dx

dx

= x2

1

dt

2

dt

=−

B

x

J

+

2

1

T

J

Transfer function:

Θ( s )

1

=

T (s)

+ B)

s ( Js

(b) Torque equations: d θ1 2

dt

2

=−

K J

1

−θ 2 ) +

1

dθ 2

K (θ 1 − θ 2 ) = B

T

J

dt

State diagram: (minimum number of integrators)

State equations: dx

1

dt

=−

K

x

B

1

dx

+ x2

dt

State equations: Let x = θ , 1

dx dt

1

=−

K B

x

1

+

K B

=−

2

x

2

x dx

2

dt

2

2

K

x

J

= θ1, =

x

1

+

1

T

J

and

x

3

dx 3

dt

State diagram:

18

= 3

1

.

dt

=

K J

x

1

K J

x

2

+

1 J

T

Transfer functions:

Θ1 ( s ) T ( s)

=

Bs + K

(

s BJs + JKs + BK 2

Θ2 (s )

)

T ( s)

=

K

(

s BJs + JKs + BK 2

)

(c) Torque equations: d θ1 2

T ( t ) = J1

dt

2

+ K (θ 1 − θ 2 )

d θ2 2

K (θ 1 − θ 2 ) = J 2

2

dt

State diagram:

State equations: state variables: dx dt

1

dx

= x2

dt

2

=−

K J

x

x

1

2

1

+

=θ2, K J

x

x

2

= dx

3

2

dt

2

x

,

dt 3

=

x

3

= θ1, dx

4

dt

4

x

=

=

4

K J

x

1

.

dt

1

1

K J

1

x

3

+

1 J

T

1

Transfer functions:

Θ1 ( s ) T ( s)

J 2s + K

Θ 2 (s )

2

=

s

2

 J1 J2 s + K ( J1 + J 2 )  2

T ( s)

=

K 2

s

 J1 J 2 s + K ( J1 + J 2 )  2

(d) Torque equations: d θm 2

T ( t) = J m

dt

2

+ K 1 (θ m − θ 1 ) + K 2 (θ m − θ 2 )

K 1 (θ m − θ 1 ) = J 1

19

d θ1 2

dt

2

K2 (θm − θ2 ) = J2

d θ2 2

dt

2

State diagram:

State equations: x = θ 1

dx

1

dt

dx

= −x2 + x3

2

dt

=

K J

1

x

m

− θ1,

dx 1

dt

1

3

x

=−

K J

2

1

=

x

1

x

,

dt

1

m

K J

2

x

3

+

4

m

=

x

,

dt

1 J

m

T

m

dx dt

4

4

= θm −θ2 ,

=

x

3

− x5

x

dx dt

5

5

=

=

2

.

dt

K J

x

2

4

2

Transfer functions:

Θ1 ( s ) T ( s) Θ 2 ( s) T (s )

K 1 (J 2 s + K 2 ) 2

=

s

2

 s 4 + ( K1 J2 J m + K2 J 1 J m + K1 J 1 J 2 + K 2 J 1 J 2 ) s 2 + K1 K 2 ( J m + J 1 + J 2 )  K2 ( J1 s + K1 ) 2

=

s

2

 s 4 + ( K1 J2 J m + K 2 J 1 J m + K1 J 1J 2 + K 2J 1J 2 ) s 2 + K1 K 2 ( J m + J 1 + J 2 ) 

(e) Torque equations: d 2θ m dt

2

=−

K1 Jm

( θ m − θ1 ) −

K2 Jm

( θ m − θ2 ) +

1 Jm

d θ1 2

T

dt

2

K1

=

J1

State diagram:

20

m

− θ1 ) −

B1 d θ1 J1 dt

d 2θ 2 dt

2

=

K2 J2

m

−θ 1 ) −

B 2 dθ 2 J 2 dt

State variables:

x

1

= θ m − θ1,

x

2

=

1

x

,

dt

=

3

m

x

,

dt

= θm −θ2 ,

4

x

=

5

.

2

dt

State equations: dx 1 dt

= −x 2 + x 3

dx

2

dt

K1

=

J

x

1

B1

J

1

dx 3

x

2

dt

1

Transfer functions:

Θ1 ( s )

(

K1 J 2 s + B2 s + K 2

=

K1

=−

2

J

x

1

K

m

J

2

)

J

Θ2 ( s )

dx

T

dt

m

=

(

dx

= x3 − x5

4

=

5

dt

K 2 J1 s + B1 s + K1 2

K2 J

x

4

2

B2 J

d θm 2

dt

2

Output equation:

+ Bm e

=

o

dθ m dt Eθ 20

∆( s )

3

m

2

1

1

2

m

1

2

2

1

2

2

1

2

1

1

m

1

1

2

m

2

1

1

2

1

2

m

2

+ K (θ m − θ L )

K ( θm − θ L ) = J L

d θL 2

dt

2

+ Bp

dθ L dt

L

π

State diagram:

Transfer function:

Θ L (s )

=

Tm ( s) Eo ( s )

=

Tm ( s)

(

)

K

(

)

s J m J L s + Bm J L + Bp J m s + Jm K + J L K + Bm B p s + Bm K  3

(

)

2

KE / 2 0π

(

)

s J m J L s + Bm J L + Bp J m s + Jm K + J L K + Bm B p s + Bm K  3

2

4-5 (a) d θ1 2

Tm ( t) = Jm

θ3 =

N1 N3 N2N4

dt θ1

2

+ T1

T2 =

T1 = N3 N4

N1 N2

T4 =

T3 =

T2

N3 N4

2

)

4-4 System equations: Tm ( t) = Jm

x

(B + B ) s + [ ( K J + K J ) J + ( K + K ) J J + B B J ] s + B K ) J + B K J + B K J ] s + K K ( J + J + J )} 4

+ [( B1K 2

1

+

T (s )

∆ ( s ) = s { J 1 J2 Jm s + J 2

4

m

∆( s )

T ( s)

x

d θ3 2

JL

dt

2

N3 N4

d θ3 2

T4 = J L

T4

dt

2

T2 = T3

θ2 =

N1 N2

θ1

2   N1 N3   d 2θ1 Tm = J m 2 + T4 =  J m +   JL  2 dt N 2N 4  N 2 N 4   dt 

d θ1 2

21

N1 N 3

2

5

(b) d θ1 2

Tm = Jm θ2 =

dt

N1

2

+ T1

2

θ1

N2

d θ2

T2 = J2 N 1N 3

θ3 =

N2N4

dt

T4 = ( J3 + J L )

+ T3

2

d θ2 2

θ1

T2 = J 2

dt

2

2

N1

dt

T4 = J 2

N4

2 d θ3    d 2θ 2 N3 Tm ( t) = Jm +  J 2 dt 2 + N ( J3 + J 4 ) dt 2  =  Jm 2 dt N2    4

d θ1

2

T1 =

2

d θ2 2

N3

+

d θ3

dt

2

+

N3 N4

(J

N1 N2

T3 =

T2

+ JL ) 3

N3 N4

d θ3 2

dt

2

2 2  d 2θ1  N1   N1 N3  +  J2 +  N N  ( J3 + J L )  dt 2  N2   2 4 

4-6 (a) Force equations:

 dy1 − dy 2    dt dt 

2  dy1 − dy 2  = M d y 2 + B dy 2  t 2 dt dt  dt dt 

f ( t) = K h ( y1 − y2 ) + Bh  (b) State variables:

x

1

=

y

1

− y2,

x

dy

=

2

K h ( y1 − y2 ) + Bh  2

dt

State equations: dx

=−

1

dt

K B

h

x

1

1

+

B

h

dx

f (t )

2

dt

h

=−

B

t

x

M

2

+

1

f (t )

M

4-7 (a) T

m

= Jm

2

d

θm

dt

2

d θm

+T1

2

Tm = Jm

Set

∂α L ∂n

dt

2

= 0.

T

=JL

d θL

2

2

θL

dt

2

d

(T

m

dt

2

T

N

1

T

2

= nT 2

(

2

)

(

θm N 1 = θ L N 2

2

nTm − n TL 2

Thus, α L =

)

− 2 nTL ) J m + n J L − 2nJL nTm − n J L = 0

n

=−

J T m

2

L

2J T

L

2

Or, n

Jm + n JL 2

2

+

J T m

L

J T

+

2

J mTL

+ 4 J m J LT m

Torque equation about the motor shaft:

=

L m

L

Relation between linear and rotational displacements:

22

J J

m L

where the + sign has been chosen.

2J T

Jm / J

n

2

= 0 , the optimal gear ratio is n

4-8 (a)

N

+ nTL = 

L m

When T

=

L m

Optimal gear ratio:

(b)

1

 J m + nJ  α + nT L  L L  n 

2

+ nJL

+TL

=0

T4

T

(b)

m

2

d

= Jm

θm

dt

2

+ Mr

2

d

2

θm

dt

+ Bm

2

m

= r θm

y

dt

Taking the Laplace transform of the equations in part (a), with zero initial conditions, we have

(

Tm ( s) = Jm + Mr

)sΘ

2

2

m

( s) + Bm sΘ m (s )

Y ( s) = rΘ m ( s)

Transfer function:

Y ( s) Tm ( s)

=

r

(

s  J m + Mr

r

)s + B

m



4-9 (a) d θm 2

Tm = Jm

dt

2

(

+ r ( T1 − T2 )

2

T1 − T2 = M

dt

d θm 2

d y

Thus, Tm = J m

2

(c) State equations: dx1 = rx3 − x2 dt

dx2

=

Tm ( s)

=

dt

2

K1 + K 2

dt

(d) Transfer function: Y ( s)

)

T1 = K2 rθ m − rθ p = K 2 ( rθ m − y )

M

+ r ( K1 + K2 )( rθ m − y )

dx3

x1

=

dt

T2 = K1 ( y − rθ m ) 2

M

− r ( K1 + K 2 ) Jm

d y dt

x1 +

= ( K1 + K2 )( rθ m − y )

2

1 Jm

Tm

r ( K1 + K 2 )

s

2

 Jm Ms + ( K1 + K2 ) ( Jm + rM ) 2

(e) Characteristic equation:

s  J m Ms + ( K1 + K2 ) ( Jm + rM )  = 0 2

2

4-10 (a) Torque equations: d θm 2

Tm ( t) = Jm

dt

2

+ Bm

dθ m dt

+ K (θ m − θ L )

K ( θm − θ L ) = J L

State diagram:

23

d θL 2

dt

2

+ BL

dθ L dt

(b) Transfer functions: Θ L ( s) Tm ( s)

=

K

Θ m ( s)

∆ ( s)

Tm ( s)

=

J L s2 + BL s + K

∆ ( s) = s J mJ L s3 + ( Bm J L + BL J m ) s 2 + ( KJ m + KJ L + Bm BL ) s + Bm K 

∆ (s )

(c) Characteristic equation:

∆( s ) = 0

T (t ) m

= T m = consta

T (s)

nt.

m

=

T

m

.

s

J L s + BL s + K 2

lim ω m ( t) = lim sΩ m ( s) = lim t →∞

(e)

s →0

s →0

ω m = ω L.

ωm

4-11 (a) State equations: dθ L =ωL

dt

dθ m dt

= ωm

L

dt

dω m dt

=

and

K J

=−

2

J m J L s + ( Bm J L + BL J m ) s + ( KJ m + KJ L + Bm BL ) s + Bm K

ωL

θm −

L

Bm Jm

3

2

do not depend on J

K J

ωm −

2

θL

L

(K

1

dt

+ K2 ) Jm

(b) State diagram:

(c) Transfer functions:

24

m

t

and J . L

= ωt

θm +

K1 Jm

θt +

dω dt

K2 Jm

t

=

θL +

K J

1

θm −

t

1 Jm

Tm

K J

1

t

θt

=

1 Bm

ΘL ( s )

=

(

K 2 J t s + K1 2

)

Θ t(s )

∆ ( s)

Tm ( s)

=

(

K1 J L s + K 2 2

)

Θ m (s )

∆ ( s)

T m( s)

J t J L s + ( K1 J L + K 2 J t ) s + K1 K 2 4

=

2

∆ ( s)

Tm ( s )

∆ ( s ) = s [ J mJ Ls + B mJ LJ t s + ( K1 J L J t + K 2J L J t + K 1Jm J L + K 2J m J t ) s 5

4

3

+ Bm J L ( K1 + K 2 ) s + K1 K2 ( J L + J t + J m ) s + BmK 1K 2 ] = 0 2

(d) Characteristic equation: ∆ ( s ) = 0 .

4-12 (a) K H (s )  −K 1  H i (s )   1 + K1 H e ( s ) + 1 i H e (s) +     B + Js  Ra + L a s  B + Js  Ra + La s  = ≅ =0 −1

Ω m (s ) TL ( s)

∆( s)

ωr = 0

∆( s)

Thus,

H e (s ) = −

H i (s )

H i (s)

Ra + La s

H e (s)

Ω m (s )

(b)

=

Ω r ( s) ∆ ( s ) = 1 + K1 H e ( s ) + = 1+

(R

a

Ω m (s ) Ω r ( s)

= TL =0

(R

(R

(R

+ La s ) ( B + Js )

K1 K b

K1 K b

+ La s )( B + Js )

+ La s )( B + Js )

+

R a + La s

d

2

θ 2

+

(R

K1 K i K b H e( s) a

+ La s ) ( ( B + Js )

K 1 Ki

(R

+ La s ) ( B + Js )

a

+ La s ) ( B + Js ) + Ki Kb + K1 Ki Kb H e ( s)

a

dt

F d a

1

= T s d 2 δ + K F d 1θ

= J α α1 = J

d

2

dt

θ 2

K

F

1

Kb H e ( s)

d

1

θ

sin

δ≅δ

− K F d 1θ = T s d 2δ

Θ( s ) − K F d 1 Θ( s ) = T s d 2 ∆ ( s )

(b)

Js

(c)

With C and P interchanged, the torque equation about C is:

2

K1 H i ( s)

+

K1 K i

dt

J

a

∆( s)

4-13 (a) Torque equation: (About the center of gravity C) 2 d θ J = T s d 2 sin δ + F d d 1 2 Thus,

K1 K i

TL =0

a

= − ( Ra + L a s )

Ts ( d1 + d 2 ) δ + Fα d 2 = J

d θ 2

dt

2

Ts ( d 1 + d 2 ) δ + K F d 2θ = J

Js Θ(s ) − K F d 2 Θ( s ) = Ts ( d1 + d 2 ) ∆( s ) 2

25

Θ( s ) ∆ (s )

=

d θ 2

dt

Ts ( d1 + d 2 ) Js − K F d 2 2

2

θe = θr −θo

4-14 (a) Cause-and-effe ct equations: dia

Ra

=−

dt

d θm

La

dt

d θo

=−

2

dt

2

JL

a

1

+

J m dt

KL

=

(e

La

Bm dθ m

2

2

1

ia +

J

− eb )

e

= Ke

a

Tm = K ii a nKL

Tm −

= K sθ e

e

( nθ

Jm

m

− θo )

Tm

T2 =

θ 2 = nθ m

n

− θo )

2

State variables:

x

1

= θo,

x

= ωo ,

2

x

=θm,

3

x

4

= ωm,

x

5

= ia

State equations: dx

1

dt dx

4

dt

dx

= x2

=−

=−

2

dt nK J

K J

L

x

x

1

n KL J

m

x

3

nK J

L

2

L

+ 1

Bm

J

m

m

x

4

L

x

dx 3

dt

L

+

Ki J

x

3

= dx

5

x

5

dt

m

4

=−

KK L

s

x

1

a

Kb L

x

4

a

Ra L

x

5

+

KK

a

L

s

a

(b) State diagram:

(c) Forward-path transfer function: Θ o ( s) KK s K i nK L = 4 3 2 2 Θe ( s ) s J mJ L La s + J L ( Ra J m + Bm J m + Bm La ) s + n K LL a J L + K L J m La + Bm R a J L s +

(

(n R K J 2

a

L

L

)

)

+ R a KL J m + B mK L L a s + K i K bK L + R a B mK L

Closed-loop transfer function:

Θo ( s) Θr ( s )

=

L

)

(R J

(n R K J

L

+ Ra KL J m + Bm KL L a s + ( K i K b K L + Ra BmK L ) s + nKK s K i K L

a

K

(

L

2

(d)

KK s K i nK L

J mJ LL a s + J 5

L

a

= ∞, θ o = θ 2 = n θ m .

+ Bm J m + B m La ) s + n K LL a J L + K L J m L a + B mR a J L s + 4

m

)

J

L

2

2

is re flecte d to m otor s ide so

State equations:

26

J

T

=

J

m

+n

2

J . L

3

θr

dω dt

m

=−

B J

m

ωm +

T

K J

i

i

dθ a

dt

T

di

= ωm

m

a

dt

=−

R L

a

i

a

a

+

KK L

s

θr −

a

KK L

s

m

a

K L

b

ωm

a

State diagram:

Forward-path transfer function:

Θ o ( s) Θe ( s )

=

KKs K i n s  J TL a s + ( Ra J T + Bm L a ) s + Ra Bm + K i K b  2

Closed-loop transfer function:

Θo ( s) Θr ( s )

=

KK sK in

JT L a s + ( R a J T + Bm La ) s + ( Ra Bm + Ki Kb ) s + KK s K i n 3

2

From part (c), when K

L

= ∞,

all t he ter ms wit hout K

L

in

Θo (s ) / Θe ( s )

and

Θ o ( s ) / Θr ( s )

can b e negl ected.

The same results as above are obtained.

4-15 (a) System equations:

f = K ii a = M T (b)

dv dt

+ B Tv

ea = Ra ia + ( La + Las )

dia dt

− Las

di s dt

0 = Rsis + ( L s + L as )

+ eb

di s dt

− L as

di a dt

Take the Laplace transform on both sides of the last three equations, with zero initial conditions, we have

Ea ( s ) = [ Ra + ( La + Las ) s ] I a ( s) − Las sI s ( s ) + K b V( s)

Ki I a ( s ) = ( MT s + BT ) V ( s )

0 = − Las sI a ( s) + [ Rs + s ( Ls + Las ) ] I s ( s ) Rearranging these equations, we get

V (s ) = Ia (s ) =

Ki M T s + BT

Y (s ) =

I a (s )

1 Ra + ( La + L as ) s

[E

a

V (s ) s

=

Ki

s ( M T s + BT )

( s ) + Las sI s ( s ) − KbV ( s ) ]

Block diagram:

27

Ia (s)

Is (s ) =

L as s

Ra + ( L a + L as ) s

I a ( s)

(c) Transfer function: Y ( s)

K i [R s + ( L s + L as ) s ]

=

s [R a + ( L a + L as ) s ][ R s + ( L s + L as ) s ]( M T s + BT ) + K i K b [R s + ( L a + L as ) s] − Lass 2

E a (s )

(M

2

T

s + BT )

4-16 (a) Cause-and-effect equations:

θe = θ r − θ L

e = K sθ e dω m

Tm = K i ia

dt

=

1

Bm

Tm −

Jm K

b

Jm

ω−

= 15 . 5 V

KL Jm

m

ea = Ke dω L

−θ L )

=

dt 15 . 5

=

/ KRPM

ia =

× 2 π / 60

1000

KL

JL

m

ea − eb Ra

−θ L )

= 0 .148

Bm

KL

eb = Kbω m

State equations:

d θL dt

= ωL

d ωL

=

dt

KL JL

θm −

KL JL

θL

d θm dt

d ωm

= ωm

=−

dt

Jm

ωm −

Jm

θL+

1 Ki Jm Ra

( KK θ s

e

− K bω m )

(b) State diagram:

(c) Forward-path transfer function: G ( s) =

J m Ra J L

K i K Ks KL s J mJ L Ra s + ( Bm Ra + K i K b ) J L s + R a K L ( J L + J 3

= 0 .03 × 1 .15 × 0 . 05 = 0 . 001725

2

Bm Ra J L

= 10 × 1 .15 × 0 .05 = 0 . 575

28

m

)s

+ K L ( Bm Ra + K i K b )

Ki K bJ L

=

21 × 0 .148

× 0 . 05 = 0 .1554

R K J a

L

L

= 1.15 × 50000 × 0 . 05 =

R K J

2875

a

L

m

= 1 .15 × 50000 × 0 . 03 = 1725

K KK K i

K L ( Bm Ra + K i K b ) = 50000(10 × 1.15 + 21 × 0.148) = 730400

s

=

L

608.7 × 10 K

21

× 1 × 50000 K = 105000

0K

6

G ( s) =

(

s s + 423.42 s + 2.6667 × 10 s + 4.2342 × 10 3

2

6

8

)

(d) Closed-loop transfer function:

M (s ) =

Θ L ( s) Θ r ( s)

=

G( s) 1 + G ( s) M (s)

K i K Ks K L

=

=

J mJ LR a s + ( B m R a + K i K b ) J L s + R a K L ( J L + J m ) s2 + K L ( Bm Ra + K iK b ) s + K i K Ks K L 4

3

6 . 087 s

+ 423

4

.42 s

3

+ 2 .6667 × 10

Characteristic equation roots: K

4-17

=1

K

s

= − 1.45

s

= − 159

s

= − 131 . 05 ±

. 88

6

s

× 10 2

8

K

+ 4.2342 × 10

8

= 2738

s

+ 6 . 087 × 10

K

s

= ± j 1000

s

= −211 . 7 ±

j 1273 . 5

s

=

s

= −617

405

±

j1223 .4

.22

(a) Block diagram:

Tr ( s )

=

KM KR

( 1 + τ s ) (1 + τ s ) + K c

s

4-19 (a) Block diagram:

29

= m

KR

3.51 20 s + 12 s + 4.51 2

K

= 5476

j 1614. 6

(b) Transfer function: TAO ( s )

8

±

j 1275

(b) Transfer function: Ω( s ) α ( s)

=

K1 K 4 e

−τ D s

Js + ( JK L + B ) s + K 2 B + K3 K 4 e 2

− τ Ds

(c) Characteristic equation:

Js + ( JK L + B ) s + K 2 B + K 3 K 4e 2

(d) Transfer function: Ω( s ) α ( s)

− τD s

=0

K1 K 4 ( 2 − τ D s )

∆ ( s)

Characteristic equation:

∆ ( s ) ≅ J τ D s + ( 2 J + JK 2τ D + Bτ D ) s + ( 2 JK 2 + 2B − τ D K 2 B − τ D K 3 K4 ) s + 2 ( K 2 B + K 3 K 4 ) = 0 3

2

4-19 (a) Transfer function: Ec ( s)

G ( s) =

=

E (s )

1 + R2C s

1 + ( R1 + R 2 ) Cs

(b) Block diagram:

(c) Forward-path transfer function: Ωm ( s)

=

E (s )

[1 + ( R

1

K (1 + R2C s )

+ R 2 ) Cs ] ( K b Ki + Ra JL s )

(d) Closed-loop transfer function: Ωm ( s) Fr ( s ) (e)

=

Gc ( s) =

[1 + ( R

1

E c ( s)

=

Kφ K (1 + R2C s )

+ R2 ) Cs] ( K b K i + Ra J L s ) + Kφ KK e N (1 + R2C s )

(1 + R C s )

E (s )

Forward-path transfer function:

Ωm ( s) E (s )

2

RCs 1

=

K (1 + R2C s )

RCs ( K b K i + Ra J L s ) 1

Closed-loop transfer function:

30

Ωm ( s) Fr ( s ) (f)

Ke

= 36

pulse s / rev

f

= 120

pulse s / sec

=

e

f

r

ω

NK

ω m = 120

Thus, N = 1. For

K φ K (1 + R2C s )

=

R1C s (K b K i + Ra J L s ) + Kφ KKe N (1 + R2C s )

= 36

/ 2 π pulse s / rad

ωm = pulse s / sec

ω m = 1800

=

= 5 . 73 pul

200 RPM

= 200(

N ( 36 / 2 π ) 200( 2 π / 60 )

RPM,

120

=

= 120

N pulse s / sec

N ( 36 / 2 π ) 1800( 2 π / 60 )

= 1080

N.

Thus,

N

= 9.

4-20 (a) Differential equations:

 dθ m − dθ L   dt dt dt   dt 2  dθ dθ L  =  J d θ L + B dθ L  + T K (θ m − θ L ) + B  m −   L 2 L  L  dt dt   dt dt  d θm 2

Ki ia = Jm

(b)

2

+ Bm

dθ m

+ K ( θm − θ L ) + B 

Take the Laplace transform of the differential equations with zero initial conditions, we get

(

)

Ki I a ( s ) = J m s + Bm s + Bs + K Θm ( s ) + ( Bs + K ) Θ L ( s )

( Bs + K ) Θ Solving for

Θm ( s )

and

2

m

ΘL ( s )

Θ m (s ) = Θ L (s ) =

(

)

( s ) − ( Bs + K ) Θ L ( s ) = J L s + BL sΘ L (s ) + TL ( s) 2

from the last two equations, we have

Ki

J m s + ( Bm + B ) s + K 2

Bs + K J L s + ( BL + B ) s + K 2

I a (s ) + Θ m (s ) −

Bs + K J m s + ( Bm + B ) s + K 2

Θ L (s )

TL ( s )

J L s + ( BL + B ) s + K 2

Signal flow graph:

(c) Transfer matrix: 2 1  Ki  J L s + ( BL + B ) s + K  Θ m ( s )  = Θ (s )  ∆ (s )  Ki ( Bs + K )  L   o

31

  Ia ( s )   2  Jm s + (B m + B ) s + K  −TL ( s )  Bs + K

∆ o ( s ) = J L Jm s + [ J L 3

(B

+ B) + J m ( BL + B ) ] s + [ BL Bm + ( BL + BM

) B + (J

3

m

+J

m

) K]s

L

+ K ( BL + B ) s

2

4-21 (a) Nonlinear differential equations: dx ( t ) dt With R

a

= 0,

φ (t ) =

=

dv ( t )

v(t )

= −k ( v ) − g ( x ) +

dt e(t )

=K

K v(t )

f

=

i (t ) f

=

K i (t ) f

f

K

= − Bv

f (t )

i (t )

(t )

f (t )

i (t )

Then,

f a

+ a

=

b

=

K

φ ( t ) ia ( t ) = i

(b) State equations:

K e (t ) i

K

2

b

K

.

2

f

dv ( t ) Thus,

v (t )

dt

= − Bv

+

(t )

K 2

K K b

i

2

e (t )

2

f

v (t )

i ( t ) as input. a

dx ( t ) dt

dv ( t )

= v (t )

= − Bv

dt

(t ) + K K

2

i (t )

i

f a

(c) State equati ons: φ ( t ) as input. f (t )

= K iK

dx ( t ) dt

2

i (t )

ia ( t )

f a

dv ( t )

= v (t )

= if

= − Bv

dt

(t )

(t ) +

φ( t )

=

K

K K

f

φ

i

2

(t )

f

4-22 (a) Force and torque equations: Broom: vertical direction:

horizontal direction:

f

f

x

=

M

Car: horizontal direction:

Eliminating f

x

=

Define the state variables as x

1

v 2

d

x (t )

f

x

d b

2

θ

= θ,

x

dx1 dt dx3 dt

= x2

dx2

= x4

dx4

=

dt

dt

(M

( L cos

b

y

θ−

x (t )

=

,

f L cos x

J

2

1

x

=

dt x and

cos x

1

b

=

1

u ( t) + M b Lx2 x1 − 3 M b gx1 / 4

(M

32

b

+ M c ) − 3M b / 4

2

b

x

4

+ M b ) gx1 − M b Lx2 x1 − u ( t )

L [ 4 (M b + M c ) / 3 − M b ]

(c) Linearization:

M L

≅ 1.

2

c

θ

3

x , and

3

2

=

θ)

dt

f L sin

dt dθ

2

and f y from the equations above, and sin x

2

d

+ Mc

2

d

2

=

2

dt

M

+ L sin θ

dt J

u(t )

− M bg =

b

(b) State equations:

K K b

2

f (t )

e(t 0

=

dx . dt

f

v (t )

∂ f1

∂ f1

=0

∂ x1 ∂ f2

∂ x2

=

c

b

∂ f2

∂f 3 ∂x 3

∂f 3

=0

M b Lx2 − 3 M b g / 4

∂ f4

∂ x1 ∂ f4

(M

+ M c ) − 3M b / 4

b

∂ f4

=0

∂ x3

∂ x2

(M

∂ f4

=0

∂ x4

=

∂ x3

=0

∂u

=0

2 M b Lx1 x2

+ M c ) − 3M b / 4

b

1

=

∂u

∂f 3

=0

∂x 4

2

=

L ( M b + M c − 3M b / 4 )

L [ 4 (M b + M c ) / 3 − M b ]

=0

∂x 2

∂ f2

=0

−1

=

∂u ∂f 3

=0

∂u

−2 M b x1 x2

=

∂ x2

∂ f1

=0

∂ x4

∂ f2

2

b

∂ f2

=0

∂ f4

b

c

=0

∂ x4 ∂x 1

∂ x3

(M + M ) g − M x = 0 L ( M + M − 3M / 4) b

∂ f1

=0

2

∂ x1

∂f 3

∂ f1

=1

(M

b

+ M c ) − 3M b / 4

Linearized state equations:

0    ∆x& 1   3 ( M b + M c ) g  ∆x&   L ( M + 4 M ) b c  2 =   ∆x& 3   0  &   − 3M b g  ∆x 4   M + 4M  b c

=

(b)

i

eq

E

=

Ki

2

y (t )

1 eq

dt dy

eq

=0

dt x

1

eq

= i, x

R

x

2 eq

2

=

y

=

y , and x

E

eq

R

=

3

E

=

eq

+

dt di ( t )

eq

E

dy

At equili brium,

R

=

dL ( y ) dy ( t )

Ri ( t ) + i ( t )

(t )

2

Define the state variables as

Then, x

=

dt

= Mg −

My ( t )

Thus,

d L ( y ) i( t )

+

Ri ( t )

0

0

L( y) = L   y 

4-23 (a) Differential equations: e(t )

0     ∆x    −3 1   0 0 0   ∆x2   L ( M b + 4 M c )  +   ∆u 0 0 1  ∆x 3  0       ∆x 4    4 0 0 0  M + 4M     b c 1

L di ( t ) y

dt

dy ( t )

= 0,

=

dt

Ri ( t ) d

= 0,

2

L y

2

y (t )

dt

2

i( t )

dy ( t ) dt

=0

K

eq

R

Mg

dy . dt

K

x

Mg

3 eq

=0

The differential equations are written in state equation form: dx dt

1

=−

R L

x x 1

2

+

x x 1

x

2

3

+

x

2

L

e

=

f

dx 1

2

dt

(c) Linearization:

33

=

x

3

=

f

dx 2

dt

2

3

=

g

K x1 M x

2 2

=

f

3

+

L di ( t ) y

dt

∂f 1 ∂x 1

=−

∂f 1 ∂e

R

x

L

=

∂f 3 ∂x 1

x

2 eq

+

2 eq

x

=

L

=−

x

3 eq

E

1

K

L

Mg

x

∂f 2

eq

∂x 1

x

L

E

∂x 2

2 K x 1 eq M

x

x x 1

x

3

2

2 eq

K

∂e

2 Rg

Mg

Eeq

K

x

1 eq

Mg

=

K

2 eq

=0

  K  0    0  

0

x

∆x& = A ∆x + B ∆e ∗

 Eeq K   RL Mg    ∗   B = 0   0      

Mg

0

=

=0

∂e ∂f 3

The linearized state equations about the equilibrium point are written as:

 Eeq K −  L Mg ∗ A = 0   2 Rg  − E eq 

∂x 3

∂f 2

Mg

eq

∂f 1

=0

=1

∂x 3

E

eq

L

2

2 Rg

=

E

+

3

∂f 2

=0

∂x 2

=

1 eq

2

∂f 3

eq

R

∂f 2

=0

2 Rg

2 eq

=−

∂x 2

Mg

R

=−

2

∂f 1

K

eq

L

2 eq

2 K x 1 eq M

E

=−

4-24 (a) Differential equations: d

M1

M

2

y1 (t ) 2

dt d

2

2

=

y 2 (t ) 2

dt

Define the state variables as x

1

=

=

M 2g

y ,

x

1

dy 1 ( t )

−B

M 1g

2

dt dy

−B

=

2

dy

(t )

2

dt 1

x

,

dt

3

2

Ki

(t )

2

y1 ( t )

2

1

(t ) y2 (t )

Ki

− =

+ Ki 2

2

x

4

2

(t )

y2 (t ) − y1 (t ) y ,

− y1 ( t )

=

dy

2

2

.

dt

The state equations are:

dx1 dt

= x2

M1

dx2 dt

Ki

= M 1 g − Bx2 − dx

At equilibrium,

1

dt

= 0,

M 1g − Solving for I, with X

1

dx

2

2

x1

dt

(x

3

= 0,

2

Ki

+

dx

2

KI

2

X1

+

3

dt

KI

(X

3

2

dx3

− x1 )

2

dt

dx

= 0,

4

dt

= 0.

= x4

Thus , x

M2

2 eq

=0

2

− X 1)

=0

2

M 2g −

dx4 dt

3

4 eq

=0

2

= 1 , we have

 M + M2  Y2 = X 3 = 1 +  1   M2 

1/2

(b) Nonlinear state equations:

34

= 0.

2

− X1)

Ki

(x

3

and x

KI

(X

= M 2 g − Bx4 −

 ( M1 + M 2) g  I=  K  

1/2

2

− x1 )

2

dx1 dt

dx2

= x2

B

= g−

dt

M1

(c) Linearization: ∂ f1 =0 ∂ x1 ∂ f2

=

∂ x1

∂ f2 ∂i

∂ f4 ∂ x1

=

=

2 KI

x2 −

3

M 1 x1

2

∂ x2

2 KI

M2 ( X3 − X1 )

3

∂ f1 ∂ x3

∂ x2

dx3

∂ f3

∂ f4

=0

∂ x3

Linearized state equations:

M

1

B

∂ f2

M1

∂ x3

∂ x2

=

= 2,

∂ f3

=0

2 KI

∂ x3

= 1,

2

g

∂ f1 ∂i

=

−2 KI

= 32 .2,

 32.2(1 + 2)  X = 96.6 X = 9.8285X  1 1 1 1  

B

= 0 .1,

1/2

I =

(

X1 =

)

X 3 = 1 + 1 + 2 X 1 = 2.732 X 1 = Y2 = 2.732 0   2 1   2 KI  1 + 3 3   M1  X1 ( X 3 − X 1 )  ∗ A = 0  2  −2 KI  3 M 2 ( X 3 − X1 ) 

1

0 −2 KI

M1

M 1 ( X 3 − X1 )

0

0 2 KI

2

M 2 ( x3 − x1 )

∂ f2 ∂ x4

3

∂ f3

= 1

∂i

∂ f4

M2

∂i

=

=0

= 0

−2 KI M2 ( X3 − X1)

= 1.

1 9.8285

2 3

2

M 2 ( X 3 − X 1)

3

=1

X 3 − X 1 = 1.732

  1  0 0    115.2 −0.05 = 1   0 0  0 − B   − 37.18  M 2 

0     1   0   2 KI  −1 +  M 1  X 12 ( X 3 − X 1 ) 2    − 6.552  ∗  B = =  0  0       − 6.552  −2 KI   2  M 2 ( X 3 − X 1) 

4-25 (a) System equations:

35

2

=0

B

K

Ki

0

−B

0

∂ x4

=−

x4 −

2

∂ f3

= 0

∂ x4

3

M2

M1 ( X3 − X1 )

∂ f4

2

M 2 ( X 3 − X1 ) M

dt

B

= g−

=0

∂ x4

∂ f3

=0

∂ x1

dt

dx4

= x4

∂ f1

=−

∂ x2

3

2

=0

∂ f2

M1 ( X 3 − X 1 )

∂ f4

2

M 1 ( x 3 − x1 )

2

 −1 1   2+ 2  M 1  X1 ( X 3 − X 1 )  2

Ki

=0

2 KI

−2 KI

i + 2

M 1x1

∂ f1

2

+

K

 −18.59 0   0 1   37.18 − 0.1 0

0

2

Tm = K i ia = ( J m + JL ) T

D

=

dω m dt

d (sec) V

+ B mω m e

ea = Ra ia + La

= r −b

b

dia

= Ksy

dt

+ K b ωm

E ( s) a

=

y = nθ m

KG ( s ) E ( s ) c

Block diagram:

(b) Forward-path transfer function: Y (s )

=

E (s )

K Kin G c ( s ) e

− TD s

s {( Ra + La s ) [( Jm + J L ) s + Bm ] + K b Ki }

Closed-loop transfer function:

Y (s ) R (s )

=

− TD s

K Kin G c ( s ) e

s ( Ra + La s ) [( Jm + J L ) s + Bm ] + K bK i s + KG c ( s )K i ne

36

− TD s

y = y ( t − TD )

Chapter 5 STATE VARIABLE ANALYSIS OF LINEAR DYNAMIC SYSTEMS 5-1 (a) State variables:

x

1

=

y,

x

2

=

dy dt

State equations:

Output equation:

 dx1

  dt   0 1   x1  0   =   +  r  dx2   −1 −4   x2  5   dt  (b) State variables:

x

1

=

y,

x

2

=

y = [1

dy

x

, dt

3

2

d

=

dt

 x1   = x1  x2 

0] 

y 2

State equations:

Output equation:

 dx1

  dt  1 0   x1    0 dx 2   = 0 0 1   x2  +    dt    dx   −1 −2.5 −1.5   x3   3  dt  x1 =

(c) State variables:

t

0

y (τ ) d τ ,

0   0 r   0.5 

x2 =

dx1

 x&   2 =  x&3   x&   4

0

dt

0

0

  x1 

 0

x

1

=

y,

x

2

=

dy ,

x

dt

3

State equations:

 x&   2 =  x&3   x&   4 5-2

dt

d y dt

2

 x1  x  2 y = [1 0 0 0]   = x1 x  3 x   4

 0 0 1 0   x   0   2+  r 0 0 0 1    x3   0  − 1 − 1 − 3 −5   x   1    4  

0

2

x4 =

,

Output equaton:

1

(d) State variables:

 x&1 

dy

x3 =

,

State equations:

 x&1 

 x1  y = [1 0 0 ]  x2  = x1    x3 

=

2

d

dt

3

y 2

x

,

4

=

d y dt

3

Output equation:

1

0

0

  x1 

 0

 x1  x  2 y = [1 0 0 0]   = x1 x  3 x   4

0 0 1 0   x2   0    +  r 0 0 1   x3   0  0  −1 −2.5 0 −1.5   x   1    4  

We shall first show that

Φ ( s ) = ( sI − A ) = We multiply both sides of the equation by

I

A

1 A

2

+ 2 + +L 2 s s 2! s ( sI − A ) , and we get I = I. Taking the inverse Laplace transform −1

37

on both sides of the equation gives the desired relationship for

5-3 (a) Characteristic equation: Eigenvalues:

s

∆( s ) =

= −0 . 5 +

− A = s2 + s + 2 = 0

sI

− 0 .5 −

j 1. 323 ,

φ( t ) .

j 1. 323

State transition matrix:

 cos1.323t + 0.378sin1.323t

φ ( t) = 

(b) Characteristic equation:

−1.512sin1.323 t

∆( s ) =

sI − A

 −0.5t e −1.069sin (1.323t − 69.3 )  0.756sin1.323t

o

= s 2 + 5s + 4 = 0

Eigenvalues:

= − 4,

s

−1

State transition matrix:

 1.333 e− t − 0.333e−4 t

φ ( t) = 

−t

 −1.333 e − 1.333 e

(c) Characteristic equation:

  −0.333e + 1.333e  −t

0.333 e − 0.333e

−4 t

−t

∆ ( s ) = ( s + 3) = 0 2

−4 t

−4 t

Eigenvalues:

= −3, − 3

s

State transition matrix:

 e −3 t φ ( t) =  0 (d) Characteristic equation:

∆( s ) =

−3 t

− 9 = 0 Eigenvalues:

2

s

  e  0

s

= −3 ,

3

s

=

State transition matrix:

 e3 t φ ( t) =  0 (e) Characteristic equation:

  e  0

−3 t

∆ ( s ) = s + 4 = 0 Eigenvalues: 2

j2,

j2

State transition matrix:

 cos2 t  − sin2 t

φ ( t) = (f) Characteristic equation:

∆( s ) =

s

3

s i n 2t 

cos2t 

+ 5 s + 8 s + 4 = 0 Eigenvalues: 2

s

= − 1, − 2 ,

−2

State transition matrix:

 e− t  φ ( t) = 0   0 (g) Characteristic equation:

∆( s ) =

s

3

0 e

0

+ 15

e  φ ( t) = 0   0

−5 t

5-4 State transition equation: x (t ) = φ (t )x( t ) +

−2 t

+ 75 s + 125 = 0

2

s

  te  −2t e  0

−2 t

te e

−5 t

−5 t

0

Eigenvalues:

s

= − 5,

− 5,

−5

  te  −5 t e  0

−5 t

∫ φ (t − τ )Br (τ )d τ t

0

(a)

38

φ (t )

for each part is given in Problem 5-3.

t

0

1  s + 1 1  0 1  1  1   ( sI − A) −1 BR( s )  = L−1       ∆ ( s )  −2 s   1 0  1 s  s+2   2 s(s + s + 2) 1 + 0.378sin1.323 t − cos1.323t  −1 = =L  t≥0   s−2   −1 + 1.134sin1.323 t + cos1.323t   s s2 + s + 2  )  ( −1

φ ( t − τ ) Br (τ ) d τ = L

(b)

∫ φ ( t − τ ) Br (τ ) dτ = L ( sI − A) t

−1

0

−1

BR ( s )  = L

−1

 1  s + 5 1  1  1  1     ∆( s)  −4 s  1  1  s 

s+6    1.5 − 1.67 + 0.167       1.5 − 1.67 e− t + 0.167e −4 t  s ( s + 1)( s + 2) −1  = L−1  s s + 1 s + 4  =  =L   −t s−4    −1 + 1.67 − 0.667   −1 + 1.67 e − 0.667 e − 4t   s( s + 1)( s + 4)   s s + 1 s + 4  (c)

 1  t −1 −1 −1  s + 3 ∫0 φ ( t − τ ) Br (τ ) dτ = L ( sI − A) BR( s)  = L   0   0  0  −1  = =L 1 t≥ −3t      0.333 (1 − e )   s ( s + 3 ) 

    0  1    1  1  s   s + 3  0

0

(d)

 1  t −1 −1 −1  s − 3   φ ( t − τ ) B r ( τ ) d τ = L ( s I − A ) B R ( s ) = L  ∫0    0    0  = =L  1    s ( s + 3 )  −1

0    0.333 (1 − e −3t )   

(e)

39

    0  1    1  1  s   s + 3 

t≥ 0

0

t≥0

 1  2 t −1 −1 −1  s + 4   φ ( t − τ ) B r ( τ ) d τ = L s I − A B R ( s ) = L ( )  ∫0    −2  s 2 + 4  2   s   2  −1 =L  = t ≥0   1   0.5sin2t   ( s 2 + 4 ) 

    0  1    s   1  s  2  s + 4  2

(f)

 1  s + 1  t −1 −1 −1   ∫0 φ ( t − τ ) Br (τ ) dτ = L ( sI − A) BR( s)  = L   0    0   0  0     1 −1 −2 t    = 0.5 (1 − e ) =L  t≥0   s( s + 2)   0      0 

0 1 s+ 2 0

     0   1  1  1   2  (s + 2 )    s  0   1     s +2  0

(g)

 1  s + 5  t −1 −1 −1  ∫0 φ ( t − τ ) Br (τ ) dτ = L  ( sI − A) BR (s )  = L   0    0 

1

( s + 5) 1 s +5 0

2

      0  1    1 0  2  ( s + 5)   s   1    1   s+5   0

        0 0   0       1 0.04 0.04 0.2   −1 −1 −5 t −5 t   =L  =L 2   s − s + 5 − ( s + 5 ) 2  =  0.04 − 0.04 e − 0.2 te  u s ( t) −5t  s ( s + 5)     0.2 − 0.2 e   1   0.2 0.2  −      s ( s + 5 )  s s +5 5-5 (a) (b)

Not a state transition matrix, since

φ( 0 ) ≠ I

(identity matrix).

Not a state transition matrix, since

φ( 0 ) ≠ I

(identity matrix).

(c) φ ( t )

is a state transition matrix, since

φ ( 0 ) = I and

40

 1 [φ ( t ) ] =  − t 1 − e (d) φ ( t )

is a state transition matrix, since

0 e

t

 = φ ( − t)

φ ( 0 ) = I , and

 e2 t [φ ( t) ]−1 =  0  0 5-6 (a) (1) Eigenvalues of A:

−1

  1 = −t  1 − et e  0

−1

− te

2 2t

2t

− te

0

e

e

2 .325 ,

t e / 2

2t

− 0 . 3376 +

  = φ ( − t) 

2t

2t

j 0 . 5623 ,

− 0 .3376 −

j 0 . 5623

(2) Transfer function relation: −1

−1

s 1  −1 X( s) = ( sI − A ) B U ( s) = 0 ∆ (s )  1

 s 2 + 3s + 2  0  U ( s) = 1  −1   ∆ (s )   −s  1 

0  0 

−1   s + 3 

s 2

s +3

1  0 

1  1      s ( s + 3 ) s  0 U ( s) = s U (s )   ∆ (s )  2  2 −2 s − 1 s  1   s 

∆( s ) = s + 3 s + 2 s + 1 3

2

(3) Output transfer function:

1 1 s  = = C ( s ) ( sI − A) B = [1 0 0 ] 3 2 U (s ) ∆( s )  2  s + 3 s + 2 s + 1  s  Y ( s)

(b) (1) Eigenvalues of A:

1

−1

− 1,

− 1.

(2) Transfer function relation:

 1   ( s + 1) 2  1   0 1 s + 1 −1  U (s ) X ( s ) = ( sI − A) BU ( s ) = U (s ) =  ∆ ( s )  0 s + 1 1   1   ( s + 1)   

∆ ( s ) = s + 2s + 1 2

(3) Output transfer function:

 1   ( s + 1) 2  Y ( s) 1 1 s+2 −1 = = C ( s ) ( sI − A) B = [1 1]  + = 2 2 U (s )  1  ( s + 1) s + 1 ( s + 1)  s + 1  (c) (1) Eigenvalues of A:

− 1,

0,

− 1.

(2) Transfer function relation:

s + 2s = 1 2

1  −1 X( s) = ( sI − A) BU ( s ) = ∆ (s ) 



0 0

s+2

1   0

1 1      s ( s + 2) ) s 0 U ( s) = s U ( s)   ∆ (s )   −s s   1  s  2

41

2

(

)

∆( s ) = s s + 2s + 1 2

(3) Output transfer function:

1 s +1 1 = C ( s ) ( sI − A) B = [1 1 0 ]  s  = = 2   U (s ) s ( s + 1) s ( s + 1)  s 2  Y ( s)

5-7

dy We write dt

=

dx

1

dt

+

dx

−1

2

dt

 dx1   dt    d x  dy  = = dt  dt   2  d y   dt2 

2

d

= x 2 + x3

dt

y

=

dx

+

2

dt

dx dt

3

= −x1 − 2 x2 − 2 x3 + u

 0 1 0   x1   0   0 1 1   x  +  0 u   2    −1 −2 −2   x3  1 

 x1   1 0 0  x =  y  =  1 1 0 x      y&   0 1 1 

(1)

 1 0 0 x =  − 1 1 0 x    1 − 1 1 

(2)

Substitute Eq. (2) into Eq. (1), we have

 −1 1 0  = A 1 x + B1u =  0 0 1  x =   dt  −1 0 −2 

dx

5-8 (a) s sI

− A = −1 1

 a1 M =  a2   1

s

−2

0

−2

0

0

a2 1 0

1 0  0

s

 0  0 u   1 

= s − 3 s + 2 = s + a 2 s + a1 s + a0 3

2

3

2

a

0

= 2,

a

1

= 0,

−1

 0 −3 1  =  − 3 1 0    1 0 0

S = B

 −2 2 0  P = SM =  0 −1 1    −4 −2 1 

42

AB

0 2 4  A B  =  1 2 6     1 1 − 1 2

a

2

= −3

(b) s sI

− A = −1

−2

0

−1

0

s

−1

1

 a1 M =  a2   1

(c) s sI

−A =

+2

0

sI

−A =

0  0

1 0

+1

3

2

3

1 0

a

2,

s

2

s

a

3

+ a 2 s + a1 s + a0 2

a

0

3 3 1  =  3 1 0    1 0 0 

s

−1 2

= 0,

−1 s

= 12 ,

= 1,

S =  B

 0 1 −1 AB A B  =  1 0 − 1    1 − 1 1  0 1  1  2

(e) −A =

0

+1

2 1 P = SM =  2 3   1 2

sI

1

a

= −3

2

a

1

= 16 ,

a

2

1 −1 0  S =  B AB A B  = 1 −2 4   1 −6 23  9 6 1 P = SM =  6 5 1     −3 1 1

3

0

0  0

=

2

−1 = s + 3 s + 3 s + 1 =

1

0

2

16 7 1  =  7 1 0    1 0 0 

=1

a2

a

= s + 7 s + 16 s + 12 = s + a 2 s + a 1 s + a 0

0

0

2

2

−1 s

0

 a1 M =  a2   1

1

a2

3

+3

s

2

 a1 M =  a2   1

2

 1 2 6 S =  B AB A B  =  1 3 8     0 0 1  0 −1 1  P = SM =  −1 0 1    1 0 0 

0

s

− 3 s + 2 = s + a 2 s + a1 s + a 0

3

 0 −3 1  =  − 3 1 0    1 0 0 

−1

1

s

0  0

0

0

(d)

1

1

+2

s

−1

s

a2

=

+3

=

s

2

+ 2 s −1 = s + a1 s + a 0 2

43

a

0

= − 1,

a

1

=2

a

1

= 3,

a

2

=3

=7

M=

 a1  1

1

 2 1 0   1 0  P = SM =

5-9 (a)

S = [B

=

AB ] =

0 1   1 −3 

 1 0  −1 1 

From Problem 5-8(a),

 0 −3 1  M = − 3 1 0     1 0 0  Then,

 C  V =  CA  =  2 CA  (b)

 0.5 1 3  Q = (MV ) =  0.5 1.5 4    −0.5 −1 −2  −1

From Problem 5-8(b),

 C  V =  CA  =  2 CA  (c)

 1 0 1 − 1 2 1     1 2 1 

16 7 1  M = 7 1 0    1 0 0   1 0 1 −1 − 1 3 1  Q = (MV ) =    2 5 1 

 0.2308 0.3077 1.0769   0.1538 0.5385 1.3846     −0.2308 −0.3077 −0.0769 

From Problem 5-8(c),

 C  V =  CA  =  2 CA 

 1 0 0 − 2 1 0    4 −4 0 

Since V is singular, the OCF transformation cannot be conducted.

(d)

From Problem 5-8(d),

 3 3 1 M =  3 1 0    1 0 0 Then,

 C  V =  CA  =  2 CA  (e)

1 0 1 − 1 1 −1     1 −2 2 

−1 1 0  Q = ( MV ) =  0 1 − 2    1 − 1 1  −1

From Problem 5-8(e),

44

M=

5-10 (a)

 2 1  1 0 

Eigenvalues of A:

T = [p1

Eigenvalues of A:

2

T = [ p1

1

2

Q = ( MV ) =

p are the eigenvectors. 3

−0.7321

1, 2.7321,

where p , p , and

1 0  1 1 

0 0.5591 0.8255  p3 ] = 0 0.7637 −0.3022    1 −0.3228 0.4766 

p2

where p , p , and

(b)

 C  CA 2  =

−0.7321

1, 2.7321,

1

V=

Then,

p2

0 0.5861 0.7546  p3 ] = 0 0.8007 −0.2762    1 0.1239 0.5952 

p are the eigenvalues. 3

(c)

Eigenvalues of A:

-3, -2, -2. A nonsingular DF transformation matrix T cannot be found.

(d)

Eigenvalues of A:

−1, −1, −1

The matrix A is already in Jordan canonical form. Thus, the DF transformation matrix T is the identity matrix I.

(e)

Eigenvalues of A:

0.4142,

−2.4142 T = [p2

p2 ] =

 0.8629 −0.2811  − 0.5054 0.9597 

5-11 (a)

1 −2  0 0 

S = [B

AB] =

S = B

 1 −1 1  2 AB A B  =  2 −2 2     3 −3 3 

S = [B

AB] =

S is singular.

(b) S is singular.

(c)

 2   2

2 +2 2  2+

2

 

S is singular.

(d)

45

−1

0 1  1 −3 

 1 −2 4  AB A B  =  0 0 0    1 −4 14 

S = B 5-12 (a)

2

S is singular.

Rewrite the differential equations as:

d θm 2

dt

2

B d θm 2

=−

J dt

2

K J

θm +

State variables: x = θ ,

x

m

1

Ki

2

J =

dia

ia dθ

=−

Kb dθ m

dt m

x

,

dt

3

La dt

Ra La

ia +

K a Ks

La

r

−θm )

= ia

State equations:

Output equation:

 dx1    dt   0     dx2  =  − K  dt   J  dx   K K  3  − a s  dt   La

1 −

B J

Kb La

     0    x1    Ki    x2 +  0  θ r J     x  K K  Ra  3   a s  −   La  La  0

y

=

1

0

0 x

(b) Forward-path transfer function:

 s  Θm ( s ) K G ( s) = = [1 0 0 ] J E (s )  0 

−1 s+

B J

Kb La

 0   Ki  − J   Ra  s+ L a 

−1

   0    KiK a  0 =  K  ∆ o ( s)  a  La 

∆ o ( s ) = J La s + ( BLa + Ra J ) s + ( KLa + Ki Kb + Ra B) s + KRa = 0 3

2

Closed-loop transfer function:

  s  Θm ( s ) K M ( s) = = [1 0 0 ]   J Θr ( s )   KaKs  La =

−1 s+

B J

Kb La K i Ka K s

 0   Ki  − J   R s+ a La 

−1

   0    K s G( s )  0 =  K K  1 + K s (s )  a s  La 

JLa s + ( BLa + Ra J ) s + ( KLa + Ki Kb + Ra B) s + K i K a K s + KRa 3

2

5-13 (a)

46

= x1

A=

 0 1  −1 0 

A = 2

− 1 0   0 −1 

A = 3

 0 − 1 1 0 

A = 4

1 0  0 1 

(1) Infinite series expansion: 3 5 t t  t2 t4  1 − + − L t − + − L  1 2 2 2! 4! 3! 5! φ ( t) = I + At + A t + L =   3 5 2 4 2!  −t + t − t + L 1 − t + t L   3! 5 ! 2! 4 ! 

 cos t = − sin t

sin t  cos t 

(2) Inverse Laplace transform: −1

1  s 1  s −1  Φ ( s ) = ( sI − A ) = 1 s  = s 2 + 1  −1 s  −1

φ (t ) =

 cos t  − sin t

sin t  cos t 

(b) A=

 −1 0   0 −2 

A = 2

1 0   0 4 

A = 3

 −1 0   0 − 8 

A = 4

 −1 0   0 16 

(1) Infinite series expansion: 2 3 4 t t t   1 − t + − + +L 0   1 2 2 2 ! 3! 4 ! φ ( t) = I + At + A t + L =   2 3 2! 4t 8t   0 1 − 2t + − +L   2! 3!

e −t = 0

  e  0

−2 t

(2) Inverse Laplace transform: Φ ( s ) = ( sI − A ) = −1

s + 1  0

 1   s +1 = s + 2   0 

   1  s + 2  0

−1

0

 e− t

 −2 t  e  0

φ (t ) = 

0

(c) A=

0 1  1 0 

A = 2

1 0  0 1 

A = 3

0 1  1 0 

A = 4

1 0  0 1 

(1) Infinite series expansion: 3 5 t t  t2 t4  1+ + +L t+ + L   1 2 2  e− t + et 2! 4! 3! 5 ! φ ( t) = I + At + A t + L =  =   −t t 3 5 2 4 2!  t + t + t L 1 + t + t + L −e + e   3! 5 ! 2 ! 4!

  e +e  −t

−e + e −t

(2) Inverse Laplace transform: −1

1 s 1   s −1  −1 Φ ( s ) = ( sI − A ) =  −1 s  = s 2 − 1 1 s  =

47

 0.5 −  s +1   −0.5 +  s + 1

 s − 1 s + 1 s − 1  0.5 0.5 0.5  + s − 1 s + 1 s − 1  0.5

− 0.5

+

0.5

t

t

 e− t + et φ ( t) = 0.5  − t t −e + e 5-14 (a)

e = K s (θ r − θ y ia = Solve for i

a

)

ea = e − es

eu − eb

eb = K b

Ra + Rs

in terms of

θy

dθ and

y

  e +e  −t

−e + e −t

t

es = Rs ia

dθ y

t

eu = Kea

Tm = K i ia = ( J m + J L )

dt

d θy 2

dt

2

, we have

dt

ia =

KKs (θr − θ y ) − Kb

dθ y dt

Rs + Rs + KRs

Differential equation:

d θy 2

dt

2

K ii a

=

Jm + J L

State variables: x = θ ,

x

y

1

dθ y   − KK sθ y + KK sθ y   −K b ( J m + J L ) (R a + R s + KRs )  dt  Ki

=

2

=

y

dt

State equations:

 dx1   0 1 0     dt    x1    θ = + − KK s K i − K b Ki − KK s Ki    r      dx2   ( J + J ) (R + R + KR ) ( J + J ) (R + R + KR )   x2   ( J + J ) (R + R + KR )   m L a s s m L a s s  s   dt   m L a s 1   x1   0   0 =    +  θr  −322.58 −80.65   x2   322.58  We can let v ( t )

= 322

.58

θr,

then the state equations are in the form of CCF.

(b) −1

−1  1  s  s + 80.65 1  = 2  322.58 s + 80.65  s + 80.65 s + 322.58  −322.58 s −0.014 0.014   −0.06 − 1.059  s + 76.42 s + 4.22 s + 76.42 + s + 4.22  =  1.0622 0.0587   4.468 − 4.468  s + 76.42 s + 4.22 s + 76.42 − s + 4.22 

( sI − A ) = −1

For a unit-step function input, u ( t ) s

=1 /

s.

322.2    1 s ( s + 76.42)( s + 4.22)  = ( sI − A)−1 B =  s  322.2   s ( s + 76.42)( s + 4.22) 

48

 1 + 0.0584 − 1.058   s s + 76.42 s + 4.22     − 4.479 + 4.479   s + 76.42 s + 4.22 

−76.42t −4.22t −0.014e + 0.01e  −0.06e −76.42 t − 1.059e −4.22 t  x (t ) =  x (0) −76.42 t −4.22t −76.42t −4.22t  − 4.468e 1.0622 e − 0.0587 e 4.468 e  1 + 0.0584 e −76.42 t − 1.058e −4.22 t  = t≥0  −76.42 t −4.22t + 4.479e  −4.479 e 

(c) Characteristic equation: (d)

∆( s ) =

s

2

+ 80 . 65 s + 322

From the state equations we see that whenever there is to increase the effective value of

Ra by (1 + K ) Rs .

=0

. 58

Ra there is ( 1 + K ) Rs .

Thus, the purpose of R is s

This improves the time constant of the system.

5-15 (a) State equations:  dx1 

0 1 0     dt    x1    θ + − KKs Ki − Kb Ki KKs Ki  = r      dx2   J ( R + R + KR ) J ( R + R + KR )   x2   J ( R + R + KR )   s s s s  s s   dt   1   x1   0   0 =   +  θ r  −818.18 −90.91   x2  818.18

Let v

= 818

.18

θr.

The equations are in the form of CCF with v as the input. −1

s −1  1  s + 90.91 ( sI − A ) =  =   818.18 s + 90.91  ( s + 10.128 ) ( s + 80.782)  −818.18 −10.128t −80.78t 0.01415 e − 0.0141e  1.143 e−10.128t − 0.142 e−80.78t   x1 (0)  x (t ) =  −10.128 t −80.78 t −10.128 t −80.78t    + 0.1433e − 0.1433 e + 1.143e − 11.58 e   x2 (0)   11.58 e−10.128 t − 11.58 e−80.78t  + t≥0 −10.128 t −80.78t  + 0.1433 e 1 − 1.1434 e  −1

(b)

(c) Characteristic equati on:

2

. 18

=0

− 10.128, −80.782

Eigenvalues:

(d)

∆ ( s ) = s + 90 . 91 s + 818

Same remark as in part (d) of Problem 5-14.

5-16 (a) Forward-path transfer function: Y ( s) 5 ( K1 + K 2s ) G ( s) = = E ( s ) s [s ( s + 4)( s + 5) + 10 ] M ( s) =

Y (s ) R (s )

=

G (s) 1 + G(s )

=

Closed-loop transfer function:

5 ( K1 + K 2s )

s + 9 s + 20 s + (10 + 5 K 2 ) s + 5 K1 4

3

2

(b) State diagram by direct decomposition:

49

1 s 

State equations:

 x&1 

 x&   2 =  x&3   x&   4

 0  0   0  −5 K 

(c) Final value:

Output equation:

  x1   0  1 0   x2   0    +   r 0 1   x3   0      −20 −9   x4   1 

1

0

0 0 − (10 + 5 K 2 )

r(t )

= u s ( t ),

R( s)

0

1

=

5-17

s →0

=

5 K1

5K2

. s

lim y( t) = lim sY ( s ) = lim t →∞

y

s →0

5 ( K1 + K 2s )

s + 9 s + 20 s + ( 10 + 5 K 2 ) s + 5 K1 4

3

2

=1

In CCF form,

 0  0  A=  M  0   − a0

1

0

0

L

0

1

0

L

M

M

M

O

0

0

0

L

− a1

− a2

− a3

L

s 0  sI − A =  M 0  a 0 sI

−A =

s

n

 0   M   1  − an −1 

−1

0

0

s

−1

0

M

M

M

0

0

0

a1

a2

a3

+ a n −1 s

n −1

 0  0   B = M   0   1 

0

+ a n−2 s

L

 L 0   O M   0 −1  L s + a n  n−2

0

+ L + a1 s + a 0

Since B has only one nonzero element which is in the last row, only the last column of going to contribute to the last row of

adj (s I − A ) B

( sI − A ) .

. The last column of

Thus, the last column of

adj (s I − A )

adj (s I − A ) B

50

is 1

adj (s I − A )

is

is obtained from the cofactors of s

s

2

L

s

n −1

'

.

0 x

5-18 (a) State variables:

x

1

=

y,

x

2

=

dy , dt

x

3

=

d

2

y 2

dt

x& ( t ) = Ax ( t ) + B r ( t )

State equ ations:

0 1 0 A= 0 0 1    −1 −3 −3 

 0 B =  0   1 

(b) State transition matrix:

s2 + 3 s + 3 s + 3 1   s −1 0  1  −1  2 Φ ( s ) = ( sI − A ) =  0 s −1  = −1 s + 3s s     ∆ (s ) 2  − s  1 3 s + 3  −3 s − 1 s   1 1 1 1 2 1  + + +   2 3 ( s + 1) 2 ( s + 1)3 ( s + 1 )3   s + 1 ( s + 1) ( s + 1)   −1 1 1 2 s = + −  2 s + 1 ( s + 1) ( s + 1 )3 ( s + 1) 3 ( s + 1) 3     2 −s −3 2 s   +  ( s + 1 )3 ( s + 1) 2 ( s + 1)3 ( s + 1)3  −1

∆ (s ) = s3 + 3 s2 + 3s + 1 = ( s + 1)

(1 + t + t 2 / 2 ) e− t  2 −t φ ( t) =  −t e / 2  ( −t + t 2 / 2 ) e −t

(t + t ) e (1 + t − t ) e −t

2

2

2

t e

3

 ( t − t / 2 ) e  (1 − 2t + t 2 / 2 ) e −t  2 −t

t e /2

−t

−t

2

−t

(d) Characteristic equation: ∆ ( s ) = s + 3 s + 3 s + 1 = 0 3

− 1, −1, −1

Eigenvalues:

5-19 (a) State variables:

2

x

1

=

y,

x

2

=

dy dt

State equations:

 dx1 (t )   dt   0 1   x1 ( t)   0   =   +   r ( t)  dx2 ( t)   − 1 − 2   x2 ( t)   1   dt  State transition matrix:

 s+ 2  ( s + 1) 2  s −1  −1  Φ ( s ) = ( sI − A ) =  1 s + 2  =  − 1  ( s + 1) 2  −1

 ( s + 1)  s  ( s + 1)2  1

2

51

 (1 + t ) e−t

φ (t ) = 

 − te

−t

  (1 − t ) e  te

−t

−t

Characteristic equation:

=

y,

−y=

x

(b) State variables:

x

1

∆ (s ) = ( s +1) = 0 2

x

2

=

y

dy

+

dt

State equations: dx dt

1

=

dy dt

=

x

2

dx

− x1

2

2

dt

d

=

2

y

dt

dy

+

2

dt

= −y −

dy

+r = −x2 +r

dt

 dx1   dt   −1 2   x1  0   =   + r  dx2   0 −1  x2  1   dt  State transition matrix:

 1 s + 1 −2   s + 1 Φ (s ) =  =  0 s + 1   0

 ( s + 1)  1   s +1  −2

−1

2

(c) Characteristic equation: ∆ (s ) = ( s +1) = 0

φ (t ) =

e −t 0 

− te e

−t

−t

  

2

5-20 (a) State transition matrix: ω  s − σ sI − A =  −ω s − σ 

( sI − A )

−1

which is the same as in part (a).

s − σ ∆( s)  ω 1

=

 s − σ  −ω

2

 cos ω t − sin ω t  σ t ( sI − A )−1  =  e  sin ω t cos ω t 

−1

φ ( t) = L

σ + jω , σ − jω

(b) Eigenvalues of A:

5-21 (a) Y (s) 1

U ( s)

=

1

Y ( s) 2

U (s) 2

s 1+ s

−1

+ 2s s

= 1

+s

−1

(b) State equations [Fig. 5-21(a)]:

0 1 0  A1 =  0 0 1     −3 −2 −1  State equations [Fig. 5-21(b)]:

 0 0 −3  A 2 =  1 0 −2     0 1 −1 

−3 −2

+3s

−3

=

1 s

3

+s +2s+3 2

−3

+2s

−2

+ 3s

−3

=

1 s

3

x& = A x + B u 1

Y ( s) 1

U (s) 1

Output equation: y = C x

1 1

1

 0 B1 =  0     1 

C1 = [1

x& = A x + B u 2

+ s + 2s +3 2

=

2

 1 B2 = 0     0 

2

0

1

0]

Output equation: y

C2 = [0

52

0

(

∆ ( s ) = s − 2σ + σ + ω

1]

2

= C 2x

2

2

)

A

Thus,

= A1 '

2

5-22 (a) State diagram:

(b)

State diagram:

5-23 (a) G (s) X (s)

=

Y (s) U ( s)

=

10 s 1 + 8 .5 s

= U ( s ) − 8 .5 s

−1

−1

−3

+ 20 . 5 s

X ( s)

X (s) −2

− 20 . 5 s

+ 15

−2

s

−3

X ( s ) − 15 s

State diagram:

State equation:

x& ( t ) = Ax ( t ) + B u ( t )

53

Y ( s)

X (s) −3

X (s)

= 10

X (s)

1 0   0  A= 0 0 1     −15 −20.5 −8.5  (b) G (s) Y ( s)

=

Y (s) U ( s)

= 10

s

−3

=

10 s

−3

+ 20

1 + 4. 5 s

−1

+ 20

−4

X ( s)

s

0  B = 0    1  s

−4

+ 3 .5 s

A and B are in CCF

X (s) −2

X (s)

X (s)

X ( s)

= −4. 5 s

−1

X ( s ) − 3 .5 s

−2

X ( s) +U ( s)

State diagram:

x& ( t ) = Ax ( t ) + B u ( t )

State equations:

0 0 A=  0 0 

 0 1 0   0 0 1   0 −3.5 −4.5  1

0

0  0  B=  0  1   

0

(c) G (s)

Y (s)

=

Y (s) U ( s)

= 5s

−2

State equations:

=

5( s s(s

X (s)

+ 1)

+ 2 )( s + 10 )

+ 5s

−3

X (s)

=

5s

−2

1 + 12 s

+ 5s

−1

X (s)

A and B are in CCF

−3

+ 20

s

X (s) −2

X (s)

= U ( s ) − 12

x& ( t ) = Ax ( t ) + B u ( t )

54

s

−1

X (s)

− 20

s

−2

X ( s)

0  0 1  A= 0 0 1    0 −20 −12 

0  B = 0    1 

A and B are in CCF

(d) G ( s) = Y ( s)

=

Y (s ) U (s ) s

−4

=

(

1

)

s ( s + 5) s + 2s + 2

X ( s)

2

X (s)

=

= U ( s) − 7s

s

−4

X (s )

−1

−2

1 + 7 s + 12 s + 10 s −1

−2

X ( s ) − 12 s

X (s)

−3

− 10

X (s ) s

−3

State diagram:

x& ( t ) = Ax ( t ) + B u ( t )

State equations:

0 0  0 1 0 0 1 0   A=  0 0 0 1  0 −10 −12 −7   

0  0  B=   0  1   

A and B are in CCF

5-24 (a) G (s)

=

Y (s) U ( s)

=

10 s

3

+ 8 . 5 s + 20 . 5 s + 15 2

=

5 . 71 s

55

+ 15

6 . 67 s

+2

+

0 . 952 s

+5

x& ( t ) = Ax ( t ) + B u ( t )

State equations:

 −1.5 0 0  A=  0 −2 0    0 −5   0

 5.71  B =  −6.67     0.952 

The matrix B is not unique. It depends on how the input and the output branches are allocated.

(b) G (s)

Y (s)

=

=

U ( s)

10( s 2

s (s

+ 2)

−4. 5

=

= 1 )( s + 3 . 5)

0 .49

+

s

+ 3 .5

s

4

+ s

State diagram:

x& ( t ) = Ax ( t ) + B u ( t )

State equation:

0  0 1 0 0 0 0 0    A= 0 0 −1 0  0 0 0 −3.5    (b) G (s)

=

Y ( s) U (s)

State equations:

=

5( s s( s

+ 1)

+ 2 )( s + 10 )

=

0  1  B=   1  1   

2 .5 s

+

0 . 313 s

+2

x& ( t ) = Ax ( t ) + B u ( t )

0  0 0  A = 0 −2 0    0 0 −10 

1  B = 1   1 

(d)

56

0 . 563 s

+ 10

+1

+

5 . 71 s

2

Y (s )

G ( s) =

U (s )

=

(

1

s ( s + 5) s + 2s + 2 2

)

=

0.1 s

0.0118 s+ 5

0.0882s + 0.235 s + 2s + 2 2

x& ( t ) = Ax ( t ) + B u ( t )

State equations:

0 0 0 −5 A=  0 0 0 0 

 0  1   −2 

0

1  1  B=   0  1   

0

0 0 −2

5-25 (a) G (s)

=

Y (s) U ( s)

=

10

+ 1. 5)( s + 2 )( s + 5)

(s

State diagram:

x& ( t ) = Ax ( t ) + B u ( t )

State equations:

 −5 1 A =  0 −2  0  0

 1   −1.5  0

0  B= 0    10 

(b) G ( s) =

Y (s ) U (s )

=

 10   s + 2   1    2  s ( s + 1)( s + 3.5)  s   s + 1   s + 3.5  10(s + 2)

2

=

State diagram:

57

x& ( t ) = Ax ( t ) + B u ( t )

State equations:

0  0 1 0 0 0 1 1   A=  0 0 −1 1  0 0 0 −3.5   

0  0  B=   0  10   

(c) G ( s) =

Y (s )

=

U (s )

 5  s + 1   1     s( s + 2)( s + 10)  s  s + 2   s + 10  59 s + 1)

=

State diagram:

x& ( t ) = Ax ( t ) + B u ( t )

State equations:

1 0  0  A = 0 − 10 −1    0 0 −2 

0  B = 0    5 

(d) G ( s) =

Y (s) U (s )

=

(

1

s ( s + 5) s + 2s + 2 2

)

State diagram:

58

x& ( t ) = Ax ( t ) + B u ( t )

State equations:

0 1 0 0 A=  0 −2 0 0 

 0  1  −5 

0

0  0  B=   0  1   

0

1 −2 0

5-26 (a) G (s)

=

Y (s) E (s)

=

10 s(s

+ 4 )( s + 5)

=

10 s 1+ 9 s

−1

−3

+ 20

X (s) s

−2

X (s)

(b) Dynamic equations:

 x&1   x&  =  2  x&3 

1  0  0 0   −10 −20

  x1  0  1   x2  + 0  r     −9   x3   1  0

y = [10

0

0] x

(c) State transition equation:

 s −1 (1 + 9s −1 + 20 s −2 ) s −2 (1 + 9 s −1 ) s −3   x1 (0)   s −3   X 1 (s )  −3 −1 −1 −2    X (s ) = 1   1  s −2  1 − 10 s s (1 + 9 s ) s x2 (0) + 2     ∆ (s )   ∆ ( s )  −1  s −2 −2 −1    s  − 10 s − 20 s s   x3 (0)   X 3 ( s )   1 2  s  s + 9 s + 20 s+9 1   x1 (0)  1  1    = − 10 s( s + 9) s  x2 (0)  + 1   ∆ c (s )   ∆ c (s )  2  s  − 10 s − 10 ( 2 s + 1 ) s   x3 (0)      ∆( s ) = 1 + 9 s

−1

+ 20

s

−2

+ 10

s

−3

∆c( s) =

59

s

3

+ 9 s + 20 s + 10 2

 1.612 0.946 0.114   −0.706 −1.117 −0.169  −0.708t  x(t ) =  −1.14 −0.669 −0.081 e +  1.692 2.678 4.056  e −2.397t      0.807 0.474 0.057   −4.056 −6.420 −0.972   

 0.0935 0.171 0.055    +  −0.551 −1.009 − 0.325 e −5.895t  x(0)     3.249 5.947 1.915  

0.1 − 0.161e−0.708t + 0.0706 e−2.397t − 0.00935e−5.895 t    −0.708 t −2.397 t −5.895 t + − 0.169e + 0.055e  0.114 e   −0.087 e−0.708t + 0.406 e− 2.397t − 0.325e− 5.895t 

t ≥0

(d) Output:

(

y ( t ) = 10 x1 (t ) = 10 1.612e

−0.708t

(

+ 10 1.141 e

− 0.706e

−0.708 t

−2.397t

− 0.169 e

+ 0.0935e

−2.397 t

−5.897 t

+ 0.0550 e

) x (0) + 10 ( 0.946e − 1.117e + 0.1711e ) x (0) ) x (0) + 1 −1.61e −0.708t + 0.706e −2.397 t − 0.0935e −5.895t −0.708 t

−2.397t

−5.895t

1

2

−5.895 t

3

5-27 (a) Closed-loop transfer function: Y (s)

10

=

R( s)

s

3

+ 9 s + 20 s + 10 2

(b) State diagram:

(c) State equations:

 x&1   x&  =  2  x&3 

(d) State transition equations: [Same answers as Problem 5-26(d)]

1  0  0 0   −10 −20

  x1  0  1   x2  + 0  r     −9   x3   1  0

(e) Output:

5-28 (a) State diagram:

(b) State equations:

60

t≥0

 x&1   x&   2 =  x&3   x&   4

 −2 20 −1 0   x1   0 −1   0 −10 1 0   x2   0 0 u     +    −0.1 0 −20 1   x3   0 0  T D   0     0 0 − 5   x4  30 0  

(c) Transfer function relations: From the system block diagram, −0.2s 0.3 30 e U ( s) 90U( s)  −1  T ( s ) + T ( s ) + + D D   ∆( s)  s + 2 ( s + 2)( s + 20) ( s + 2)( s + 5)( s + 20) (s + 5)( s + 20) 

1

Y ( s) =

∆( s ) = 1 + Y ( s)

−0 . 2 s

+ 2 )( s + 20

(s

− ( s + 19 . 7 )

=

+ 2 )( s + 20 ) + 0 .1e

(s

−( s + 20

Ω( s ) =

−0 . 2 s

)

+ 2 )( s + 20 ) + 0 .1e

(s

5-29 (a)

0 . 1e

−0 . 2 s

=

+ 2 )( s + 20 ) + 0 .1 e

(s

(s

) T

D

30 e

(s)+

+ 5)

(s

T (s) D

+ 2 )( s + 20 −0 . 2 s

(s

(s

+ 5)

(s

)

+ 90(

s

+ 2 )U ( s )

+ 2 )( s + 20 ) + 0 . 1e 30 e

+

− 0. 2 s

−0 . 2 s

−0 . 2 s

U ( s)

+ 2 )( s + 20 ) + 0 .1e

There should not be any incoming branches to a state variable node other than the s

−1

−0 . 2 s

branch. Thus, we

should create a new node as shown in the following state diagram.

Notice that there is a loop with gain −1 after all the s

(b) State equations: dx

1

dt

=

17

x

2

1

+

1

x

2

dx 2

2

dt

=

15 2

x

1

1 2

x

2

+

1

r

R (s )

= 2.

Output equation:

Ks + 5 s + 1

(

( s + 1) s + 11s + 2 2

( s + 1) ( s

)

2

y

= 6 .5 x 1 + 0 .5 x 2

K = 1: Y (s) R( s)

=

s s

3

2

+ 12

+ 5s +1 s

2

+ 13 s + 2

State diagram:

61

)

+ 11s + 2 = 0

Roots of characteristic equation: −1, −0.185, −-10.82. These are not functions of K.

(c) When

(b) Characteristic equation: 2

=

branches are deleted, so

2

5-30 (a) Transfer function: Y (s )

−1

(d) When

K = 4:

Y (s ) R (s )

4s + 5s + 1

(s + 1)(4 s + 1)

2

=

( s + 1) ( s

2

+ 11s + 2

=

) ( s + 1) (s

2

+ 11s + 2

)

=

4s + 1 s + 11s + 2 2

State diagram:

(e) Y (s) R( s)

=

Ks (s

2

+ 5s =1

+ 1 )( s + 0 .185)(

s

+ 10 .82

When K = 4, 2.1914, 0.4536, pole-zero cancellation occurs. )

5-31 (a) Gp ( s) =

Y ( s) U (s )

=

1

( 1 + 0.5s ) (1 + 0.2s + 0.02s

2

)

=

100 s + 12 s + 70 s + 100 3

2

State diagram by direct decomposition:

State equations:

 x&1   x&  =  2  x&3 

 0  0   −100

  x1  0  0 1   x2  + 0  u     −70 −12   x3  1  1

0

(b) Characteristic equation of closed-loop system:

Roots of characteristic equation:

62

s

3

+ 12

s

+ 70 s + 200 = 0

2

− 5. 88 ,

− 3 . 06 +

j 4. 965 ,

− 3 . 06 −

5-32 (a) Gp ( s) =

Y ( s) U (s )

1 − 0.066 s

(1 + 0.5s ) (1 + 0.133s + 0.0067s

2

)

=

−20( s − 15) s + 22 s + 190 s + 300 3

2

State diagram by direct decomposition:

State equations:

 x&1   x&  =  2  x&3 

 0  0   −300

  x1   0  0 1   x2  +  0      −190 −22   x3   1  1

0

Characteristic equation of closed-loop system: s

3

+ 22

s

2

+ 170

5-33 (a) State variables:

x

s

+ 600 = 0

1

= ωm

State equations: dω dt

m

=−

K K b

i

and x

−12, −5 + j5, −5 −j5

2

+ K bRa

JR

a

Roots of characteristic equation:

= ωD

+

K

D

J

ωD +

KK JR

i

e

a

(b) State diagram:

63

dω dt

D

=

K J

D R

ωm −

K J

D R

ωD

j 4. 965

(c) Open-loop transfer function: Ωm ( s)

=

E (s )

KK i ( J R s + K D )

JJ RR a s + ( K b J R Ki + K DR a J R + K D JRa ) s + K DK b Ki 2

Closed-loop transfer function:

Ωm ( s) Ω r ( s)

=

Ks KKi ( J R s + KD )

JJ R Ra s + ( Kb J R Ki + KD R a J R + KD JRa + Ks K Ki J R ) s + KD K b Ki + Ks K Ki KD 2

(d) Characteristic equation of closed-loop system: ∆ ( s ) = JJ R R a s + ( K D J R K i + K D R a J R + K D JRa + K s KK i J R ) s + K D K b K i + K s K Ki KD = 0 2

∆( s ) = Characteristic equation roots:

−19.8,

s

2

+ 1037

+ 20131

.2

=0

−1017.2

5-34 (a) State equations: x& ( t ) = Ax ( t ) + B r ( t ) A=

s

 −b d   −2 1   c − a =  2 − 1 

B=

 0  1 

S = [B

AB ] =

0 1  1 −1 

Since S is nonsingular, the system is controllable.

(b) S = [B

AB ] =

0 d   1 − a 

AB

1 −1 1  A B  = 1 −1 1    1 −1 1 

The system is controllable for d

≠ 0.

5-35 (a) S = B

2

S is singular. The system is uncontrollable.

(b)

64

1 − 1 1  AB A B  = 1 − 2 4    1 −3 9 

S = B

2

S is nonsingular. The system is controllable.

5-36 (a) State equations: x& ( t ) = Ax ( t ) + B u ( t )

A=

 −2 3   1 0 

B=

Output equation:

V = C

'

'

AC

'

y

=

1  1 

S = [B

= Cx

0 x

1

1 −2   =  0 3   

1 1  1 1 

C = 1

S is singular. The system is uncontrollable.

0

V is nonsingular. The system is observable.

(b) Transfer function: Y (s)

AB ] =

s

=

U ( s)

s

2

+3

+2s−3

1

= s

−1

Since there is pole-zero cancellation in the input-output transfer function, the system is either uncontrollable or unobservable or both. In this case, the state variables are already defined, and the system is uncontrollable as found out in part (a).

5-37 (a) α = 1, (b)

2 , or 4

.

These values of

α will cause pole-zero cancellation in the transfer function.

The transfer function is expanded by partial fraction expansion, Y (s) α −1 α R( s)

=

3( s

+ 1)

2( s

By parallel decomposition, the state equations are: x& ( t )

 −1 0 0  A =  0 −2 0   0 −4   0 The system is uncontrollable for

(c)

α = 1,

or

−2 + 2)

+

α −4 6( s

+4)

= Ax ( t ) + B r ( t ) ,

α −1  B = α − 2    α − 4 

output equation: y ( t )

D=

= C x ( t ).

1 − 1  3 2

1 6 

α = 2, or α = 4.

Define the state variables so that

 −1 0 0  A =  0 −2 0    0 0 − 4

The system is unobservable for

1  3   −1 B=   2  −1     6 

α = 1, or α = 2, or α = 4.

65

D = [α − 1

α −2

α − 4]

5-38 S = [B

AB] =

b  1 b ab − 1 

S = ab − 1 − b ≠ 0 2

The boundary of the region of controllability is described by ab

− 1 − b = 0. 2

Regions of controllability:

5-39 S = [B

AB] =

 b1 b1 + b 2  b b 2   2

The system is completely controllable when b

V = C

'

'

AC

'

S = 0 when b1 b2 − b1 b2 − b2 = 0,or b2 = 0 2

≠ 0.

2

d2   d1  =    d2 d 1 + d 2 

The system is completely observable when d

≠ 0.

2

5-40 (a) State equations: dh 1 K nN K = ( qi − qo ) = I θm − o h dt A A A State variable:

x

1

= h,

x

=θm,

x

x& = Ax + B e

State equations:

 −K o  A  A=  0   0 

2

K I nN A 0 0

dθ m

dt 3

=

dθ dt

m

d ωm m

=−

dt

Ki Kb JRa

ωm +

Ki K a JRa

ei

J

= Jm +n

= ωm

i

   1 =  KK − i b JRa  0

V = 0 when d 1 ≠ 0 .

0  −1 0.016  0  0 1    0 0 −11.767 

State diagram:

66

   0    B =  0 = K K   i a  JRa 

 0   0    8333.33

2

J

L

(b) Characteristic equation of A: s+ sI − A =

Ko

− K I nN

A

A

0

s

0

0

Eigenvalues of A:

0,

0

s+

 

= s s +

−1 Ki K b

  s + Ki K b  = s ( s +1)( s + 11.767)   A  JRa 

Ko

JRa

−1, −11.767.

(c) Controllability: S = B

0 133.33   0  A B  = 0 8333.33 − 98058    8333.33 −98058 1153848 2

AB

S ≠ 0.

The system is controllable.

(d) Observability: (1) C =

1

0

V = C

(2) C =

0

(3) C =

0

'

1

V = C

'

'

'

'

'

'

−1 1  1  ( A ) C  = 0 0.016 −0.016    0 0.016  0 ' 2

'

V is nonsingular. The system is observable.

:

AC

1

'

'

AC

0

0

V = C

:

0

0 0 0    ( A ) C  = 1 0 0   0 1 −11.767  ' 2

'

V is singular. The system is unobservable.

:

AC

5-41 (a) Characteristic equation:

0 0  0  ( A ) C  = 0 0 0     1 −11.767 138.46 ' 2

'

∆( s ) = s I − A

Roots of characteristic equation: −5.0912,

=

s

4

− 25 . 92

s

5.0912, 0, 0

(b) Controllability:

67

2

=0

V is singular. The system is unobservable.

S = B

AB

∗2

A B

− 0.0732 0 − 1.8973  0  −0.0732 0 −1.8973 0  ∗3 ∗  A B  =   0 0.0976 0 0.1728   0.0976  0 0.1728 0  

S is nonsingular. Thus, A ,

B

is controllable.

(c) Observability: (1)

C

=

1

0

V = C

∗'

0

0

∗'

A C

∗'

∗' 2

(A ) C

∗'

1 0 ∗' 3 ∗' ( A ) C  =  0 0 

0

25.92

1

0

0

0

0

0

 25.92   0   0  0

S is singular. The system is unobservable. ∗

(2) C = 0

1

V = C

0

∗'

0

∗'

A C

∗'

∗'

2

(A ) C

∗'

0 671.85  0 25.92 1 0 25.92 0  ∗' 3 ∗'  (A ) C  =  0 0 0 0  0  0 0 0  

S is singular. The system is unobservable.

(3) C = 0

0

1

V = C

∗'

0

∗'

A C

∗'

∗'

2

(A ) C

∗'

0 0 ∗' 3 ∗' (A ) C  =  1 0 

S is nonsingular. The system is observable.

(4) C = 0

0

0

1

68

0

− 2.36

0

0

0

0

1

0

 − 2.36   0   0  0

V = C

∗'

∗'

A C

∗'

∗' 2

(A ) C

0 −61.17  0 −2.36 0 0 −2.36 0  ∗' 3 ∗'  ( A ) C  =  0 0 0 0  1  0 0 0  

∗'

S is singular. The system is unobservable.

5-42

The controllability matrix is

0 − 384  0 −1 0 −16  −1 0 −16 0 −384 0    0 0 0 16 0 512  S=  0 512 0  0 0 16  0 1 0 0 0 0    0 0 0   1 0 0

Rank of S is 6. The system is controllable.

5-43 (a) Transfer function: Θv ( s ) R (s )

=

J vs

2

(J

KI H s + K Ps + K I + K N 2

G

)

State diagram by direct decomposition:

x& ( t ) = Ax ( t ) + B r ( t )

State equations:

0

1

0 0  A = 0 0  0 0 

 0   1   − KP  J G 

0

0

1 0 − ( KI + KN

(b) Characteristic equation:

JG

Jvs

2

(J

)

 0  0 B=   0  1  

)

s + K P s + K I + KN = 0 2

G

69

5-44 (a) State equations: x& ( t ) = Ax ( t ) + B u 1 ( t )

A=

 −3 1   0 − 2 

B=

S is nonsingular. A ,

 0 1 

S = [B

(b)

With feedback, u 2

= − kc 2 ,

'

0 1  1 −2 

B is controllable.

Output equation: y 2 = C x

V = C

AB ] =

'

AC

'

C = −1

 −1 3   =   1 −3 

the state equation is: x& ( t )

 −3 − 2 k A =  1+ g   0

=

V is singular. The system is unobservable.

Ax ( t )

 1+ k   − 2  1

1

+ B u1 ( t ) . 1  0  S= 1+k   1 −2 

 0 B= 1 

S is nonsingular for all finite values of k. The system is controllable. State diagram:

y2 = Cx

Output equation:

V =  D '

 −1  1 + k  −1 1 + K ' ' A D  =   1  1 + k

1

C=

1 + k 

 (1 + k )   3 + 2k  − 2 (1 + k )  3 + 2k

2

V is singular for any k. The system with feedback is unobservable.

5-45 (a) S = [B V = C (b) u = − k 1

1 2   2 − 7  1 − 1  ' ' A C  = 1 −2 

AB ] = '

S is nonsingular. System is controllable.

V is nonsingular. System is observable.

k2 x

 0 1   k1  −  −1 −3   2 k1

A c = A − BK = 

70

1 − k2    −k1 = 2 k2   − 1 − 2 k1 − 3 − 2 k 2  k2

S = [B

A cB ] =

For controllabillity, k 2

V = C For observability, V

'

≠−

1 − k1 − 2 k2 + 2   2 − 7 − 2k − 4k   1 2

11 2 '

A cC

'

− 1 − 1 − 3 k1   =    1 −2 − 3 k 2 

= −1 + 3k 1 − 3 k 2 ≠ 0

71

S = − 11 − 2 k2 ≠ 0

Chapter 6 6-1 (a)

STABILITY OF LINEAR CONTROL SYSTEMS = 0 , − 1. 5 +

Poles are at s

j 1. 6583 ,

− 1. 5 −

One poles at s = 0. Marginally stable .

j 1. 6583

(b) Poles are at s = − 5, − j 2 , j 2 (c) Poles are at s = − 0 .8688 , 0 .4344 + j 2 . 3593 , 0 .4344 − j 2 . 3593 (d) Poles are at s = − 5, − 1 + j , − 1 − j (e) Poles are at s = − 1.3387 , 1. 6634 + j 2 . 164, 1. 6634 − j 2 .164 (f) Poles are at s = − 22 . 8487 ± j 22 . 6376 , 21 . 3487 ± j 22 . 6023 6-2 (a)

s

+ 25

3

s

+ 10 s + 450 = 0

2

Two poles on j ω axis. Marginally stabl e . Two poles in RHP. Unstable . All poles in the LHP. Stable . Two poles in RHP. Unstable . Two poles in RHP. Unstable .

Roots: − 25 . 31 , 0 .1537 + j 4.214, 0 .1537 − 4.214

Routh Tabulation: s

3

s

2

s

1

1

10

25

450

− 450

250

Two sign changes in the first column. Two roots in RHP.

= −8

0

25

(b)

s

s

0

3

+ 25

450

+ 10 s + 50 = 0

2

s

Roots: − 24. 6769 , − 0 .1616 + j 1.4142 , − 0 .1616 − j 1.4142

Routh Tabulation: s

3

s

2

s

1

s

0

1

10

25

50

− 50

250

No sign changes in the first column. No roots in RHP.

=8

0

25

(c)

s

50

+ 25

3

s

2

+ 250

s

+ 10 = 0

Roots: − 0 . 0402 , − 12 .48 + j 9 . 6566 , − j 9 . 6566

Routh Tabulation: s

3

s

2

s

1

s

0

1

250

25

10

− 10

6250

= 249

No sign changes in the first column. No roots in RHP. .6

0

25

(d)

2s

4

10

+ 10

s

3

+ 5 . 5 s + 5 . 5 s + 10 = 0 2

Roots: − 4.466 , − 1.116 , 0 .2888 + j 0 . 9611 , 0 .2888 − j 0 . 9611

Routh Tabulation:

71

s

4

s

3

s

2

55

1

10 24.2

s

2

5 .5

10

5 .5

− 11

= 4.4

− 100

10

10

= − 75 . 8

4.4 s

0

10

Two sign changes in the first column. Two roots in RHP.

(e)

s

6

+ 2 s + 8 s + 15 5

4

s

3

+ 20

s

2

+ 16 s + 16 = 0

Roots: − 1.222 ± j 0 .8169 , 0 . 0447 ± j 1.153 , 0 .1776 ± j 2 .352

Routh Tabulation: s

6

s

5

s

4

s

3

s

2

s

1

16

1

8

20

2

15

16

− 15

40

= 0 .5

− 16

2

2

− 33

− 48

−396 + 24

= 11 .27

−33 − 541 .1 + 528

= −1.16

16

= 12

16

0

11 .27 s

0

0

Four sign changes in the first column. Four roots in RHP.

(f)

s

4

+ 2 s + 10 3

s

2

+ 20 s + 5 = 0

Roots: −0 .29 , − 1. 788 , 0 . 039 + j 3 .105 , 0 . 039 − j 3 .105

Routh Tabulation: s

4

s

3

s

2

20

1

10

2

20

− 20

=0

5

5

2

6-3 (a)

s

4

s

2

s

1

s

0

ε 20

5

ε − 10 ε

≅−

Replac e 0 in last row by

10

ε

Two sign changes in first column. Two roots in RHP.

5

+ 25 s + 15 3

s

2

ε

+ 20 s + K = 0

Routh Tabulation:

72

s

4

s

3

s

2

1

K

15 25

20

− 20

375

= 14.2

K

25 s

− 25

284

1

K

= 20 − 1. 76

K

20

− 1. 76

K

>0

>0

K

or K

< 11 . 36

14.2 s

0

K

Thus, the system is stable for 0 < K < 11.36. When K = 11.36, the system is marginally stable. The auxiliary equation equation is A ( s ) = 14.2 s frequency of oscillation is 0.894 rad/sec.

(b)

s

2

+ 11 . 36 = 0 .

The solution of A(s) = 0 is s

2

= − 0 .8 .

The

+ Ks + 2 s + ( K + 1) s + 10 = 0

4

3

2

Routh Tabulation: s

4

s

3

s

2

1

2

K

K

− K −1

2K

K

=

K

−1

10

+1

>0

K K

10

K

−9 K − 1

>1

2

s

1

s

0

−9 K −1 > 0 2

−1

K 10

The conditions for stability are: K > 0, K > 1, and − 9 K − 1 > 0 . Since K is always positive, the last condition cannot be met by any real value of K. Thus, the system is unstable for all values of K. 2

(c)

s

2

+ ( K + 2 ) s + 2 Ks + 10 = 0

3

2

Routh Tabulation: s

3

s

2

s

1

K 2K

1

2K

+2

10

2

+ 4 K − 10 K

s

0

+2

K

> −2

K

2

+ 2 K −5 > 0

10

The conditions for stability are: K > −2 and K + 2 K − 5 > 0 or (K +3.4495)(K − 1.4495) > 0, or K > 1.4495. Thus, the condition for stability is K > 1.4495. When K = 1.4495 the system is 2

marginally stable. The auxiliary equation is A ( s ) The frequency of oscillation is 1.7026 rad/sec.

(d)

s

3

+ 20

s

2

+ 5 s + 10

K

= 3.4495

s

=0

Routh Tabulation:

73

2

+ 10 = 0 .

The solution is s

2

= −2 .899

.

s

3

s

2

s

1

1

5 10 K

20

− 10

100

K

= 5 − 0 .5 K

5

− 0 .5 K > 0

K

>0

or K

< 10

20 s

0

10 K

The conditions for stability are: K > 0 and K < 10. Thus, 0 < K < 10. When K = 10, the system is

= 20

marginally stable. The auxiliary equation is A ( s ) equation is s

(e)

s

= −5 .

2

3

2

+ 100 = 0 .

The solution of the auxiliary

The frequency of oscillation is 2.236 rad/sec.

+ Ks + 5 s + 10 s + 10

4

s

2

K

=0

Routh Tabulation: s

4

s

3

s

2

5

K

10

K

10 K

5K

− 10

5K

10 K

1

>0 − 10 > 0

K

− 100

50 K s

K 5K

1

− 10

K

>2

2

=

− 10

or K

50 K

− 100 − 10 5K

K

3

5K

− 10

− 10 − K > 0 3

K s

0

10 K

K

The conditions for stability are: K > 0, K > 2, and 5 K

β K

+ 2 . 9055

γε

K

2

− 2 . 9055

K

+ 3 .4419

ϕ

2 and K < is unstable for all values of K.

(f)

s

4

− 10 − K > 0 . 3

>0 The last condition is written as

The second-order term is positive for all values of K.

−2.9055.

Since these are contradictory, the system

+ 12 . 5 s + s + 5 s + K = 0 3

2

Routh Tabulation: s

4

s

3

s

2

1

1

12 . 5

5

−5

12 . 5

= 0 .6

K

K

12 . 5 s

1

3

− 12 . 5 K

= 5 − 20 .83

K

5

− 20 . 83 K > 0

or K

< 0 .24

0 .6 s K K >0 The condition for stability is 0 < K < 0.24. When K = 0.24 the system is marginally stable. The auxiliary 0

equation is A ( s ) = 0 . 6 s oscillation is 0.632 rad/sec.

2

6-4

The characteristic equation is Ts

3

+ 0 .24 = 0 .

The solution of the auxiliary equation is s

+ ( 2T +1) s + ( 2 + K ) s + 5 K = 0 2

Routh Tabulation:

74

2

= − 0 .4.

The frequency of

s

3

T

s

2

2T

s

1

( 2T

+1

0

T

>0

5K

T

> −1 /

+ 1 )( K + 2 ) − 5 KT 2T

s

+2

K

K (1 − 3 T )

+1

5K

T > 0, K > 0, and K

4T

+2

3T

−1

+ 24

Ks

<

T-versus-K parameter plane is shown below.

6-5 (a)

Characteristic equation: s

5

+ 600

s

4

+ 50000

s

3

+ Ks

2

+ 4T + 2 > 0

>0

K

The conditions for stability are:

2

. The regions of stability in the

+ 80

K

=0

Routh Tabulation: s

5

s

4

s

3

3

s

2

214080

1

50000

24 K

600

K

80 K

× 10

7

−K

14320 K

600

3 s

1

0

< 3 × 10

7

600 00 K

−K

2

80 K

× 10 − K 7

− 7 .2 × 10

+ 3 .113256 × 10

16

600( 214080 s

K

00

11

− 14400

K

−K

K

< 214080

K

00

2

K

2

− 2 .162 × 10

)

80 K

K

Conditions for stability:

75

>0

7

K

+ 5 × 10

12

0

row:

Thus, the final condition for stability is: When K

5

(K

or

+ ( K + 2 ) s + 30 2

Ks

+ 200

=0

K

Routh tabulation: s

3

s

2

s

1

s

0

30 K

1

+2

K

2

− 140

K

+2

30 K

200 K K

200 K

Stability Condition:

Characteristic equation: s

3

> −2

K

> 4. 6667

K

>0

K > 4.6667

When K = 4.6667, the auxiliary equation is A ( s ) The frequency of oscillation is 11.832 rad/sec.

(c)

K

+ 30

s

+ 200

2

= 6 . 6667

s

2

+ 933

. 333

=0.

The solution is s

2

+K =0

s

Routh tabulation: s

3

s

2

s

1

1

200

30

K

−K

6000

K

< 6000

K

>0

30 s

0

K

Stabililty Condition:

0

<

< 6000

K

When K = 6000, the auxiliary equation is A ( s ) The frequency of oscillation is 14.142 rad/sec.

(d)

Characteristic equation:

s

3

= 30

s

2

+ 6000 = 0 .

The solution is s

2

= − 200

.

+ 2 s + ( K + 3) s + K + 1 = 0 2

Routh tabulation: s

3

s

2

s

1

K

+3

1

K

2

K +1

+5

K

> −5

K

> −1

30 s

0

K +1

Stability condition:

K>

−1.

When K =

−1 the zero element occurs in the first element of the

76

= − 140

.

0

= s + ( k 2 − 1 ) s + 20 − 2 k1 − k 2 = 0 2

or k

2

>1

− 2 k1 − k 2 > 0

or

k

< 20 − 2 k 1

2

Parameter plane:

6-7 Characteristic equation of closed-loop system: s −1 0 sI − A + BK = 0 k1

= s + ( k 3 + 3 ) s + ( k 2 + 4 ) s + k1 = 0

s

−1

k2 + 4

s + k3 + 3

3

2

Routh Tabulation:

s

3

s

2

s

1

s

k2 + 4

1 k3 + 3

(k

3

k3 +3>0 or k3 > − 3

k1

+ 3) ( k 2 + 4 ) − k 1

(k

k3 + 3 0

+ 3 )( k + 4) − k > 0 2

1

k >0

k

1

1

Stability Requirements:

k 3 > − 3, 6-8 (a)

3

(k

k 1 > 0,

3

+ 3) ( k 2 + 4 ) − k 1 > 0

Since A is a diagonal matrix with distinct eigenvalues, the states are decoupled from each other. The second row of B is zero; thus, the second state variable, x is uncontrollable. Since the uncontrollable 2

77

state has the eigenvalue at −3 which is stable, and the unstable state x with the eigenvalue at −2 is 3

controllable, the system is stabilizable.

(b) 6-9

Since the uncontrollable state x has an unstable eigenvalue at 1, the system is no stabilizable. 1

The closed-loop transfer function of the sysetm is

Y (s)

=

R (s )

1000 s + 15.6 s + ( 56 + 100 K t ) s + 1000 3

s

The characteristic equation is:

2

3

+ 15 . 6 s + ( 56 + 100 2

K )s t

+ 1000 = 0

Routh Tabulation: s

3

s

2

s

1

1 15 . 6 873 . 6

+ 100

56

+ 1560

K

t

1000 Kt

− 1000

1560 K t

15 .6 s

0

.4

>0

1000

Stability Requirements:

6-10

− 126

K

t

> 0 . 081

The closed-loop transfer function is Y (s) R( s)

=

K(s s

3

s

The characteristic equation:

+ Ks 3

+ Ks

+ 2 )( s + α )

+ ( 2 K + αK − 1 ) s + 2 αK

2

2

+ ( 2 K + α K − 1) s + 2 αK = 0

Routh Tabulation: s

3

s

2

s

1

2 αK

K (2

+ αK − 1

2K

1

K

+ α ) K − K − 2 αK

>0

2

(2 + α )K

K s

0

2 αK

Stability Requirements: α > 0 , K -versus- α Parameter Plane:

α >0 K

> 0,

K

>

1

+ 2α

2

78

.

− 1 − 2α > 0

6-11 (a)

Only the attitude sensor loop is in operation: K

Θ( s) Θr ( s ) If KK

s

If KK

s

t

= 0. The system transfer function is:

G (s) p

=

1+ K G ( s) s

=

p

K s

2

− α + KK

s

>α,

the characteristic equation roots are on the imaginary axis, and the missible will oscillate.

≤α,

the characteristic equaton roots are at the origin or in the right-half plane, and the system

is unstable. The missile will tumble end over end.

(b)

Both loops are in operation: The system transfer function is

Θ( s) Θr ( s ) KK

For stability: When K

=0

t

G (s) p

1 + K sG t

> 0,

t

and KK

=

KK

>α,

s

(s)

p

s

+ K sG p ( s )

=

K s

2

+ KK

t

s

+ KK s − α

−α > 0 .

the characteristic equation roots are on the imaginary axis, and the missile

will oscillate back and forth. For any KK − α if KK < 0, the characteristic equation roots are in the right-half plane, and the system s

If KK

t

t

is unstable. The missile will tumble end over end. > 0 , and KK < α , the characteristic equation roots are in the right-half plane, and the system is t

unstable. The missile will tumble end over end.

6-12

Let s

1

= s + α,

then when s

= −α,

s

1

= 0.

This transforms the

s = −α axis in the s-plane onto the imaginary

axis of the s -plane. 1

(a)

F (s)

=

Or

s

+ 5s +3 = 0

s

2

2

+ 3 s1 − 1 = 0

1

Le t s

= s1 − 1

2

s1

Routh Tabulation:

s s

We get

1

−1) + 5 ( s1 −1) + 3 = 0 2

−1

1

1

(s

3

1

−1

0 1

Since there is one sign change in the first column of the Routh tabulation, there is one root in the region to the right of s = −1 in the s-plane. The roots are at −3.3028 and 0.3028.

(b)

F (s)

=

s

3

+ 3s + 3s +1 = 0 2

Let s

=

s

1

−1

We get

79

( s1 −1)3

+ 3( s1 −1) + 3 ( s1 −1) + 1 = 0 2

(c)

Or

s

F (s)

=

= 0.

3 1

s

Or

3

s

3 1

The three roots in the s -plane are all at s 1

+ 4 s + 3 s + 10 = 0 2

=

Let s

s

1

1

= 0.

Thus, F(s) has three roots at s =

We get

1

−2

1

10

(s

−1

1

−1.

−1) + 4 ( s1 −1) + 3 ( s1 −1) +10 = 0 3

2

+ s 1 − 2 s1 + 10 = 0 2

s s

Routh Tabulation:

3 1 2 1

− 12

1

s1 s

0

10

1

Since there are two sign changes in the first column of the Routh tabulation, F(s) has two roots in the region to the right of s = −1 in the s-plane. The roots are at −3.8897, −0.0552 + j1.605, and −0.0552

(d)

F (s) Or

=

s

+4s +4s +4 = 0

3

s

− j1.6025. 2

3 1

Let s

=

s

1

−1

(s

We get

1

−1) + 4 ( s1 −1) + 4 ( s1 −1) + 4 = 0 3

2

+ s1 − s1 + 3 = 0 2

s s

Routh Tabulation:

3 1 2 1 1

s1 s

1

−1

1

3

−4

0

3

1

Since there are two sign changes in the first column of the Routh tabulation, F(s) has two roots in the region to the right of s = −1 in the s-plane. The roots are at −3.1304, −0.4348 + j1.0434, and −0.4348

−j1.04348.

6-13 (a) Block diagram:

(b) Open-loop transfer function: G ( s) =

H ( s)

=

E (s )

K a K i nK I N

s (R a Js + K i K b ) ( As + K o )

=

16.667 N s( s + 1)( s + 11.767)

Closed-loop transfer function: H (s) R( s)

G (s)

= 1

+G ( s)

16 .667 N

= s

3

+ 12 . 767

(c) Characteristic equation:

80

s

2

+ 11 . 767

s

+ 16 . 667

N

s

+ 12 . 767

3

2

s

+ 11 . 767

s

+ 16 . 667

=0

N

Routh Tabulation: s

3

s

2

s

1

1

11 . 767 16 . 667 N

12 .767 150 .22

− 16 . 667

N

− 16.667

1 50.22

N

>0

or

N

NA

>0

12 . 767 s

0

N

16 .667N

Stability condition:

0

<

N

>0

0

0 . 706 A

+3 > 0

24.92 A

+ 105

.9

− 15

N > 0

250N

From the s row,

(b)

+3

.9

A

35 .3

N < 1.66 + 7.06/A

When A

→∞

N

max

→ 1. 66

For A = 50, the characteristic equation is

3 s + ( 35.3 + 0.06 K o ) s + 0.706 Ko s + 250 N = 0 3

2

Routh tabulation

81

Thus, N

max

= 1.

− 588

o

. 33

2

N < 0.0000 5653 K o

K

250 N

N

+ 0 . 03323

>0

N = 10, A = 50. The characteristic equation is

s + ( 35.3 + 0.06 K o ) s + 0.706 Ko s + 50 KI = 0 3

2

Routh Tabulution: s

3

s

2

s

1

0 . 706 K

1 35 . 3

+ 0 . 06 2

0 . 04236 K o

0

50 K

6-15 (a) Block diagram:

50 K

o

+ 24. 92

Ko

+ 0 . 06

Ko

35 .3 s

K

− 50

o

K

I

KI

I

o

> −588

. 33

KI

< 0 . 000847

K

>0

I

2

2Ko

+ 0 .498

(b) Characteristic equation: 2 Ms + K s + K + K = 0 D s P 500 s

82

2

+ K D s + 500 + K P = 0

Ko

K

o

(c)

For stability, K

D

> 0,

0 .5

+ K P > 0.

Thu s,

K

P

> − 0 .5

Stability Region:

6-16 State diagram:

∆ = 1 + s + Ks

2

Characteristic equation: s

Stability requirement:

2

+s+K =0

K>0

83

Chapter 7

TIME-DOMAIN ANALYSIS OF CONTROL SYSTEMS

7-1 (a) ζ ≥ 0 . 707

(c) ζ ≤ 0 . 5

7-2 (a)

Type 0

7-3 (a)

K

(b)

K

(c)

K

(d)

K

(e)

K

(f)

K

ωn ≥ 2

1

≤ ω n ≤ 5 rad

(b)

/ sec

Type 0

(c)

(d)

Type 1

Type 2

≤ ζ ≤ 0 . 707

(b)

0

(d)

0 .5

(e)

Type 3

ωn ≤ 2

≤ ζ ≤ 0 . 707

(f)

ω n ≤ 0 . 5 rad

Type 3

= lim

G ( s)

= 1000

Kv

= lim

sG ( s )

=0

Ka

= lim

s G (s)

p

= lim

G ( s)

=∞

Kv

= lim

sG ( s )

=1

Ka

= lim

s G (s)

p

= lim

G ( s)

=∞

Kv

= lim

sG ( s )

=K

Ka

= lim

s G (s)

p

= lim

G ( s)

=∞

Kv

= lim

sG ( s )

=∞

Ka

= lim

s G ( s)

p

= lim

G ( s)

=∞

Kv

= lim

sG ( s )

=1

Ka

= lim

s G (s)

p

= lim

G ( s)

=∞

Kv

= lim

sG ( s )

=∞

Ka

= lim

s G (s)

p

s→0

s→0

s→0

s→0

s→0

s→0

s →0

s →0

s→0

s →0

s →0 s →0

84

s→0

s→0

s→0

s→0

s→0 s→0

2

=0

2

=0

2

=0

2

=1

2

=0

2

=

K

/ sec

7-4 (a)

Input Error Constants Steady -state Error ________________________________________________________________________________ u (t )

K

tu ( t )

K

2

K

s

s

t u (t ) / 2 s

(b)

p

= 1000

v

=0

a

=0

1 1001

∞ ∞

Input Error Constants Steady -state Error ________________________________________________________________________________ u (t )

K

tu ( t )

K

2

K

s

s

t u (t ) / 2 s

p

=∞

0

v

=1

1

a

=0

(c) Input

Error Constants Steady -state Error ________________________________________________________________________________ u (t )

K

tu ( t )

K

2

K

s

s

t u (t ) / 2 s

p

=∞

0

v

=K

1/ K

a

=0

The above results are valid if the value of K corresponds to a stable closed-loop system.

(d)

The closed-loop system is unstable. It is meaningless to conduct a steady-state error analysis.

(e)

Input Error Constants Steady -state Error ________________________________________________________________________________ u (t )

K

tu ( t )

K

2

K

s

s

t u (t ) / 2 s

(f)

p

=∞

0

v

=1

1

a

=0

Input Error Constants Steady -state Error ________________________________________________________________________________ u (t )

K

tu ( t )

K

2

K

s

s

t u (t ) / 2 s

p

=∞

0

v

=∞

0

a

=K

1/ K

The closed-loop system is stable for all positive values of K. Thus the above results are valid.

7-5 (a)

K

H

=

H (0)

=1

M (s) a

0

=

= 3,

G (s) 1+ G ( s)H ( s) a

1

= 3,

a

2

s

=

=

3

s

+1

+ 2 s + 3s +3 = 1, b 1 = 1. 0 2

b

2,

Unit-step Input:

ess = Unit-ramp input: a

(b)

K

H

=

H (0)

 2 =3 

0

− b0 K H = 3 − 1 = 2 ≠ 0.

0

− b0 K H = 2 ≠ 0

Unit-parabolic Input: a

 b0 KH 1 − a KH  0 1

and

a

Thus

1

e

ss

= ∞.

− b1 K H = 1 ≠ 0.

Thus

e

ss

= ∞.

=5 M (s)

=

G (s) 1+ G ( s)H ( s)

=

1 s

2

+5s + 5

85

a

0

= 5,

a

1

= 5,

b

0

= 1,

b

1

= 0.

Unit-step Input:

ess = Unit-ramp Input:

 b0K H 1 − a KH  0

= 0:

i e

ss

=

a a

− b1 K H

1

e

(c)

K

H

=

H (0)

ss

1

5 1 −  = 0 5 5

− b0 K H = 0

0

a K

5

=

i

= 1:

s

+5

a

1

− b1 K H = 5 ≠ 0

1

=

25

H

0

Unit-parabolic Input:

 = 

1

5

=∞

=1/5 M (s)

=

G (s) 1+ G ( s)H ( s) a

0

= 1,

a

=

+ 15 s + 50 s + s + 1 a = 50 , a = 15 , b = 5 , 2 3 0

s

= 1,

1

4

3

2

The system is stable. b

1

=1

Unit-step Input:

ess = Unit-ramp Input:

 b0K H 1 − a KH  0

= 0:

i e

ss

=

a a

− b0 K H = 0

0

− b1 K H

1

a K

e

(d)

K

H

=

H (0)

ss

=

= 1:

i

1−1 / 5

a

1

− b1 K H =

4 /5

≠0

=4

1/5

H

0

Unit-parabolic Input:

  5/5  = 5  1 − 1  = 0 

1

=∞

= 10 M (s)

G (s)

=

1+ G ( s)H ( s)

a

0

= 10 ,

a

=

= 5,

1

1 s

+ 12 s + 5 s + 10 a = 12 , b = 1, 2 0 3

The system is stable.

2

b

= 0,

1

b

=0

2

Unit-step Input:

ess = Unit-ramp Input:

 b0K H 1 − a KH  0

= 0:

i e

ss

=

a a

1

− b0 K H = 0

0

− b1 K H a K 0

Unit-parabolic Input: e

7-6 (a)

M (s)

=

s s

4

+ 16

s

3

+4

ss

=

5

i

= 1:

a

1

− b1 K H = 5 ≠ 0

= 0 . 05

100

H

=∞

+ 48 s + 4 s + 4 a = 4, a = 4, 0 1 2

 1  10   = 10  1 − 10  = 0 

1

K a

=1

H

2

= 48 ,

The system is stable. a

3

= 16 ,

b

0

= 4,

b

1

= 1,

b

2

= 0,

Unit-step Input:

ess = Unit-ramp input: i

= 0:

 b0K H 1 − a KH  0 1

a

0

  4  =  1 − 4  = 0 

−b0 KH = 0

i

= 1:

86

a

1

− b1 K H = 4 − 1 = 3 ≠ 0

b

3

=0

e

=

ss

a

− b1 K H

1

a K 0

Unit-parabolic Input: e

(b)

M (s)

s

3

3

=

4

H

4

=∞

+ 3)

K(s

=

ss

−1

4

=

K

+ 3s + ( K + 2)s + 3K a = 3K , a = K + 2, 0 1 2

a

=1

H

The system is stable for K

= 3,

2

b

0

=3K,

> 0.

=K

b

1

Unit-step Input:

ess = Unit-ramp Input:

 b0K H 1 − a KH  0

= 0:

i e

=

ss

  3K   =  1 − 3 K  = 0 

1

a a

0

− b0 K H = 0

− b1 K H

1

a K 0

Unit-parabolic Input: e

ss

+2 −K

K

=

= 1:

i

=

3K

H

a

1

− b1 K H = K + 2 − K = 2 ≠ 0

2 3K

=∞

The above results are valid for K > 0.

(c)

M (s)

=

s s

4

+ 15

s

3

+5 + 50 s + 10 s a = 0 , a = 10 , 0 1 2

a

2

10 s

=

H ( s)

s+5 50 , a

=

K

3

H

= lim

H (s)

s →0

= 15 ,

b

0

= 5,

s b

1

=2 =1

Unit-step Input:

ess =

 a 2 − b1K H  a KH  1

Unit-ramp Input: e

ss

=∞

ss

=∞

Unit-parabolic Input: e

(d)

M (s)

=

s

4

+ 17

s

3

K(s

+ 5)

+ 60

s

a

2

K

+ 5 Ks + 5 K

= 5K ,

0

 1  50 − 1 × 2   = 2  10  = 2.4 

1

a

1

= 5K ,

a

2

H

=1

The system is stable for 0 < K < 204.

= 60 ,

a

3

= 17 ,

b

0

= 5K,

b

1

=

K

Unit-step Input:

ess = Unit-ramp Input: i

= 0:

e

=

ss

 b0K H 1 − a KH  0 1

a a

1

0

−b0 KH = 0

− b1 K H a K 0

  5K   =  1 − 5 K  = 0 

=

H

Unit-parabolic Input: e

ss

5K

i

−K

5K

= 1:

=

4 5

=∞

The results are valid for 0 < K < 204.

87

a

1

− b1 K H = 5 K − K = 4 K ≠ 0

7-7 G (s)

=

Y (s)

=

E ( s)

KG 1

(b) (c)

r(t )

r(t )

r(t )

K

= u s ( t ):

e

= tu s ( t ): =t

2

e

u ( t ) / 2:

e

s

ss

=

=

+ K tG p( s)

Error constants:

(a)

( s ) 20 s

p

1

p

= ∞,

=

ss

=

ss

K

1 1+ K 1

20 s ( 1 + 0 .2 s

=

5K 1 + 100 K

+ 100 K

,

Type-1 system. K ) t

a

=0

t

=0 p

1 + 100 K

=

K

v

100 K

t

5K

v

=∞

K

a

7-8 G p (s)

G (s)

100

= (1

+ 0 .1 s )( 1 + 0 . 5 s )

(b) (c)

r(t )

r(t )

K

= u s ( t ):

e

= tu s ( t ): =t

2

KG

=

E (s)

+ 0 .1 s )( 1 + 0 . 5 s ) + 100

Error constants:

r(t )

Y (s)

=

20 s 1

p

(s)

+ K tG p ( s )

100 K

= 20 s ( 1

(a)

G (s)

e

u ( t ) / 2:

e

s

ss

=

1 K

p

ss

ss

= ∞,

= =

K

1 1+ K 1 K

=

5K 1 + 100 K

,

K

a

=0

t

=0 p

= v

v

Kt

1 + 100 K

t

5K

=∞ a

Since the system is of the third order, the values of K and K

must be constrained so that the system is

t

stable. The characteristic equation is

s + 12 s + ( 20 + 2000 K t ) s + 100 K = 0 3

2

Routh Tabulation: s

3

s

2

s

1

1

20

+ 24000

K

t

100 K

12 240

+ 2000

Kt

− 100

K

12 s

0

100 K

Stability Conditions:

K>0

12 ( 1+ 100 K t ) − 5 K > 0 or

1 + 100 K t

Thus, the minimum steady-state error that can be obtained with a unit-ramp input is 1/12.

88

5K

>

1 12

7-9 (a)

From Figure 3P-19,

Θ o ( s)

=

Θr ( s )

Θo ( s) Θr ( s )

=

K1 K 2

1+ 1+

K1 K 2 Ra + La s

Ra + La s

+

K i K b + KK1 K i K t

(R

a

+ La s ) ( Bt + J t s )

K i K b + KK1 K i K t

+

(R

a

+ La s ) ( Bt + J t s )

+

KK s K 1K i N

s ( Ra + La s )( Bt + J t s )

s [( Ra + La s ) ( Bt + Jt s ) + K1 K2 ( Bt + Jt s ) + Ki Kb + KK1 K i K t ]

L a J t s + ( L a Bt + Ra J t + K 1 K 2 J t ) s + ( Ra Bt + K i K b + K Ki K1K t + K 1 K 2 Bt ) s + KK s K 1K i N 3

2

θ r ( t ) = u s ( t ),

Θr ( s ) =

1

lim s Θ ( s ) e

s →0

s

=0

Provided that all the poles of s Θ ( s ) are all in the left -half s-plane. e

(b)

For a unit-ramp input,

Θr ( s ) = 1 / e

ss

=

lim

t →∞

2

s .

θ e (t ) =

lim s Θ ( s ) e

s →0

R B

=

a

t

+ K 1 K 2 B t + K i K b + KK KK

s

1

K K i

t

K K N i

1

if the limit is valid.

7-10 (a) Forward-path transfer function:

[n(t) = 0]:

K (1 + 0.02s ) G ( s) =

Y ( s)

=

E (s )

Error Constants:

K

For a unit-ramp input, r ( t )

2 K (1 + 0.02s ) s ( s + 25) = 2 K Kt s s s + 25 s + KKt 1+ 2 s ( s + 25)

(

= ∞,

p

= tu s ( t ),

K

R( s)

=

1

=

v

1 s

2

K ,

K

,

a

)

Type-1 system.

=0

t

e

ss

=

lim e ( t )

t→ ∞

= lim

s →0

sE ( s )

=

1 K

=

K

t

v

Routh Tabulation:

s s

3

2

1

s

1

KK t + 0.02 K

25

K

25 K ( Kt + 0.02) − K 25

s

0

K K >0

Stability Conditions:

(b)

With r(t) = 0, n ( t )

= u s ( t ),

N (s)

=1/

25 ( Kt + 0.02 ) − K > 0 or K t > 0.02

s.

System Transfer Function with N( s) as Input:

K Y (s ) N (s )

K s ( s + 25) = 3 2 K (1 + 0.02 s) K Kt s s + 25 s + K ( K t + 0.02 ) s + K 1+ 2 + 2 s ( s + 25) s ( s + 25) 2

=

89

Steady -State O utput due to n ( t):

=

y ss

7-11 (a)

n(t )

= 0,

r (t )

lim y ( t )

t →∞

= lim

=1

sY ( s )

s→0

if the limit is valid.

= tu s ( t ).

Forward-path Transfer function:

G ( s) =

Y ( s) E (s )

K ( s + α )( s + 3)

=

(

2

Ramp-error constant:

Kv

= lim

e

=

Characteristic equation: s + Ks Routh Tabulation: 3

s

3

s

2

s

1

s

0

ss

3K

s →0

1 K

sG ( s )

=− v

= −3 K α

1 3K

v

+ αK − 1 3 αK

K K (3K

Type-1 system.

+ [ K ( 3 + α ) − 1] s + 3α K = 0

2

1

)

s s −1

n =0

+ αK − 1 ) − 3 αK K 3α K

+ αK − 1 − 3α > 0

3K

Stability Conditions:

K

or

1+ 3K

>

3

αK > 0 (b)

When r(t) = 0, n ( t )

= u s ( t ),

N (s)

=1/

s.

K ( s + 3) Y (s )

Transfer Function between n ( t) and y( t):

N (s ) Steady -State Output due to n ( t):

=

y ss

lim y ( t )

t →∞

= 1+

r =0

= lim

s→0

2 Ks ( s + 3) s −1 = K ( s + α )( s + 3) s 3 + Ks 2 + [ K ( s + α ) − 1]s + 3α K

(

s s −1 2

)

=0

sY ( s )

7-12

if the limit is valid.

− πζ

Percen t maxi mum ov

=e

ershoo t = 0 .25

1 −ζ

2

Thus

πζ Solving for

ζ

Peak T ime

1−ζ

2

= − ln0.25 = 1.386

from the last equation, we have t

max

=

π ωn

1−ζ

ζ

= 0 .01

(

π ζ = 1.922 1 − ζ 2

2

2

)

= 0.404. sec.

π

ωn =

Thus,

2

1 − ( 0 .404 )

0 . 01

Transfer Function of the Second-order Prototype System: Y (s) R( s)

ωn 2

=

s

2

+ 2ζω n s + ω n 2

7-13 Closed-Loop Transfer Function:

=

117916 s

2

+ 277

.3 s

+ 117916

Characteristic equation:

90

= 343 2

.4

Y (s )

=

R (s )

25 K

s + ( 5 + 500 Kt ) s + 25 K = 0 2

s + ( 5 + 500 Kt ) s + 25 K 2

For a second-order prototype system, when the maximum overshoot is 4.3%,

ωn =

2 ζω

25 K ,

n

= 5 + 500

K

ζ = 0 . 707

= 1.414

t

.

25 K

Rise Time: [Eq. (7-104)] t

r

=

1 − 0 .4167

ζ + 2 . 917 ζ

K K

With

= =

=

ωn ωn 2

Thus,

2

( 10 . 82 )

=

25 4. 68

K

= 0 .2

ωn

sec

ω n = 10 . 82

Thu s

2

= 4. 68

25 and

2 .164

t

5

= 0 . 0206

,

+ 500

K

t

= 1.414 ω n = 15 . 3

the sy stem t ransfe

Y (s)

t

=

10 . 3

= 0 . 0206

500

r func tion i s

117

=

R( s)

K

Thus

s

2

+ 15 . 3 s + 117

Unit-step Response: y = 0.1 at t = 0.047 sec. y = 0.9 at t = 0.244 sec. t = 0 .244 − 0 . 047 = 0 .197 r

y

7-14 Closed-loop Transfer Function: Y (s )

=

R (s )

25 K

2

πζ 1−ζ

, we get

ζ

( 4. 32% max.

s + ( 5 + 500 Kt ) s + 25 K = 0

s + ( 5 + 500 Kt ) s + 25 K

ζ

= 0 . 0432

Characteristic Equation:

2

When Maximum overshoot = 10%,

Solving for

max

= − ln0.1 = 2.3

(

π ζ = 5.3 1 − ζ 2

2

2

2

)

= 0.59.

The Natural undamped frequency is

ωn =

25 K

Thus,

5 + 500 K

t

= 2 ζω n = 1.18 ω n

Rise Time: [Eq. (7-114)] r

=

K

=

t

1 − 0 .4167

ζ + 2 . 917 ζ ωn

ωn

2

= 0 .1 =

1. 7696 sec.

ωn

Th us

ω n = 17 . 7

2

= 12 . 58

25 With K = 12.58 and K

t

=

sec.

K

t

=

15 . 88

= 0 . 0318 500 0 . 0318 , the system transfer function is Thus

Y (s) R( s)

313

= s

2

+ 20 .88 s + 314.

5

Unit-step Response: y = 0.1 when t = 0.028 sec. y = 0.9 when t = 0.131 sec.

91

overs hoot)

t

y

7-15

= 0 . 131 − 0 . 028 = 0 .103

r

= 1.1

max

( 10%

sec.

max.

overs hoot )

Closed-Loop Transfer Function:

Characteristic Equation:

Y (s)

s + ( 5 + 500 Kt ) s + 25 K = 0

R( s)

=

25 K s

2

+ ( 5 + 500

2

+ 25 K

K )s t

πζ

When Maximum overshoot = 20%,

1−ζ Solving for

ζ

, we get

ζ = 0 .456

The Natural undamped frequency

= − ln0.2 = 1.61

(

π ζ = 2.59 1 − ζ 2

2

2

2

)

.

ωn =

25 K

5

+ 500

K

t

= 2 ζω n = 0 . 912 ω n

Rise Time: [Eq. (7-114)] t

r

=

1 − 0 .4167

ζ + 2 . 917 ζ ωn

2

= 0 . 05 =

1.4165 sec.

ωn

Thus,

ωn

ωn =

1.4165

=

28 .33

0 . 05

2

= 32 .1 5 + 500 K = 0 . 912 ω = 25 . 84 Thus, t n 25 With K = 32.1 and K = 0 . 0417 , the system transfer function is K

=

K

t

= 0 . 0417

t

Y (s) R( s)

802 . 59

=

s

2

+ 25 . 84 s + 802

. 59

Unit-step Response: y = 0.1 when t = 0.0178 sec. y = 0.9 when t = 0.072 sec. t = 0 . 072 − 0 . 0178 = 0 . 0542

sec.

r

y

7-16 Closed-Loop Transfer Function: Y (s ) 25 K = 2 R ( s ) s + ( 5 + 500 Kt ) s + 25 K From Eq. (7-102), Delay time t

d

When Maximum overshoot = 4.3%,

1.1

max

= 1.2

With K = 20.12 and K

max.

overs hoot )

Characteristic Equation:

s + ( 5 + 500 Kt ) s + 25 K = 0 2

+ 0 .125 ζ + 0 .469 ζ

2

= 0 .1

ωn

ζ = 0 . 707

t

.

=

d

1.423

ωn

sec.

= 0 .1

sec.

Thus

 ω n  =  14.23  = 8.1 5 + 500 K = 2ζω = 1.414 ω = 20.12 K =    t n n  5   5  2

( 20%

ω n = 14.23

2

t

= 0 . 0302

, the system transfer function is Y (s) R( s)

202 . 5

= s

2

+ 20 .1 s + 202

Unit-Step Response:

92

.5

Thus

K

t

=

15 .12 500

= 0 . 0302

When y = 0.5, t = 0.1005 sec. Thus, t = 0 .1005 sec. d

y

7-17 Closed-Loop Transfer Function: Y (s ) 25 K = 2 R ( s ) s + ( 5 + 500 Kt ) s + 25 K From Eq. (7-102), Delay time t

d

1.1

( 4. 3%

max.

overs hoot )

s + ( 5 + 500 Kt ) s + 25 K = 0 2

+ 0 .125 ζ + 0 .469 ζ

2

= 0 . 05 =

ωn

1. 337 Thus,

ωn

ωn =

1. 337

= 26 . 74

0 . 05

2

K =

With K = 28.6 and K

= 1. 043

Characteristic Equation:

 ω n  =  26.74  = 28.6     5   5  2

max

t

= 0 . 0531

,

5

+ 500

K

t

= 2 ζω n = 2 × 0 . 59 × 26 . 74 = 31 . 55

the sy stem t ransfe

Y (s)

K

t

= 0 . 0531

r func tion i s

715

=

R( s)

Thus

s

2

+ 31 . 55 s + 715

Unit-Step Response: y = 0.5 when t = 0.0505 sec. Thus, t = 0 . 0505 sec. d

y

7-18 Closed-Loop Transfer Fu nction: Y (s ) 25 K = 2 R ( s ) s + ( 5 + 500 Kt ) s + 25 K For Maximum overshoot = 0.2, ζ = 0.456 From Eq. (7-102), Delay time t

d

=

1.1

max

= 1.1007

( 10 . 07%

max.

overs hoot )

Characteristic Equation:

s + ( 5 + 500 Kt ) s + 25 K = 0 2

.

+ 0 .125 ζ + 0 .469 ζ

2

=

ωn

1.2545

= 0 . 01

ωn

sec.

 ω n  = 15737.7 = 629.5  25  5  2

Natural Undamped Frequency 5

+ 500

K

t

ωn =

1.2545

t

= 0.2188

.45

0 . 01

= 2 ζω n = 2 × 0 .456 × 125

With K = 629.5 and K

= 125 .45

= 114.41

K =

Thus, K

t

= 0.2188

, the system transfer function is Y (s) R( s)

=

15737 . 7 s

2

+ 114.41

Unit-step Response:

s

+ 15737

.7

y = 0.5 when t = 0.0101 sec.

93

Thus, t y

7-19 Closed-Loop Transfer Function: Y (s ) 25 K = 2 R ( s ) s + ( 5 + 500 Kt ) s + 25 K ζ = 0 .6 2 ζω = 5 + 500 n From Eq. (7-109), settling time t

K

t

=

3 .2

s

1.2 ω

n

−5

= 1.2

sec.

( 20% ma

x. overs hoot )

Characteristic Equation:

s + ( 5 + 5000 Kt ) s + 25 K = 0 2

K

=

ζω n

max

= 0 . 0101

d

t

= 1.2 ω n 3.2

0 .6 ω

= 0 .1

sec. Thus,

ωn =

n

ωn

3 .2

= 53 . 33

0 . 06

2

= 0 .118

=

K

500

= 113

. 76

25

System Transfer Function: Y (s) R( s)

=

2844 s

2

+ 64 s + 2844

Unit-step Response:

y(t) reaches 1.00 and never exceeds this value at t = 0.098 sec. Thus, t = 0 . 098 sec. s

7-20 (a) Closed-Loop Transfer Function: Y (s ) 25 K = 2 R ( s ) s + ( 5 + 500 Kt ) s + 25 K For maximum overshoot = 0.1, ζ = 0 . 59 . Settling time:

t

s

=

K

3 .2

ζω n t

=

3 .2

=

1.18

0 .59

ωn

ωn −5

Characteristic Equation:

s + ( 5 + 500 Kt ) s + 25 K = 0 2

5

+ 500

= 0 . 05

K

t

= 2 ζω n = 2 × 0 . 59 ω n = 1.18 ω n ωn =

sec.

ωn

3 .2 0 . 05

× 0 . 59

= 108

.47

2

= 0 .246

K

=

500

= 470

. 63

25

System Transfer Function: Y (s) R( s)

=

11765 . 74 s

2

+ 128

s

+ 11765

. 74

Unit-Step Response: y(t) reaches 1.05 and never exceeds this value at t = 0.048 sec. Thus, t = 0 . 048 sec. s

94

(b)

ζ = 0 .456

For maximum overshoot = 0.2, Settling time t

s

=

3 .2

ζω n

.

3 .2

=

0 .456 K

t

=

5

+ 500

= 0 . 01

ωn

0 . 912

K

t

= 2 ζω n = 0 . 912 ω n ωn =

sec.

ωn −5

3 .2

× 0 . 01

0 .456

= 701 . 75

= 1.27

500

System Transfer Function: Y (s) R( s)

=

492453 s

2

+ 640

s

+ 492453

Unit-Step Response: y(t) reaches 1.05 and never exceeds this value at t = 0.0074 sec. Thus, t = 0 . 0074 sec. This is less s

than the calculated value of 0.01 sec.

7-21 Closed-Loop Transfer Function: Y (s ) 25 K = 2 R ( s ) s + ( 5 + 500 Kt ) s + 25 K Damping ratio

ζ = 0 . 707

Characteristic Equation:

s + ( 5 + 500 Kt ) s + 25 K = 0 2

Settling time t

.

s

=

4. 5ζ

ωn

=

3 .1815

ωn

= 0 .1

sec.

Thus,

ω n = 31 .815 ωn 2

5

+ 500

K

t

= 2 ζω n = 44. 986

Thus , K

t

= 0 . 08

K

=

= 40 .488

System Transfer Function: Y (s) R( s)

Unit-Step Response:

=

1012 .2 s

2

+ 44.

986 s

+ 1012

.2

The unit-step response reaches 0.95 at t = 0.092 sec. which is the measured t . s

95

7-22 (a)

When

ζ = 0 . 5 , the rise time is t

r

1 − 0 .4167

ζ + 2 . 917 ζ ωn

2

=

1. 521

=1

ωn

sec.

Thus

ω n = 1. 521

The second-order term of the characteristic equation is written s

2

+ 2 ζω n s + ω n = 2

The characteristic equation of the system is

For zero remainders,

28 .48 a

=

2

+ 1. 521 s + 2 . 313 = 0

s

3

+ ( a + 30

)s

+ 1. 521 s + 2 . 313

2

s

Dividing the characteristic equation by

s

45 . 63

Thus,

a

+ 30

2

as

+K =0

, we have

= 1. 6

K

= 65 .874 + 2 . 313

a

= 69 . 58

Forward-Path Transfer Function: G (s)

69 . 58

= s( s

+ 1.6 )( s + 30

)

Unit-Step Response: y = 0.1 when t = 0.355 sec. y = 0.9 when t = 1.43 sec. Rise Time: t = 1.43 − 0 . 355 r

(b)

seconds

The system is type 1.

(i)

For a unit-step input, e

(ii)

For a unit-ramp input,

= 1. 075

ss

= 0. K

= lim

v

s→0

sG ( s )

=

K 30 a

=

60 . 58 30

× 1. 6

7-23 (a) Characteristic Equation: s

3

+ 3s + (2 + K ) s − K = 0 2

Apply the Routh-Hurwitz criterion to find the range of K for stability.

Routh Tabulation:

96

= 1.45

e

ss

=

1 K

= 0 .69 v

sec.

s

3

s

2

s

1

s

0

6

+K

1

2

3

−K

+4K 3

−K

Stability Condition:

-1.5 < K < 0 This simplifies the search for K for two equal roots. When K = −0.27806, the characteristic equation roots are: −0.347,

−0.347, and −2.3054. (b) Unit-Step Response: ( K = − 0.27806)

(c) Unit-Step Response ( K = − 1)

The step responses in (a) and (b) all have a negative undershoot for small values of t. This is due to the zero of G(s) that lies in the right-half s-plane.

7-24 (a)

The state equations of the closed-loop system are: dx

= − x1 + 5 x 2

dx

= −6 x 1 dt dt The characteristic equation of the closed-loop system is 1

∆=

2

s +1

−5

6 + k1

s + k2

− k1 x 1 − k 2 x 2 + r

= s + ( 1 + k2 ) s + ( 30 + 5 k1 + k 2 ) = 0 2

97

For

(b)

For

ω n = 10

ζ = 0 . 707

, 2ζω

2

For

ω n = 10

1

2

k

2

1

+ k 2 = 70

.

1.414

ζ = 0 . 707 an d

k 2 = 59 + 10 k1 2

Thus

, 1

+ k 2 = 2ζω n = 14. 14

k

Th us

2

= 13 .14

The closed-loop transfer function is

Y (s)

=

R (s )

5 s + ( k 2 + 1) s + ( 30 + 5 k1 + k 2 ) 2

lim y ( t )

For a unit-step input,

t →∞

(e)

k

= 30 + 5k1 + k 2

= 11 . 37 .

1

Thu s 5 k

.

2

2

and

1

ωn = 1 +

Th us

+ k 2 = 100

Solving for k , we hav e

(d)

2

(1 + k )

+ 5 k 1 + k 2 = ω n = 100

= 1+ k2.

n

ωn = (c)

30

= lim

sY ( s )

s→0

5

=

=

5 s + 14.14 s + 100 2

= 0 . 05

100

For zero steady-state error due to a unit-step input, 30

+ 5 k1 + k 2 = 5

Thus

5k

1

+ k 2 = −25

Parameter Plane k versus k : 1

2

7-25 (a) Closed-Loop Transfer Function Y (s) 100 ( K P + K D s ) = 2 R ( s ) s + 100 K D s + 100 K P The system is stable for K > 0 and P (b)

For

ζ = 1,

2ζω

ω n = 10

n

= 100 K

K

P

(c)

See parameter plane in part (g) .

(d)

See parameter plane in part (g) .

D

(b) Characteristic Equation: s K

D

> 0.

n

= 100

2

+ 100

K

.

T hus

KD

= 20

98

KP

K

D

= 0 .2

K

P

D

s

+ 100

K

P

=0

(e)

K

Parabolic error constant

= 1000

a

−2

sec

K a = lim s G ( s ) = lim100 ( K P + K D s ) = 100 K P = 1000

Thus K P = 10

2

s →0

(f)

s →0

Natural undamped frequency

ω n = 50

ω n = 10 (g)

When K

P

K

= 50

P

K

Th us

=

P

25

= 0, G (s)

=

100 K s

D

s

2

100 K

=

D

(pole-zero cancellation)

s

7-26 (a) Forward-path Transfer Function: Y ( s)

G ( s) = When r ( t )

(b)

= tu s ( t ),

=

E (s ) K

= lim

v

KKi

s [ Js(1 + Ts ) + K i K t ]

s→0

sG ( s )

=

K K

e

=

ss

10 K

(

s 0.001 s + 0.01 s + 10 K t =

t

2

1 K

=

K

)

t

K

v

When r(t) = 0

Y (s ) Td ( s) For T ( s ) d

(c)

=

1 s

=

1 + Ts

=

s [ Js(1 + Ts) + Ki Kt ] + KKi lim y ( t ) t→ ∞

=

lim sY ( s ) s →0

=

1 + 0.1s

(

1 if the system is stable. 10 K

The characteristic equation of the closed-loop system is 0 . 001 s

3

+ 0 . 01 s + 0 .1 s + 10 2

K

=0

The system is unstable for K > 0.1. So we can set K to just less than 0.1. Then, the minimum value of the steady-state value of y(t) is 1 + =1 − 10 K K =0 .1 However, with this value of K, the system response will be very oscillatory. The maximum overshoot will be nearly 100%.

(d)

For K = 0.1, the characteristic equation is 0 . 001 s

3

+ 0 . 01 s + 10 2

K s t

For the two complex roots to have real parts of

+1 = 0

s

or

3

+ 10

s

2

+ 10

4

K s t

+ 1000 = 0

−2/5. we let the characteristic equation be written as

99

)

s 0.001 s + 0.01 s + 10 Kt + 10 K 2

( s + a) ( s

+ 5 = 10 = − a = −5

a

Then,

The three roots are: s

7-27 (a)

K

= 10000

t

)

+ 5s + b = 0

2

s + ( s + 5)s + (5a+ b) s+ ab = 0 3

or

2

=5 ab = 1000 b = 200 s = − a = −5 s = −2 . 5 ± j13 . 92

a

+ b = 10

5a

4

K

t

K

t

= 0 . 0225

The Forward-Path Transfer Function:

9 × 10 K 12

G ( s) =

(

s s + 5000 s + 1.067 ×10 s + 50.5 × 10 s + 5.724 × 10 4

3

7

2

9

12

)

9 × 10 K 12

=

s ( s + 116)( s + 4883)( s + 41.68 + j3178.3)( s + 41.68 − j 3178.3)

Routh Tabulation: s

5

s

4

s

3

s

1

1

s

0

× 10

5.7 2. 895

× 10

16 . 6

5

5 . 72

× 10 + 1. 579 × 10 8

13

+ 8 .473 × 10 29

12

7

K

+ 1. 579 × 10

9

7

50.5

× 10

9

12

− 2 . 8422 × 10

K

− 2981

− 58303

.14 K

9

5.7 24 9

− 1. 8 × 10

12

× 10

9

12

9

K

× 10

× 10

12

12

K

0

K

2

K

K

K

1

2

× 10

K

From the s row, the condition of stability is 165710 or

× 10

500 0

2

s

1.067

0 − 3000 . 57 ) < 0 2

K

0 < K < 3000.56

The critical value of K for stability is 3000.56. With this value of K, the roots of the characteristic equation are: −4916.9, −41.57 + j3113.3, −41.57 + j3113.3, −j752.68, and j752.68

(b)

K

= 1000

L

oz-in/rad. The forward-path transfer function is

9 × 10 K 11

G ( s) =

(

s s + 5000 s + 1.582 × 10 s + 5.05 ×10 s + 5.724 × 10 4

3

6

2

9

11

)

9 × 10 K 11

=

s (1 + 116.06)( s + 4882.8)( s + 56.248 + j1005)( s + 56.248 − j1005)

(c) Characteristic Equation of the Closed-Loop System: s

5

+ 5000

s

+ 1. 582 × 10

4

6

3

s

+ 5 . 05 × 10

9

s

2

+ 5 . 724 × 10

11

+ 9 × 10

s

11

K

=0

Routh Tabulation: s

5

s

4

s

3

s

2

1 500 0 5.72 4.6503

× 10

5

× 10 + 1. 5734 × 10 7

5 .724 6

K

1 .582

× 10

6

5. 05

× 10

9

× 10

11

9

100

5.7 24 9

− 1.8 × 10

× 10

11

K

8

K

× 10

× 10 0

11

11

K

s

26 . 618

1

× 10

18

+ 377

15

− 2 . 832 × 10

K

× 10 + 1. 5734 × 10 7

4. 6503 s

× 10

.43

0

× 10

9

11

1

2

K

− 1332

. 73 K

K

2

K

K

From the s row, the condition of stability is Or,

6

14

× 10

26 . 618

− 93990 < 0

(K

or

− 1400

4

)( K

+ 3774. 3 K − 2 .832 + 67 .14 ) < 0

K

2

>0

Stability Condition: 0 < K < 1400 The critical value of K for stability is 1400. With this value of K, the characteristic equation root are: −4885.1, −57.465 + j676, −57.465 − j676, j748.44, and −j748.44

(c)

K

L

= ∞.

Forward-Path Transfer Function:

G ( s) = =

nK s K i K s  L a J T s + ( Ra J T + R mL a ) s + R a B m + K i K b  891100 K

(

s s + 5000 s + 566700 2

=

)

891100 K s ( s + 116)( s + 4884)

Characteristic Equ ation of the Closed-Loop System: s

3

+ 5000

s

2

+ 566700

s

+ 891100

K

=0

Routh Tabulation: s

3

s

2

s

1

s

0

1

5667 00

5000

89 1100 K

− 178

5 66700

.22 K

89 91100 K

From the s row, the condition of K for stability is 566700 − 178.22K > 0. 1

Stability Condition :

0 < K < 3179.78

The critical value of K for stability is 3179.78. With K = 3179.78, the characteristic equation roots are

−5000, When the motor shaft is flexible, K becomes stiffer, K K

L

=∞,

L

L

−j752.79.

j752.79, and

is finite, two of the open-loop poles are complex. As the shaft

increases, and the imaginary parts of the open-loop poles also increase. When

the shaft is rigid, the poles of the forward-path transfer function are all real. Similar effects

are observed for the roots of the characteristic equation with respect to the value of K . L

7-28 (a) G (s) c

=1

G (s)

=

+ 2)

100( s s

2

K

−1

p

= lim

G (s)

s→0

= − 200

When d(t) = 0, the steady-state error due to a unit-step input is e

ss

=

1 1+ K

1

= p

1

− 200

=−

1 199

101

JT = J m + n J L 2

2

= − 0 . 005025

(b) Gc (s ) = (c)

s +α

G (s ) =

s

100( s + 2)( s + α )

(

s s −1 2

Kp = ∞

)

α=5

m aximum

overs hoot

= 5.6%

α

= 50

ma ximum

oversh oot

= 22%

α

= 500

max imum o versho ot

e ss = 0

= 54.6%

As the value of α increases, the maximum overshoot increases because the damping effect of the zero at s = −α becomes less effective.

Unit-Step Responses:

(d)

r(t )

=0

and

G (s) c

= 1.

d (t )

= u s(t )

D (s)

1

=

s

System Transfer Function: ( r = 0) Y ( s) D (s)

= r =0

+ 2)

100( s s

3

+ 100

s

2

+ ( 199 + 100 α ) s + 200 α

Output Due to Unit-Step Input: Y ( s)

100( s

= s s

y

(e)

r(t )

= 0,

d (t )

ss

=

3

+ 100

lim y ( t ) t →∞

s

+ ( 199 + 100 α ) s + 200 α

2

= lim

sY ( s )

s→0

= us(t ) G (s) c

=

s

+ 2)

=

200 200

α

=

1

α

+α s

System Transfer Function [ r( t) = 0] Y (s) D ( s)

= r =0

100 s ( s s

3

+ 100

s

2

+ 20

+ ( 199 + 100 α ) s + 200 α

102

D (s)

=

1 s

y ss

=

= lim

sY ( s )

100 s ( s

+ 2)

lim y ( t )

t →∞

s→0

=0

(f) Y (s)

α=5

=

D ( s ) r =0 Y (s)

α = 50

D ( s)

D (s)

s

3

s

3

= r =0

Y ( s)

α = 5000

s

3

= r =0

+ 100

s

2

+ 699

100 s ( s

+ 100

s

2

+ 2)

+ 5199

100 s ( s

+ 100

s

2

+ 1000

s

s

+ 10000

+ 2)

+ 50199

s

+ 100000

Unit-Step Responses:

(g)

As the value of α increases, the output response y(t) due to r(t) becomes more oscillatory, and the overshoot is larger. As the value of α increases, the amplitude of the output response y(t) due to d(t) becomes smaller and more oscillatory.

7-29 (a) Forward-Path Transfer function: G (s)

=

H (s) E (s)

=

Characteristic Equation:

10 N s( s

+ 1 )( s + 10

N

s(s

)

N=1: Characteristic Equation:

s

2

s

+ 1)

+ s +1 = 0

+s+N =0

2

ζ = 0 .5

ωn = 1

−πζ

Maximu

m over shoot

= e

1 −ζ

2

= 0 .163

N=10: Characteristic Equation:

s

2

(16.3%)

Peak time t

+ s = 10 = 0

max

=

π ωn

1

ζ = 0 . 158

−ζ

= 3 . 628

ω n = 10

= 1.006

sec.

−πζ

Maximum overshoot

=e

(b) Unit-Step Response:

1 −ζ

2

= 0 . 605

(60.5%)

Peak time t

N=1

103

max

=

π ωn

1

−ζ

2

sec.

2

Second-order System Maximum overshoot Peak time

Third-order System

0.163 3.628 sec.

0.206 3.628 sec.

Unit-Step Response: N = 10

Second-order System Maximum overshoot Peak time

Third-order System

0.605 1.006 sec.

0.926 1.13

sec.

7-30 Unit-Step Responses:

When T

z

is small, the effect is lower overshoot due to improved damping. When T

overshoot becomes very large due to the derivative effect. T

z

z

is very large, the

improves the rise time, since 1

derivative control or a high-pass filter.

104

+Tz s

is a

7-31 Unit-Step Responses

> 0 . 707

less stable. When T p

1

=−

The effect of adding the pole at s

T

to G(s) is to increase the rise time and the overshoot. The system is

p

, the closed-loop system is stable.

7-32 (a) N=1 Closed-loop Transfer Function: M

H

Y ( s)

=

(s)

10

=

R( s)

s

1

=

+ 11 s + 10 s + 10

3

2

1

+ s + 1. 1 s + 0 .1 s 2

3

Second-Order Approximating system: M

L

(s)

=

1 1+ d s

+ d2s

1

M M d e f f

1

2

= m1 =

f

2

d

=

2m

2

(s)

H L

(s)

= m2

l

1

=

1

1+ s

=1

+ d2s

+ 1.1 s + 0 .1 s 2

l

2

= 1.1

l

=

3

6

= 2 l6 − 2 l 1 l 5 + 2 l 2 l 4 − l 3 = − l3 = − 0 . 01

f

2

f M

L

(s)

= 1.2

6

e

2

= −0 . 01

d

1

1

= 1

+ 0 . 9 s + 1. 005

s

2

2

+ l2 s + l3s 2

3

4

4

=

f

=

2l

= 2 m 4 − 2 m 1m 3 + m 2 = m 2 = d 2 2

4

2

− 2 l1 l 3 + l 2 = −2 × 1 × 0 .1 + ( 1.1 ) = 1. 01 2

4

2

2d

− d1

2

2

f

= 2 d 2 − 1.2 = 0 .81 =

s

+ 0 . 8955

s

= 1. 01

Thus d

0 . 995 2

4

+ 0 . 995

1

e

= 1. 01 = d 2

2

4

Thus d

= 0 .9 G L ( s)

=

0 . 995 s( s

+ 0 .895)

Roots of Characteristic Equation: Third-order System

−10.11 −0.4457 + j0.8892

2

2

= 1.2 = 2

+m2s

= 0 .1

3

2

2

1+ l s

e

= 2 l 2 − l1 = 2 × 1.1 − 1 = 1.2 f

1

1

2

2

Thus,

1+ m s

2

− m1 = 2 d 2 − d1 2

2

1+ d s

2

−0.4457 − j0.8892

Second-order System

−0.4478 + j0.8914 −0.4478 − j0.8914

105

2

= 1. 005

The real root at −10.11 is dropped by the second-order approximating system, and the two complex roots are slightly preturbed.

Unit-Step Responses:

(b)

N = 2:

Closed-loop Transfer Function: M

l

= 0 .5

1

e d

l

= d2 = 2

4

=

2 1

M

L

2d

(s)

2

f

2

H

(s)

20

=

s

= 0 . 55

l

3

3

=

+ 11 s + 10 s + 20 2

= 0 . 05

e

2

=

4

2d

+ 0 . 55

− d1 = 2

2

= −2 l1 l 3 + l 2 = −0 . 05 + 0 . 3025 = 0 .2525

1 1

1 + 0 .5 s

2

− e 2 = 2 × 0 . 5025 − 0 .85 = 0 .155

=

1

+ 0 . 3937

s

+ 0 . 5025

s

2

d

1

s

2

+ 0 . 7834

s

2

+ 0 . 05

2

=

2l

Thus

s

3

− l1 = 2 × 0 . 55 − ( 0 . 5) 2

2

d

2 2

=

f

4

+ 1. 99

G L(s)

1. 99

= s(s

Roots of Characteristic Equations: Third-order System

− 10.213

−0.3937 + j1.343 −0.3937 − j1.343

= 0 .2525

= 0 . 3937

1. 99

=

f

s

Second-order System

−0.3917 + j1.3552 −0.3917 − j1.3552

The real root at −10.213 is dropped by the second-order approximating system, and the two complex roots are slightly preturbed.

Unit-Step Responses:

106

+ 0 . 7835)

2

= 0 .85 d

2

= 0 . 5025

(c)

N=3 Closed-Loop Transfer Function: M

l e

= 0 . 3333

1

= 2 d 2 − d1 = 2

2

Thus, d

H

=

2 2

f

4

(s)

l f

=

L

(s)

s

+ 11 s + 10 s + 30 l

2

+ 0 . 333

s

+ 0 . 3667

d 1

2 1

= 2 d2 −

+ 0 .2186

s

f

= 0 .2186

+ 0 . 335

2

s

2

+ 0 . 0333

s

3

s

2

= d2 = 2

4

f

= −2 l1 l 3 + l 2 = 0 .1122 2

4

= 0 . 335

2

2 . 985

=

e

= 2 × 0 . 335 − 0 . 6222 = 0 . 0477 d

2

s

= 0 . 0333

= 2 l 2 − l1 = 0 . 7333 − 0 .1111 = 0 . 6222

1 1

1

2

= 0 . 1122

=

3

1

=

2

= 0 . 3667

2

d

M

30 3

+ 0 . 6524

s

+ 2 .985

G L(s)

2 . 985

= s( s

+ 0 . 6525)

Roots of Characteristic Equation: Third-order System

Second-order System

−10.312 −0.3438 + j1.6707 −0.3438 − j1.6707

−0.3262 + j1.6966 −0.3262 − j1.6966

The real root at −10.312 is dropped by the second-order approximating system, and the complex roots are slightly preturbed.

Unit-Step Responses:

(d)

N= 4

Closed-Loop Transfer Function:

107

M

40

=

(s)

H

s

3

+ 11 s + 10 s + 40 l

e

2

=

Thus,

d

− d1 = 2

2d

2 2

2

= M

f

L

= 0 .25

1

d

2

+ 0 .1225

s

l

=

3

=

2 1

+ 0 . 025

2

s

s

3

= 0 . 025

4

d

2

+ 0 .275

=d2 =

4

= 0 .2513

+ 0 .2513

s

1 + 0 .25 s

e

1 1

1

= 0 .275

2

= 2 l 2 − l 1 = 0 .4875

2

=

(s)

l

2

f

= 0 . 06313

4

=

2

2d

4

= −2 l1 l 3 + l 2 = 0 . 06313

f

2

f 2

2

= 0 . 5025 − 0 .4875 = 0 . 015

3 . 98 s

+ 0 .4874

2

s

G

+ 3 .98

L

(s)

=

Roots of Characteristic Equation: Third-order System

Second-order System

− 10.408 −0.2958 + j1.9379 −/2958 − j1.9379

−0.2437 + j1.98 −0.2437 − j1.98

The real root at −10.408 is dropped by the second-order approximating system, but the complex roots are preturbed. As the value of N increases, the gain of the system is increased, and the roots are more preturbed.

Unit-Step Responses:

(e)

N= 5

Closed-loop Transfer Function: M

l e e

H

1

(s)

= 0 .2

2

= 2 d1 − d 2 =

4

= d2 =

2

2

f

=

f

50 s

3

+ 11 s + 10 s + 50 2

l

2

= 0 .22

1

= 1 l

3

= 0 . 02

= 2 l 2 − l1 = 0 .44 − 0 .04 = 0 .4 2

2

= −2 l1 l 3 + l 2 = −0 . 008 + 0 . 0484 = 0 . 0404 2

4

+ 0 .2 s + 0 .22

108

s

2

+ 0 . 02

s

3

d

1

= 0 .1225

3 . 98 s( s

+ 0 .4874

)

Thus, d d M

L

(s)

2 2 2 1

=

f

= 0 .0404

=

2d

4

+ 0 . 0447

1

s

= 0 .201

2

− e 2 = 0 .402 − 0 .4 = 0 . 002

2

1

=

d

+ 0 .201

s

2

=

d

1

1 s

2

+ 0 .2225

s

= 0 . 04472 G L(s)

+ 4. 975

4. 975

=

s(s

+ 0 .2225)

Roots of Characteristic Equation: Third-order System

Second-order System

− 10.501 −0.2494 + j2.678 −0.2494 −j2.678

−0.1113 + j2.2277 −0.1113 − j2.2277

The real root at −10.501 is dropped by the second-order approximating system, and the complex roots are changed, especially the real parts.

Unit-Step Responses:

7-33 (a)

K=1

Forward-path Transfer Function:

G ( s) =

891100

(

s s + 5000 s + 566700 2

)

=

891100 s ( s + 116)( s + 4884)

Closed-Loop Transfer Function : M

H

(s)

891100

= l

s

1

e e

3

+ 5000

s

+ 566700

2

= 0 . 636

l

2

= 2 d 2 − d1 =

4

= d2 =

2

2

f

2

f

s

+ 891100

= 5. 611 × 10

−3

+ 0 . 636 l

s

+ 5. 611 × 10

= 1.122 × 10

3

−3

Thus,

109

−3

s

+ 1.1222 × 10

2

−6

s

3

−6

− 0 .4045 = −0 . 3933

= − 2 l1 l3 + l 2 = −2 × 0 . 636 × 1.1222 × 10 2

4

1

= 2 l 2 − l1 = 1.1222 × 10 2

2

1

=

−6

+ ( 5 . 611 × 10

−3

)

2

= 0 . 000030

06

2

d

2

d

2

M

L

(s)

1

= 0 . 000030 = 2 d2 −

f

d

06 2

1

+ 0 . 6358

s

= 0 .005482

= 0 . 01096 + 0 . 3933 = 0 .4042

1

=

2

+ 0 . 005482

s

=

2

d

1

= 0 . 6358

182 .4 s

+ 115

2

+ 182

. 97 s

G L ( s)

.4

182 .4

=

s( s

+ 115

. 97 )

Roots of Characteristic Equations: Third-order System

Second-order System

−1.595 −114.4 −4884

−1.5948 −114.38

The real root at −4884 is dropped by the second-order approximating system. the other two roots are hardly preturbed.

Unit-Step Responses

(b)

K = 100

M

(s)

H

Closed-loop Transfer Function:

891100

=

s

3

+ 5000 l

e e

s

00

+ 566700 + 891100

2

= 0 .00636

1

2

= 2 d 2 − d1 =

4

= d2 =

2

2

f

f

l

2 2 2

d

M

L

(s)

1

=

f

4

1

+ 0 . 007403

−5

= 5. 611 × 10

−5

l

= 1. 1222 × 10

−8

−5

3

− 4. 045 × 10

4

f

−5 −8

= 2 l 4 − 2 l1 l 3 + l 2 = − 2 × 0 . 00636 × 1.1222 × 10 2

2

−9

= 0 . 000109

d 6

s

+ 0 . 000054

8s

2

− 0 . 000071

1

=

+ 5. 611 × 10

1 + 0 . 00636 s

= 2 l 2 − l1 = 11 .222 × 10

= 3 . 0056 × 10

= 2 d2 −

00

1

2

2

Thus, d

2

=

2

= 0 . 000054 8

s

2

+ 135

= 0 . 000054

.1 s

2

+ 1.1222 × 10

= 0 . 000071

+ 18248

8

d

1

G

+ ( 5 . 611 × 10

−5

)

2

(s)

=

18248 s( s

+ 135

. 1)

Roots of Characteristic Equations: Third-order System

Second-order System

110

s

3

= 3 . 0056 × 10

= 0 . 007403

L

−8

8

8

18248

=

s

−9

−4887.8 −56.106 + j122.81 −56.106 − j122.81

−67.55 + j114.98 −67.55 − j114.98

Unit-Step Responses

(c)

K = 1000 Closed-loop Transfer Function: M

H

(s)

891100

=

s

+ 5000

3

l e e

s

2

+ 566700

= 2 d 2 − d1 =

4

= d2 =

2

f

f

d

M

L

(s)

2 2 2 1

=

f

=

2d

4

l

2

1

−6

= 5 . 611 × 10

−6

l

3

f

2

= 0 . 000010

d 965

− 0 . 000010

+ 0 . 000382

9s

+ 0 . 000005

482 s

2

=

818

−6

= 1.1222 × 10

−9

− 0 .000000

2

−11

+ 5 . 611 × 10

s

= − 2 l1 l3 + l 2 = −2 × 0 . 000636 × 1.1222 × 10

4

1

=

1 + 0 . 000636

000

= 2 l 2 − l1 = 11 .222 × 10

= 3 . 0055 × 10 2

+ 891100

1

=

2

2

Thus, d

s

= 0 . 000636

1

2

2

000

2

404 −9

= 0 . 000010

+ ( 5 . 611 × 10

= 0 . 000005

482

= 0 . 000000

147

d

182415 .177 s

2

+ 69 . 8555

s

+ 182415

.177

1

s

−6

)

2

= 3 . 0056 × 10

= 0 . 000382

−34.928 + j425.67

111

3

G L (s)

−34.928 − j425.67

−11

9

=

182415 .177 s( s

Second-order System

− 4921.6 −39.178 + j423.7 −39.178 − j423.7

s

82

Roots of Characteristic Equations: Third-order System

−9

+ 1.1122 × 10

2

+ 69 . 8555)

Unit-step Re sponses

7-34

Forward-path Transfer Function G (s)

=

K(s s(s

Closed-loop Transfer Function

− 1)

M (s)

+ 1 )( s + 2 )

Second-order System: M

L

(s)

=

M L (s )

−s +1 s

+3s + s +1

3

2

+ c1 s

1

1+ d s

+ d2s

1

M H (s )

=

2

( − s + 1) (1 + d s + d s ) 1 + ( d − 1) s + ( d − d ) s − d s = ( s + 3s + s + 1) ( 1+ c s ) 1 + ( c + 1) s + ( c + 3) s + ( 3c + 1) s + c s 2

=

1

3

2

2

1

2

= 1 + c1

1

m

1

l

= d1 −1

3

1

2

2

1

l

2

m

1

2

= 3 + c1

2

= d 2 − d1

l

3

= 1 + 3 c1

3

= −d 2

m

3

1

1

l

1

= c1

4

e2 = f 2 = 2m 2 − m1 = 2 ( d 2 − d 1 ) − ( d 1 − 1) = 2 d 2− d 1 − 1 2

2

2

= 2l2 − l1 = 2 ( 3 + c1 )− ( 1+ c1 ) = 5 − c1 2

2

2

e4 = f 4 = 2m 4 − 2m1 m3 + m 2 = −2 ( d 1 − 1) ( −d 2 ) + ( d 2 − d1 ) = d 2 − 2d 2 + d 1 2

2

2

2

= 2l4 − 2l1l3 + l2 = − 2 ( 1 + c1 ) (1 + 3c1 ) + ( 3 + c1 ) = 7 − 2c1 − 5c1 2

2

2

e6 = f 6 = 2m 6 − 2 m1m 5 + 2 m 2 m4 − m 3 = −m 3 = − ( −d 2 ) = −d 2 2

2

2

2

= 2l6 − 2l1l5 + 2l 2l4 − l3 = 2l2 l4 − l3 = 2 ( 3 + c1 ) c1 − (1 + 3c1 ) = − 1 − 7c1 2

2

2

Simultaneous equations to be solved:

Solutions:

112

2

4

c1

= −1. 0408

= 1 + 7 c1

d

= 0 . 971

− 2 d 2 + d 1 = 5 − c1

d

2d

− d1 = 6 −c1

2

d d

M

L

2 2

(s)

2

2

2

2

2 2

− 1. 0408

1

=

2

s

+ 0 . 971 s + 2 . 93

1

s

−0 .3552(

=

2

s

s

+ 0 . 3314

2

− 0 . 9608 s

1

2

)

= 2 . 93 G L(s)

+ 0 .3413

−0 . 3552(

=

s( s

s

− 0 . 9608

+ 0 . 69655)

Roots of Characteristic Equations: Third-orde r System

Second-order System

−2.7693 −0.1154 + j0.5897 −0.1154 − j0.5897

−0.1657 + j0.5602 −0.1657 −j0.5602

Unit-step Responses

7-35 (a)

K = 10 Closed-loop Transfer Function: M

(s)

H

=

10 s

+ 23

4

s

3

+ 62

s

2

+ 40 s + 10

=

1 1+ 4 s

+ 6 .2 s + 2 . 3 s + 0 .1 s 2

3

Second-order System Approximation: M

L

(s)

=

1 1+ d s 1

l e e e

2

4

6

=

f

=

f

=

f

2

=4

1

=

2d

l

2

= 6 .2

2

l

3

= 2 .3

l

4

− d 1 = 2 l 2 − l1 = 12 .4 − 16 = −3 . 6 2

2

+ d2s

2

4

= d 2 = 2 l 4 − 2 l1 l 3 + l 2 = 0 .2 − 2 × 4 × 2 . 3 + ( 6 .2 ) = 20 .24

6

= 2 d 6 − 2 d 1 d 5 + 2 d 2 d 4 − d 3 = − d 3 = 2 l 2 l 4 − l3 = − 4. 05

2

2

2

2

2

2

Thus, d d

2 2 2 1

=

20 .24

=

2d

2

d f

2

= 9 + 3. 6 = 12 . 6

2

d

113

1

= 4. 5 = 3 . 55

= 0 .1

4

)

M

L

1

=

(s)

+ 3 .55 s + 4. 5 s

1

0 .2222

=

2

s

2

+ 0 . 7888

s

G L(s)

+ 0 .2222

0 .2222

=

s( s

+ 0 . 7888

Roots of Characteristic Equations: Fourth -order system

Second-order system

−2.21 −20 −0.3957 + j0.264 −0.3957 −j0.264 (b)

−0.3944 + j0.258 −0.3944 −j0.258

K = 10 Third-order System Approximation:

M

L

(s)

=

1+ d s 1

e e

2

4

=

f

=

f

2

=

2d

M

1

+ d 2s + d3s 2

3

M

− d 1 = −3 . 6 2

2

d

L

(s) (s)

1

= 1

= −f6 =

2 3

+ d1 s + d 2 s + d 3 s 2

2

3

+ 4 s + 6 .2 s + 2 . 3 s + 0 .1 s 2

4. 05

= 2 d 4 − 2 d 1 d 3 + d 2 = − 2 d 1 d 3 + d 2 = − 4. 025 2

4

H

Thus

d

+d2 = 2

1

f

4

3

d

3

4

= 2 . 0125

= 20 .24

Thus, d

= 0 . 5 d 1 − 1.8 2

2

Solving for d ,

d

1

1

= 3 .9528

2

d ,

2

− 3 . 0422

= 0 .25 d 1 − 1. 8 d 1 + 3 .24 = 20 .24 + 4. 025 4

− 0 .4553 +

,

d

Selecting the positive and real solution, we have

M

L

(s)

2

1

= 3 . 9528

1

= 1

+ 3. 9528

s

+ 6 . 0123

s

2

GL ( s) =

+ 2 . 0125

(

s

3

=

− 0 .4553 −

j 2334, d

.

1

j 2334.

= 0 . 5 d 1 − 1.8 = 6 . 0123 2

2

0 .4969 s

3

+ 2 . 9875

0.4969

s s + 2.9875 s + 1.964 2

d

s

2

+ 1. 964

s

+ 0 .4969

)

Roots of Characteristic Equations: Fourth -order System

−2.21 −20 −0.39565 + j0.264 −0.39565 − j0.264

Third-order System

−2.1963 −0.3956 + j0.264 −2.1963 − j0.264

Unit-step Response

114

)

(c)

K = 40 Closed-loop Transfer Function: M

(s)

H

40

=

s

l

1

4

+ 23

s

+ 62

3

=1

l

s

2

+ 40 s + 40

= 1. 55

2

=

l

1 1+ s

+ 1. 55

= 0 . 575

3

s l

2

4

+ 0 . 575

s

+ 0 . 025

3

s

4

= 0 . 025

Second-order System Approximation: M

L

(s)

1

=

1+ d s 1

e e

2

4

=

f

=

f

2

2

= 2 d2 − d1 =

4

= d 2 = 2 l 4 − 2 l1 l3 + l 2 = 0 . 05 − 1.15 + 2 .4025 = 1. 3025

2

2l

− l 1 = 3 .1 − 1 =

+ d2s

2

2

2

2 .1

2

Thus, d d

M

(s)

L

= 1. 3025

2 2

d

= 2 d2 −

f

+ 1.1413

2

2 1

2

1

= 1

+ 0 .4273

s

s

= 2 × 1.1413 − 2 . 1 = 0 .1825

= 1.1413

2

d

1

= 0 .4273

0 .8762

=

s

+ 0 . 3744

2

s

G L ( s)

+ 0 .8762

=

0 .8762 s( s

+ 0 . 3744

Roots of Characteristic Equations: Fourth -order System

Second-order System

−2.5692 −19.994 −0.2183 + j0.855 −0.2183 − j0.855

−0.1872 + j0.9172

−0.1872 − j0.9172

Third-order system Approximation: M

L

(s)

1

= 1

e2

=

f2

= 2 d2 − d1 =

e4

=

f4

= −2 d 1 d 3 + d 2 = −1. 0062

e

=

f

= − d 3 = 2 l 2 l 4 − l3 = − 0 .2531

6

2

2l2

2

− l 1 = 3 .1 − 1 = 2

2

2

6

+ d 1s + d 2 s + d 3 s

d1

3

2 .1

+ d 2 = 2 l 4 − 2 l1 l 3 + l 2 = 1. 3025 2

2

2

Equations to be Solved Simultaneously: 2d d

2 2

− d 1 = 2 .1 2

2

= 0 .25

d

4 1

d

2 2

= −1. 00623

+ 1. 05 d 1 + 1.1025 2

d

1

= 1. 3025

Th us

115

d

4 1

+ 4.2

Thus d

2 1

d

− 4. 0249

= 0 .5 d 1 + 1. 05 2

2

d

1

− 0 .8 = 0

)

d

The roots of the last equation are:

1

= − 0 .1688

Selecting the positive real solution, we have d d

= 1. 00623

2 2

1

1

0 . 9525 ,

= 0 .9525

+ 1.3025 = 2 .261

= 0 .2531

2 3

M

d

d

,

L

(s)

1

= 1

+ 0 . 9525

GL ( s) =

s

(

+ 1. 5037

s

2

+ 0 . 5031

s

1.9876

s s + 2.9886 s + 1.8932 2

d

T hus

d

)

Roots of Characteristic Equati ons: Fourth -order System

−2.5692 −19.994 −0.2183 + j0.855 −0.2183 − j0.855 Unit-step Responses

116

=

j 2 .196 ,

− 0 . 392 −

.

Th us

=

3

− 0 .392 +

2

3

= 1. 5037 = 0 . 5031 1. 9876

s

3

+ 2 . 9886

s

2

+ 1. 8932

s

+ 1.9876

1.9876 s ( s + 2.0772)( s + 0.9114)

Third-order System

−2.552 −0.2183 ± j 0 .8551

j 2 .196

Chapter 8 8-1 (a)

P (s)

ROOT LOCUS TECHNIQUE

= s + 4 s + 4 s +8 s 4

3

2

Q (s)

= s +1

0,

Finite zeros of Q( s):

−1

−3.5098, −0.24512 ± j1.4897

Finite zeros of P( s):

Asymptotes: K > 0: Intersect of Asymptotes:

P( s)

= s + 5s + s 3

2

Q (s)

Finite zeros of P( s) : Finite zeros of Q( s):

,

180

=s

2

Q (s)

Finite zeros of P( s): Finite zeros of Q( s): Asymptotes:

(d)

P (s)

−1

o

90

3

= s + 2 s + 3s 5

4

,

270

o

,

240

o

= −1

K < 0:

3

±

−1

o

0 ,

180

o

= −2

j 1. 5874

o

(

−1± 1, −1, −3 0,

2

K < 0:

0

o

) ( s + 3)

j 1.414

There are no asymptotes, since the number of zeros of P( s) and Q( s) are equal.

3

Q (s) 0,

0,

− 1.5 ±

= s +3s + 5 2

−1 ±

0,

= s + 2 s + 10 2

Finite zeros of P( s): Finite zeros of Q( s):

Q (s)

60

o

,

180

o

,

300

o

K < 0:

−1 − 1 − ( −1. 5) − ( −1. 5) 5

−2

=

o

0 ,

120

o

,

240

,

240

o

1 3

= s +5

− 1.0398 ± −5

Asymptotes: K > 0: Intersect of Asymptotes:

j 1.414

j 1. 6583

σ1 = 4

o

−4. 7913 − 0 .2087 − ( − 1)

0 . 083156 180

Asymptotes: K > 0: Intersect of Asymptotes:

P (s)

120

Q(s ) = s − 1

Finite zeros of P( s): Finite zeros of Q( s):

(f)

−1

0 ,

2

0, 0 − 3.156 ,

2

Finite zeros of Q( s): Asymptotes:

P (s)

o

K < 0:

= s + 3s + 2s +8

K > 0:

Finite zeros of P( s):

(e)

o

−4.7912, −0.20871

0,

= s + 2s + 3s 3

300

4

σ1 = P( s)

,

= s +1

Asymptotes: K > 0: Intersect of Asymptotes:

(c)

o

−3 . 5 − 0 .24512 − 0 .24512 − ( −1 )

σ1 = (b)

o

60

j 1.4426 ,

60

o

,

180

1. 0398 o

,

300

120

± o

j1.4426 K < 0:

o

0 ,

120

o

o

−1. 0398 − 1. 0398 + 1. 0398 + 1. 0398 − ( − 5)

σ1 =

4

8-2 (a) Angles of departure and arrival. K > 0:

−θ 1 − θ 2 − θ 3 + θ 4 = −180 o

o

o

= −180

o

o

=0

o

−θ 1 − θ 2 − θ 3 + θ 4 = −180

o

−θ 1 − 90 − 45 + 90 o

−θ 1 − 90 − 45 + 90 θ 1 = −45

o

o

θ 1 = 135 K < 0:

o

o

(b) Angles of departure and arrival. K > 0: K < 0:

o

−θ 1 − 135

o

− 90 + 90

o

o

=0

o

θ 1 = −135

(c) Angle of departure: K > 0:

−θ 1 − θ 2 − θ 3 + θ 4 = −180 o

−θ 1 − 135 θ 1 = −90

− 90

o

− 45

o

o

= −180

o

o

(d) Angle of departure K > 0:

−θ 1 − θ 2 − θ 3 − θ 4 = −180 −θ 1 − 135

o

θ 1 = −180

− 135

o

− 90

o

o

= −180

o

o

121

−1

=

−5 3

(e) Angle of arrival K < 0:

θ 1 + θ 6 − θ 2 − θ 3 − θ 4 − θ 5 = −360 o

8-3 (a)

θ 1 + 90 − 135

o

θ 1 = −108

o

.435

− 135

o

o

o

− 45 − 26 . 565

o

= −360

(b)

(c)

(d)

122

o

8-4 (a) Breakaway-point Equation:

2s

5

+ 20

s

4

+ 74

− 0 . 7275

Breakaway Points:

(b) Breakaway-point Equation: 3 s + 22 6

s

+ 65

5

+ 110

− 2 . 3887

,

+ 100

4

s

(c) Breakaway-point Equation: 3 s + 54 6

s

+ 347

5

− 2 .5 ,

Breakaway Points:

(d) Breakaway-point Equation: − s − 8 s − 19 6

5

+ 86

3

s

8-5 (a) =

G ( s)H ( s)

90

o

K(s s(s

+ 925

s

+ 8 s + 94

s

.5 s

s

2

+ 44 s + 12 = 0

4

3

+ 867

.2 s

+ 120

s

2

− 781 .25 s − 1953 = 0

1. 09 4

s

3

− 0 . 6428

Breakaway Points:

,

2

+ 48 = 0

2 .1208

+8)

+ 5)( s + 6 )

and

o

270

Intersect of Asymptotes:

σ1 =

0

K < 0:

− 5 − 6 − ( −8 ) 3

Breakaway-point Equation: 2s

Breakaway Points: Root Locus Diagram:

+ 48 s = 0

2

s

− 1, − 2 . 5

Breakaway Points:

Asymptotes: K > 0:

3

s

3

+ 35

− 2 .2178

s ,

2

−1

+ 176

s

123

o

= − 1.5

+ 240 = 0

− 5 . 5724,

8-5 (b)

0

− 9 . 7098

and

180

o

G ( s)H ( s)

Asymptotes: K > 0: Intersect of Asymptotes:

K

=

45

o

s(s ,

+ 1 )( s + 3)( s + 4 )

135

o

,

225

σ1 =

G ( s)H ( s)

=

+4)

2

+ 2) o

Asymptotes: K > 0: 60 , Intersect of Asymptotes:

o

K < 0:

0 ,

K < 0:

0 ,

90

o

,

180

o

,

= −2

4 2

+ 38 s + 12 = 0 − 2 , − 3 . 5811

2

180

o

,

300

σ1 =

0

o

+ 0 − 2 − 2 − ( −4 ) 4

Breakaway-point Equation: Breakaway Points:

o

315

−1 −3 − 4

3

K(s s (s

,

4 s + 24 s −0 .4189 ,

Breakaway-point Equation: Breakaway Points: Root Locus Diagram:

8-5 (c)

0

o

3s 0,

4

−1

=0

+ 24 s + 52 s + 32 s = 0 − 1. 085 , − 2 , − 4. 915 3

2

Root Locus Diagram:

124

o

120

o

,

240

o

270

o

8-5 (d) G ( s)H ( s)

=

K(s s( s

2

+ 2)

+ 2s + 2) o

Asymptotes: K > 0: 90 , Intersect of Asymptotes:

270

o

K < 0:

σ1 = 3

0

− 1 − j − 1 − j − ( −2 ) 3

−1

o

0 ,

180

o

=0

+8 s +8 s + 4 = 0

Breakaway-point Equation:

2s

Breakaway Points:

− 2 .8393

2

The other two solutions are not breakaway points.

Root Locus Diagram

125

8-5 (e) G ( s) H( s) =

(

K ( s + 5)

s s + 2s + 2 2

o

Asymptotes: K > 0: 90 , Intersect of Asymptotes:

)

o

270

σ1 =

0

K < 0:

− 1 − j − 1 − j − ( −5) 3

Breakaway-point Equation: 2 s + 17 s − 7.2091 3

Breakaway Points:

2

−1

(

K

s ( s + 4 ) s + 2s + 2 2

180

o

= 1. 5

+ 20 s + 10 = 0 The other two solutions are not breakaway points.

8-5 (f) G ( s) H( s) =

o

0 ,

) 126

o

Asymptotes: K > 0: 45 , Intersect of Asymptotes:

135

o

,

σ1 =

o

225 0

315

o

o

K < 0:

− 1 − j −1 + j − 4

0 ,

3

G ( s)H ( s)

Asymptotes:

=

2

s (s

K > 0:

+ 4)

2

+ 8)

2

90

o

,

270

o

2

Intesect of Asymptotes:

Breakaway-point Equation: Breakaway Points:

0,

s

5

,

180

o

+ 20 s + 8 = 0 The other solutons are not breakaway points.

o

K < 0:

σ1 =

o

= −1. 5

8-5 (g) K(s

90

4

4 s + 18 s −3 . 0922

Breakaway-point Equation: Breakaway Point:

,

0

+ 20

0 ,

180

o

+ 0 − 8 − 8 − ( −4 ) − ( −4 ) 4 s

4

+ 160

s

−2 3

−4, −8, −4 − j4, −4 + j4

127

+ 640

s

2

+ 1040

s

=0

,

270

o

8-5 (h) G ( s)H ( s)

=

K 2

s (s o

Asymptotes: K > 0: 45 , Intersect of Asymptotes:

+ 8)

2

135

o

,

225

o

σ1 = Breakaway-point Equation: Breakaway Point:

s 0,

3

,

315

−8 − 8

o

K < 0:

= −4

4

+ 12 s + 32 −4, −8 2

=0

s

8-5 (i) 128

o

0 ,

90

o

,

180

o

,

270

o

G ( s) H( s) =

(

K s + 8 s + 20 2

s ( s + 8) 2

o

Asymptotes: K > 0: 90 , Intersect of Asymptotes:

270

)

2

o

o

K < 0:

σ1 = 5

0 ,

−8 − 8 − ( − 4 ) − ( −4 ) 4

+ 20

Breakaway-point Equation:

s

Breakaway Points:

− 4,

o

180

s

4

+ 128

−2 s

−8, − 4 +

3

+ 736

j 4. 9 ,

s

2

= −4 + 1280

−4 −

s

=0

j 4. 9

(j) G ( s) H( s) =

Ks

(s

2

2

−4

)

Since the number of finite poles and zeros of G ( s ) H ( s ) are the same, there are no asymptotes.

Breakaway-point Equation: 8 s = 0 Breakaway Points:

s=0

129

8-5 (k) G ( s) H( s) = Asymptotes:

(s

K > 0:

(

K s −4 2

2

)( s

+1 90

Intersect of Asymptotes: σ

1

Breakaway-point Equation: Breakaway Points:

o

,

=

2

)

+4 270

)

o

−2 + 2 4

−2 6

s 0,

K < 0:

o

0 ,

180

o

=0 − 8 s − 24 4

3 .2132 ,

=0 − 3 .2132

s

2

8-5 (l) 130

,

j 1. 5246 ,

j1. 5246

K ( s − 1) 2

G ( s) H( s) =

(s

2

Asymptotes: K > 0:

)( s

+1 90

o

,

2

+4 270

)

o

Intersect of Asymptotes:

σ1 = 5

−1 + 1 4

s

Breakaway Points:

− 2 . 07 ,

G ( s)H ( s)

=

K(s

3

j 1.47 ,

j 1.47

=0

3

2 . 07 ,

+ 1 )( s + 2 )( s + 3) s (s

0 ,

− 2s − 9 s = 0

Breakaway-point Equation:

(m)

−2

o

K < 0:

− 1) 131

180

o

Asymptotes: K > 0:

180

o

K < 0:

Breakaway-point Equation: s + 12 s + 27 s 6

5

G ( s)H ( s)

Asymptotes:

=

K (s 3

s (s

K > 0:

+ 5)( s + 40

+ 250 60

+ 2 s − 18 3

s

o

2

− 1.21, −2.4, −9.07,

Breakaway Points:

(n)

4

0

o

,

)( s

o

,

0.683,

0,

0

)

+ 1000

180

=0

)

300

o

K < 0:

132

o

0 ,

120

o

,

240

o

Intersect of asymptotes:

σ1 = Breakaway-point Equation: 3750 s

8-5 (o)

Asymptotes:

=

K(s s( s

K > 0:

+ 0 + 0 − 250 − 1000 − ( −5) − ( −40 5

+ 335000

− 7.288, −712.2,

Breakaway Points:

G ( s)H ( s)

6

0

0,

s

5

−2

+ 5 .247 × 10

8

s

4

0

− 1)

o

,

270

o

K < 0:

Intersect of Asymptotes:

133

o

0 ,

= − 401 . 67

+ 2 . 9375 × 10

+ 1 )( s + 2 ) 90

)

180

o

10

s

3

+ 1. 875 × 10

11

s

2

=0

σ1 =

8-6 (a)

3

−1 − 2 − 1 3

−1

−3 s −1 = 0

Breakaway-point Equation:

s

Breakaway Points;

−0.3473, −1.532,

Q(s ) = s + 5

(

= −2

1.879

)

P (s ) = s s + 3s + 2 = s (s +1)( s + 2) 2

o

Asymptotes: K > 0: 90 , Intersect of Asymptotes:

270

o

K < 0:

σ1 = Breakaway-point Equation:

s

3

−1 − 2 − ( −5) 3

−1

=1

+ 9 s + 15 s + 5 = 0 2

134

o

0 ,

180

o

−0.4475, −1.609,

Breakaway Points:

8-6 (b)

Q (s)

= s +3

Asymptotes:

P (s)

=

ε

s s

K > 0:

2

90

+ s+2 o

,

−6.9434

ϕ

270

o

K < 0:

Intersect of Asymptotes:

σ1 = 3

−1 − ( −3) 3

−1

o

0 ,

180

o

=1

+5s + 3s + 3 = 0

Breakaway-point Equation:

s

Breakaway Points:

− 4.4798

2

The other solutions are not breakaway points.

135

8-6 (c)

Q (s)

= 5s

Asymptotes:

P (s)

=

s

K > 0:

2

+ 10 180

o

K < 0: 2

− 50 = 0

Breakaway-point Equation:

5s

Breakaway Points:

− 3.162,

3.162

136

0

o

(

8-6 (d) Q(s ) = s s + s + 2 2

)

Asymptotes: K > 0:

P( s) = s + 3s + s + 5s +10 4

180

3

2

o

K < 0: 6

Breakaway-point Equation:

s

Breakaway Points:

−2,

0

o

+ 2 s + 8 s + 2 s − 33 5

4

1.784.

3

s

2

− 20 s − 20 = 0

The other solutions are not breakaway points.

137

(

) ( s + 2)

8-6 (e) Q(s ) = s −1 2

(

P( s) = s s + 2 s + 2 2

)

Since Q ( s ) and P ( s ) are of the same order, there are no asymptotes. 3

+ 12

Breakaway-point Equation:

6s

Breakaway Points:

−1.3848

s

2

+ 8s + 4 = 0

138

8-6 (f)

Q(s ) = (s + 1)(s + 4) Asymptotes:

K > 0:

(

P (s ) = s s − 2 180

2

)

o

K < 0: 4

+ 10

Breakaway-point equations:

s

Breakaway Points:

−8.334,

s

3

+ 14

0 s

2

0.623

139

o

−8 = 0

8-6 (g) Q(s ) = s + 4s + 5 2

Asymptotes:

P( s) = s

K > 0:

90

o

,

2

(s

+ 8 s + 16

2

270

o

σ1 = 5

s

Breakaway Points:

0,

o

K < 0:

Intersect of Asymptotes:

Breakaway-point Equation:

)

−8 − ( −4 )

+ 10

4 s

− 2,

4

s

180

o

= −2

−2 + 42

0 ,

3

− 4, 140

+ 92 −2 +

s

2

+ 80

j 2 .45 ,

s

=0 −2 −

j 2 .45

(

8-6 (h) Q(s ) = s − 2 2

)( s + 4)

(

P (s ) = s s + 2s + 2 2

)

Since Q ( s ) and P ( s ) are of the same order, there are no asymptotes.

Breakaway Points:

− 2,

6.95

141

8-6 (i)

Q (s)

Asymptotes:

= ( s + 2 )( s + 0 . 5) K > 0:

180

P( s)

o

Breakaway-point Equation: Breakaway Points:

=s

−4.0205,

ε −ϕ s

2

1

K < 0: s

4

0

o

+5s + 4s −1 = 0 3

2

0.40245 The other solutions are not breakaway points.

142

8-6 (j)

Q(s ) = 2 s + 5 Asymptote s:

K > 0:

P (s ) = s 60

o

,

180

2

o

(s

2

,

300

Intersect of Asymptotes;

σ1 = Breakaway-point Equation:

6s

4

)

+2s +1 = s

0

o

2

( s + 1)

K < 0:

+ 0 − 1 − 1 − ( − 2 . 5)

+ 28

4 s

3

−1

+ 32 143

s

2

2

o

0 ,

=

0 .5 3

+ 10

s

=0

120

o

,

= 0 .167

240

o

Breakaway Points:

8-7 (a) Asymptotes:

K > 0:

−0.5316, −1, −3.135

0,

45

o

,

135

o

,

225

Intersect of Asymptotes:

σ1 = 5

+ 65

4s

Breakaway Points:

−0.6325,

ζ = 0 . 707

,

,

315

o

−2 − 2 − 5 − 6 − ( −4 )

Breakaway-point Equation:

When

o

5 s

4

−1

+ 396 −5.511

K = 13.07

144

s

3

+ 1100

= − 2 . 75 s

2

+ 1312

(on the RL)

s

+ 480 = 0

8-7 (b) Asymptotes:

K > 0:

45

o

,

135

o

,

225

Intersect of Asymptotes:

σ1 = Breakaway-point Equation: When ζ = 0 . 707 , K = 61.5

4s

3

0

o

,

315

o

− 2 − 5 − 10

= −4.25

4

+ 51 s + 160 2

145

s

+ 100 = 0

8-7 (c) Asymptotes:

K > 0:

180

o

Breakaway-point Equation: Breakaway Points:

−1.727

When

K = 9.65

ζ = 0 . 707

,

s

4

+ 4 s + 10 3

s

+ 300

2

(on the RL)

146

s

+ 500 = 0

8-7 (d)

K > 0:

90

o

,

270

o

Intersect of Asymptotes:

σ1 = When

ζ = 0 . 707

,

−2 − 2 − 5 − 6 4

−2

K = 8.4

147

= −7 . 5

8-8 (a) Asymptotes:

K > 0:

60

o

,

180

o

,

300

Intersect of Asymptotes:

σ1 =

0

o

− 10 − 20

= −10

3

Breakaway-point Equation: 3 s + 60 s + 200 = 0 2

148

Breakaway Point: (RL)

−4.2265,

K = 384.9

(b) Asymptotes:

K > 0:

45

o

,

135

o

,

225

Intersect of Asymptotes:

σ1 = Breakaway-point Equation: 4 s + 27 s Breakaway Points: (RL) −0.4258 3

2

0

o

,

315

−1 − 3 − 5

o

= −2 .25

4

+ 46 s + 15 = 0 K = 2.879 ,

149

−4.2537

K = 12.95

150

8-10

P (s)

= s + 25 3

Asymptotes:

s

2

+ 2 s + 100 K

t

> 0:

Q (s) 90

o

,

270

= 100

Intersect of Asymptotes:

σ1 = 3

−25 − 0

s

Breakaway Points:

−2.2037,

8-11 Characteristic e quation:

s

3

3

−1

= −12 . 5

+ 12 . 5 s − 50 = 0

Breakaway-point Equation: (RL)

s

o

2

−12.162

+5s + K ts + K = 0 2

151

(a) Kt = 0 :

P (s ) = s

2

( s + 5)

Q (s) = 1 o

Asymptotes: K > 0: 60 , Intersect of Asymptotes:

180

o

,

σ1 = Breakaway-point Equation:

8-11 (b)

P (s)

3s

= s + 5 s + 10 = 0 3

2

o

Asymptotes: K > 0: 90 , Intersect of Asymptotes:

2

−5 − 0

= −1. 667

3

+ 10

Q (s)

270

o

300

s

=

=0

s

o

152

Breakaway Points:

0,

−3.333

σ1 =

−5 − 0 2

−1

=0

Breakaway-point Equation: 2 s + 5 s − 10 = 0 There are no breakaway points on RL. 3

8-12

P( s)

= s + 116 2

. 84 s

Asymptotes:

+ 1843 J

L

= 0:

Q (s) 180

(RL)

2

2 . 05 s ( s

+ 5)

o

Breakaway-point Equation: − 2 . 05 s Breakaway Points:

=

4

− 479 0,

s

3

− 12532

−204.18

s

153

2

− 37782

s

=0

(

)

8-13 (a) P (s ) = s s − 1 2

Asymptotes:

Q (s ) = ( s+ 5)( s + 3) K > 0:

180 4

o

+ 16

Breakaway-point Equation:

s

Breakaway Points:

0.5239,

(RL)

s

3

+ 46

s

2

−12.254

154

− 15 = 0

(

)

8-13 (b) P (s ) = s s + 10s + 29 2

Q (s ) = 10(s + 3) o

Asymptote s: K > 0: 90 , Intersect of Asymptotes:

270

o

σ1 =

0

− 10 − ( −3) 3

−1

Breakaway-point Equation: 20 s + 190 s There are no breakaway points on the RL. 3

2

+ 600

155

= −3 . 5 s

+ 870 = 0

8-14 (a)

P( s)

= s ( s + 12 . 5)( s + 1 )

Q (s)

Asymptotes: N > 0: Intersect of Asymptotes:

= 83 .333 60

o

,

180

o

,

300

σ1 = + 27 −0.4896

Breakaway-point Equation: 3 s Breakaway Point:

(RL)

2

− 12 . 5 − 1

0

s _12 .5

o

3

=0

156

= − 4. 5

8-14 (b)

P( s)

= s + 12 . 5 s + 833 2

A > 0:

180

.333

Q (s)

= 0 .02

2

s (s

+ 12 . 5)

o

Breakaway-point Equation: 0 . 02 s Breakaway Points: (RL) 0

4

+ 0 . 5 s + 53 .125 3

157

s

2

+ 416

. 67 s

=0

8-14 (c)

P (s)

= s + 12 . 5 s + 1666 . 67 = ( s + 17 . 78 Q ( s ) = 0 . 02 s ( s + 12 . 5) 3

2

Asymptotes:

K

o

> 0:

180

)( s

− 2 . 64 +

j 9 . 3)( s

− 2 . 64 −

j 9 . 3)

o

Breakaway-point Equation: 0 . 02 s + 0 . 5 s + 3 .125 s Breakaway Point: (RL) −5.797 4

3

158

2

− 66 .67 s − 416

. 67

=0

8-15 (a)

A

= K o = 100

:

P (s)

Asymptotes: N > 0: Intersect of Asymptotes:

= s ( s + 12 . 5)( s + 1 ) 60

o

σ1 = 2

180 0

Q (s)

300

− 1 − 12 . 5

o

= − 4. 5

3

+ 27 s + 12 . 5 = 0

Breakaway-point Equation:

3s

Breakaway Points:

−0.4896

(RL)

o

159

= 41 . 67

8-15 (b)

P( s)

= s + 12 . 5 + 1666

Q (s)

2

= 0 . 02

Asymptotes:

2

s (s

. 67

= ( s + 6 .25 +

j 40 . 34 )( s

+ 6 .25 −

j 40 . 34 )

+ 12 . 5)

A > 0:

180

o

Breakaway-point Equation:

0 . 02 s

Breakaway Points:

0

(RL)

4

+ 0 . 5 s + 103 3

160

.13 s

2

+ 833

. 33 s

=0

8-15 (c)

P (s)

= s + 12 . 5 s + 833

Q (s)

3

2

. 33

= ( s + 15 . 83)(

s

− 1. 663 +

j 7 . 063)( s

− 1. 663 −

j 7 .063)

= 0 . 01 s ( s + 12 . 5)

Asymptotes:

K

o

> 0:

180

o

Breakaway-point Equation:

0 . 01 s

Breakaway Point:

−5.37

(RL)

4

+ 0 .15

s

3

+ 1. 5625

161

s

2

− 16 . 67 s − 104.

17

=0

8-16 (a)

P (s)

=s

2

(s

+ 1 )( s + 5)

Q (s)

=1

Asymptotes: K > 0: Intersect of Asymptotes:

o

45

,

135

o

,

σ1 = Breakaway-point Equation:

4s

3

+ 18

s

2

o

225 0

315

+0 −1− 5

+ 10

162

,

o

= −1. 5

4 s

=0

Breakaway point: (RL) 0, −3.851

(b)

P (s)

=s

2

(s

+ 1 )( s + 5)

Q (s)

Asymptotes: K > 0: Intersect of Asymptotes:

= 5s +1 60

o

,

σ1 =

o

180 0

,

300

o

+ 0 − 1 − 5 − ( −0 .2 ) 4

−1

=−

5. 8

= −1. 93

3

Breakaway-point Equation: 15 s + 64 s + 43 s + 10 s = 0 Breakaway Points: (RL) −3.5026 4

8-17

P( s) Q (s)

=s = 10

2

(s

+ 1 )( s + 5) + 10 = ( s + 4. 893)(

3

s

+ 1.896

2

)( s

− 0 . 394 +

j 0 .96 )( s

− 0 . 394 +

j 0 . 96 )

s

Asymptotes:

T

d

> 0:

60

o

,

180

o

,

Intersection of Asymptotes:

300

o

σ1 =

−4. 893 − 1. 896 + 0 . 3944 + 0 . 3944

There are no breakaway points on the RL.

163

4

−1

= −2

8-18 (a)

K = 1:

P (s)

=s

3

(s

+ 117

Asymptotes:

.23)( s K

L

+ 4882

90

Intersect of Asymptotes:

σ1 = Breakaway Point: (RL)

8-18 (b)

K = 1000:

P( s)

=s

Q (s)

= 1010(

3

(s

Asymptotes:

K

L

> 0:

90

+ 5000

s s

+ 4921

,

= 1010(

s

+ 1. 5948

)( s

+ 114.41

)( s

+ 4884

o

,

270

−117

.23

− 4882

.8

+ 1. 5948 + 114.41 + 4884 5

.23)( s

3

o

o

−3

= −0 .126

0

+ 117

= 1010(

Q (s)

.8 )

> 0:

270

s

+ 4882 2

. 6 )( s

.8 )

+ 5 . 6673 × 10 + 39 .18 +

5

s

+ 891089

j 423 . 7 )( s

110 )

+ 39 .18 − 423

.7 )

o

Intersect of Asymptotes:

σ1 =

−117

.23

− 4882

.8

+ 4921 5

164

−3

.6

+ 39 .18 + 39 .18

= −0 . 033

)

Breakaway-point Equation:

+ 5.279 × 10 Breakaway points: (RL) 0, −87.576 2020 s

8-19

7

+ 2 .02 × 10

7

s

6

Characteristic Equation:

s

3

10

+ 5000

s

s

2

+ 1. 5977 × 10

5

+ 572

, 400 s

P( s)

3

≅ ( s + 1. 5945)(

2

s

+ 115

Q ( s)

.6 )

Breakaway-point Equation: 1200 s

2

≅ 10 .24

+ 3775

s

s

s

s

4

+ 1. 8655 × 10

16

,000

+JL

ε

+ 115

. 6 )( s

+ 4882

+ 900

= s + 5000 s + 572 , 400 s + 900 ,000 = ( s + 1. 5945)( Since the pole at −5000 is very close to the zero at −4882.8, P ( s ) and P( s)

13

10 s

3

s

3

+ 1. 54455 × 10

+ 50 ,000 .8 )

s

2

ϕ=

Q (s)

Q ( s ) can be approximated as:

2

=0

Breakaway Points: (RL): 0, −3.146

165

18

s

2

=0

0

= 10

2

s (s

+ 5000

)

8-20 (a) α = 12 :

P ( s)

=

2

s (s

+ 12

= s +1

Q (s)

) o

Asymptotes: K > 0: 90 , Intersect of Asymptotes:

270

o

K < 0:

σ1 = Breakaway-point Equation:

8-20 (b) α = 4 :

P( s)

=s

2

(s

+4) o

Asymptotes: K > 0: 90 , Intersect of Asymptotes:

Q (s) 270

3

2s

o

+ 0 − 12 − ( −1 ) 3

−1

+ 15 s + 24 2

s

=0

180

o

= −5 . 5 Breakaway Points:

0,

−2.314, −5.186

= s +1 o

K < 0: 0 ,

σ1 = Breakaway-point Equation: 2 s + 7 s 3

0

o

0 ,

2

0

o

180

+ 0 − 4 − ( − 10 3

+8 s = 0

−1

= − 1. 5

Breakaway Points: K > 0 0. None for K < 0.

166

(c) Breakaway-point Equation:

2s

2

+ ( α + 3) s + 2 s = 0

Solutions:

s

=−

α +3 4

±

+ 3) − 16 α 2

, s

4

For one nonzero breakaway point, the quantity under the square-root sign must equal zero. Thus,

α − 10 α + 9 = 0 , α = 1 or α = 9 . 2

cancellation in the equivalent G ( s ). When

8-21 (a)

P (s)

=s

2

(s

+ 3)

Q ( s)

α = 9,

The a nswer

is

α = 9.

= s +α

Breakaway-point Equation: 2 s + 3 (1 + α ) s + 6 α = 0 The roots of the breakaway-point equation are: 3

s

=

−3 ( 1 + α ) 4

±

The

α = 1 solution represents pole-zero = − 3. σ 1 = −4.

the nonzero breakaway point is at s

9( 1

+ α ) − 48 α 2

4

167

=0

For no breakaway point other than at s = 0 , set

9( 1 + α )

2

− 48 α < 0 or - 0 . 333 < α < 3

Root Locus Diagram with No Breakaway Point other than at s = 0.

8-21 (b) One breakaway point other than at s = 0: α = 0 . 333

,

168

Breaka way po int at s

= −1.

8-21 (d) Two breakaway points :

α > 3:

169

8-22

Let the angle of the vector drawn from the zero at s

=

j12

to a point s on the root locuss near the zero 1

θ. θ1 =

Let angle

of th e vect or dra wn fro m the

pole a t j 10 to s .

θ2 =

angle

of th e vect or dra wn fro

pole a t 0 to s .

θ3 =

angle

of th e vect or dra wn fro m the

pole a t

j 10 to s1 .

θ4 =

angle

of th e vect or dra wn fro m the

zero a t

j 12 to s .

be

m the

1

1

1

Then the angle conditions on the root loci are:

θ = θ 1 − θ 2 −θ 3 + θ 4 = θ 1 = θ 2 = θ 3 = θ 4 = 90

odd m ultipl es of o

Thus,

θ

180 o

= 0

The root loci shown in (b) are the correct ones.

170

o

Chapter 9

9-1 (a)

FREQUENCY DOMAIN ANALYSIS

ωn =

K=5

(b)

= 2 .24

5

= 4. 62

21 . 39

ω n = 10

K = 100

= 2 . 944

( 9 . 38 dB)

r

= 15 . 34

( 23 . 71 dB)

r

= 4.17

r

=1

r

= 1. 57

r

= ∞ ( unstab

9-2 (a)

M

(b)

M

(c)

M

(d)

M

(e)

M

(f)

M

(g)

M

(h)

M

r

1−ς

= 3 .27

2

( 9 . 8 dB)

r

= 4.12

( 12 . 3 dB)

d / sec

Thus, minimum

Minimum BW =

2

ω n = 17 . 7 ωn

M

/ sec

r

( ( 1 − 2ζ

2

)+

1

= 2ζ

= 1. 618

BW = 2.44 r ad / sec

BW = 2.07 r ad / sec

/ sec

ζ

1 − 0 .416

BW = 5.16 r ad / sec

= 0.59

ζ + 2 . 917 ζ

2

= 0 .1

ωn Maxi mum

M

4ζ − 4ζ + 2 4

2

)

r

1/2

9-4

sec

= 1. 05 = 20.56 rad/sec

−πζ

Maximu

m over shoot

= 0.2

1

=1 −ζ

ω r = 9 .45 ra

BW = 1.12 r ad / sec

d / sec

T hus,

=

r

ωr = 0

BW = 9.18 r ad / sec

ω r = 3 . 5 rad

t

=1

r

BW = 0.46 r ad / sec

ω r = 0 .82

= 0.1

r

BW = 6.223

= 1. 05

1−ζ

M

ω r = 6 .25 ra

ω r = 1.25 ra

m over shoot

= 0 . 327

BW = 4.495

9-3 1

= 0 . 707

/ sec

ωr =4

ω r = 1. 5 rad

le)

= 3 . 09

r

M

2

6.54 =

( 3 . 918 dB)

=

6.54 =

20

ωr =0

r

M

ζ

ζ

( 12 .4 dB)

( 0 dB)

Maximu

= 1.46

9.24

ωr = ωn (c)

6.54 = 4.48

ωn =

K = 21.39

ζ

Thus,

0 .2

=e

1 −ζ

171

2

ζ = 0.456

d / sec

M

r

1

= 2ζ

= 1.232

1−ζ

Maximum M

t

2

= 1.232

r

r

=

1 − 0 .416

ζ + 2 . 917 ζ ωn

Minimum BW =

( (1 − 2ζ

9-5

2

)+

2

= 0 .2

Thus, minimum

4ζ − 4ζ + 2 4

2

)

1/2

ω n = 14. 168

= 18.7

− πζ

Maximum overshoot = 0.3

M

r

1

= 2ζ

= 1.496

1−ζ

Maximum M

r

=

t

2

1.496

=

r

1 − 0 .416

1 −ζ

=e

Thus, 0 . 3

2

ζ + 2 . 917 ζ

ζ = 0 . 358 2

ωn

Minimum BW =

( (1 − 2ζ

2

)+

= 0 .2

Thus, minimum

4ζ − 4ζ + 2 4

2

9-6

)

1/2

ω n = 6 .1246

= 1.4106

−πζ

M

r

= 1.4 =

ω r = 3 rad t

max

1

=

Thus, 2ζ

1−ζ

/ sec

= ωn

π ωn

1

−ζ

= 0.387

1 −ζ

Maximum overshoot = e

2

= 0.2675

1

− 2ζ

2

= 0 .8367 ω n

π 3 . 586

1

− ( 0 . 387

= 0 .95 )

sec

ωn =

3

= 3 . 586

At

ω = 0,

M

2

Unit-step Response:

9-7 T BW (rad/sec) Mr ________________________________________________________________ 1.14 1.17 1.26 1.63

1.54 1.09 1.00 1.09

172

0 . 8367

This indicates that the steady-state value of the unit-step response is 0.9.

0 0.5 1.0 2.0

(26.75%)

2

= 2

ζ

= 0 .9 .

3.0 1.96 4.0 2.26 5.0 2.52 _________________________________________________________________

1.29 1.46 1.63

9-8 T BW (rad/sec) Mr _________________________________________________________________ 0 1.14 0.5 1.00 1.0 0.90 2.0 0.74 3.0 0.63 4.0 0.55 5.0 0.50 _________________________________________________________________

1.54 2.32 2.65 2.91 3.18 3.37 3.62

9-9 (a) 20

L( s )

=

When

ω = 0: ∠L ( j ω ) = −90 o

s ( 1 + 0 . 1 s )( 1

L ( jω ) =

1

+ 0 .5 s )

L( jω )

(

−0.6ω + jω 1 − 0.05ω

− 0 .05 ω = 0 2

Thus ,

ω = ±4.47

P

=0

=∞

When

2

)

=

(

0.36ω + ω 4

2

o

o

2

(1 − 0.05ω ) 2

L ( j 4.47 )

Φ 11 = 270 = ( Z − 0.5 Pω − P ) 180 = ( Z − 0.5 )180 o

ω = ∞: ∠L ( jω ) = −270 o

20  −0.6ω − jω 1 − 0.05ω 2

20 2

= 1,

) 

L( jω )

Setting Im L ( jω )

2

= −1. 667

Thus, Z =

360 180

o o

=2

The closed-loop system is unstable. The characteristic equation has two roots in the right-half s-plane.

Nyquist Plot of L ( jω ):

173

=0 =0

(b) L( s )

=

10 s ( 1 + 0 . 1 s )( 1

Based on the analysis conducted in part (a), the intersect of the negative

+ 0 .5 s )

real axis by the L ( jω ) plot is at

Φ 11 = −90

o

=

δ− Z

−0.8333, and the corresponding ω

0 . 5 Pω

−P

ι

180

o

= 180

− 90

Z

Thus,

Z

= 0 . The closed-loop system is stable.

Nyquist Plot of L ( jω ):

(c) L( s )

=

When When

100( 1 + s ) s ( 1 + 0 . 1 s )( 1

ω = 0 : ∠L ( j 0 ) = −90 o

ω

ω = ∞: ∠L ( jω ) = −270

L( j0 ) o

= 1,

P

+ 0 .2 s )( 1 + 0 . 5 s )

=∞

L( jω )

When

=0

P

= 0.

ω = ∞: When

174

∠L ( j∞ ) = −270

o

L ( j ∞)

ω = ∞: ∠L ( jω ) = −270

o

=0 L( jω )

=0

L ( jω ) =

( 0.01ω

100(1 + jω )

Setting Im L ( jω ) Thus,

− 0.8ω

4

=0

ω = 64. 55

2

) + jω (1 − 0.17ω ) 2

0 .01 ω

ω = ± 8 . 03

2

4

=

(

) − jω (1 − 0.17ω )  ) + ω ( 1 − 0.17ω )

100(1 + jω )  0.01ω − 0.8 ω

( 0.01ω

4

4

− 0.8ω

− 0 .8 ω 2 − 1 + 0 .17 ω 2 = 0

2

2

2

2

2

2

2

ω 4 − 63 ω 2 − 100 = 0

 100  ( 0.01ω 4 − 0.8ω 2 ) + ω 2 (1 − 0.17ω 2 )   L ( j 8.03) =   = − 10  ( 0.01ω 2 − 0.8ω 2 )2 + ω 2 (1 − 0.17ω 2 ) 2    ω=8.03 Φ 11 = 270 = ( Z − 0.5 Pω − P ) 180 = ( Z − 0.5 )180 o

o

o

Thus, Z = 2

The closed-loop system is

unstable. The characteristic equation has two roots in the right-half s-plane.

Nyquist Plot of L ( jω ):

(d) 10

L( s )

=

When

ω = 0: ∠L ( j ω ) = −180 o

P

ω

s ( 1 + 0 .2 s )( 1 + 0 . 5 s ) 2

L ( jω ) =

( 0.1ω

10 4

Setting Im L ( jω ) origin where

L ( jω )

ω = ∞.

−ω

2

) − j 0.7ω

= 0 , ω = ∞.

3

=

=2

P

=∞

(

=0

10 0.1ω − ω + j0.7ω

( 0.1ω

4

4

2

−ω

ω = ∞: ∠L ( jω ) = −360 o

When

2

)

2

+ 0.49ω

3

)

6

The Nyquist plot of L ( jω ) does not intersect the real axis except at the

175

L ( jω )

=0

Φ 11 = ( Z − 0.5 Pω − P )180 = ( Z − 1) 180 o

Thus, Z = 2 .

o

The closed-loop system is unstable. The characteristic equation has two roots in the right-half s-plane. Nyquist Plot of L ( jω ):

9-9 (e) L(s ) =

When

(

3( s + 2)

s s + 3s + 1 3

ω = 0 : ∠L ( j 0 ) = −90 o

L ( jω ) =

3( jω + 2) 4

− 3ω

ω − 3ω − 2 = 0 4

Pω = 1

)

2

2

) + jω or

=

P=2

L( j0 )

=∞

(

3( jω + 2)  ω − 3ω

(4

ω = 3 . 56 2

4

ω = ∞: ∠L ( j ∞) = −270 o

When 4

− 3ω

2

)

2

) − jω 

2

ω = ± 1.89

o

o

Setting Im L ( jω )

2

rad/sec. L ( j 1.89 )

Φ 11 = ( Z − 0.5 Pω − P )180 = ( Z − 2.5 ) 180 = − 90

o

=3

Thus, Z = 2

The closed-loop system is unstable. The characteristic equation has two roots in the right-half s-plane.

Nyquist Plot of L ( jω ):

176

L ( j∞)

=0 = 0,

9-9 (f) L(s ) =

When

0.1

(

2

ω = 0 : ∠L ( j 0 ) = −90 o

L ( jω ) = ω =∞

Pω = 1

)

s ( s + 1) s + s + 1

or

L( j0 )

0.1

4

− 2ω

2

) + jω (1 − 2ω )

ω = 0 .5 2

2

ω = ±0 . 707

=

P=0

=∞

When

(

4

− 2ω

Φ 11 = ( Z − 0.5 Pω − P )180 = ( Z − 0.5 ) 180 = − 90 Nyquist Plot of L ( jω ): o

9-9 (g) L(s ) =

100

(

s ( s + 1) s + 2 2

)

Pω = 3

2

2

P=0

177

o

L ( j∞)

=0

2

2

2

L ( j 0 . 707 )

o

) − jω (1 − 2ω )  ) + ω ( 1 − 2ω )

0.1  ω − 2ω 4

ω = ∞: ∠L ( j ∞) = −360 o

2

2

Settiing Im L ( jω )

=0

= −0 .1333

Thus, Z = 0

The closed-loop system is stable.

When

ω = 0 : ∠L ( j 0 ) = −90 o

The phase of L ( jω ) is discontinuous at

(

Φ11 = 35.27 + 270 − 215.27 o

o

=∞

L( j0 )

o

ω = 1.414

) = 90

o

ω = ∞: ∠L ( j ∞) = −360 o

When

L ( j∞)

=0

Φ11 = ( Z − 1.5) 180 = 90 o

o

Thus, P

11

=

360 180

o o

=

The closed-loop system is unstable. The characteristic equation has two roots in the right-half s-plane. Nyquist Plot of L ( jω ):

9-9 (h) L( s )

When

=

10( s s(s

+ 10

)

+ 1 )( s + 100

P

ω

)

=1

ω = 0 : ∠L ( j 0 ) = −90 o L ( jω ) =

P

=∞

L( j0 )

−101ω + jω (100 − ω )

Setting Im L ( jω )

2

= 0, ω = 0

ω = ∞: ∠L ( j ∞) = −180 o

When

=

(

10( jω + 10) − 101ω − jω 100 − ω 2

10( jω + 10) 2

=0

10201ω + ω 4

2

(100 − ω ) 2

2

) 

2

is the only solution. Thus, the Nyquist plot of L ( jω ) does not intersect

the real axis, except at the origin.

Φ 11 = ( Z − 0.5 Pω − P )180 = ( Z − 0.5 ) 180 = − 90 o

o

The closed-loop system is stable.

Nyquist Plot of L ( jω ):

178

o

Thus, Z = 0 .

L ( j∞)

=0

2

9-10 (a) L( s )

K

= s(s

+ 2 )( s + 10

For stability, Z = 0.

)

=1

P

o

Φ11 = −0 . 5 Pω × 180

=0

= −90

o

This means that the (−1, j0) point must not be

enclosed by the Nyquist plot, or

0 < 0 .004167 K Nyquist Plot of L ( jω ):

9-10 (b) L( s )

K(s

= s(s

< 1.

+ 1)

+ 2 )( s + 5)( s + 15)

For stability, Z = 0.

Φ11 = −0 . 5 × 180

by the Nyquist plot, or 0

o

< 0 . 000517

Thus,

0

< K < 240

=1

P

=0

= −90 9K

o

This means that the (−1, j0) must not be enclosed

0

Φ 11 = 180

Thus, the system is unstable for all K > 0. For K < 0, the critical point is (1, j0), all K < 0. Thus, the system is unstable for all values of K.

Nyquist Plot of L ( jω ):

9-11 (a) G (s)

=

K (s

+ 5)

2

P

ω

=0

P

=0

180

o

− 180

, not

Φ11 = 0

o

for

o

.

∠G ( j 0 ) = 0

o

(K

> 0)

∠G ( 0 ) = 180

o

∠G ( j ∞ ) = 0

o

(K

0)

(K < 0)

G ( j ∞)

=0

For stability, Z = 0.

Φ 11 = − ( 0.5 Pω + P ) 180 = 0 o

0 K

<

0)

∠G ( 0 ) = 180

o

(K

< 0)

K

=

G ( j0 )

125 G ( j ∞)

= −270

o

∠G ( j ∞ ) = 270

(K > 0)

o

(K < 0)

G ( j ∞)

=0

For stability, Z = 0.

Φ 11 = − ( 0.5 Pω + P ) 180 = 0 o

Φ11 = 0

0 < K < 1000 K K

> 1000 < − 125

−125 < K < 0

o

Stable

Φ 11 = 360

o

Φ11 = 180

o

Φ11 = 0

o

Unstable Unstable

o

Stable

The system is stable for − 125 < K < 0.

9-11 (c) G (s)

=

K (s

+ 5)

4

∠G ( j 0 ) = 0

o

=P =0

(K

> 0)

∠G ( 0 ) = 180

o

(K

< 0)

G ( j0 )

K

=

625 G ( j∞ )

=0

o

(K > 0)

∠G ( j ∞ ) = 180

181

o

(K < 0)

G ( j ∞)

=0

For stability, Z = 0.

Φ 11 = − ( 0.5 Pω + P ) 180 = 0 o

0

<

K

< 2500

Φ11 = 0

> 2500

K K

< − 625

−625 < K < 0

o

o

Stable

Φ 11 = 360

o

Φ11 = 180

o

Unstable Unstable

o

Φ11 = 0

Stable

The system is stable for − 625 < K < 2500.

9-12

(

)

(

)

s s + 2s + s +1 + K s + s + 1 = 0 3

2

Leq ( s) =

(

2

K s + s +1

(

2

)

s s + 2s + s + 1

Leq ( jω ) =

3

2

K

Setting Im L

eq

4

(1 − ω

−ω

( jω )

2

Pω = 1

)

P =0

) + jω  = K  − ( ω ) + jω (1 − 2ω ) (ω 2

2

L eq ( j 0 )

4

4

−ω

2

ω 4 − 2ω 2 + 1 = 0 Thus,

ω = ±1

L eq ( j 1)

= −K

For stability,

Φ11 = − ( 0 . 5 Pω + P )180

o

= − 90

o

When K = 1 the system is marginally stable. K K

>0 −1

2 s

K

1

− 2K +1

2

+1

K s

0

=

− 1)

(K

2

+1

K

K

K>0

When K = 1 the coefficients of the s row are all zero. The auxiliary equation is s + 1 = 0 The solutions are ω = ± 1 rad/sec. Thus the Nyquist plot of L eq ( j ω ) intersects the −1 point when K = 1, when ω = ± 1 1

2

∞, except at K = 1.

rad/sec. The system is stable for 0 < K <

9-13 Parabolic error constant

K a = lim s G ( s ) = lim10 ( K P + K D s ) = 10 K P = 100 s →0

Characteristic Equation:

G

eq

( s)

=

10 K s

2

D

Thus K P = 10

2

s →0

s

s

P

ω

+ 100

+ 10

2

=2

P

K

D

s

+ 100 = 0

=0

For stability, o

Φ 11 = − ( 0 . 5 Pω + P )180

The system is stable for 0 < K

9-14 (a) 1

o

= −180

< ∞.

D

The characteristic equation is

+G ( s) −G ( s) − 2

G (s)

2

G eq ( s )

= −2

G (s)

G eq ( j ω ) =

G

eq

( j0)

=

= 1− 2 2

G (s)

=

−2 K (s

( 400 − 120ω K

2

200

∠180

o

=0

2

+ 4)

2

2

(s

−2 K 2

+ω G

4

eq

+ 5)

2

=0

P

2

) + jω ( 360 − 18ω ) 2

( j ∞)

= 0 ∠180

o

183

=

=0

−2 K

2

2 2 2 ( 400 − 120ω + ω ) − jω ( 360 − 18ω ) 

( 400 − 120ω

Setting Im G

eq

( jω )

2

+ω =0

2

) + ω ( 360 − 18ω ) 2

2

2

ω =0

ω

and

=

± 4.47 r

G

eq

=

( j 4.47 )

K

2

800

For stability,

Φ 11 = − ( 0.5 Pω + P ) 180 = 0 o

<

K

or

< 200

2

The system is stable for K

o

200

Characteristic Equation: s

4

+ 18

s

+ 121

3

s

2

+ 360

s

+ 400 − 2 K

=0

2

Routh Tabulation s

4

s

3

s s

2

1

1

121

18

360

101

−2K

400

- 36 K

29160

400

− 2K

2

2

2

29160

+ 36 K

2

101 s

0

400

−2K

2

K

K

Thus for stability,

<

=

Nyquist Plot

83 . 33 N s(s

< 200

200

9-15 (a) G (s)

2

+ 2 )( s + 11 . 767

)

For stability, N < 3.89 N < 3 since N must be an integer.

Thus

(b) G (s)

2500

= s ( 0 . 06 s

+ 0 . 706

)( As

+ 100

Characteristic Equation: 0 . 06 As

3

)

+ ( 6 + 0 . 706

2

A )s

184

+ 70 . 6 s + 2500 = 0

>0

G

eq

=

(s)

+ 0 . 706

As

2

( 0 . 06 s

6s

2

+ 70 . 6 s + 2500

)

Since G eq ( s ) has more zeros than poles, we should sketch the Nyquist

plot of 1 / G eq ( s ) for stability study.

( 2500 − 6ω ) + j70.6 ω = ( 2500 − 6ω ) + j70.6ω  ( −0.706ω + j 0.06ω ) = ( jω ) A ( −0.706 ω − j0.06ω ) A ( 0.498ω + 0.0036ω ) 2

1

G eq

2

2

3

o

= ∞∠ − 180

1 / G eq ( j 0 )

4

1 / G eq ( j∞ )

− 0 . 36 ω = 0

ω = ±16 . 68

2

100 .156

2

= 0 ∠ − 90

6

o

Setting 1

eq

( j16 . 68 )

=

3

 =0  G eq ( jω )  1

Im  −4.23 A

For stability,

Φ 11 = − ( 0.5 Pω + P ) 180 = −180 o

For A

> 4.23

For 0

<

A

Φ11 = 180

< 4.23

o

o

Unstable o

Φ11 = −180

Stable

The system is stable for 0 < A < 4.23.

(c) G (s)

=

2500 s ( 0 . 06 s

+ 0 . 706

+Ko)

)( 50 s

Characteristic Equation:

G

eq

( s)

=

K s ( 0 . 06 s o

3s

3

s ( 0 . 06 s

+ 0 . 706

)

P

ω

+ 35 . 3 s + 2500 2

K o ( −0.06ω + 0.706 jω )

=0

3

G eq ( j ω ) =

( 2500 − 35.3ω ) − j3ω 2

( jω )

=0

= −0 . 004

Ko

Setting Im G G eq ( j 7 .18 )

eq

ω + 138 4

3

+ 0 . 706

=

P

)( 50 s

+ K o ) + 2500 = 0

=0

(

G eq ( j 0 )

= 0 ∠ 90

o

G eq ( j∞ )

)  ( 2500 − 35.5ω ) + j 3ω ( 2500 − 35.3ω ) + 9ω

K o − 0.06ω + 0.706 jω

.45 ω

2

2

2

2

− 9805

. 55

185

= 0 ∠ − 90

=0

ω = 51 . 6 2

2

3

6

ω = ± 7 .18



o

For stability,

Φ 11 = − ( 0.5 Pω + P ) 180 = 0 o

For stability, 0 < K

9-16 (a)

K

o

< 217

o

.4

= 0:

t

G ( s) =

Y ( s) E (s )

=

(

10000 K

s s + 10 s + 10000 Kt 2

)

=

10000 K s ( s + 10) 2

The (−1, j0) point is enclosed for all values of K. The system is unstable

for all values of K.

(b)

G ( s) =

K

(

t

= 0 . 01 :

10000 K

s s + 10 s + 100 2

2

)

G( jω ) =

(

10000 K − 10ω − jω 100 − ω 100ω + ω 4

2

(100 − ω )

186

2

2

2

) 

Setting Im G ( j ω )

ω = ±10

= 0 ω 2 = 100 G ( j 10 )

= −10

K

The system is stable for 0 < K < 0.1

(c)

G ( s) =

K

(

t

= 0 .1:

(

10000 K  −10ω − jω 1000 − ω 2

10000K

s s + 10 s + 1000 2

G ( jω ) =

)

Setting Im G ( j ω )

100ω + ω 4

= 0 ω 2 = 100 ω = ±31 . 6

(1000 − ω )

2

2

G ( j 31 . 6 )

2

) 

2

= −K

For stability, 0 < K < 1

9-17

The characteristic equation for K = 10 is: s G eq ( s )

=

3

+ 10

10 ,000 K t s s

G eq ( j ω ) =

3

+ 10

s

2

+ 100

=0

K s t

P

+ 100

3

=

=0

,000

=2

(

10,000 K t  −ω + jω 10,000 − 10ω 4

10,000 K t jω 100,000 − 10 ω − jω

+ 10 ,000

2

,000

2

s

(10,000 − 10ω ) 2

187

2

6

2

)

Setting Im G

eq

( jω )

=0

ω = 0 , ω = 10 ,000 2

ω = ±100

G eq ( j 100 )

= −Kt

For stability,

Φ 11 = − ( 0.5 Pω + P ) 180 = −360 o

o

The system is stable for K > 0. t

9-18 (a) Let G ( s )

G1 ( s ) =

(

= G 1 ( s )e

− Td s

Then

100

s s + 10 s + 100 2

)

100 Let

(

− 10ω + jω 100 − ω 2

100ω + ω 4

Thus

∠G1 ( j 1) = − tan ω Td

)

(100 − ω ) 2

−1

2

100

or

100ω 4 + ω 2 (100 − ω 2 )2   

1/2

= 10,000

=

84.23

o

6

4

2

)

− 180 π o

180 π = 1.47

180

Thus the maximum time delay for stability is T

(b)

T

d

=1

= 1.47 sec.

d

sec.

G ( s) =

(

100 Ke

−s

s s + 10 s + 100 2

At the intersect on the negative real axis,

)

G( jω ) =

ω = 1.42

=1

ω − 100ω + 10,000ω − 10,000 = 0

 100 − ω 2  o  −10ω  = 264.23   ω=1

( 264.23 =

ω =1

=1

ω are ω = ± 1 rad/sec.

The real solution for

Equating

2

2

100 Ke

(

− jω

− 10ω + jω 100 − ω

188

2

2

)

G ( j 1.42 )

= −0 . 7107

K.

The system is stable for 0 < K < 1.407

9-19 (a)

K = 0.1

G ( s) =

10 e

(

− Tds

s s + 10 s + 100 2

)

= G1 (s )e

10 Let

(

− 10ω + jω 100 − ω 2

=1

)

10

or

100 ω 4 + ω 2 (100 − ω 2 ) 2   

1/2

ω − 100 ω + 10 ,000 ω − 100 = 0 6

Thus

∠G1 ( j 0.1) = − tan

4

−1

ω Td

ω =0.1

=

2

ω

The real solutions for

( 269.43

)

− 180 π o

180

o

= 1.56

We have

T

d

= 15 . 6

sec.

We have the maximum time delay for stability is 15.6 sec.

9-19 (b) T d = 0 .1 sec.

G ( s) =

(

100 Ke

−0.1 s

s s + 10 s + 100 2

)

G( jω ) =

100 Ke

rad/sec. G ( j 6 . 76 )

= −0 .1706

−0.1 jω

(

− 10ω + jω 100 − ω

At the intersect on the negative real axis,

ω = 6 . 76

is

 100 − ω 2  o = 269.43  −10ω    ω =0.1 o

Equate

2

−Td s

K

189

2

2

)

=1

ω = ± 0 .1

The system is stable for 0 < K < 5.86

9-20 (a)

The transfer function (gain) for the sensor-amplifier combination is 10 V/0.1 in = 100 V/in. The velocity of flow of the solution is v

=

10 in

3

/ sec

= 100

in/sec

0 .1 in The time delay between the valve and the sensor is T G (s)

=

100 Ke s

2

d

=D

/ v sec. The loop transfer function is

− Td s

+ 10 s + 100

190

191

9-21 (a)

The transfer function (gain) for the sensor-amplifier combination is 1 V/0.1 in = 10 V/in. The velocity of flow of the solutions is v

=

10 in

The time delay between the valve and sensor is T G (s)

(b)

=

2

/ sec

= 100 in / sec 0 .1 in = D / v sec. The loop transfer function is

d

10 Ke s

3

−T d s

+ 10 s + 100

K = 10:

G ( s) = G1 (s ) e

Thus,

G ( jω ) =

100

(

Setting

− Td s

100 − ω

2

)

+ j10ω

ω − 100 ω = 0 4

2

∠G1 ( j 10) = − tan

=1

−1

100 e

− jω Td

(100 − ω ) + j10ω 2

(100 − ω ) 2

Real s olutio ns:

ω

2

+ 100ω = 10,000

= 0,

2

ω

=

/ sec

 10ω  = − 90 o    100 − ω 2  ω =10

Thus, 10 T

d

Maximum D

(c)

π

=

d

o

180

Thus, T

90

=

π o

=

π

2

= 0 .157

sec

20

=

vT

d

= 100 × 0 .157 = 15 . 7

in

D = 10 in. T

d

=

D

=

10

= 0 .1

sec

G (s)

=

10 Ke

−0 . 1 s

v 100 s + 10 s + 100 The Nyquist plot of G ( jω ) intersects the negative real axis at ω = 12 . 09 rad/sec. G ( j ) For stability, the maximum value of K is 12.94 . 2

9-22 (a)

192

= −0 . 0773

K

G H ( s) =

8

(

s s + 6 s + 12 2

M

r

G L ( s)

)

= 1, ω r = 0

2 . 31

=

s( s

+ 2 . 936

BW = 1.02 r ad / sec

)

M

= 1, ω r = 0

r

BW = 1.03 r ad / sec

(b) G H ( s) =

M

r

(

0.909

s 1 + 0.5455 s + 0.0455s

= 1, ω r = 0

2

G L ( s)

)

M

BW = 1.4 ra d / sec

=

0 . 995 s ( 1 + 0 .4975 s )

= 1, ω r = 0

r

BW = 1.41 r ad / sec

(c) G H ( s) =

M

r

(

0.5

s 1 + 0.75 s + 0.25 s

= 1, ω r = 0

2

G L ( s)

)

BW = 0.87 r ad / sec

M

=

0 . 707 s ( 1 + 0 . 3536 s )

= 1, ω r = 0

r

BW = 0.91 r ad / sec

(d) G H ( s) =

M

r

90.3

(

−7

s 1 + 0.00283 s + 8.3056 × 10 s

= 1, ω r = 0

BW = 119.74

2

G L (s)

)

M

92 . 94

= s (1

+ 0 . 002594

= 1, ω r = 0

r

s)

rad / sec B W = 118.76

(e) G H ( s) =

M

r

180.6

(

−7

s 1 + 0.00283 s + 8.3056 × 10 s

= 1, ω r = 0

BW = 270.55

)

G L (s)

M

)

G

2

=

189 . 54 s (1

+ 0 . 002644

= 1, ω r = 0

r

s)

BW = 268.11

(f) G H ( s) =

M

r

(

1245.52 −7

s 1 + 0.00283 s + 8.3056 × 10 s

= 2 . 96 , ω r = 666

2

M

BW = 1054.4 1 rad / sec

9-23 (a)

M

r

= 2 . 06 , ω r = 9 . 33 rad

L

r

(s)

=

2617 . 56 s (1

= 3 . 74,

+ 0 . 0053

s)

ω r = 700

BW = 1128.7 4 rad / sec

/ sec,

BW = 15.2 r ad / sec

(b)

193

M

r

ωr

1

= 2ζ

= 2 . 06

1−ζ

1 − 2ζ

2

ζ − ζ + 0 . 0589 = 0 4

ωn

/ sec

2

GL ( s) = 9-24 (a) (b)

M

M

r

r

s ( s + 2ζω n )

= 2 . 96 , ω r = 666 1

= 2ζ

ωr

=

1 − 2ζ

s( s + 4.987)

=

9 . 33

=

s(1 + 0.2005 s)

=

is

ζ

= 0.25.

BW = 1054.4 1 rad / sec

ζ − ζ + 0 . 0285 = 0 4

= 9 . 974

19.94

2

= 666

2

ωn

s ( s + 2ζω n )

=

ωn =

Thus

The solution for

472227.43 s( s + 235.7)

=

666 . 67

ζ < 0 . 707

is

ζ

= 0.1715

2003.5 s(1 + 0.00424 s)

5 s ( 1 + 0 . 5 s )( 1

+ 0 .1 s )

10 s ( 1 + 0 . 5 s )( 1

= 687

0 . 97

9-25 (b) G (s)

ζ < 0 . 707

0 .9354

9-25 (a) G (s)

The solution for

2

2

GL ( s) =

ωn =

Thus

99.48

= 2 . 96

1−ζ

2

2

+ 0 .1 s )

194

(c) G (s)

=

500 (s

+ 1.2 )( s + 4 )( s + 10

)

(d) G (s)

10( s

= s(s

+ 1)

+ 2 )( s + 10

)

9-25 (e)

195

G ( s) =

0.5

(

s s + s +1 2

)

(f) G ( s) =

(

100 e

−s

s s + 10 s + 50 2

)

(g) G ( s) =

(

100 e

−s

s s + 10 s + 100 2

)

9-25 (h)

196

G ( s) =

10( s + 5)

(

s s + 5s + 5 2

)

9-26 (a) =

G (s)

K s ( 1 + 0 .1 s )( 1 + 0 . 5 s )

The Bode plot is done with K = 1. GM = 21.58 dB For GM = 20 dB, K must be reduced by −1.58 dB. Thus K = 0.8337 o

o

PM = 60 .42 . For PM = 45 K should be increased by 5.6 dB. Or, K = 1.91

(b) G (s)

K(s

=

+ 1)

s ( 1 + 0 .1 s )( 1 + 0 .2 s )( 1 The Bode plot is done with K = 1. GM = 19.98 dB. For GM = 20 dB, K ≅ 1. o

+ 0 .5 s )

o

PM = 86 . 9 . For PM = 45 K should be increased by 8.9 dB. Or, K = 2.79.

9-26 (c) G ( s)

K

= (s

+ 3)

3

The Bode plot is done with K = 1. GM = 46.69 dB PM = infinity.

197

For GM = 20 dB K can be increased by 26.69 dB or K = 21.6. For PM = 45 deg. K can be increased by 28.71 dB, or K = 27.26.

(d) =

G (s)

K (s

+ 3)

4

The Bode plot is done with K = 1. GM = 50.21 dB PM = infinity. For GM = 20 dB K can be increased by 30.21 dB or K = 32.4 For PM = 45 deg. K can be increased by 38.24 dB, or K = 81.66

(e) G ( s) =

Ke

(

−s

s 1 + 0.1s + 0.01s

2

)

The Bode plot is done with K = 1. GM=2.97 dB PM = 26.58 deg For GM = 20 dB K must be decreased by −17.03 dB or K = 0.141. For PM = 45 deg. K must be decreased by −2.92 dB or K = 0.71.

9-26 (f) G ( s) =

K (1 + 0.5s )

(

s s + s +1 2

)

The Bode plot is done with K = 1. GM = 6.26 dB PM = 22.24 deg For GM = 20 dB K must be decreased by − 13.74 dB or K = 0.2055. For PM = 45 deg K must be decreased by − 3.55 dB or K = 0.665.

9-27 (a)

198

G (s)

=

10 K s ( 1 + 0 .1 s )( 1 + 0 . 5 s )

The gain-phase plot is done with K = 1. GM = 1.58 dB PM = 3.95 deg. For GM = 10 dB, K must be decreased by − 8.42 dB or K = 0.38. For PM = 45 deg, K must be decreased by −14 dB, or K = 0.2. For M = 1.2 , K must be decreased to 0.16. r

(b) G (s)

=

5K ( s

+ 1)

s ( 1 + 0 .1 s )( 1 + 0 .2 s )( 1

+ 0 .5 s )

The Gain-phase plot is done with K = 1. GM = 6 dB PM = 22.31 deg. For GM = 10 dB, K must be decreased by −4 dB or K = 0.631. For PM = 45 deg, K must be decrease by −5 dB. For M = 1.2 , K must be decreased to 0.48. r

9-27 (c) G ( s) =

10 K

(

s 1 + 0.1s + 0.01s

2

)

The gain-phase plot is done for K = 1. GM = 0 dB M =∞ r

PM = 0 deg For GM = 10 dB, K must be decreased by −10 dB or K = 0.316. For PM = 45 deg, K must be decreased by − 5.3 dB, or K = 0.543. For M = 1.2 , K must be decreased to r

0.2213.

(d)

199

G ( s) =

Ke

(

−s

s 1 + 0.1s + 0.01s

2

)

The gain-phase plot is done for K = 1. GM = 2.97 dB M = 3 . 09 r

PM = 26.58 deg For GM = 10 dB, K must be decreased by − 7.03 dB, K = 0.445. For PM = 45 deg, K must be decreased by − 2.92 dB, or K = 0.71. For M = 1.2 , K = 0 . 61 . r

(b) (c)

Gain crossover frequency = 2.09 PM = 115.85 deg Phase crossover frequency = 20.31 rad/sec

GM = 21.13 dB

Gain crossover frequency = 6.63 rad/sec Phase crossover frequency = 20.31 rad/sec

PM = 72.08 deg GM = 15.11 dB

Gain crossover frequency = 19.1 rad/sec Phase crossover frequency = 20.31 rad/sec

PM = 4.07 deg GM = 1.13 dB

×

(d)

For GM = 40 dB, reduce gain by (40 − 21.13) dB = 18.7 dB, or gain = 0.116

(e)

For PM = 45 deg, the magntude curve reads −10 dB. This means that the loop gain can be increased by 10 dB from the nominal value. Or gain = 3.16 × nominal value.

(f)

The system is type 1, since the slope of G ( jω ) is

(g)

GM = 12.7 dB. PM = 109.85 deg.

(h)

−20 dB/decade as ω → 0.

The gain crossover frequency is 2.09 rad/sec. The phase margin is 115.85 deg. Set

ωT d = 2 . 09 T d = Thus, the maximum time delay is T

9-29 (a)

nominal value.

d

o

115 .85 180

o

= 0 .9674

π

= 2 .022

sec.

The gain is increased to four times its nominal value. The magnitude curve is raised by 12.04 dB. Gain crossover frequency = 10 rad/sec PM = 46 deg Phase crossover frequency = 20.31 rad/sec GM = 9.09 dB

(b)

The GM that corresponds to the nominal gain is 21.13 dB. To change the GM to 20 dB we need to increase the gain by 1.13 dB, or 1.139 times the nominal gain.

(c)

The GM is 21.13 dB. The forward-path gain for stability is 21.13 dB, or 11.39.

(d)

The PM for the nominal gain is 115.85 deg. For PM = 60 deg, the gain crossover frequency must be

(e)

moved to approximately 8.5 rad/sec, at which point the gain is −10 dB. Thus, the gain must be increased by 10 dB, or by a factor of 3.162. With the gain at twice its nominal value, the system is stable. Since the system is type 1, the steady-state error due to a step input is 0.

200

(f) (g)

With the gain at 20 times its nominal value, the system is unstable. Thus the steady-state error would be infinite. With a pure time delay of 0.1 sec, the magnitude curve is not changed, but the the phase curve is subject to a negative phase of

− 0 .1ω

PM

= 115

.85

− 0 .1 × gain c

rossov er fre quency

= 115.85

− 0 .209 = 115

The new phase crossover frequency is approximately 9 rad/sec, where the original phase curve is reduced by −0.9 rad or −51.5 deg. The magnitude of the gain curve at this frequency is −10 dB. Thus, the gain margin is 10 dB.

(h)

When the gain is set at 10 times its nominal value, the magnitude curve is raised by 20 dB. The new gain crossover frequency is approximately 17 rad/sec. The phase at this frequency is Thus, setting

ωT d = 17 T d =

30

o

180

π o

= 0 . 5236

Thus

9-30 (a) Bode Plot:

201

T

d

−30 deg.

= 0 .0308

sec.

. 64 deg

For stability: 166 (44.4 dB) < K < 7079 (77 dB) Phase crossover frequencies: 7 rad/sec and 85 rad/sec Nyquist Plot:

9-30 (b) Root Loci.

202

9-31 (a) Nquist Plot

203

9-31 Bode Plot

9-31 (b) Root Loci

9-32 Bode Diagram

204

When K = 1, GM = 68.75 dB, PM = 90 deg. The critical value of K for stability is 2738.

9-33 (a) Forward-path transfer function:

205

G (s)

=

where

ΘL ( s ) E (s)

=

K G (s) a

p

=

K K ( Bs a

i

+K

)

∆o

(

∆ o = 0.12 s ( s + 0.0325 ) s + 2.5675 s + 6667

(

2

= s 0.12 s + 0.312 s + 80.05 s + 26 G ( s) =

3

(

2

)

43.33( s + 500)

s s + 2.6 s + 667.12 s + 216.67 3

2

(b) Bode Diagram:

Gain crossover frequency = 5.85 rad/sec

PM = 2.65 deg.

Phase crossover frequency = 11.81 rad/sec

GM = 10.51 dB

9-33 (c) Closed-loop Frequency Response:

206

)

)

M

9-34 (a)

r

= 17 . 72 , ω r = 5 . 75

BW = 9.53 r ad / sec

When K = 1, the gain crossover frequency is 8 rad/sec.

(b) When K = 1, the phase crossover frequency is 20 rad/sec.

(c)

When K = 1, GM = 10 dB.

(d)

When K = 1, PM = 57 deg.

(e)

When K = 1, M

(f)

When K = 1,

(g)

When K = 1, BW = 15 rad/sec.

(h)

When K =

(i)

When K = 10 dB (3.16), the system is marginally stable. The frequency of oscillation is 20 rad/sec.

r

= 1.2.

ωr = 3

−10 dB (0.316), GM = 20 dB

207

(j) 9-35

The system is type 1, since the gain-phase plot of G ( jω ) approaches infinity at −90 deg. Thus, the steady-state error due to a unit-step input is zero.

When K = 5 dB, the gain-phase plot of G ( jω ) is raised by 5 dB.

(a)

The gain crossover frequency is 10 rad/sec.

(b)

The phase crossover frequency is 20 rad/sec.

(c)

GM = 5 dB.

(d)

PM = 34.5 deg.

(e)

M

r

=2

(f) ω r = 15 (g) (h)

When K =

9-36 (a)

−30 dB, the GM is 40 dB.

The phase margin with K = 1 and T produces a phase lag of Thus,

−17 deg.

T

(b)

d

=

sec is approximately 57 deg. For a PM of 40 deg, the time delay

The gain crossover frequency is 8 rad/sec.

o

ωT d = 17

=0

d

=

17

o

= 0 . 0371

The phase margin with K = 5 dB and T

o

ωT d = 4. 5 =

ω = 10

4. 5

o

180

d

=0

o

o

π o

= 0 . 9948

T

d

=

0 . 9948

= 0 .1244

sec

8

is approximately 34.5 deg. For a PM of 30 deg, the time The gain crossover frequency is 10 rad/sec. Thus,

= 0 . 0785

o

57

180

−4.5 deg.

π

=

−57 deg.

T

Thus

d

=

0 . 0785

= 0 .00785

sec

10

−34.5 deg.

ωT d = 34. 9-38

sec

With K = 5 dB, for marginal stability, the time delay must produce a phase lag of Thus at

ω

T hus

With K = 1, for marginal stability, the time delay must produce a phase lag of Thus, at ω = 8 rad/sec,

delay must produce a phase lag of

(b)

8

ωT d = 57 9-37 (a)

= 0 .2967

o

180

0 .2967

π

5

o

=

34. 5 180

o

π

o

= 0 . 602

For a GM of 5 dB, the time delay must produce a phase lag of

ωT d = 34.

5

o

=

34. 5 180

o

π

o

T

Thus

−34.5 deg at ω = 10

= 0 . 602

208

d

=

0 . 602

= 0 .0602

sec

10

Thus

T

d

=

0 . 602 10

= 0 .0602

sec

9-39 (a) Forward-path Transfer Function: G (s)

=

Y (s) E ( s)

=

e

−2 s

( 1 + 10 s )( 1

+ 25

s)

From the Bode diagram, phase crossover frequency = 0.21 rad/sec GM = 21.55 dB gain crossover frequency = 0 rad/sec

PM = infinite

(b) G ( s) =

1

(1 + 10s )( 1 + 25s ) ( 1 + 2 s + 2 s

From the Bode diagram, phase crossover frequency = 0.26 rad/sec

2

)

GM = 25 dB

gain crossover frequency = 0 rad/sec

PM = infinite

(c) G ( s) =

1− s

(1 + s ) (1 + 10 s )( 1 + 2s )

From the Bode diagram, phase crossover frequency = 0.26 rad/sec GM = 25.44 dB gain crossover frequency = 0 rad/sec PM = infinite

9-39 (continued) Bode diagrams for all three parts.

209

9-40 (a) Forward-path Transfer Function:

210

G (s)

=

e

−s

( 1 + 10 s )( 1 + 25 s ) From the Bode diagram, phase crossover frequency = 0.37 rad/sec GM = 31.08 dB gain crossover frequency = 0 rad/sec PM = infinite

(b) G ( s) =

1

(1 + 10s ) ( 1 + 25s ) (1 + s + 0.5s

From the Bode diagram, phase crossover frequency = 0.367 rad/sec gain crossover frequency = 0 rad/sec

(c) G (s)

=

2

)

GM = 30.72 dB PM = infinite

( 1 − 0 .5 s )

( 1 + 10 s )( 1 + 25 s )( 1 + 0 .5 s ) From the Bode diagram, phase crossover frequency = 0.3731 rad/sec gain crossover frequency = 0 rad/sec

9-41 Sensitivity Plot:

211

GM = 31.18 dB PM = infinite

S

M G

= 17 .15 max

ω max = 5 .75

212

Chapter 10

DESIGN OF CONTROL SYSTEMS

10-1 Forward-path Transfer Function: G (s)

M ( s)

= 1

−M

For type 1 system, 200 a Ramp-error constant:

(s)

=

K s

3

−K = 0 K

v

+ ( 20 + a ) s + ( 200 + 20 2

K

=

sG ( s )

s →0

200

+ 20

The forward-path transfer function is

G ( s) =

(

a

−K

K = 200a

Thus

= lim

+ 200

a) s

a

200 a

=

200

+ 20 a

=5

a = 10

Thus

K = 2000

The controller transfer function is

2000

s s + 30 s + 400 2

Gc ( s) =

G ( s)

)

Gp (s )

(

20 s + 10 s + 100

=

(s

2

2

+ 30 s + 400

)

)

The maximum overshoot of the unit-step response is 0 percent.

10-2 Forward-path Transfer Function: G (s)

M ( s)

= 1

−M

For type 1 system, 200 a Ramp-error constant:

(s)

=

K s

3

−K = 0 K

v

+ ( 20 + a ) s + ( 200 + 20 2

sG ( s )

s →0

K

= 200

The forward-path transfer function is

G ( s) =

(

a

−K

K = 200a

Thus

= lim

+ 200

a) s

+ 20

a

200 a

=

200

+ 20 a

=9

a = 90

Thus

K = 18000

The controller transfer function is

18000

Gc ( s) =

)

s s + 110 s + 2000 2

G ( s) Gp (s )

=

(

) ( s + 110 s + 2000 )

180 s + 10 s + 100 2

2

The maximum overshoot of the unit-step response is 4.3 percent. From the expression for the ramp-error constant, we see that as a or K goes to infinity, K Thus the maximum value of K

v

v

approaches 10.

that can be realized is 10. The difficulties with very large values of K and

a are that a high-gain amplifier is needed and unrealistic circuit parameters are needed for the controller.

10-3 (a) Ramp-error Constant: K v = lim s

1000 ( K P + K D s ) s ( s + 10)

s →0

=

1000K P 10

(

= 100 K P = 1000

Thus

K

P

= 10

)

Characteristic Equation: s + 10 + 1000 K D s + 1000 K P = 0 2

ωn = Thus

1000 K P

=

10000

K

=

D

90

= 100

2 ζω

= 0 . 09

1000

213

n

= 10 + 1000

K

D

= 2 × 0 . 5 × 100 = 100

10-3 (b)

For K

2 ζω

(c)

= 1000

v

For K

2 ζω

ζ = 0 . 707

and

n

= 10 + 1000

v

= 1000

and

ζ

n

= 10 + 1000

K

K

D

, and from part (a), ω

= 2 × 0 . 707 × 100 = 141

= 1.0 , and from part (a), ω

D

n

n

= 100

.4

K

Thus

= 100

D

=

D

=

131 .4

= 0 . 1314

1000

= 2 × 1 × 100 = 200

K

Thus

190

= 0 . 19

1000

10-4 The ramp-error constant: 1000 ( K P + K D s )

K v = lim s

s ( s + 10)

s →0

= 100 K P = 10,000

The forward-path transfer function is: G ( s) =

KD 0 0.2 0.4 0.6 0.8 1.0

PM (deg)

1000 (100 + K D s ) s( s + 10) Mr 13.5 1.817 1.291 1.226 1.092 1.06

GM

∞ ∞ ∞ ∞ ∞ ∞

1.814 36.58 62.52 75.9 81.92 84.88

K P = 100

Thus

Max overshoot (%)

493 525 615 753 916 1090

The phase margin increases and the maximum overshoot decreases monotonically as K

D

46.6 41.1 22 13.3 8.8 6.2

increases.

10-5 (a) Forward-path Transfer Function: G ( s) = G c ( s ) G p ( s) =

Ramp Error Constant:

e

ss

=

1 K

= v

0 . 0802 KK

K

v

4500 K ( K D + K P s ) s ( s + 361.2) = lim

s→0

≤ 0 . 001

sG ( s )

Thus

=

4500 KK

P

= 12 .458

KK

361 .2 KK

P

≥ 80 .2

Let

K

P

P

=1

and

K

= 80 .2

P

Attributes of Unit-step Response: KD 0 0.0005 0.0010 0.0015 0.0016 0.0017 0.0018 0.0020 Select K ≥ 0 . 0017

tr (sec)

ts (sec)

0.00221 0.00242 0.00245 0.0024 0.00239 0.00238 0.00236 0.00233

0.0166 0.00812 0.00775 0.0065 0.00597 0.00287 0.0029 0.00283

D

(b) BW must be less than 850 rad/sec.

214

Max Overshoot (%) 37.1 21.5 12.2 6.4 5.6 4.8 4.0 2.8

KD

GM

10-6

D

Mr

48.45 62.04 73.5 75.46 77.33 78.22 79.07

1.276 1.105 1.033 1.025 1.018 1.015 1.012

827 812 827 834 842 847 852

∞ ∞ ∞ ∞ ∞ ∞ ∞

0.0005 0.0010 0.0015 0.0016 0.0017 0.00175 0.0018 Select K

PM (deg)

≅ 0 . 00175

. A larger K

D

would yield a BW larger than 850 rad/sec.

The forward-path Transfer Function: N = 20

G ( s) =

200 ( K P + K D s ) s ( s + 1)( s + 10)

To stabilize the system, we can reduce the forward-path gain. Since the system is type 1, reducing the gain does not affect the steady-state liquid level to a step input. Let K = 0 . 05

G ( s) =

200 ( 0.05 + K D s )

P

s ( s + 1)( s + 10)

Unit-step Response Attributes: KD

When K

10-7 (a)

For e

ss

D

ts

(sec)

Max Overshoot (%)

0.01 5.159 12.7 0.02 4.57 7.1 0.03 2.35 3.2 0.04 2.526 0.8 0.05 2.721 0 0.06 3.039 0 0.10 4.317 0 = 0 . 05 the rise time is 2.721 sec, and the step response has no overshoot.

= 1, K v = lim sG ( s ) = lim s s →0

s →0

200 ( K P + K D s ) s ( s + 1)( s + 10)

Forward-path Transfer Function:

G ( s) =

= 20 K P = 1

Thus

K

P

= 0 . 05

200 ( 0.05 + K D s ) s ( s + 1)( s + 10)

Attributes of Frequency Response: KD 0 0.01 0.02 0.05 0.09 0.10 0.11 0.20 0.30 0.50

PM (deg)

GM (deg)

47.4 56.11 64.25 84.32 93.80 93.49 92.71 81.49 71.42 58.55

For maximum phase margin, the value of K

20.83

1.24 1.09 1.02 1.00 1.00 1.00 1.00 1.00 1.00 1.03

∞ ∞ ∞ ∞ ∞ ∞ ∞ ∞ ∞

D

Mr

is 0.09. PM = 93.80 deg. GM =

215

1.32 1.24 1.18 1.12 1.42 1.59 1.80 4.66 7.79 12.36

∞, M

r

= 1,

(b) Sensitivity Plots: M

The PD control reduces the peak value of the sensitivity function S

G

( jω )

10-8 (a) Forward-path Transfer Function:

G ( s) =

Thus

(b)

K

 

100  K P +

KI s

  

For

s + 10 s + 100 2

I

K v = 10,

K v = lim sG ( s ) = lim s s →0

s →0

100 ( K P s + K I )

(

s s + 10 s + 100

= 10 .

Let the complex roots of the characteristic equation be written as

= −σ +

s

(

j 15 an d s

= −σ −

j 15 .

)

s + 2σ s + σ + 225 = 0 2

The quadratic portion of the characteristic equation is

2

s + 10 s + (100 + 100 KP ) s + 1000 = 0 3

The characteristic equation of the system is

2

The quadratic equation must satisfy the characteristic equation. Using long division and solve for zero remainder condition.

s + (10 − 2 σ ) s + 2σ s + σ + 225 s + 10 s + (100 + 100 K P ) s + 1000 2

2

2

3

2

(

)

s + 2σ s + σ + 225 s 3

2

2

( ) + ( 20 σ − 4σ ) s + (10 − 2 σ ) ( s

(10 − 2 σ ) s + 100 KP − σ − 125 s + 1000 2

(10 − 2 σ ) s

2

2

2

216

2

+ 225

)

)

= K I = 10

(100 K For zero remainder, 2 σ

3

)

+ 3σ − 20σ − 125 s + 2σ − 10σ 2

P

3

2

+ 450σ −1250

− 10 σ 2 + 450 σ − 1250 = 0 + 3 σ − 20 σ − 125 = 0 2

and

100 K

The real solution of Eq. (1) is

σ = 2 .8555 K

P

The characteristic equation roots are:

(1)

P

=

125

(2)

. From Eq. (2),

+ 20 σ − 3 σ

2

= 1. 5765

100 s

= − 2 .8555 +

− 2 . 8555 −

j 15 ,

j 15 , and

s

(c) Root Contours: G eq ( s ) =

100 K P s s + 10 s + 100 s + 1000 3

2

=

Root Contours:

217

100 K P s

( s + 10) ( s

2

+ 100

)

= −10 + 2 σ = −4.289

10-9 (a) Forward-path Transfer Function:

G ( s) =

 

100  K P +

KI s

  

s + 10 s + 100 2

For

K v = 10,

K v = lim sG ( s ) = lim s s →0

s →0

100 ( K P s + K I )

(

s s + 10 s + 100 2

)

= K I = 10

Thus the forward-path transfer function becomes

G ( s) =

10 (1 + 0.1K P s )

(

s 1 + 0.1s + 0.01s

2

)

Attributes of the Frequency Response: KP 0.1

PM (deg) 5.51

GM (dB) 1.21

218

Mr 10.05

When K

0.5 0.6 0.7 0.8 0.9 1.0

22.59 25.44 27.70 29.40 30.56 31.25

1.5

31.19

1.1

31.51

1.2

31.40

P

= 1.1

and K

I

= 10 ,

K

6.38 8.25 10.77 14.15 20.10

∞ ∞ ∞ ∞

v

= 10 ,

2.24 1.94 1.74 1.88 1.97 2.00

15.81 16.11 16.38 16.62 17.33 18.01

1.81

20.43

2.00

18.59

1.97

19.08

the phase margin is 31.51 deg., and is maximum.

The corresponding roots of the characteristic equation roots are:

−5 .4,

− 2 .3 +

j 13 .41 , and

− 2 .3 −

j 13 .41

Referring these roots to the root contours in Problem 10-8(c), the complex roots corresponds to a relative damping ratio that is near optimal.

(b) Sensitivity Function:

In the present case, the system with the PI controller has a higher maximum value for the sensitivity function.

10-10 (a) Forward-path Transfer Function: G ( s) = For K

v

= 100

100 ( K P s + K I )

(

s s + 10 s + 100 2

)

,

K v = lim sG ( s ) = lim s s →0

(b)

The characteristic equation is

s →0

100 ( K P s + K I )

(

s s + 10 s + 100 2

)

= K I = 100

s + 10 s + (100 + 100 KP ) s + 100 KI = 0 3

2

Routh Tabulation:

219

Thus

K

I

= 100

.

s

3

s

2

s

1

s

0

1

100 + 100 K

10

10,000

− 900

100 K P

P

For stability, 100 K

P

− 900 > 0

Thus

K

P

>9

0

10,0 00

Root Contours: G eq ( s )

(c)

K

I

=

100 K s

3

+ 10

s

2

P

+ 100

s s

+ 10 ,000

=

100 K P s (s

+ 23 . 65)(

s

− 6 .825 +

j 19 .4 )( s

100 ( K P s + 100 )

(

s s + 10 s + 100 2

)

The following maximum overshoots of the system are computed for various values of K

ymax

j19 .4 )

= 100 G ( s) =

KP

− 6 . 825 −

15

20

1.794

1.779

When K P = 25,

22 1.7788

24 1.7785

25

26

30

1.7756

1.779

1.782

minimum ymax = 1.7756

10-11 (a) Forward-path Transfer Function:

220

P

. 40 1.795

100

1000

1.844

1.859

G ( s) =

100 ( K P s + K I )

(

s s + 10 s + 100 2

For K

)

v

(

=

100 K

= 10 ,

I

100

K

I

= 10

)

s3 + 10 s2 + 100 K + 1 s + 1000 = 0 P

(b) Characteristic Equation: Routh Tabulation: s

3

s

2

s

1

s

0

1

100 + 100 K

10

1000

100 K

P

For stability,

KP > 0

0

P

1000

Root Contours: G

(c)

eq

(s)

=

100 K s

3

+ 10

s

2

P

s

+ 100

s

The maximum overshoots of the system for different values of K

+ 1000

P

ranging from 0.5 to 20 are

computed and tabulated below. KP ymax

0.5

1.0

1.393

1.275

1.6

1.7

1.8

1.9

1.2317

1.2416

1.2424

1.2441

When K P = 1.7,

2.0

3.0

5.0

1.246

1.28

1.372

maximum ymax = 1.2416

10-12 Gc (s ) = K P + K D s + where K

P

KI

=

s K

K Ds + K P s + K I 2

=

P2

s + K D1K I 2

 

= (1 + K D1 s )  K P2 + K

D

= K D1K P 2

Forward-path Transfer Function:

221

  s 

K I2

K

I

=

K

I2

10 1.514

20 1.642

G ( s) = G c ( s ) G p ( s) = Thus K

I

100 ( 1 + K D1 s ) ( K P2 s + K I 2 )

(

s s + 10 s + 100 2

For stability, K

s s + 10 s + 100 2

> 9.

Select K

s + 10 s + (100 + 100 KP 2 ) s + 10,000 = 0 3

)

P2

= 10

K

P

=

K

P2

D1

= 0 .2 ,

2

for fast rise time.

1000 ( 1 + K D 1s ) ( s + 10 )

G ( s) = When K

= K I 2 = 100

Characteristic Equation:

100 ( K P 2 s + 100 )

P2

sG ( s )

=0)

D1

Forward-path Transfer Function:

(

s→0

= K I 2 = 100

Consider only the PI controller, (with K

G ( s) =

= lim

Kv

)

(

s s + 10 s + 100 2

)

K

the rise time and overshoot requirements are satisfied.

D

=

K

D1

K

P2

= 0 .2 × 10 = 2

+ K D 1 K I 2 = 10 + 0 .2 × 100 = 30 G (s) c

= 30 + 2 s +

100 s

Unit-step Response

10-13 Process Transfer Function: Gp ( s) =

Y ( s) U (s )

=

e

−0.2s

1 + 0.25 s

1

(1 + 0.25s ) (1 + 0.2s + 0.02s

2

)

(a) PI Controller: KP + G ( s) = G c ( s ) G p ( s) ≅

For K

v

= 2,

K

G ( s) =

Thus

v

KI

s 2 1 + 0.2 s + 0.02 s

(1 + 0.25s ) ( = lim

s→0

sG ( s )

=

4

200 ( 2 + K P s )

(

200 K

× 50

s ( s + 4 ) s + 10s + 50 2

I

=

K

I

)

=

200 ( K P s + K I )

(

s ( s + 4 ) s + 10s + 50 2

=2

Thus

)

The following values of the attributes of the unit-step response are computed for the system with various values for K . P

KP

ts

Max overshoot (%)

(sec)

222

ts

(sec)

) K

I

=2

0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 1.0

19.5 13.8 8.8 4.6 1.0 0 0 0 0

0.61 0.617 0.615 0.606 0.5905 0.568 0.541 0.5078 0.44

223

2.08 1.966 1.733 0.898 0.878 0.851 1.464 1.603 1.618

224

225

Unit-step Response.

226

The unit-step response shows a maximum overshoot of 26%. Although the relative damping ratio of the complex roots is 0.707, the real pole of the third-order system transfer function is at −0.667 which adds to the overshoot.

(c) G ( s) = G c ( s ) G p ( s) =

For K

v

= 100

, K

I2

=

K

For a small overshoot, K

= 100

I

G ( s) =

D1

0.00667 (1 + K D1 s ) ( K P2 s + K I 2 )

(

s s + 0.00667 2

. Let us select K

P2

)

= 50 .

Then

0.00667 (1 + K D1 s ) ( 50 s + 100 )

(

s s + 0.00667 2

)

must be relatively large. When K

D1

= 100

, the maximum overshoot is

approximately 4.5%. Thus, KP

=

K

K

D

=

K

I

= 100

K

P2 D1

+ K D 1 K I 2 = 50 + 100 × 100 = 10050 K

P2

= 100 × 50 = 5000

System Characteristic Equation: s + 33 . 35 s + 67 . 04 s + 0 .667 = 0 3

Roots:

2

− 0.01,

Unit-step Response.

227

− 2.138,

− 31.2

10-16 (a) G (s) p

=

Z (s)

=

F (s)

1 2

Ms

=

+Ks

1 150 s

2

=

+1

0 . 00667 s

+ 0 . 00667

2

The transfer function G p ( s ) has poles on the j ω axis. The natural undamped frequency is

ω n = 0 . 0816

(b) PID Controller:

= lim

Kv

s →0

sG ( s )

=

ζ = 1 and ω n = 1

(

2

s s + 0.00667 2

= 100

KI

T hus

KI

)

)

= 100

s + 0.00667 K D s + 0.00667 (1 + K P ) s + 0.00667 K I = 0 3

Characteristic Equation: For

(

0.00667 K D s + K P s + K I

G ( s) = G c ( s ) G p ( s) =

2

rad/sec, the second-order term of the characteristic equation is s

2

+ 2 s + 1.

Dividing the characteristic equation by the seond-order term.

s + ( 0.00667 K D − 2 ) s + 2s + 1 s + 0.00667 K D s + ( 0.00667 + 0.00667 K P ) s + 0.00667 K I 2

3

2

s + 2s + s 3

2

( 0.00667 K ( 0.00667 K ( 0.00667 K

D

− 2 ) s + ( 0.00667 KP − 0.99333 ) s + 0.00667 K I

D

− 2 ) s + ( 0.01334 K D − 4 ) s + 0.00667 K D − 2

P

− 0.01334 K D + 3.00667 ) s + 0.00667 K I − 0.00667 K D + 2

2

2

For zero remainder, 0 . 00667 K

−0 . 00667

K

P D

− 0 . 01334

K

+ 0 . 00667

K

D I

+ 3. 00667 = 0 +2 =0

From Eq. (2),

228

(1) (2)

0 . 00667 K

D

= 0 . 00667

K

P

= 0 . 01334

K

From Eq. (1), 0 . 00667 K

I

+ 2 = 2 . 667

D

− 3 . 00667 = 2 .3273

K

Thus

D

= 399 K

Th us

. 85

P

= 348

. 93

Forward-path Transfer Function:

(

0.00667 399.85 s + 348.93 s + 100

G ( s) =

2

(

s s + 0.00667 2

)

)

Characteristic Equation:

s 3 + 2.667 s 2 + 2.334 s + 0.667 = ( s + 1) Roots:

− 1,

− 1,

2

( s + 0.667 ) = 0

− 0.667

Unit-step Response.

The maximum overshoot is 20%.

10-17 (a) Process Transfer Function: G

p

( s)

=

4

G ( s) = G c ( s ) G p ( s) =

2

s

Characteristic Equation: K

P

= 0 .25

a nd

Forward-path Transfer Function

K

D

s

2

+ 4K Ds + 4 K P =

= 0 . 3535

Unit-step Response.

229

s

2

+ 1.414

s

4 ( KP + KD s) s +1 = 0

2

for

ζ = 0 .707

,

ωn =1

Maximum overshoot = 20.8%

(b)

Select a relatively large value for K The closed-loop zero at s

D

= −K P

and a small value for K / K

D

P

dynamics are governed by the other closed-loop poles. Let K The following results show that the value of K Kp

For BW

P

= 10

D

and use small values of K

tr

.

= 1, we can again select

ts

(sec)

0 0 0

r

P

is not critical as long as it is small.

Max overshoot (%)

0.1 0.05 0.2

(c)

so that the closed-loop poles are real.

is very close to one of the closed-loop poles, and the system

(sec)

0.0549 0.0549 0.0549

K

D

= 10

and a small value for K

0.0707 0.0707 0.0707

P

.

The following frequency-domain results substantiate the design. Kp

PM (deg)

Mr

0.1 0.05 0.2

89.99 89.99 89.99

1 1 1

10-18 (a) Forward-path Transfer Function: G ( s) = G c ( s ) G p ( s) =

2

40 40 40

Characteristic Equation:

10,000 ( K P + K D s ) s

s

( s + 10)

Routh Tabulation:

230

3

+ 10

s

2

+ 10 ,000

K

D

s

+ 10 ,000

K

P

=0

s

3

s

2

s

1

s

0

1

10 ,000 K

10

10, 000 K

10,000 K

D

10,000

− 1000 K

K

D

P

The system is stable for K

0

P

P

(b) Root Locus Diagram: 10,000 K P

G ( s) =

s

Root Contours: 0 ≤ K

G

eq

D

< ∞,

(s)

=

2

( s + 10 )

K

= 0 . 001 ,

P

10 ,000 K s

3

+ 10

s

2

D

0 . 002 , 0 .005 , 0 . 01 . s

+ 10 ,000

231

K

P

P

>0

and

K

D

> 0 .1 K P

(c)

The root contours show that for small values of K means that if we choose K

P

P

the design is insensitive to the variation ofK

to be between 0.001 and 0.005, the value of K

for a relative damping ratio of 0.707. Let K

P

= 0 . 001

and K

D

= 0 . 005

D

P

. This

can be chosen to be 0.005 . G (s) c

= 0 . 001 + 0 . 005

The forward-path transfer function becomes

10 (1 + 5 s )

G ( s) =

s

( s + 10 )

−0.2, which is very close to s = 0, G(s) can be approximated as:

Since the zero of G ( s ) is at s =

G (s) ≅ For the second-order system,

2

ζ = 0 . 707 t

r

=

50 s ( s + 10 ) . Using Eq. (7-104), the rise time is obtained as

1 − 0 .4167

ζ + 2 . 917 ζ ωn

Unit-step Response:

232

2

= 0 . 306

sec

s.

(d) Frequency-domain Characteristics: 10 (1 + 5 s )

G ( s) =

s PM (deg)

2

( s + 10 )

GM (dB)

63

Mr

1.041

7.156

10-19

 0  25.92 ∗ A =  0  −2.36 

1

0

0

−1 0 0   s  −25.92 s 0 0  ∗  sI − A =   0 0 s − 1  2.36  0 0 s 

0 0 0  0 0 1  0 0 0

s

0

0

∆ = sI − A = s 0 s − 1 + ∗

0

0

 s3  2 1 25.92 s ∗ −1  ( sI − A ) =  ∆ − 2.36 s  2  −2.36 s

s

−25.92

0

0

s

−1 = s

2.36

0

s

0 2

(s

2

− 25.92

  s 0 0  3 2 −2.36 s − 25.92 s s − 25.92   3 −2.36 s 0 s − 25.92s  2

s

0

3

233

0

)

( sI − A )

−1

 −0.0732 s 2   0   −0.0732  1  −0.0732 s 3  ∗ −1  =  B = ( sI − A )   0  ∆  0.0976 s 2 − 2.357     0.0976  3    0.0976 s − 2.357 s 

(

Y ( s) = D sI − A

)

−1

B = [0

0

(

0 ] sI − A

1

)

−1

B=

s

2

(s

3

3

10-20

1

c

=

K

P

+

)

D

which

.

Let us first attempt to compensate the system with a PI controller. G (s)

− 25.92

)

2

The system cannot be stabilized by the PD controller, since the s and the s terms involve K D

2

2

s + 0.0976 s + ( 0.0976 KP − 25.92 ) s − 2.357 K D s − 2.357 KP = 0 4

Characteristic Equation:

require opposite signs for K

(

0.0976 s − 24.15

K

I

G ( s) = Gc ( s ) Gp ( s) =

Then

s

100 ( K P s + K I )

(

s s + 10 s + 100 2

)

Since the system with the PI controller is now a type 1 system, the steady-state error of the system due to a step input will be zero as long as the values of K and K are chosen so that the system is stable. Let us choose the ramp-error constant K

= 100

v

performance characteristics are obtained with K KP 10 20 30 40 50 100

I

P

I

. Then, K

= 100

I

= 100

. The following frequency-domain

and various value of K

PM (deg)

P

ranging from 10 to 100.

GM (dB)

∞ ∞ ∞ ∞ ∞ ∞

1.60 6.76 7.15 6.90 6.56 5.18

Mr

29.70 7.62 7.41 8.28 8.45 11.04

50.13 69.90 85.40 98.50 106.56 160.00

The maximum phase margin that can be achieved with the PI controller is only 7.15 deg when K

Thus, the overshoot requirement cannot be satisfied with the PI controller alone. Next, we try a PID controller.

Gc (s ) = K P + K Ds +

KI

=

(1 + K s ) (K D1

s

Based on the PI-controller design, let us select K becomes

G ( s) =

P2

= 30 .

P2

s + KI2 )

=

P

= 30 .

(1 + K s ) (K D1

s

P2

s + 100 )

s

Then the forward-path transfer function

100 ( 30 s + 100 )( 1 + K D1s )

(

s s + 10 s + 100 2

)

The following attributes of the frequency-domain performance of the system with the PID controller are obtained for various values of K ranging from 0.05 to 0.4. D1

K D1 0.05 0.10 0.20 0.30

PM (deg) 85.0 89.4 90.2 90.2

GM (dB)

∞ ∞ ∞ ∞

234

Mr

1.04 1.00 1.00 1.00

164.3 303.8 598.6 897.0

0.40

90.2

We see that for values of K

D1

K

D1

P

=

= 0 .2, K

1201.0

greater than 0.2, the phase margin no longer increases, but the

bandwidth increases with the increase in K K

1.00

P2

K

I

=

D1

K

. Thus we choose

I2

= 100

,

K

D

=

K

D1

K

P2

= 0 .2 × 30 = 6 ,

+ K D 1 K I 2 = 30 + 0 .2 × 100 = 50

The transfer function of the PID controller is

G (s) c

= 50 + 6 s +

100

s The unit-step response is show below. The maximum overshoot is zero, and the rise time is 0.0172 sec.

10-21 (a) Gp (s ) =

4 s

2

G ( s) = Gc ( s ) G p ( s ) =

4 (1 + aTs ) s (1 + Ts )

Root Contours: ( T is fixed and a varies)

235

2

G

eq

(s)

=

4 aTs Ts

3

+ s +4 2

small value for T and a large and a = 100.

Select a value for a. Let T = 0.02

G (s) c

=

1+ 2 s

G ( s) =

1 + 0 . 02 s

400 ( s + 0.5) s

The characteristic equation is The roots are: The system transfer function is

s

3

+ 50

s

2

+ 400

s

2

( s + 50)

+ 200 = 0

−0.5355, −9.3, −40.17 Y (s)

=

R (s )

400 ( s + 0.5 )

( s + 0.5355) ( s + 9.3) ( s + 40.17 )

Since the zero at −0.5 is very close to the pole at −0.5355, the system can be approximated by a secondorder system,

Y (s)

=

R (s )

373.48

( s + 9.3)( s + 40.17 )

The unit-step response is shown below. The attributes of the response are: Maximum overshoot = 5%

t

s

= 0 . 6225

Unit-step Response.

236

sec

t

r

= 0 .2173

sec

The following attributes of the frequency-domain performance are obtained for the system with the phase-lead controller. PM = 77.4 deg GM = infinite Mr = 1.05 BW = 9.976 rad/sec

10-21 (b)

The Bode plot of the uncompensated forward-path transfer function is shown below. The diagram shows that the uncompensated system is marginally stable. The phase of G ( jω ) is

ωm

frequencies. For the phase-lead controller we need to place

−180 deg at all

at the new gain crossover frequency

to realize the desired phase margin which has a theoretical maximum of 90 deg. For a desired phase margin of 80 deg,

The gain of the controller is 20 log

10

1 + sin 80

o

1 − sin 80

o

=

a

= 42

a

= 130

dB. The new gain crossover frequency is at

G ( jω )

42

=−

= −21

dB

2 Or

G ( jω )

1 T 1 aT

=

=

4

ω m

2

= 0 . 0877

=

Gc (s ) =

1 + aTs 1 + Ts

× 6 . 75 = 77

130

= 0 . 592

Thus

=

Thus

aT

ω = 2

Thus

ω = 6 . 75

45 . 61

T

= 0 . 013

= 1. 69

1 + 1.702 s

4 (1 + 1.702s )

G( s) =

1 + 0.0131s

s

Bode Plot.

237

2

(1 + 0.0131s )

10-22 (a) Forward-path Transfer Functi on: 1000 (1 + aTs )

 

1000a  s +

1

 

aT  G ( s) = G c ( s ) GP ( s) = = s ( s + 10 ) (1 + Ts )  1 s ( s + 10 )  s +   T

238

Set 1/aT = 10 so that the pole of G ( s ) at s = becomes s

ωn =

2 ζω

1000 a

n

=

1

2

+

=2

T

−10 is cancelled. 1

s

T

+ 1000

a

c

=

=0

1000 a

a = 40

Thus

Controller Transfer Function: G (s)

The characteristic equation of the system

and

T = 0.0025

Forward-path Transfer Function:

1 + 0 . 01 s

G ( s) =

1 + 0 . 0025 s

40,000 s ( s + 400 )

The attributes of the unit-step response of the compensated system are: t

Maximum overshoot = 0

= 0 . 0168

r

t

sec

s

= 0 . 02367

sec

(b) Frequency-domain Design The Bode plot of the uncompensated forward-path transfer function is made below.

G (s) =

1000 The attributes of the system are PM = 17.96 deg, GM = infinite.

s ( s + 10 )

Mr = 3.117, and BW = 48.53 rad/sec. To realize a phase margin of 75 deg, we need more than 57 deg of additional phase. Let us add an additional 10 deg for safety. Thus, the value of φ for the phase-lead controller is chosen to be m

67 deg. The value of a is calculated from

The gain of the controller is 20 log

10

1 + sin 67

o

1 − sin 67

o

a

=

a

= 20

log

10

= 24. 16

24. 16

= 27 . 66

dB. The new gain crossover frequency

is at G ( jω

From the Bode plot 1 T

=

aT

=

ωm '

'

m

)

=−

27 . 66

= − 13 . 83

dB

2

is found to be 70 rad/sec. Thus,

24. 16

× 70 = 344 =

1 + aTs

Thus

G (s)

PM = 75.19 deg

GM = infinite

c

T

or

=

1

= 0 . 0029

+ 0 . 0702

aT

= 0 . 0702

s

1 + Ts 1 + 0 . 0029 s The compensated system has the following frequency-domain attributes: Mr = 1.024 BW = 91.85 rad/sec

The attributes of the unit-step response are: Rise time tr = 0.02278 sec Settling time ts = 0.02828 sec

239

Maximum overshoot = 3.3%

10-23 (a) Forward-path Transfer Function: ( N = 10) G ( s) = G c ( s ) G p ( s) =

200 ( 1 + aTs )

s ( s + 1) ( s + 10 ) ( 1 + Ts )

Starting with a = 1000, we vary T first to stabilize the system. The following time-domain attributes are obtained by varying the value of T.

240

T

tr

Max Overshoot (%)

ts

0.0001

59.4

0.370

5.205

0.0002 0.0003 0.0004 0.0005 0.0006 0.0007 0.0008 0.0009 0.0010

41.5 29.9 22.7 18.5 16.3 15.4 15.4 15.5 16.7

0.293 0.315 0.282 0.254 0.230 0.210 0.192 0.182 0.163

2.911 1.83 1.178 1.013 0.844 0.699 0.620 0.533 0.525

The maximum overshoot is at a minimum when T = 0.0007 or T = 0.0008. The maximum overshoot is 15.4%.

Unit-step Response. ( T = 0.0008 sec a = 1000)

10-23 (b) Frequency-domain Design. Similar to the design in part (a), we set a = 1000, and vary the value of T between 0.0001 and 0.001. The attributes of the frequency-domain characteristics are given below. T 0.0001 0.0002 0.0003 0.0004 0.0005 0.0006 0.0007

PM (deg)

GM

17.95 31.99 42.77 49.78 53.39 54.69 54.62

60.00 63.53 58.62 54.53 51.16 48.32 45.87

Mr

(dB)

241

3.194 1.854 1.448 1.272 1.183 1.138 1.121

BW (rad/sec) 4.849 5.285 5.941 6.821 7.817 8.869 9.913

0.0008 0.0009 0.0010

53.83 52.68 51.38

43.72 41.81 40.09

1.125 1.140 1.162

The phase margin is at a maximum of 54.69 deg when T = 0.0006. The performance worsens if the value of a is less than 1000.

10-24 (a) Bode Plot.

The attributes of the frequency response are: PM = 4.07 deg

Mr = 23.24

GM = 1.34 dB

10-24 (b) Single-stage Phase-lead Controller. G ( s) =

6 (1 + aTs )

s (1 + 0.2s ) (1 + 0.5s ) (1 + Ts )

We first set a = 1000, and vary T. The following attributes of the frequency-domain characteristics are obtained. T 0.0050 0.0010 0.0007 0.0006

PM (deg) 17.77 43.70 47.53 48.27

242

Mr 3.21 1.34 1.24 1.22

10.92 11.88 12.79

0.0005 0.0004 0.0002 0.0001

48.06 46.01 32.08 19.57

1.23 1.29 1.81 2.97

The phase margin is maximum at 48.27 deg when T = 0.0006. Next, we set T = 0.0006 and reduce a from 1000. We can show that the phase margin is not very sensitive to the variation of a when a is near 1000. The optimal value of a is around 980, and the corresponding phase margin is 48.34 deg. With a = 980 and T = 0.0006, the attributes of the unit-step response are: Maximum overshoot = 18.8%

tr = 0.262 sec

(c) Two-stage Phase-lead Controller. ( a = 980,

G( s) =

ts = 0.851 sec

T = 0.0006)

(

)

6 ( 1 + 0.588s ) 1 + bT s 2 s (1 + 0.2s )(1 + 0.5 s ) (1 + 0.0006 s ) 1 + T s 2

(

)

Again, let b = 1000, and vary T . The following results are obtained in the frequency domain. 2

T2

Mr

PM (deg)

0.0010 93.81 1.00 0.0009 94.89 1.00 0.0008 96.02 1.00 0.0007 97.21 1.00 0.0006 98.43 1.00 0.0005 99.61 1.00 0.0004 100.40 1.00 0.0003 99.34 1.00 0.0002 91.98 1.00 0.0001 73.86 1.00 Reducing the value of b from 1000 reduces the phase margin. Thus, the maximum phase margin of 100.4 deg is obtained with b = 1000 and T 2 = 0.0004. The transfer function of the two-stage phaselead controller is

Gc ( s) =

( 1 + 0.588s ) (1 + 0.4s ) (1 + 0.0006s ) (1 + 0.0004s )

(c) Unit-step Responses.

243

10-25 (a)

The loop transfer function of the system is

G ( s) H( s) =

1 + R2C s

10 K pK aK e Ns (1 + 0.05 s ) s

The characteristic equation is For root locus plot with R

2

3

+ 20

RCs 1

s

2

=

68.76 s (1 + 0.05s )

+ 6 .876 × 10

−4

6.876 × 10 R2 s s + 20 s + 687.6 3

2

+ 687

2s .6

=0

as the variable parameter, we have −4

G eq ( s ) =

R s

−6

1 + R2 × 10 s

2

−4

=

6.876 × 10 R2 s

( s + 21.5) ( s − 0.745 +

Root Locus Plot.

244

j 5.61) ( s − 0.745 − j 5.61 )

When R

2

= 2 . 65 × 10

5

, the roots are at

−6 . 02 ±

j 7 . 08 , and the relative damping ratio is 0.65 which is

maximum. The unit-step response is plotted at the end together with those of parts (b) and (c).

(b) Phase-lead Controller. G ( s) H( s) =

68.76 (1 + aTs )

s (1 + 0.05 s ) (1 + Ts )

(

)

(

)

Characteristic Equation: Ts + 1 + 20 T s + 20 + 1375.2 aT s + 1375.2 = 0 3

2

With T = 0.01, the characteristic equation becomes

3 2 s + 120 s + ( 2000 + 1375.2 a ) s + 137520 = 0 The last equation is conditioned for a root contour plot with a as the variable parameter. Thus 1375 .2 as G eq ( s ) = 3 2 s + 120 s + 2000 s + 137 ,520 From the root contour plot on the next page we see that when a = 3.4 the characteristic equation roots are at − 39 .2 , − 40 .4 + j 43 . 3, and − 40 .4 − j 43 . 3 , and the relative damping ratio is maximum and is 0.682.

Root Contour Plot ( a varies).

245

Unit-step Responses.

10-25 (c) Frequency-domain Design of Phase-lead Controller. For a phase margin of 60 deg, a = 4.373 and T = 0.00923. The transfer function of the controller is G (s) c

=

1 + aTs 1

+ Ts

=

1

+ 0 . 04036

s

1 + 0 . 00923 s

246

10-26 (a) Time-domain Design of Phase-lag Controller. Process Transfer Function:

Gp ( s) =

200 s ( s + 1 )( s + 10 )

For the uncompensated system, the two complex characteristic equation roots are at s

= − 0 .475 +

j 0 .471

and −0 .475 − j 0 .471 which correspond to a relative damping ratio of 0.707, when the forward path gain is 4.5 (as against 200). Thus, the value of a of the phase-lag controller is chosen to be a

=

4. 5

= 0 . 0225

Select T = 1000

which is a large number.

200 Then G (s) c

=

1 + aTs 1 + Ts

=

1

+ 22 . 5 s

G ( s) = Gc ( s ) Gp ( s) =

1 + 1000 s

Unit-step Response.

Maximum overshoot = 13.6

tr = 3.238 sec

ts = 18.86 sec

Bode Plot (with phase-lag controller, a = 0.0225, T = 1000)

247

4.5 ( s + 0.0889) )

s ( s + 1 )( s + 10 )( s + 0.001)

PM = 59 deg.

Mr = 1.1

GM = 27.34 dB

10-26 (b) Frequency-domain Design of Phase-lag Controller. For PM = 60 deg, we choose a = 0.02178 and T = 1130.55. The transfer function of the phase-lag controller is 1 + 24. 62 s G (s) = GM = 27.66 dB Mr = 1.093 BW = 0.619 rad/sec c 1 + 1130 . 55 s

Unit-step Response.

Max overshoot = 12.6%, tr = 3.297 sec

ts = 18.18 sec

10-27 (a) Time-domain Design of Phase-lead Controller Forward-path Transfer Function.

G ( s) = Gc ( s ) Gp ( s) =

K (1 + aTs ) s ( s + 5)

2

(1 + Ts )

248

K

v

= lim

s →0

sG ( s )

=

K 25

= 10

Thus

K = 250

With K = 250, the system without compensation is marginally stable. For a > 1, select a small value for T and a large value for a. Let a = 1000. The following results are obtained for various values of T ranging from 0.0001 to 0.001. When T = 0.0004, the maximum overshoot is near minimum at 23%. T 0.0010 0.0005 0.0004 0.0003 0.0002 0.0001

tr

ts

Max Overshoot (%)

(sec)

(sec)

33.5 23.8 23.0 24.4 30.6 47.8

0.0905 0.1295 0.1471 0.1689 0.1981 0.2326

0.808 0.6869 0.7711 0.8765 1.096 2.399

As it turns out a = 1000 is near optimal. A higher or lower value for a will give larger overshoot.

Unit-step Response.

(b) Frequency-domain Design of Phase-lead Controller 250 ( 1 + aTs )

G ( s) = s

2

( s + 5 ) (1 + Ts ) 2

Setting a = 1000, and varying T, the following attributes are obtained. T 0.00050 0.00040 0.00035 0.00030 0.00020

Mr

PM (deg) 41.15 42.85 43.30 43.10 38.60

1.418 1.369 1.355 1.361 1.513

BW (rad/sec) 16.05 14.15 13.16 12.12 10.04

When a = 1000, the best value of T for a maximum phase margin is 0.00035, and PM = 43.3 deg. As it turns out varying the value of a from 1000 does not improve the phase margin. Thus the transfer function of the controller is

249

G (s) c

=

1 + aTs 1

1

=

+ Ts

+ 0 . 35

s

+ 0 . 00035

1

G ( s) =

and s

250 ( 1+ 0.35s ) s ( s + 5)

2

(1 + 0.00035s )

10-27 (c) Time-domain Design of Phase-lag Controller Without compensation, the relative damping is critical when K = 18.5. Then, the value of a is chosen to be 18 . 5 a = = 0 . 074 250 We can use this value of a as a reference, and conduct the design around this point. The value of T is preferrably to be large. However, if T is too large, rise and settling times will suffer. The following performance attributes of the unit-step response are obtained for various values of a and T. a

T

0.105 0.100 0.095 0.090 0.090 0.090 0.090 0.090 0.090 0.090

Max Overshoot (%)

tr

ts

2.6 2.9 2.6 2.5 2.1 1.9 1.7 1.4 0.9 0.7

1.272 1.348 1.422 1.522 1.532 1.538 1.543 1.550 1.560 1.566

1.82 1.82 1.82 2.02 2.02 2.02 2.02 2.22 2.22 2.22

500 500 500 500 600 700 800 1000 2000 3000

As seen from the results, when a = 0.09 and for T ≥ 2000, the maximum overshoot is less than 1% and the settling time is less than 2.5 sec. We choose T = 2000 and a = 0.09. The corresponding frequency-domain characteristics are: PM = 69.84 deg

Mr = 1.004

GM = 20.9 dB

(d) Frequency-domain Design of Phase-lag Controller G ( s) =

250 ( 1 + aTs ) s ( s + 5)

2

a 1)

1 + Ts

By selecting a small value for T, the value of a becomes the critical design parameter in this case. If a is too small, the overshoot will be excessive. If the value of a is too large, the oscillation in the step response will be objectionable. By trial and error, the best value of a is selected to be 6, and T = 0.001. The following performance attributes are obtained for the unit-step response. tr = 0.01262 sec

Maximum overshoot = 0%

ts = 0.1818 sec

However, the step response still has oscillations due to the compliance in the motor shaft. The unitstep response of the phase-lead compensated system is shown below, together with that of the uncompensated system.

(b) Phase-lead and Second-order Controller The poles of the process G p ( s ) are at −161.3, order term is

+ 262

2

s

s

+ 2 ,624, G

The value of

ωn

is set to

Let the two poles of G

c1

c1

( s)

=

s

2

=

+ 262 s

2

s

2

2

417 .1

+ 2ζ pω n s + ω n

and

s

+ 2 ,624,

s

2

= 1620

= − 1620

G c1 ( s )

The second-

417 .1 . Let the second-order controller transfer function be

2 ,624, 417 .1

( s ) be at s

−131+ j1614.7 and −131 − j1614.7.

− 1620

. Then,

ζ p = 405

+ 262

s

+ 2 ,624,

417 .1

+ 3240

s

+ 2 ,624,

417 .1

6.087 × 10

10

G ( s) = G c ( s ) G c1 ( s ) G p ( s) =

(

.

(1 + 0.006 s )

)

s ( s + 161.3 ) s + 3240 s + 2,624,417.1 ( 1 + 0.001s )

The unit-step response is shown below, and the attributes are:

252

2

tr = 0.01012 sec

Maximum overshoot = 0.2

ts = 0.01414 sec

The step response does not have any ripples.

Unit-step Responses

10-29 (a) System Equations. e

a

= R a i a + K bω m

T

= K i ia

m

Tm = Jm

dω m dt

+ Bmωm + KL ( θ m − θL ) + BL ( ωm − ω L )

dω L

K L ( θ m − θ L ) + BL ( ω m − ω L ) = J L

dt

State Equations in Vector-matrix Form:

 dθ L   dt   0    K  dω L   − L  dt   J L  dθ  =  0  m   dt   K L  dω m   J    m  dt 

1

0

BL

KL

JL

JL

0

0

BL Jm

KL Jm

  θ   0  BL  L   0   JL  ω L   + 0   ea  θ  1 m     Ka   B + Bl Ki K b  ω m  − m −  R J    a m Jm J m Ra  0

253

State Diagram:

Transfer Functions:

Ωm ( s)

=

Ea ( s ) Ω L ( s)

=

E a (s ) Ωm ( s)

=

Ea ( s ) Ω L ( s) E a (s )

=

(

)

Ki s + BL s + KL / Ra 2

J m J L s + ( K e J L + BL J L + BL J m ) s + ( J m K L + J L K L + K e BL ) s + K L K e 2

2

K i ( B Ls + K L ) / R a

J m J L s + ( K e J L + BL J L + B L J m ) s + ( J m K L + J L K L + K e BL ) s + K L K e 3

2

(

133.33 s + 10 s + 3000 2

)

s + 318.15 s + 60694.13 s + 58240 3

2

=

(

133.33 s + 10s + 3000 2

( s + 0.9644) ( s + 158.59 +

j187.71) ( s + 158.59 − j187.71)

1333.33 ( s + 300 )

( s + 0.9644 ) ( s + 158.59 +

j187.71 ) ( s + 158.59 − j187.71)

(b) Design of PI Controller.

G ( s) =

Kv

Ω L (s )

= lim

E (s )

s→0

With K I

 

1333.33 K P  s + =

sG ( s )

KP 20 18

KI

s ( s + 0.9644 ) s + 317.186s + 60388.23

=

= 14 .56 ,

(

  ( s + 300 ) KP 

1333 .33 0 . 9644

× 300

× 60388

2

K

I

. 23

= 6 . 87

KI

= 100

) Thus

KI

= 14 .56

we study the effects of varying K P . The following results are obtained. tr

ts

(sec)

(sec)

0.00932 0.01041

0.02778 0.01263

254

Max Overshoot (%) 4.2 0.7

)

17 16 15 10 With K I

= 14 .56

0.01113 0.01184 0.01303 0.02756

0.01515 0.01515 0.01768 0.04040

0 0 0 0.6

and K P ranging from 15 to 17, the design specifications are satisfied.

Unit-step Response:

(c) Frequency-domain Design of PI Controller ( KI = 14.56) G ( s) =

(

1333.33 ( K P s + 14.56 )( s + 300 )

s s + 318.15 s + 60694.13s + 58240 3

2

The following results are obtained by setting K I KP

PM (deg)

20 18 17 16 15 10 8 7 6 5

65.93 69.76 71.54 73.26 74.89 81.11 82.66 83.14 83.29 82.88

GM (dB)

∞ ∞ ∞ ∞ ∞ ∞ ∞ ∞ ∞ ∞

= 14 .56

)

and varying the value of K P .

Mr

Max Overshoot (%)

1.000 1.000 1.000 1.000 1.000 1.005 1.012 1.017 1.025 1.038

266.1 243 229 211.6 190.3 84.92 63.33 54.19 45.81 38.12

4.2 0.7 0 0 0 0.6 1.3 1.9 2.7 4.1

255

tr

ts

(sec)

(sec)

0.00932 0.01041 0.01113 0.01184 0.01313 0.0294 0.04848 0.03952 0.04697 0.05457

0.02778 0.01263 0.01515 0.01515 0.01768 0.0404 0.03492 0.05253 0.0606 0.0606

From these results we see that the phase margin is at a maximum of 83.29 degrees when K P = 6 . However, the maximum overshoot of the unit-step response is 2.7%, and M r is slightly greater than one. In part (b), the optimal value of K P from the standpoint of minimum value of the maximum overshoot is between 15 and 17. Thus, the phase margin criterion is not a good indicator in the present case.

10-30 (a) Forward-path Transfer Function Gp ( s) =

K Θ m ( s) Tm ( s )

=

(

(

( = + 2100 s + 10,000 ) s ( s + 4.937 ) ( s

100 K s + 10 s + 100

s s + 20 s 3

2

2

)

10,000 s + 10 s + 100 2

2

)

+ 15.06 s + 2025.6

)

The unit-step response is plotted as shown below. The attributes of the response are: tr = 0.01345 sec

Maximum overshoot = 57%

ts = 0.4949 sec

(b) Design of the Second-order Notch Controller The complex zeros of the notch controller are to cancel the complex poles of the process transfer function. Thus

G (s) c

=

s

2

s

2

+ 15 . 06 s + 2025 + 90 ζ p s + 2025

.6 .6

( G ( s) = G ( s ) G ( s) = s ( s + 4.937 ) ( s

10,000 s + 10 s + 100

and

c

p

The following results are obtained for the unit-step response when various values of The maximum overshoot is at a minimum of 4.1% when ζ

p

= 1.222

ζp

are used.

. The unit-step response is

plotted below, along with that of the uncompensated system.

ζp

2 ζω

2.444 2.333 2.222 1.667 1.333 1.222 1.111 1.000

200 210 200 150 120 110 100 90

Max Overshoot (%)

n

7.3 6.9 6.5 4.9 4.3 4.1 5.8 9.8

Unit-step Response

256

2

2

)

+ 90ζ p s + 2025.6

)

10-30 (c) Frequency-domain Design of the Notch Controller The forward-path transfer function of the uncompensated system is

( G ( s) = s ( s + 4.937 ) ( s

10000 s + 10 s + 100 2

2

)

+ 15.06 s + 2025.6

)

The Bode plot of G ( jω ) is constructed in the following. We see that the peak value of G ( jω ) is approximately 22 dB. Thus, the notch controller should provide an attentuation of at the resonant frequency of 45 rad/sec. Using Eq. (10-155), we have

G ( j 45) c

ζz

=

ζp

=

0 .167

ζp

Notch Controller Transfer Function

G (s) c

=

s

+ 15 .06 s + 2025

s

2

2

+ 189

.216 s

= 0 . 0794

Thus

ζ p = 2 .1024

Forward-path Transfer Function

.6

+ 2025

−22 dB or 0.0794

G (s ) =

.6

Bode Plots

257

( s ( s + 4.937 ) (s

10,000 s + 10 s + 100 2

2

)

+ 189.22s + 2025.6

)

Attributes of the frequency response: PM = 80.37 deg

M

GM = infinite

r

= 1. 097

Attributes of the frequency response of the system designed in part (b): PM = 59.64 deg

GM = infinite

M

r

= 1. 048

10-31 (a) Process Transfer Function Gp ( s) =

(

500 ( s + 10 )

s s + 10 s + 1000 2

)

The Bode plot is constructed below. The frequency-domain attributes of the uncompensated system are: PM = 30 deg

GM = infinite

M

r

= 1.86

The unit-step response is oscillatory.

(b) Design of the Notch Controller

258

and

For the uncompensated process, the complex poles have the following constants:

ωn =

1000

= 31 .6

2 ζω

n

= 10

Thus

ζ

= 0.158

The transfer function of the notch controller is

G (s) c

=

s

2

+ 2ζ zω n s + ω n

s

2

2

+ 2ζ p s + ω n 2

For the zeros of G ( s ) to cancel the complex poles of G p ( s ) , c

ζ z = ζ = 0 .158

.

From the Bode plot, we see that to bring down the peak resonance of G ( j ω n ) in order to smooth out the magnitude curve, the notch controller should provide approximately −26 dB of attenuation. Thus, using Eq. (10-155),

ζz ζp

−26

= 10

= 0 . 05

20

ζp =

Thu s

0 .158

= 3 .1525

0 . 05

The transfer function of the notch controller is

G (s) c

=

s s

2

2

+ 10 s + 1000

+ 199

. 08 s

G ( s) = G c ( s ) G p ( s) =

+ 1000

(

500 ( s + 10 )

s s + 199.08 s + 1000 2

The attributes of the compensated system are: PM = 72.38 deg

GM = infinite

Maximum overshoot = 3.4%

t

r

= 0 . 3868

M

r

=1

sec

Bode Plots

259

s

= 0 .4848

sec

)

Step Responses

10-31 (c) Time-domain design of the Notch Controller

260

With

ζ z = 0 .158

ω n = 31 . 6 ,

and

the forward-path transfer function of the compensated system is

G ( s) = G c ( s ) G p ( s) =

500 ( s + 10 )

(

s s + 63.2ζ p s + 1000 2

)

The following attributes of the unit-step response are obtained by varying the value of

ζp

2 ζω

1.582 1.741 1.899 2.057 2.215 2.500 3.318 When

ζ p = 2 .5

tr

Max Overshoot (%)

n

100 110 120 130 140 158.25 209.7

ζ p. ts

(sec)

0 0 0 0 0.2 0.9 4.1

(sec)

0.4292 0.4172 0.4074 0.3998 0.3941 0.3879 0.3884

0.5859 0.5657 0.5455 0.5253 0.5152 0.4840 0.4848

the maximum overshoot is 0.9%, the rise time is 0.3879 sec and the setting

time is 0.4840 sec. These performance attributes are within the required specifications.

10-32

Let the transfer function of the controller be

(

20,000 s + 10 s + 50

Gc ( s) =

2

( s + 1000 )

)

2

Then, the forward-path transfer function becomes

For G

For

cf

(s)

= 1,

K

± 20% variation in K,

= lim

v

sG ( s )

s→0

K

we let

G cf ( s ) =

10

(

and

2

s s + 10 s + 100 6

2

K

= 50

8

10

= 4000

min

=

(

20,000 K s + 10 s + 50

G ( s) = Gc ( s ) Gp ( s) =

K

max

50 ( s + 1 )

)

) ( s + 1000)

2

Thus t he nom inal K

= 6000

= 5000

. To cancel the complex closed-loop poles,

where the (s + 1) term is added to reduce the rise time.

s + 10 s + 50 2

Closed-loop Transfer Function:

Y (s) R (s )

10 K ( s + 1) 6

=

(

s s + 10 s + 100 2

)( s + 1000 ) + 20,000K ( s 2

2

+ 10 s + 50

)

Characteristic Equation: K = 4000:

Roots:

s

5

+ 2010

− 97 . 7 ,

s + 1,020 , 100 s + 9 . 02 × 10 s + 9 × 10 − 648 . 9 , − 1252 . 7 , − 5 . 35 + j 4. 6635 , 4

3

Max overshoot K = 5000:

Roots:

s

5

+ 2010

−132

.46 ,

s

4

≅ 6.7%

+ 1,02010

587 .44,

Max overshoot

7

≅ 4%

2

+ 4 × 10 = 0 − 5 . 35 − j 4. 6635

8

9

Rise time < 0.04 sec

0 s + 1.1 × 10 s + 1.1 × 10 s + 5 × 10 − 1279 . 6 , − 5 .272 + j 4. 7353 , − 5 .272 3

8

2

9

Rise time < 0.04 sec

261

9

=0 − j 4. 7353

K = 6000

Roots:

s

5

+ 2010

−176

s

. 77 ,

, 100 s + 1. 3 × 10 s + 1.3 × 10 s + 6 × 10 = 0 − 519 . 37 , − 1303 .4, − 5 .223 + j 4. 7818 , − 5 .223 − j 4. 7818 4

+ 1,020

Max overshoot

3

8

≅ 2.5%

2

9

9

Rise time < 0.04 sec

Thus all the required specifications stay within the required tolerances when the value of K varies by plus and minus 20%.

Unit-step Responses

10-33

Let the transfer function of the controller be

Gc ( s) =

(

200 s + 10 s + 50 2

( s + 100 )

)

2

The forward-path transfer function becomes

G ( s) = G c ( s ) G p ( s) =

(

200,000 K s + 10 s + 50 2

s ( s + a ) ( s + 100 )

)

2

For a = 10, K

v

= lim

sG ( s )

s →0

=

10

7

10

K 5

= 100

K

= 100

K

Thus

=1

Characteristic Equations: ( K = 1) a = 10:

Roots:

+ 210 s + 2 .12 × 10 s + 2 .1 × 10 s + 10 = 0 −4. 978 + j 4. 78 , − 4. 978 − j 4. 78 , − 100 + j 447 s

4

3

5

2

6

7

.16

− 100 −

s + 208 s + 2 .116 × 10 s + 2 . 08 × 10 s + 10 = 0 Roots: − 4. 939 + j 4. 828 , − 4. 939 − j 4. 828 , − 99 . 06 + j 446 .97 , − 99 . 06 4

a = 8:

a = 12:

s

4

3

+ 212

s

3

5

2

6

+ 2 .124 × 10 + 2 .12 × 10 5

262

6

s

+ 10

j 447 .16

7

7

=0

j 446 . 97

Roots: − 5. 017 + j 4. 73 ,

− 5 . 017 −

Unit-step Responses:

j 4. 73 ,

− 100

. 98

+

j 447 . 36 ,

− 100

. 98

j 447 . 36

All three responses for a = 8, a = 10, and 12 are similar.

10-34 Forward-path Transfer Function: G ( s) =

Y ( s)

=

E (s )

K s ( s + 1 )( s + 10 ) + K Kt s

Kv = lim sG( s ) = s →0

K =1 10 + KKt

Characteristic Equation:

s + 11s + (10 + KKt ) s + K = s + 11 s + Ks+ K = 0

For root loci,

G eq ( s ) =

3

2

3

K ( s + 1) s

2

( s + 11)

263

2

Root Locus Plot ( K varies)

The root loci show that a relative damping ratio of 0.707 can be realized by two values of K. K = 22 and 59.3. As stipulated by the problem, we select K = 59.3.

10-35 Forward-path Transfer Function: G ( s) =

10 K s ( s + 1) ( s + 10 ) + 10 K t s

K

v

= lim

s →0

sG ( s )

264

10 K

= 10

+ 10

K

= t

K 1+ K

=1 t

Thus

K

t

=

K

−1

s ( s + 1 )( s + 10 ) + 10 Kt + 10 K = s + 11 s +10 Ks + 10 K = 0 3

Characteristic Equation: When K = 5.93 and K s

3

The roots are:

t

=

K

− 1 = 4. 93

+ 11 s + 10 . 046 2

− 10 . 046

s

, the characteristic equation becomes

+ 4. 6 = 0

− 0 .47723 +

,

2

− 0 .47723 −

j 0 .47976 ,

j 0 .47976

10-36 Forward-path Transfer Function: G ( s) =

K (1 + aTs )

(

s ( (1 + Ts ) s + 10 s + KK t 2

K

)

v

= lim

s →0

=

sG ( s )

1 K

= 100

Thu s

t

Let T = 0.01 and a = 100. The characteristic equation of the system is written:

(

)

s + 110 s + 1000s + K 0.001s + 101s +100 = 0 4

3

2

2

To construct the root contours as K varies, we form the following equivalent forward-path transfer function:

G eq ( s ) =

(

0.001 K s + 101,000 s + 100,000 s

2

2

( s + 10) ( s + 100 )

) = 0.001K ( s + 1)( s + 50499 ) s

2

( s + 10) ( s + 100)

From the root contour diagram we see that two sets of solutions exist for a damping ratio of 0.707. These are: K = 20: Complex roots: − 1.158 + j 1.155 , − 1.158 − j1 .155 K = 44.6:

Complex roots: − 4. 0957 + j 4. 0957 ,

The unit-step responses of the system for K = 20 and 44.6 are shown below.

Unit-step Responses:

265

− 4. 0957 −

j 4. 0957

K

t

= 0 . 01

10-37 Forward-path Transfer Function: G ( s) =

K s K K1 Ki N s  J t L a s + ( Ra J t + L a Bt + K 1 K 2 J t ) s + R a Bt + K 1 K 2 Bt + K b K i + K K1 Ki Kt  2

1.5 × 10 K 7

G ( s) =

Ramp Error Constant:

Thus

(

s s + 3408.33s + 1,204,000 + 1.5 × 10 KK t K

2

v

8

= lim

s →0

sG ( s )

1.204

+ 150

1.204

15 K

=

KK

t

= 0 .15

+ 150

KK

)

= 100 t

K

The forward-path transfer function becomes

1.5 × 10 K 7

G ( s) =

(

s s + 3408.33s + 150,000 K

Characteristic Equation:

2

s

3

+ 3408

. 33 s

+ 150

)

,000 Ks

+ 1. 5 × 10

7

K

=0

When K = 38.667 the roots of the characteristic equation are at

− 0 .1065

,

− 1. 651 +

− 1.651 −

j 1. 65 ,

j1. 65

The forward-path transfer function becomes

G ( s) =

(

5.8 × 10

8

s s + 3408.33 s + 5.8 × 10 2

266

6

)

≅ 0 . 707

for the complex roots)

Unit-step Response

Unit-step response attributes:

10-38

Maximum overshoot = 0

Rise time = 0.0208 sec

Settling time = 0.0283 sec

(a) Disturburnce -to-Output Transfer Function Y (s)

==

TL ( s) For T ( s ) L

r =0

= 1/

2 (1 + 0.1s )

c

=1

s

t →∞

= lim

sY ( s )

s →0

=

1

≤ 0 . 01 Thus 10 K Performance of Uncompensated System. K = 10, G ( s ) = 1 lim y ( t )

10-38 (b)

G (s)

s (1 + 0.01 s)(1 + 0.1 s ) + 20 K

K

≥ 10

c

G ( s) =

200 s (1 + 0.01s ) (1 + 0.1s )

The Bode diagram of G ( jω ) is shown below. The system is unstable. The attributes of the frequency response are:

PM = −9.65 deg

GM =

−5.19 dB.

(c) Single-stage Phase-lead Controller Design To realize a phase margin of 30 degrees, a = 14 and T = 0.00348. G (s) c

=

1 + aTs 1

+ Ts

=

1 + 0 . 0487 s 1

+ 0 . 00348

267

s

The Bode diagram of the phase-lead compensated system is shown below. The performance attributes are: PM = 30 deg GM = 10.66 dB M = 1. 95 BW = 131.6 rad/sec. r

(d) Two-stage Phase-lead Controller Design Starting with the forward-path transfer function

G ( s) =

200 ( 1+ 0.0487 s )

s (1 + 0.1s ) (1 + 0.01s ) (1 + 0.00348 s )

The problem becomes that of designing a single-stage phase-lead controller. For a phase margin or 55 degrees, a = 7.385 and T = 0.00263. The transfer function of the second-stage controller is G

Thus

G ( s) =

c1

( s)

=

1

+ aTs

1

+ Ts

=

1

+ 0 . 01845

s

1

+ 0 . 00263

s

200 ( 1+ 0.0487 s ) (1 + 0.01845 s )

s (1 + 0.1s ) (1 + 0.01s ) (1 + 0.00348s )( 1 + 0.00263s )

The Bode diagram is shown on the following page. The following frequency-response attributes are obtained: PM = 55 deg

GM = 12.94 dB

M

r

= 1.11

Bode Plot [parts (b), (c), and (d)]

268

10-39 (a) Two-stage Phase-lead Controller Design. The uncompensated system is unstable. PM = −43.25 deg and GM = −18.66 dB. With a single-stage phase-lead controller, the maximum phase margin that can be realized affectively is 12 degrees. Setting the desired PM at 11 deg, we have the parameters of the single-stage phaselead controller as a = 128.2 and T = 0 . 00472 . The transfer function of the single-stage controller 1

G

is

c1

( s)

=

1

+ aT 1 s

1

+ T1 s

1

=

+ 0 . 6057

s

1 + 0 . 00472

s

Starting with the single-stage-controller compensated system, the second stage of the phase-lead controller is designed to realize a phase margin of 60 degrees. The parameters of the second-stage controller are: b = 16.1 and T = 0 . 0066 . Thus, 2

G

c2

(s)

=

1 + bT s 2

1+T s

=

2

G (s) c

= G c 1 ( s )G c 2 ( s ) =

1 + 0 .106 s 1 + 0 . 0066 s

1 + 0 . 6057 s

1

+ 0 .106

s

1 + 0 . 00472 s 1 + 0 . 0066 s

Forward-path Transfer Function:

269

1,236,598.6 ( s + 1.651 )( s + 9.39 )

G ( s) = G c1 ( s ) G c2 ( s ) G p ( s) =

s ( s + 2)( s + 5 ) ( s + 211.86 ) ( s + 151.5 )

Attributes of the frequency response of the compensated system are: GM = 19.1 dB

M

PM = 60 deg

= 1. 08

r

The unit-step response is plotted below. The time-response attributes are: t

Maximum overshoot = 10.2%

s

= 0 .1212

t

sec

r

= 0 . 037

sec

(b) Single-stage Phase-lag Controller Design. With a single-stage phase-lag controller, for a phase margin of 60 degrees, a = 0.0108 and T = 1483.8. The controller transfer function is 1 + 16 . 08 s G (s) = c 1 + 1483 . 8 s The forward-path transfer function is G (s)

= G c ( s )G p ( s ) =

β+ γ β+ γβ+ γβ+ γ 6 .5 s

s s

s

2

0 . 0662

5

s

0 . 000674

From the Bode plot, the following frequency-response attributes are obtained: PM = 60 deg

M

GM = 20.27 dB

r

= 1. 09

The unit-step response has a long rise time and settling time. The attributes are: Maximum overshoot = 12.5%

t

s

= 12 . 6

t

sec

r

=

2 .126 sec

(c) Lead-lag Controller Design. For the lead-lag controller, we first design the phase-lag portion for a 40-degree phase margin. The result is a = 0.0238 and T = 350 . The transfer function of the controller is 1

G

c1

( s)

=

(s)

=

1

+ 8 . 333

s

1 + 350 s The phase-lead portion is designed to yield a total phase margin of 60 degrees. The result is b = 4.8 and T = 0 .2245 . The transfer function of the phase-lead controller is 2

G

c2

1

+ 1.076

s

1 + 0 .2245 s The forward-path transfer function of the lead-lag compensated system is

G ( s) = Frequency-response attributes: Unit-step response attributes:

68.63 ( s + 0.12 )( s + 0.929 )

s ( s + 2 ) ( s + 5 ) ( s + 0.00286 ) ( s + 4.454 ) PM = 60 deg GM = 13.07 dB M = 1. 05 BW = 3.83 rad/sec r Maximum overshoot = 5.9%

270

t

s

= 1. 512

sec

t

r

= 0 . 7882

sec

Unit-step Responses.

10-40 (a)

The uncompensated system has the following frequency-domain attributes: PM = 3.87 deg The Bode plot of G

p

M

GM = 1 dB

r

= 7 .73

( j ω ) shows that the phase curve drops off sharply, so that the phase-lead

controller would not be very effective. Consider a single-stage phase-lag controller. The phase margin of 60 degrees is realized if the gain crossover is moved from 2.8 rad/sec to 0.8 rad/sec. The attenuation of the phase-lag controller at high frequencies is approximately −15 dB. Choosing an attenuation of

−17.5 dB, 20 log

we calculate the value of a from a = −17 . 5 dB Thus a = 0.1334

10

The upper corner frequency of the phase-lag controller is chosen to be at 1/aT = 0.064 rad/sec. Thus, 1/T = 0.00854 or T = 117.13. The transfer function of the phase-lag controller is 1 + 15 .63 s G (s) = c 1 + 117 .13 The forward-path transfer function is

G ( s) = G c ( s ) G p ( s) =

5 ( 1 + 15.63 s ) (1 − 0.05 s )

s (1 + 0.1 s ) (1 + 0.5 s ) (1 + 117.13 s ) (1 + 0.05 s )

From the Bode plot of G ( jω ) , the following frequency-domain attributes are obtained: PM = 60 deg

GM = 18.2 dB

M

r

= 1. 08

The unit-step response attributes are: maximum overshoot = 10.7%

t

s

= 10 .1

sec

t

Bode Plots

271

r

= 2 .186

sec

10-40 (b)

Using the exact expression of the time delay, the same design holds. The time and frequency domain attributes are not much affected.

272

10-41 (a) Uncompensated System. Forward-path Transfer Function: G ( s) = The Bode plot of G ( jω ) is shown below.

10

(1 + s ) (1 + 10 s )( 1 + 2s ) (1 + 5s )

The performance attributes are: PM = −10.64 deg The uncompensated system is unstable.

GM =

−2.26 dB

(b) PI Controller Design.

(

10 K p s + K I

Forward-path Transfer Function: G ( s) = Ramp-error Constant: K v = lim sG s →0

G ( s) =

)

s (1 + s ) (1 + 10 s )( 1 + 2s ) (1 + 5s ) ( s ) = 10 K I = 0 .1 Thu s K I = 0 . 01 0.1 (1 + 100K P s )

s (1 + s ) (1 + 10 s )( 1 + 2s ) (1 + 5s )

The following frequency-domain attributes are obtained for various values of K

P

.

KP

PM (deg)

GM (dB)

Mr

0.01 0.02 0.05 0.10 0.12 0.15 0.16 0.17 0.20

24.5 28.24 38.84 50.63 52.87 53.28 52.83 51.75 49.08

5.92 7.43 11.76 12.80 12.23 11.22 10.88 10.38 9.58

2.54 2.15 1.52 1.17 1.13 1.14 1.16 1.18 1.29

The phase margin is maximum at 53.28 degrees when K

P

= 0 .15 .

The forward-path transfer function of the compensated system is

G ( s) =

0.1 (1 + 15 s )

s (1 + s ) (1 + 10 s )( 1 + 5 s ) ( 1 + 2 s )

The attributes of the frequency response are: PM = 53.28 deg

M

GM = 11.22 dB

= 1.14

r

The attributes of the unit-step response are: Maximum overshoot = 14.1%

t

r

= 10 . 68

sec

Bode Plots

273

t

s

= 48

sec

BW (rad/sec) 0.13 0.13 0.14 0.17 0.18 0.21 0.22 0.23 0.25

10-41 (c) Time-domain Design of PI Controller.

274

By setting K

= 0 . 01

I

and varying K

we found that the value of K

P

P

that minimizes the maximum

overshoot of the unit-step response is also 0.15. Thus, the unit-step response obtained in part (b) is still applicable for this case.

10-42 Closed-loop System Transfer Function. Y (s)

1

=

s + ( 4 + k 3 ) s + ( 3 + k 2 + k3 ) s + k1 3

R (s )

2

For zero steady-state error to a step input, k

=1.

1 2

we divide the characteristic polynomial by s

s + ( 2 + k2 )

For the complex roots to be located at −1 +j and −1

+2s+ 2

− j,

and solve for zero remainder.

s + 2 s + 2 s + ( 4 + k3 ) s + ( 3 + k2 + k3 ) s + 1 2

3

2

s + 3

2s

+ 2s

2

( 2 + k ) s + (1 + k + k ) s + 1 ( 2 + k ) s + ( 4 + 2k ) s + 4 + 2k 2

3

2

3

2

3

3

( -3+k

2

− k 3 ) s − 3 − 2 k3

− 3 − 2k3 = 0

For zero remainder,

−0.5.

k

Thus

−3 + k 2 − k 3 = 0

The third root is at steady-state error.

3

3

= −1. 5 k

Thus

2

= 1. 5

Not all the roots can be arbitrarily assigned, due to the requirement on the

10-43 (a) Open-loop Transfer Function. G ( s) =

X 1 (s )

=

E (s )

k3 s  s + ( 4 + k 2 ) s + 3 + k 1 + k 2  2

Since the system is type 1, the steady-state error due to a step input is zero for all values of k , k , and k 1

2

3

that correspond to a stable system. The characteristic equation of the closed-loop system is

s + ( 4 + k2 ) s + ( 3 + k1 +k 2 )s + k 3 = 0 3

For the roots to be at

−1 + j, −1 − j,

2

and

−10, the equation should be: 3 2 s + 12 s + 22 s + 20 = 0

Equating like coefficients of the last two equations, we have 4 + k = 12 2

3

+ k 1 + k 2 = 22 k

3

= 20

k

Thus

k

Thus

(b) Open-loop Transfer Function. Y (s ) Gc ( s) 20 = = 2 E ( s ) ( s + 1) ( s + 3 ) s s + 12 s + 22

(

Thus

)

10-44 (a)

275

Thus

2 1

=8 = 11

k3 = 20

Gc ( s) =

20 ( s + 1 ) ( s + 3 )

(

s s + 12 s + 22 2

)

 0  25.92 ∗ A =  0  −2.36 

0

1

0

0

0

0

0

0

1

0

0

0

 0   −0.0732  ∗  B =  0   0.0976   

 

The feedback gains, from k 1 to k 4 :

−2.4071E+03

−4.3631E+02

The A 0.0000E+00

−1.5028E+02

− B ∗K

0.0000E+00 2/3258E+02

−1.0182E+02

matrix of the closed-loop system

1.0000E+00

−3.1938E+01

−8.4852E+01

0.0000E+00 0.0000E+00 −7.4534E+00 0.0000E+00 1.0000E+00 8.2816E+00 9.9379E+00

−6.2112E+00

0.0000E+00 4.2584E+01

The B vector 0.0000E+00

−7.3200E−02

0.0000E+00 9.7600E−02

Time Responses:

10-44 (b)

The feedback gains, from k 1 to k 2 :

−9.9238E+03 The A

−1.6872E+03 ∗

− B ∗K

−1.3576E+03

matrix of the closed-loop system

276

−8.1458E+02

0.0000E+00

−7.0051E+02

1.0000E+00

−1.2350E+02

0.0000E+00 9.6621E+02

0.0000E+00 0.0000E+00 −5.9627E+01 0.0000E+00 1.0000E+00 1.3251E+02 7.9503E+01

−9.9379E+01

0.0000E+00 1.6467E+02

The B vector 0.0000E+00

−7.3200E−02

0.0000E+00 9.7600E−02

Time Responses:

10-45

The solutions are obtained by using the pole-placement design option of the linsys program in the ACSP software package.

(a)

The feedback gains, from k 1 to k 2 :

−6.4840E+01

−5.6067E+00

The A 0.0000E+00

−3.0950E+02

0.0000E+00

−4.6190E+02

− B ∗K

2.0341E+01

2.2708E+00

matrix of the closed-loop system

1.0000E+00 0.0000E+00 0.0000E+00 1.1463E+02 1.4874E+01 0.0000E+00 0.0000E+00 1.0000E+00 −3.6724E+01 1.7043E+02 1.477eE+01

−3.6774E+01

The B vector 0.0000E+00

−6.5500E+00

0.0000E+00

277

−6.5500E+00 (b) Time Responses: ∆x ( 0 ) =

0 .1

0

0

0

∆x ( 0 ) =

0 .1

0

0

0 , the initial position of

With the initial states

'

'

∆x 1 or ∆y1 is preturbed downward ∆x 3 = ∆y 2 is

from its stable equilibrium position. The steel ball is initially pulled toward the magnet, so

negative at first. Finally, the feedback control pulls both bodies back to the equilibrium position. With the initial states

∆x ( 0 ) =

0

0

0 .1

'

0 , the initial position of

∆x 3 or ∆y 2

is preturbed

downward from its stable equilibrium. For t > 0, the ball is going to be attracted up by the magnet toward the equilibrium position. The magnet will initially be attracted toward the fixed iron plate, and then settles to the stable equilibrium position. Since the steel ball has a small mass, it will move more actively.

10-46 (a) Block Diagram of System.

278

u = −k1 x1 + k 2 ∫ ( − x1 + w1 ) dt State Equations of Closed-loop System:

 dx1   dt   −2 − k1 1   x1   0  =   x  +  k dx − k 0  2 2 2  2  dt 

1   w1  0   w2 

Characteristic Equation:

sI − A = For s =

−10, −10, X (s)

sI

=

X ( s) =

−1

k2

s

−A =

X 1 (s)

W1 ( s )

s + 2 + k1

=

(

=

s2

1+ 2 s

s

−1

−2

+ 18

W2 (s)

=

200 + W2s

s s + 20 s + 200 2

2

+ 20 s + 200 = 0

200 W 1 ( s ) s

1

= s + ( 2 + k1 ) s + k2 = 0

)

s

+ s −1W 2 ( s ) −1

W2 s

Thus

+ 200

s

k1

= 18

200 W 1 ( s )

−2

=

=

const ant

W2

s

2

k2

and

+ sW 2 ( s )

+ 20 s + 200

lim x( t) = lim sX ( s) = 1 t →∞

10-46 (b) With PI C ontroller: Block Diagram of System:

279

s →0

= 200

Set K P

=

2 and

KI

= 200 .

X (s ) =

(K

P

s + K I ) W1 ( s ) + sW2 ( s ) s + 20 s + 200 2

Time Responses:

280

=

( 2s + 200 ) W ( s ) + sW ( s ) 1

s + 20 s + 200 2

2

Chapter 11

THE VIRTUAL LAB

Part 1) Solution to Lab questions within Chapter 11 11-5-1 Open Loop Speed 1. Open loop speed response using SIMLab: a. +5 V input:

The form of response is like the one that we expected; a second order system response with overshoot and oscilla tion. Considering an amplifier gain of 2 and K b = 0.1 , the desired set point should be set to 2.5 and as seen in the figure, the final value is approximately 50 rad/sec which is armature voltage divided by K b . To find the above response the systems parameters are extracted from 11-3-1 of the text and B is calculated from 11-3 by having τ m as:

τm =

Ra J m , Ra B + k b k m

B=

Ra J m − k b k m τ m = 0.000792kg ⋅ m 2 / sec Ra τ m

b. +15 V input:

282

c. –10 V input:

2. Study of the effect of viscous friction:

283

The above figure is plotted for three different friction coefficients (0, 0.001, 0.005) for 5 V armature input. As seen in figure, two important effects are observed as the viscous coefficient is increased. First, the final steady state velocity is decreased and second the response has less oscillation. Both of these effects could be predicted from Eq. (11.1) by increasing damping ratio ζ.

284

As the overall inertia of the system is increased by 0.005 / 5.2 2 and becomes 1.8493 × 10 −3 kg.m2 , the mechanical time constant is substantially increased and we can assume the first order model for the motor (ignoring the electrical sub-system) and as a result of this the response is more like an exponential form. The above results are plotted for 5 V armature input.

285

4. Reduce the speed by 50% by increasing viscous friction:

As seen in above figure, if we set B=0.0075 N.s/m the output speed drop by half comparing with the case that B=0 N.s/m. The above results are plotted for 5 V armature input.

286

5. Study of the effect of disturbance:

Repeating experiment 3 for B=0.001 N.s/m and T L =0.05 N.m result in above figure. As seen, the effect of disturbance on the speed of open loop system is like the effect of higher viscous friction and caused to decrease the steady state value of speed.

287

6. Using speed response to estimate motor and load inertia:

Using first order model we are able to identify system parameters based on unit step response of the system. In above plot we repeated the experiments 3 with B=0.001 and set point voltage equal to 1 V. The final value of the speed can be read from the curve and it is 8.8, using the definition of system time constant and the cursor we read 63.2% of speed final value 5.57 occurs at 0.22 sec, which is the system time constant. Considering Eq. (11-3), and using the given value for the rest of parameters, the inertia of the motor and load can be calculated as:

J=

τ m ( Ra B + K m K b ) 0.22(1.35 × 0.001 + 0.01) = = 1.8496 × 10 − 3 kg.m2 Ra 1.35

We also can use the open loop speed response to estimate B by letting the speed to coast down when it gets to the steady state situation and then measuring the required time to get to zero speed. Based on this time and energy conservation principle and knowing the rest of parameters we are able to calculate B. However, this method of identification gives us limited information about the system parameters and we need to measure some parameters directly from motor such as Ra , K m , K b and so on. So far, no current or voltage saturation limit is considered for all simulations using SIMLab software.

288

7. Open loop speed response using Virtual Lab: a. +5 V:

289

b. +15 V:

c. –10 V:

290

Comparing these results with the part 1, the final values are approximately the same but the shape of responses is closed to the first order system behavior. Then the system time constant is obviously different and it can be identified from open loop response. The effect of nonlinearities such as saturation can be seen in +15 V input with appearing a straight line at the beginning of the response and also the effects of noise and friction on the response can be observed in above curves by reducing input voltage for example, the following response is plotted for a 0.1 V step input:

291

8. Identifying the system based on open loop response:

Open loop response of the motor to a unit step input voltage is plotted in above figure. Using the definition of time constant and final value of the system, a first order model can be found as:

G( s ) =

9 , 0.23s + 1

where the time constant (0.23) is found at 5.68 rad/sec (63.2% of the final value).

292

11-5-2 Open Loop Sine Input 9. Sine input to SIMLab and Virtual Lab (1 V. amplitude, and 0.5, 5, and 50 rad/sec frequencies) a. 0.5 rad/sec (SIMLab):

293

294

295

10. Sine input to SIMLab and Virtual Lab (20 V. amplitude) a. 0.5 rad/sec (SIMLab):

296

297

298

In both experiments 9 and 10, no saturation considered for voltage and current in SIMLab software. If we use the calculation of phase and magnitude in both SIMLab and Virtual Lab we will find that as input frequency increases the magnitude of the output decreases and phase lag increases. Because of existing saturations this phenomenon is more sever in the Virtual Lab experiment (10.f). In this experiments we observe that M = 0.288 and ϕ = −93.82 o for ω = 50.

299

11-5-3 Speed Control 11. Apply step inputs (SIMLab) In this section no saturation is considered either for current or for voltage.

a. +5 V:

300

b. +15 V:

c. -10 V:

301

b. +15 V:

302

c. -10 V:

13. Study of the effect of viscous friction:

303

As seen in above figure, two different values for B are selected, zero and 0.0075. We could change the final speed by 50% in open loop system. The same values selected for closed loop speed control but as seen in the figure the final value of speeds stayed the same for both cases. It means that closed loop system is robust against changing in system’s parameters. For this case, the gain of proportional controller and speed set point are 10 and 5 rad/sec, respectively.

14. Study of the effect of disturbance:

Repeating part 5 in section 11-5-1 for B=0.001 and T L =0.05 N.m result in above figure. As seen, the effect of disturbance on the speed of closed loop system is not substantial like the one on the open loop system in part 5, and again it is shown the robustness of closed loop system against disturbance. Also, to study the effects of conversion factor see below figure, which is plotted for two different C.F. and the set point is 5 V.

304

By decreasing the C.F. from 1 to 0.2, the final va lue of the speed increases by a factor of 5.

15. Apply step inputs (Virtual Lab) a. +5 V:

305

b. +15 V:

c. –10 V:

306

As seen the responses of Virtual Lab software, they are clearly different from the same results of SIMLab software. The nonlinearities such as friction and saturation cause these differences. For example, the chattering phenomenon and flatness of the response at the beginning can be considered as some results of nonlinear elements in Virtual Lab software.

11-5-4 Position Control 16. 160 o step input (SIMLab)

307

17. –0.1 N.m step disturbance

18. Examine the effect of integral control

308

In above figure, an integral gain of 1 is considered for all curves. Comparing this plot with the previous one without integral gain, results in less steady state error for the case of controller with integral part.

309

20. Set B=0:

21. Study the effect of saturation

310

The above figure is obtained in the same conditions of part 20 but in this case we considered ± 10 V. and ± 4 A. as the saturation values for voltage and current, respectively. As seen in the figure, for higher proportional gains the effect of saturations appears by reducing the frequency and damping property of the system.

22. Comments on Eq. 11-13 After neglecting of electrical time constant, the second order closed loop transfer function of position control obtained in Eq. 11-13. In experiments 19 through 21 we observe an under damp response of a second order system. According to the equation, as the proportional gain increases, the damped frequency must be increased and this fact is verified in experiments 19 through 21. Experiments16 through 18 exhibits an over damped second order system responses.

23. In following, we repeat parts 16 and 18 using Virtual Lab:

Study the effect of integral gain of 5:

311

312

Ch. 11 Problem Solutions Part 2) Solution to Problems in Chapter 11 11-1. In order to find the current of the motor, the motor constant has to be separated from the electrical component of the motor.

The response of the motor when 5V of step input is applied is:

a) The steady state speed: 41.67rad/sec b) It takes 0.0678 second to reach 63% of the steady state speed (26.25rad/sec). This is the time constant of the motor. c) The maximum current: 2.228A 313

11.2

The steady state speed at 5V step input is 50rad/sec. a) It takes 0.0797 seconds to reach 63% of the steady state speed (31.5rad/sec). b) The maximum current: 2.226A c) 100rad/sec

314

11-3

a) b) c) d)

50rad/sec 0.0795 seconds 2.5A. The current When Jm is increased by a factor of 2, it takes 0.159 seconds to reach 63% of its steady state speed, which is exactly twice the original time period . This means that the time constant has been doubled.

315

11-4 Part 1: Repeat problem 11-1 with TL = -0.1Nm

a) It changes from 41.67 rad/sec to 25 rad/sec. b) First, the speed of 63% of the steady state has to be calculated. 41.67 - (41.67 - 25) × 0.63 = 31.17 rad/sec. The motor achieves this speed 0.0629 seconds after the load torque is applied c) 2.228A. It does not change Part 2: Repeat problem 11-2 with TL = -0.1Nm a) It changes from 50 rad/sec to 30 rad/sec. b) The speed of 63% of the steady state becomes 50 - (50 - 30) × 0.63 = 37.4 rad/sec. The motor achieves this speed 0.0756 seconds after the load torque is applied c) 2.226A. It does not change.

316

Part 3: Repeat problem 11-3 with TL = -0.1Nm

a) It changes from 50 rad/sec to 30 rad/sec. b) 50 - (50 - 30) × 0.63 = 37.4 rad/s The motor achieves this speed 0.0795 seconds after the load torque is applied. This is the same as problem 11-3. c) 2.5A. It does not change d) As TL increases in magnitude, the steady state velocity decreases and steady state current increases; however, the time constant does not change in all three cases.

317

11-5 The steady state speed is 4.716 rad/sec when the amplifier input voltage is 5V:

11-6

a) 6.25 rad/sec. b) 63% of the steady state speed: 6.25 × 0.63 = 3.938 rad/sec It takes 0.0249 seconds to reach 63% of its steady state speed. c) The maximum current drawn by the motor is 1 Ampere.

318

11-7 a) 9.434 rad/sec. b) 63% of the steady state speed: 9.434 × 0.63 = 5.943 rad/sec It takes 0.00375 seconds to reach 63% of its steady state speed. c) The maximum current drawn by the motor is 10 Amperes. d) When there is no saturation, higher Kp value reduces the steady state error and decreases the rise time. If there is saturation, the rise time does not decrease as much as it without saturation. Also, if there is saturation and Kp value is too high, chattering phenomenon may appear. 11-8

a) The steady state becomes zero. The torque generated by the motor is 0.1 Nm. b) 6.25 - (6.25 - 0) × 0.63 = 2.31 rad/sec. It takes 0.0249 seconds to reach 63% of its new steady state speed. It is the same time period to reach 63% of its steady state speed without the load torque (compare with the answer for the Problem 11-6 b). 11-9 The SIMLab model becomes

The sensor gain and the speed input are reduced by a factor of 5. In order to get the same result as Proble m 11-6, the Kp value has to increase by a factor of 5. Therefore, Kp = 0.5. The following graphs illustrate the speed and current when the input is 2 rad/sec and Kp = 0.5. 319

11-10

320

11-11

a) The steady state position is very close to 1 radian. b) 1.377 radians. c) 0.148 seconds. It has less steady state error and a faster rise time than Problem 11-10, but has larger overshoot.

321

11-12 Different proportional gains and the ir corresponding responses are shown on the following graph.

As the proportional gain gets higher, the motor has a faster response time and lower steady state error, but if it the gain is too high, the motor overshoot increases. If the system requires that there be no overshoot, Kp = 0.2 is the best value. If the system allows for overshoot, the best proportional gain is dependant on how much overshoot the system can have. For instance, if the system allows for a 30% overshoot, Kp = 1 is the best value.

322

11-13 Let Kp = 1 is the best value.

As the derivative gain increases, overshoot decreases, but rise time increases.

323

11-14

324

11-15 There could be many possible answers for this problem. One possible answer would be Kp = 100 Ki= 10 Kd= 1.4

The Percent Overshoot in this case is 3.8%.

325

11-16 0.1 Hz

0.2 Hz

326

0.5 Hz

1 Hz

327

2Hz

5Hz

328

10Hz

50Hz

329

As frequency increases, the phase shift of the input and output also increase. Also, the amplitude of the output starts to decrease when the frequency increases above 0.5Hz. 11-17

As proportional gain increases, the steady state error decreases.

330

11-18

Considering fast response time and low overshoot, Kp =1 is considered to be the best value.

331

11-19 It was found that the best Kp = 1

As Kd value increases, the overshoot decreases and the rise time increases.

332

Appendix H

H-1 (a) L( s )

5( s

=

s(s

GENERAL NYQUIST CRITERION

− 2)

+ 1 )( s − 1 )

When ω = 0 : ∠L ( j 0 ) = −90

o

L ( jω ) =

5( jω − 2)

(

− jω 1 + ω

The Nyquist plot of

2

)

=

P

L( j0 ) o

Whenω = ∞: ∠L ( j ∞) = −180

=1

=∞

L ( j∞)

− 5(ω + 2 j )

(

ω 1+ω

L ( jω )

2

=1

)

=0

L ( jω )

When

= 0 , ω = ∞.

does not intersect the real axis except at the origin.

Φ 11 = ( Z − 0.5 Pω − P )180 = ( Z − 1.5 ) 180 = − 90 o

o

o

Thus, Z=1.

The closed-loop system is unstable. The characteristic equation has 1 root in the right-half s-plane. Nyquist Plot of

(b)

L ( jω )

50

= s( s

+ 5)( s − 1 )

L ( jω ):

When ω = 0 : ∠L ( j 0 ) = 90

=1 o

When ω = ∞: ∠L ( j ∞) = −270

P

L( j0) o

=1

=∞

L ( j∞)

333

=0

50

L ( jω ) =

2

(

−4ω − j ω 5 + ω 2

(

50  −4ω + jω 5 + ω

2

)

=

16ω + ω 4

2

2

(5+ ω ) 2

)

2

For Im L ( jω ) = 0 , ω = ∞. Thus, the Nyquist plot of L(s) intersects the real axis only at the origin where ω = ∞. Φ 11 = ( Z − 0.5 Pω − P )180 = ( Z − 1.5 ) 180 = 90 o

o

o

Thus, Z = 2

The closed-loop system is unstable. The characteristic equation has two roots in the right-half s-plane. The Nyquist plot of

(c) L ( s ) =

(

3( s + 2)

s s + 3s + 1 3

L ( jω ):

Pω = 1

)

When ω = 0 : ∠L ( j 0 ) = −90

o

When ω = ∞: ∠L ( j ∞) = −270

L ( jω ) =

3( jω + 2) 4

− 3ω

ω − 3ω − 2 = 0 4

2

2

) + jω or

=

P=2

L( j0 ) o

=∞

L ( j∞)

=0

(

3( jω + 2)  ω − 3ω

(4

ω = 3 . 56 2

4

4

− 3ω

2

)

2

2

) − jω 

Φ 11 = ( Z − 0.5 Pω − P )180 = ( Z − 2.5 ) 180 = − 90 o

o

Setting

2

o

L ( j 1.89 )

Im L ( jω )

= 0,

=3

Thus, Z = 2

The closed-loop system is unstable. The characteristic equation has two roots in the right-half s-plane.

334

Nyquist Plot of

(d) L ( s ) =

100

(

s ( s + 1) s + 2 2

L ( jω ):

Pω = 3

)

When ω = 0 : ∠L ( j 0 ) = −90

o

P=0 L( j0 )

=∞

o

When ω = ∞: ∠L ( j ∞) = −360 L ( j∞) = 0 The phase of L ( jω ) is discontinuous at ω = 1.414 rad/sec.

(

Φ11 = 35.27 + 270 − 215.27 o

o

o

) = 90

o

Φ11 = ( Z − 1.5) 180 = 90 o

o

Thus,

P

11

=

360 180

o o

=

2

The closed-loop system is unstable. The characteristic equation has two roots in the right-half s-plane. Nyquist Plot of

L ( jω ):

335

s − 5s + 2 2

(e)

L(s ) =

(

s s + 2 s + 2 s + 10 3

2

When ω = 0 : ∠L ( j 0 ) = −90

o

When ω = ∞: ∠L ( j ∞) = −180

)

Pω = 1

L( j0 )

=∞

o

L ( j∞)

P=2

=0

( 2 − ω ) − j5ω  ( ω − 2 ω ) − jω (10 − 2ω )  ( 2 − ω ) − j5ω L ( jω ) = = ( ω − 2ω ) + jω (10 − 2ω ) (ω − 2ω ) + ω (10 − 2ω ) 2

4

2

2

2

4

( 2 − ω )(10 − 2ω ) + 5( ω

Setting L( jω) = 0,

4

2

2

4

− 2ω

or ω − 3 .43ω + 2 .86 = 0 2 Thus, ω = 1.43 o r 2.0 1. ω = ±1.2 rad / sec or L ( j1.2 ) = − 0 . 7 L ( j1.42 ) = −0 . 34 4

2

2

2

2

2

2

2

2

)= 0

2

Φ 11 = ( Z − 0.5 Pω − P )180 = ( Z − 2.5 )180 = −90 o

o

o

/ sec

Thus, Z = 2 .

The closed-loop system is unstable. The characteristic equation has two roots in the right-half s-plane. L ( jω ):

Nyquist Plot of

(f)

L(s ) =

(

) ( s + 2 ) = −0.1s − 0.2 s + 0.1s + 0.2 s ( s + s + 1) s ( s + s + 1)

− s −1 2

3

2

2

When ω = 0 : ∠L ( j 0 ) = −90

2

o

When ω = ∞: ∠L ( j∞ ) = 180

L( j0 ) o

=∞

L( j∞)

336

= 0 .1

Pω = 1

P=0

( 0.2 + 0.2ω ) + 0.1 jω ( 1 + ω ) = (1+ ω ) ( 0.2 + j 0.1ω ) − ω − jω ( 1− ω )  L ( jω ) = −ω + jω (1 − ω ) ω + ω ( 1− ω ) 2

2

2

2

2

2

4

Setting Im L ( jω ) = 0 , ω = 2 . 2

2

Thus,

Φ 11 = ( Z − 0.5 Pω − P )180 = ( Z − 0.5 ) 180 = − 90 o

2

o

H-2 (a)

L(s ) =

L ( jω ):

K ( s − 2)

(

s s −1 2

)

Pω = 1

P =1

For stability, Z = 0. Φ 11 = − ( 0.5 Pω + P ) 180 = −270 o

o

For K > 0, Φ 11 = −90 . For K < 0, Φ 11 = +90

o

≠ −270

o

The system is unstable. o

The system is unstable.

Thus the system is unstable for all K.

337

2

L ( j 1.414 )

Thus, Z = 0

o

The closed-loop system is stable. Nyquist Plot of

2

= − 0 .3

Nyquist plot

(b)

L( s )

K

= s(s

+ 10

)( s

P

ω

− 2)

=1

P

=1

For stability, Z = 0. Φ 11 = − ( 0.5 Pω + P ) 180 = −1.5 × 180 = − 270 o

For K > 0, Φ 11 = 90

o

For K < 0, Φ 11 = −90 Nyquist Plot of

o

The system is unstable.

. o

o

≠ −270

o

.

The system is unstable for all values of K .

L ( jω ):

338

(c) L ( s ) =

(

K ( s + 1)

s s + 3s + 1 3

P

)

ω

=1

P

=

2

For stability, Z = 0. Φ 11 = − ( 0.5 Pω + P− 1 ) 180 = − 450 o

For K > 0, Φ 11 = −90

o

For K < 0, Φ 11 = +90

o

o

The system is unstable.

.

when K > −1.

Φ11 = −270

o

when K < −1.

The Nyquist plot of L ( jω ) crosses the real axis at K, and the phase crossover frequency is 1.817 rad/sec. The system is unstable for all values of K. Nyquist Plot of

(d) L ( s ) =

(

L ( jω ):

K s − 5s + 2

(

2

)

s s + 2 s + 2 s + 10 3

2

P

)

ω

=1

P

=

2

For stability, Z = 0. Φ 11 = − ( 0.5 Pω + P ) 180 = −2.5 × 180 = − 450 o

The Nyquist plot of

L ( jω )

o

o

intersects the real axis at the following points:

ω = ∞: L ( j 0 ) = 0 ω = 1.42 rad / sec: L ( j 1.42 ) = − 0 . 34 ω = 1.2 rad / sec: L ( j 1.4 ) = −0 . 7 K

K

339

0

<

K

1.43

<

2 . 94

0

+ 5 K − 2 .5 > 0 + 5 .45)(

K

− 0 .458

)

>0

For system stability, K > 0.458.

(b) s 3 + K ( s 3 + 2s 2 +1 ) = 0 Leq ( s) =

(

)

K s + 2s + 1 3

s

When ω = 0:

∠L eq ( j 0 ) = − 270

When ω = ∞: Leq ( jω ) = 1

− 2ω ω

Pω = 3

3

o

∠L eq ( j ∞ ) = 0

(

K  1 − 2ω

2

− jω

) − jω

3

o

P =0 L

L

eq

eq

=∞

( j0 )

( j ∞)

=K

 K ω 3 + j ( 1 − 2ω 2 ) = 3

3

=0

ω = ∞ , ω = ±0 . 707 rad/sec

always equal to K for all values of ω . For stability, Φ = − ( 0.5 P + P ) 180 = −270 o

o

ω

11

Φ 11 = −90

K>0 K < 0 Φ 11 = 90

o

o

Unstable

Unstable

The system is unstable for all values of K. Routh Tabulation: s

3

1+ K

0

s

2

2K

K

s

1

−1 − K

K

>0

K

< −1

K

>0

2K o

K

Im L

eq

( jω )

=0

2

L eq ( j 0 . 707 )

The Nyquist plot is a straight line, since the equation of

s

Setting

ω

3

The system is unstable for all values of K .

345

=K

L eq ( j ω )

shows that its real part is

(c) s ( s + 1) ( s2 + 4) + K ( s 2 + 1) = 0 Leq ( s) =

(

K s +1 2

(

)

Pω = 3

)

s ( s + 1) s + 4 2

P =0

For stability, Φ 11 = − ( 0.5 Pω + P ) 180 = −270 o

<

0

K

0

> −3

K

Φ 11 = −90

o

Φ 11 = +90

o

Φ11 = −270

o

Unstable Unstable o

Unstable

The system is unstable for all values of K .

Routh Tabulation

s

4

s

3

s

2

s

1

1

3

ε

3 + K

−3− K ε

−3 − K

−( 3 + K ) − 2 K ε −( 3 + K ) s

o

ε

2 K

2K

K

< −3

2

≅1

2K

K >0

The conditions contradict. The system is unstable for all values of K.

347

= 0 ∠ − 270

o

Appendix I

I-1

DISCRETE-DATA CONTROL SYSTEMS

(a)

(b) F (z) =

ze−3

(z−e ) −3

F (z ) =

2

(c)

(d) z z   F (z ) = − − (2− jω ) − (2+ jω )   2 j  z −e z −e  1

I-2

−2 j  ze 2 j  ze −   2 −2 j 2 2 j ( z − e 2 j ) z − e ( )  

1

F (z ) =

(z + e ) (z − e ) −2

ze

−2

3

−2

(a) F (z ) =

− j 2T  ze j 2 T  ze −   2 2 2 j  ( z − e j 2T ) ( z − e − j 2T )  

T

(b) F (z)

−1

=1 − z

+z

−2

−3

−z

+z

−4

−L

−1

F (z )

(1 + z ) F ( z ) = 1

=

z

−1

Add Eqs. (1) and (2). We have

I-3

z

(1)

−1

−z

−2

+z

F (z)

Or,

−3

−z

=z

−4

(z

+L

+ 1)

(a) Z

2 −5 T 2 −10T T e z T e z  1  ( −1) 2 ∂ 2  z  = = + 2 −aT 2  ( s + 5) 3  2 ! ∂ a  z − e  a = 5 2 ( z − e −5T ) ( z − e −5T )3

(b) F (s)

F (z)

1

= =

3

s (s z z

+ 1)

−1

=

1

s

Tz (z

1 s

+

− 1)

2

=

10

1

+

2

T

2

3

s

z(z

2( z

1

s

+ 1)

− 1)

3

+1 z

z

−T

−e

(c) F (s)

10

= s(s

F (z ) =

+ 5)

2

10 z 25( z − 1)

25 s

10

(

25 ( s

10 z

25 z − e

−5 T

+ 5) −

2

− (s

+ 5)

2 Tze

2

−5T

) (z −e ) −5 T

2

(d) F (s ) =

(

5

s s +2 2

)

=

2.5 s

2.5 s

F (z)

s +2 2

348

=

2 .5 z z

−1

2 .5 z ( z z

2

− cos ωT )

− 2 z cos ωT + 1

(2)

I-4 (a) F (z)

=

10 z (z

− 1 )( z − 0 .2 )

=

12 . 5 z z

−1

)

=

(b) F (z ) =

z

( z − 1) ( z

2

+ z +1

f ( k ) = 0.333 − (0.1667 + j 0.2887) e

 2 kπ  3

= 0.333 − 0.333cos 

12 . 5 z

f (k )

− 0 .2

z

= 12 . 5 1 − ( 0 .2 )

k

k

= 0 , 1,

2,

L

( 0.1667 + j 0.2887 ) z ( 0.1667 − j 0.2887) z − 3( z − 1 ) ( z + 0.5 + j0.866) ( z + 0.5 − j0.866 ) z

− j 2 kπ 3

− (0.1667 − j 0.2887)e

 − 0.576sin  2 kπ     3

j 2 kπ 3

 = 0.333 + 0.666cos  2 kπ + 240 0  k = 0,1, 2 , L      3 

(c) F (z)

=

z (z

− 1 )( z + 0 . 85)

=

0 . 541 z z

−1

0 . 541 z z

f (k )

+ 0 .85

= 0 .541

1

− ( −0 . 85)

k

k

= 0 , 1, 2 , L

k

= 0 , 1, 2 , L

(d) F (z)

=

10 (z

− 1 )( z − 0 . 5)

=

20 z

−1

20 z

f (0)

− 0 .5

=0

f (k )

=

20 1

− ( 0 .5)

k −1

I-5 (a)

(

lim f ( k ) = lim 1 − z k →∞

z →1

−1

) F ( z ) = lim z z →1

0.368 2

− 1.364 z + 0.732

Expand F(z) into a power series of k 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15

z

−1

f(k) 0.000000 0.000000 0.368000 0.869952 1.285238 1.484260 1.451736 1.261688 1.026271 0.844277 0.768362 0.798033 0.894075 1.003357 1.082114

. k 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30

349

=1

f(k) 1.109545 1.089310 1.041630 0.991406 0.957803 0.948732 0.960956 0.984269 1.007121 1.021225 1.023736 1.016836 1.005586 0.995292 0.989487

(b) F (z)

=

(

10 z (z

The function 1 − z

− 1 )( z + 1 )

−1

) F ( z) has a pole at z = −1 , so the final-value

theorem cannot be applied. The response f(k) oscillates between 0 and 10 as shown below. k 1 2 3 4 5 6 7 8 9 10

f(k) 0.000000 10.000000 0.000000 10.000000 0.000000 10.000000 0.000000 10.000000 0.000000 10.000000

(c) lim

k→∞

f (k )

= lim

ε− ϕ

z →1

1

z

−1

F (z)

k 11 12 13 14 15 16 17 18 19 20

= lim

z →1

F(z) is expanded into a power series of k 1 2 3 4 5 6 7 8 9 10

z (z z

− 0 . 5)

−1

=

f(k) 0.000000 10.000000 0.000000 10.000000 0.000000 10.000000 0.000000 10.000000 0.000000 10.000000

2

by long division.

f(k) 1.000000 1.500000 1.750000 1.875000 1.937500 1.968750 1.984375 1.992188 1.996094 1.998047

k 11 12 13 14 15 16 17 18 19 20

f(k) 1.999023 1.999512 1.999756 1.999878 1.999939 1.999969 1.999985 1.999992 1.999996 1.999998

(d) F (z)

=

z (z

− 1 )( z − 1. 5)

F (z)

has a pole outside the unit circle.

The final-value theorem cannot be applied to z

−1

k 1 2 3 4 5 6

F ( z ) . F ( z ) is

f (k )

is unbounded.

expanded into a power series of

. f(k) 0.000000 1.000000 2.500000 4.750000 8.125000 13.187500

k 11 12 13 14 15 16

350

f(k) 113.33078 170.995117 257.492676 387.239014 581.858521 873.787781

7 8 9 10

I-6 (a)

x( k

20.781250 32.171875 49.257813 74.886719

17 18 19 20

+ 2 ) − x ( k + 1) + 0 .1 x ( k ) = u s ( k )

x (0)

=

1311.681641 1968.522461 2953.783691 4431.675781

=0.

x (1 )

Taking the z-transform on both

sides, 2

z X (z)

−z

X ( z) =

2

x(0)

− zx ( 1) − zX

( z − 1) ( z

z 2

− z + 0.1

10 z

=

)

z

+ zx ( 0 ) + 0 .1 X ( z ) =

(z)

+

z −1

1.455 z

z

z − 0.1127

−1 11.455z z − 0.8873

Taking the inverse z-

transform, x( k )

(b)

x( k

= 10 + 1.455

( 0 .1127 )

+ 2) − x (k ) = 0

k

= 1,

x (0)

z X ( z ) − z x(0) − zx(1) − X ( z ) = 0 2

− 11 .455

=0

x (1 )

X ( z) =

2

( 0 . 8873)

z

k

k

= 0 , 1, 2 , L

Taking the z-transform on both sides, we have

 kπ    2 

2

k = 0,1,2, L

x ( k ) = cos 

z −1 2

I-7 (a) P (1 )

= (1+ r) P (0) − u

P( 2)

= ( 1 + r ) P ( 1) − u

P (1 )

=

amoun t owed

P(0 )

=

P0

where

=

after

the f irst p eriod.

amoun t borr owed i nitial ly

M P (k

+ 1 ) = (1 + r ) P ( k ) − u

=

u

amoun t paid

each

period

inclu ding p rincip al and

inter est.

(b) By direct substitution, P (2)

= ( 1 + r ) P (1 ) − u = (1 + r )

2

P (0)

−(1+ r)u − u

P ( 3)

= (1 + r ) P ( 2 ) − u = ( 1 + r )

3

P (0)

− (1 + r )

P (N )

= (1 + r )

N

P (0)

−u

M

(1+ r)

N −1

+ (1 + r )

2

− (1 + r ) u − u

u

N −2

+ L + (1 + r ) + 1 = 0

Solving for u from the last equation, we have u

=

(1+ r) N −1

N

P

∑ (1 + r )

0

=

(1 (1

k

+r )

N

+ r)

N

P r 0

−1

k =0

(c) Taking the z-transform on both sides of the difference equation, we have ∞

∑ k =0

P( k

+ 1) z

−k

= (1 + r )

∑ P( k ) z k =0

351

−k

uz z

−1

P( k

+ 1 ) = (1 + r ) P ( k ) − u ,

zP ( z ) − zP

Or,

= (1 + r ) P ( z ) −

0

uz

−1

z

P (z)

Thus,

=

zP z

0

−(1+ r)

uz (z

− 1)

z

− (1 + r )

Taking the inverse z-transform and setting k = N, we have N

= P 0 (1 + r )

P (N )

+

u r

u (1 r

+ r)

N

=0

Solving for u from the last equation, we have u

(1

=

(1

(d) For u

P

0

=

= \$15 (1

+r )

N

+ r)

N

(1

I-8

= 0 . 01 ,

,000 , r P0 r

−1

=

48

N

+ r)

N

= 48

and N

( 1. 01 )

+ r)

P0 r

−1

month s,

( 15000 )( 0 . 01 ) 48

( 1. 01 )

−1

=

( 1. 612 )( 15000 )( 0 . 01 ) 1. 612

−1

= \$395

.15

(a) G( z) =

5.556

G( z ) =

20 z

z −1

5.556 z − 0.1

(b) z − 0.5

20z z − 0.8

+

80 z z −1

(c) G( z ) =

4z z −1

4z z − 0.5

2z

( z − 0.5 )

2

(d) G( z ) =

I-9

lim

k→∞

f (k )

= lim

2z

z − 1 (z − 0.5) 2

ε− ϕ

z →1

1

2z 2

z

−1

F (z)

z

= lim

z →1

(z

− 0 . 5)

F(z) is expanded into a power series of k 1 2 3 4 5 6 7 8 9 10

= z

2 −1

f(k) 1.000000 1.500000 1.750000 1.875000 1.937500 1.968750 1.984375 1.992188 1.996094 1.998047

by long division. k 11 12 13 14 15 16 17 18 19 20

(d) 352

f(k) 1.999023 1.999512 1.999756 1.999878 1.999939 1.999969 1.999985 1.999992 1.999996 1.999998

F (z)

=

z (z

F (z)

− 1 )( z − 1. 5)

has a pole outside the unit circle.

f (k )

is unbounded.

The finalvalue theorem cannot be applied to k 1 2 3 4 5 6 7 8 9 10

2-18 (a)

x( k

F ( z ) . F ( z ) is

f(k) 0.000000 1.000000 2.500000 4.750000 8.125000 13.187500 20.781250 32.171875 49.257813 74.886719

expanded into a power series of

k 11 12 13 14 15 16 17 18 19 20

+ 2 ) − x ( k + 1) + 0 .1 x ( k ) = u s ( k )

x (0)

=

x (1 )

z

−1

.

f(k) 113.33078 170.995117 257.492676 387.239014 581.858521 873.787781 1311.681641 1968.522461 2953.783691 4431.675781

= 0 . Taking the z-transform on both

sides, 2

z X (z) X (z )

−z

=

2

x(0)

β− γε z

z

1

− zx ( 1) − zX z

=

ϕ

− z + 0 .1

2

10 z z

z

+ zx ( 0 ) + 0 .1 X ( z ) =

(z)

+

−1

1.455 z z

z −1 11 .455 z

− 0 .1127

Taking the inverse z-

− 0 . 8873

z

transform, x( k )

(b)

x( k

= 10 + 1.455

k

( 0 .1127 )

+ 2) − x (k ) = 0

− 11 .455

x (0)

= 1,

( 0 . 8873)

x (1 )

k

= 0 , 1, 2 , L

k

=0

Taking the z-transform on both sides, we

have 2

z X (z)

−z

2

x ( 0 ) − zx ( 1 ) − X ( z )

=0

X (z)

=

z

z

2

2

−1

x (k )

= cos

Φ πΙ Γ Ηϑ Κ k

2

k

= 0 , 1, 2 , L

2-19 (a) P (1 )

= (1+ r) P (0) − u

P( 2)

= ( 1 + r ) P ( 1) − u

where

P (1 )

=

amoun t owed

P(0 )

=

P0

=

after

the f irst p eriod.

amoun t borr owed i nitial ly

M P (k

+ 1 ) = (1 + r ) P ( k ) − u

u

=

amoun t paid

each

period

inclu ding p rincip al and

(b) By direct substitution, P (2)

= ( 1 + r ) P (1 ) − u = (1 + r )

2

P (0)

−(1+ r)u − u

P ( 3)

= (1 + r ) P ( 2 ) − u = ( 1 + r )

3

P (0)

− (1 + r )

P (N )

= (1 + r )

N

P (0)

−u

M

(1+ r)

N −1

+ (1 + r )

353

2

N −2

u

− (1 + r ) u − u

+ L + (1 + r ) + 1 = 0

inter est.

Solving for u from the last equation, we have u

=

(1+ r)

N

P

0

N −1

∑ (1 + r )

=

(1

+r )

N

+ r)

N

(1

k

P r 0

−1

k =0

(c) Taking the z-transform on both sides of the difference equation,

P( k

+ 1 ) = (1 + r ) P ( k ) − u ,

we have ∞

P( k

+ 1) z

−k

= (1 + r )

k =0

∑ P( k ) z

−k

P (z)

=

uz

z

k =0

−1

Or, zP ( z ) − zP

= (1 + r ) P ( z ) −

0

uz z

Thus,

−1

zP z

0

−(1+ r)

uz (z

− 1)

z

− (1 + r )

Taking the inverse z-transform and setting k = N, we have P (N )

= P 0 (1 + r )

N

+

u r

u (1 r

+ r)

N

=0

Solving for u from the last equation, we have u

(d) For u

P

0

=

= \$15 (1 (1

,000 , r

+r )

N

+ r)

N

P0 r

−1

= 0 . 01 , =

( 1. 01 )

=

and N 48

(1

+ r)

N

(1

+ r)

N

= 48

P0 r

−1

month s,

( 15000 )( 0 . 01 )

( 1. 01 )

48

−1

I-9

I-10

354

=

( 1. 612 )( 15000 )( 0 . 01 ) 1. 612

−1

= \$395

.15

I-11

355

(a) Discrete state equations:

I-12

A=

0 1  −2 −3    −1

φ ( t) = L

B=

0  1   

 s −1    2 s + 3

 2 e − t − e −2 t  ( sI − A ) −1  =  − t −2 t  −2 e + 2 e

−1

−1

=

1 s + 3 1 ( s +1)( s + 2)  −2 s 

  2 e − T − e −2 T φ ( T ) =  −2 e− T + 2 e−2 T −t −2 t  − e + 2e   0.2325   0.6 φ (T ) =  −0.465 −0.0972 

T = 1 sec.

θ ( t) = L

( sI − A )

( sI − A ) =  −t

e −e

− 2t

−t −2 t 1 1   e − e  ( sI − A)−1 B  = L−1  =    −t −2 t   ( s + 1)( s + 2) s  − e + 2 e   0.2325  T = 1 sec. θ (T ) =  − 0.0972

θ (T ) =

  −e + 2 e  e

−T

−e

−T

−2 T

−2 T

 e − T − e −2 T  − e −T + 2 e −2 T   

(b)  2 e −NT − e −2 NT φ ( NT ) =  − NT −2 NT  − 2e + e

e −e

−NT − NT

 −2 NT  + 2e  −e

−2 NT

 2 e −N − e −2 N = −N −2 N  − 2e + e

e −e

−N −N

−e

−2 N

+ 2e

−2 N

  

N −1

x ( NT ) = φ ( NT )x (0) +

∑φ [ ( N − k − 1) T ]θ (T )u (kT ) k=0

(a) Discrete state equations:

I-13

A=

0

1

0 0 

(b)

1 1  s s2  −1 B= sI − A ) = sI − A ) = = ( (  1  0 s  s 0 s   1  0 s  1 t  1 T  1 0.001 φ ( t) = φ (T ) =  0 1  0 1  =  0 1  1  s 2  t  T  0.001  −1 −1 −1 θ ( t) = L ( sI − A) B  = L   = θ (T ) =      1  =  1   1  1   s  0 

φ ( NT ) =

s

1  0

−1 

NT 

 1 0.001N  =  1  0 1 

N −1

x ( NT ) = φ ( NT )x (0) +

∑φ [ ( N − k − 1) T ]θ (T )u (kT ) k=0

356

1 s

1

1 ( N − k − 1) T  1 0.001( N − k − 1)   0  = 0  1 1  0.001( N − k )  φ [( N − k − 1) T ]θ (T ) =   1

φ [( N − k − 1) T ] =

I-14

(a) Transfer function: X( z ) U (z )

 z − 0.6  0.465

= [ zI − φ (T ) ] θ (T ) = −1

X( z )

=

U (z )

−1

1  z + 0.0972 0.2325   0.2325    0.2325  = z + 0.0972   − 0.0972 ∆ ( z )  −0.465 z − 0.6   − 0.0972 −0.2325

0.2325 z    ∆( z )  −0.0972( z + 0.6673) 1

∆ ( z ) = z − 0.5028 z + 0.04978 2

(b) Characteristic equation: ∆ ( z ) = z − 0 . 5028 2

I-15

z

+ 0 . 04978 = 0

(a) Transfer function: X( z ) U (z )

= [ zI − φ (T ) ] θ (T ) = −1

1  0.001z   z − 1 0.001  0.001 =     ∆ (z )  0 z − 1   1  ∆ ( z )  z − 1  1

∆( z ) = ( z − 1)

2

(b) Characteristic equation: ∆( z ) = ( z −1) = 0 2

I-16

State diagram:

∆( z ) = z

−1

−z

−1

−6z

−2

−z

Characteristic equation:

−2

2z

+ 5z

−2

− 10

z

−3

− 5z

+ 15 = 0

357

−3

= −2 z

−2

− 15

z

−3

=0

I-17

State diagram:

State equations: x ( k + 1) = 2 x ( k ) + x ( k ) 1 2 3

Output equation:

x (k

+ 1) = x 3 ( k )

x (k

+ 1) = −0 . 1 x 1 ( k ) − 0 .2 x 2 ( k ) − 0 .1 x 3 ( k ) + r ( k )

2

3

y (k )

= x1( k )

Transfer function: Y (z) R( z )

=

+z

−3

+ 0 .2 z

−2

z 1 + 0 .1 z

−1

−2

−3

+ 0 .2 z

z

=

z

3

+2

+ 0 .1 z + 0 .2 z + 0 .2 2

I-18 Open-loop transfer function: G( z ) =

Y ( z)

(

= 1− z

E( z)

−1

) Z  s ( s + 1)  = z − 0.368 1

0.632

Closed-loop transfer function: Y (z) R( z )

=

G (z) 1+ G ( z)

=

0 . 632 z

− 0 .264

Discrete-data state equation: x [( k

I-19

+ 1 ) T ] = 0 .264

x ( kT )

+ 0 . 632

r ( kT )

= z + 1. 5 z − 1 = 0 This is a second-order system, n = 2. F(1) = 1.5 > 0 F(−1) = −1.5 < 0 Thus, for n = 2 = even, F(−1) < 0. The system is unstable. The characteristic equation roots are at z = 0.5 and z = -2. F(z) has one root outside the unit circle.

(a)

2

F (z)

Let

z

=

w

+1

w

−1

. The characteristic equation becomes

( w + 1)

Or

2

(

)

2  w + 1  + 1.5  w + 1  − 1 = 0      w−1   w−1

+ 1.5 w − 1 − ( w − 1) = 0 or 1.5 w + 4 w − 1.5 = 0 2

2

2

The last coefficient of the last equation is negative. Thus, the system is unstable.

(b)

F (z)

= z + z + 3 z + 0 .2 = 0 This is a third-order system, n = 3. 3

2

358

z

Let

=

w

+1

w

−1

. The characteristic equation becomes

 w + 1  +  w + 1  + 3  w + 1  + 0.2 = 0        w−1   w−1   w−1  3

2

Or + 1 ) + ( w + 1 ) ( w − 1 ) + 3 ( w + 1 )( w − 1 ) + 0 .2( w − 1 ) = 0 or 5 .2 w + 0 .4 w − 0 .4 w + 2 . 8 = 0 Since there is a negative sign in the characteristic equation in w, the equation has at least one root in the right-half w-plane. (w

3

2

2

3

3

2

Routh Tabulation: w

3

w

2

w

1

w

0

5 .2

− 0 .4

0 .4

2 .8

− 36 .8 2 .8

Since there are two sign changes in the first column of the Routh tabulation, F(z) has two roots outside the unit circle. The three roots are at: −0.0681, −0.466 + j1.649 and −0.466 − j1.649.

(c)

= z − 1.2 z − 2 z + 3 = 0 w +1 Let z = . The characteristic equation becomes w −1 F (z)

3

2

3 2  w + 1  − 1.2  w + 1  − 2  w + 1  + 3 = 0        w−1   w −1  w−1 

0 .8 w

Or

3

− 5 .2 w + 15 .2 w − 2 .8 = 0 2

Routh Tabulation: w

3

w

2

w

1

w

0

0 .8

15 .2

− 5.2

− 2 .8

14. 77

− 2 .8

There are three sign changes in the first column of the Routh tabulation. Thus, F(z) has three roots outside the unit circle. The three roots are at −1.491, 1.3455 + j0.4492, and 1.3455 − j0.4492.

(d)

= z − z − 2 z + 0 .5 = 0 w +1 Let z = . The characteristic equation becomes w −1 F (z)

3

2

 w + 1  −  w + 1  − 2  w + 1  + 0.5 = 0        w−1   w−1   w−1  3

Or

2

− 1.5 w + 2 . 5 w + 7 . 5 w − 0 . 5 = 0 3

2

359

Routh Tabulation: 3 w − 1. 5 w

2

w

1

w

0

7 .5

− 0 .5

2 .5 7 .2

− 0 .5

Since there are two sign changes in the first column of the Routh tabulation, F(z) has three zeros outside the unit circle. The three roots are at z = 1.91, −1.1397, and 0.2297.

I-20 Taking the z-transform of the state equation, we have zX ( z )

= ( 0 . 368 − 0 . 632 X (z)

Or,

K

=

R (z )

z z

The characteristic equation is Stability Condition: 0 . 368

I-21

z

K ) X ( z ) + KR ( z )

− 0 .368 + 0 . 632 − 0 .368 + 0 . 632

− 0 . 632

K

0

< 0 . 07

(a) Forward-path transfer function:

(

G ho Gp ( z ) = 1 − z

−1

) Z  s

 = 1 − z −1 K Z  0.667 − 0.444 + 0.444  = K (0.00476 z + 0.004527) )  s 2   ( 2 s s + 1.5  z − 1.8607 z + 0.8607  ( s + 1.5)  K

2

Characteristic equation: For T = 0.1 sec.

z

2

z

2

− 1.8607 z + 0 . 8607 + 0 .00476 Kz + 0 . 004527 + ( 0 . 00476 K − 1.8607 ) z + 0 .8607 + 0 . 004527

360

K K

=0 =0

Let

z

=

w

+1

w

−1

. The characteristic equation becomes

+ ( 0 .2786 − 0 .009054 K ) w + 3 .7214 − 0 . 000233 K = 0 For stability, all the coefficients of the characteristic equation must be of the same sign. Thus, the conditons for stability are: 0.2786 − 0.009054K > 0 or K < 30.77 3.7214 − 0.000233K > 0 or K < 15971.67 Thus, for stability, 0 < K < 30.77 T = 0.5 sec. Forward-path transfer function: K ( 0 . 09883 z + 0 . 07705) G G (z ) = ho p 2 z − 1.4724 z + 0 .47237 w

2

2

+ ( 0 . 09883

0 . 009287

(b)

Characteristic equation: z

Let

z

=

w

+1

w

−1

.

K

− 1.4724

)z

+ ( 0 . 07705

K

+ 0 .47237

)

=0

The characteristic equation becomes + ( 1. 05526 − 0 .1541

2

0 .17588 Kw

+ 2 .94477 − 0 . 02178

K )w

K

=0

For stability, all the coefficients of the characteristic equation must be of the same sign. Thus, the conditions for stability are:

− 0 . 1541 − 0 . 02178

1. 05526 2 . 9447

Stability condition:

(c) T = 1 sec.

0

<

K K

> 0 or > 0 or

K K

< 6 . 8479 < 135 .2

< 6 . 8479

K

Forward-path transfer function: G

ho

G

p

(z )

=

+ 0 .19652

K ( 0 .3214 z z

2

− 1.2231

)

+ 0 .22313

z

Characteristic equation: z

Let

z

=

w

+1

w

−1

.

2

+ ( 0 . 3214

K

− 1.2231

)z

+ ( 0 .22313 + 0 .19652

=0

K )

The characteristic equation becomes

0 . 5179 Kw

2

+ ( 1. 55374 − 0 . 393

K )w

+ 2 .4462 − 0 .12488

K

=0

For stability, all the coefficients of the characteristic equation must be of the same sign. Thus, the conditions for stability are :

1. 55374 2 .4462

Stability condition:

I-23

0

<

K

− 0 . 393 K > 0 − 0 .12488 K > 0

K

or or

< 3 . 9535

(a)

Roots: -1.397 -0.3136 + j0.5167 Unstable System.

-0.3136 – j0.5167

(b)

Roots: 0.3425 -0.6712 + j1.0046 Unstable System.

-0.06712 – j1.0046

361

< 3 . 9535 < 19 . 588

K

I-24

(c)

Roots: -0.4302 Unstable System.

-0.7849 + j1.307

(d)

Roots: 0.5 -0.8115 Stable System.

-0.7849 – j1.307

-0.0992 + j0.7708

-0.0992 – j0.7708

(a) Forward-path Transfer Function: T = 0.1 sec.  5  = 1 − z −1 Z  2.5 − 1.25 + 1.25  = 0.02341z + 0.021904 −1 G ( z ) = (1 − z ) Z  2 )  s 2 s s + 2.5  z 2 − 1.8187z + 0.8187  (  s ( s + 2)  Closed-loop Transfer Function: Y (z) R( z )

=

G (z) 1+ G ( z)

=

0 . 0234 z z

2

+ 0 . 021904

− 1. 7953

z

+ 0 .8406

(b) Unit-Step Response y(kT)

(c) Forward-path Transfer Function: T = 0.05 sec.  5  = 0.0060468 z + 0.0058484 −1 G ( z ) = (1 − z ) Z  2   s ( s + 2)  z 2 − 1.9048z + 0.9048 Closed-loop Transfer Function: Y (z) R( z )

=

0 . 006046 z

2

z

− 1. 8988

362

+ 0 . 005848 z

+ 0 . 91069

4

Unit-step Response y(kT)

(a)

I-25

(

Y ( z) = 1 − z

−1

(

= 10

Y ( z)

G( z) =

−1

∗ p

= lim

z →1

K

∗ a

G (z)

=

1 2

T

)T

2

K T

E (z)

t

z

−1

(

U ( z ) = 10 E( z) − Kt 1 − z U (z)

z ( z + 1) 10( z −1) z − 1 + K tT

3

10( z z

) Z  s

1

− 1)

−1+ K tT

2

U ( z)  

E (z)

E ( z)

5T ( z + 1)

( z − 1) ( z − 1 + K T ) t

lim ( z

K

− 1)

2

=

v

G (z)

1 T

lim ( z z →1

−1 )G ( z ) =

10 T

2

K T

2

t

=

10 K

t

=0

(b) Forward-path Transfer Function:

Closed-loop Transfer Function:

5T ( z + 1) 2

G( z) =

=

U (z)

Th us

−1

2

=

=∞

z →1

3

2( z − 1)

E( z) Error Constants: K

U ( z)  

1

U (z)

Y ( z) = 1 − z

) Z  s

Y ( z)

( z − 1) ( z − 1 + K T )

R (z )

t

5T ( z + 1) 2

=

(

) (

)

z + K tT + 5T − 2 z + 5T − K tT + 1 2

2

2

(c) Characteristic Equation: T = 0.1 sec.

(

) (

)

F (z ) = z + Kt T + 5T − 2 z + 5T − K tT + 1 = z + ( 0.1Kt − 1.95) z + ( 1.05 − 0.1Kt ) = 0 2

2

2

363

2

For stability, from Eq. (6-62), where Thus,

a

0

= 1. 05 − 0 .1 K t F (1 ) a0

= 0 .1 > 0

=

> 0,

F (1 ) a

1

F ( − 1)

0 .5 K

t

> 0,

a0

= 4 − 0 .2 K t > 0

− 0 .1 K t < a 1 = 1

1. 05

Stability Condition:

(d) Unit-step Response:

F ( −1 )

or

0.5 < K t

< K t < 20

=5 Y (z)

=

R( z)

(e) Unit-step Response

< a2

= 0 .1 K t − 1. 95

K

t

0 . 05 ( z z

2

z

2

+ 1)

− 1.45 z + 0 . 55

=1

Y (z) R( z )

=

0 . 05 ( z

+1)

364

− 1. 85 z + 0 . 95

or

< 20 . 5

K

t

< 20

I-26 (a) T = 0.1 sec.

  s + 37.06 s + 141.2 E (z )  ( s + 37.06 z + 141.2 )  Ωm ( z) −1  − 30 + 14.2 + 29.7 + 0.418  = (1 − z ) Z   2 E( z) s s + 4.31 s + 32.7   s 1000( z + 20)

Gp (s ) =

Ω m ( z)

(

= 1− z

2

(

= 1− z

=

−1

) 

z

+

 z −1

3 . 368 z 3

−30 z

− 1. 6876

( z −1)

+ 1. 7117

2

z

14.2 z

2

z

+ 0 . 71215

29.7 z

+

2

z

+

z − 0.65

− 0 . 3064

=

− 0 . 024576

−1

100( s + 20)

)Z  s

2

  z − 0.038  0.418 z

+ 1. 7117

z

− 0 . 3064

− 1 )( z − 0 . 65)(

z

− 0 . 038

3 . 368 z (z

2

(b) Closed-loop Transfer Function: T = 0.1 sec. Ωm ( z ) R (z )

=

3. 368 z

+ 1. 7117

2

+ 1. 6805

z

3

3

+ 1. 6805

2

z

z

− 0 . 3074

+ 2 .424

z

− 0 . 331

Characteristic Equation: z

z

2

+ 2 .424

z

− 0 . 331 = 0

Roots of Characteristic Equation: z

= 0 .125

,

− 0 . 903 +

The complex roots are outside the unit circle

365

j1. 3545 , z

− 0 . 903 −

j 1. 3545

= 1, so the system is unstable.

)

(c) T = 0.01 sec.

Forward-path Transfer Function

Ωm ( z ) E (z)

Ωm ( z )

Closed-loop Transfer Function

R (z )

= =

− 2 . 6785

3

z

z 2

z

0 . 047354

− 2 . 6312

3

z

+ 0 . 005974

2

0 . 04735 z

z

− 0 . 69032

z

+ 0 . 005875

− 0 . 03663

3z

+ 2 . 3748

2

− 0 . 03663

3z

+ 2 . 3689

2

z

− 0 . 72695

Characteristic Equation: − 2 . 6312 Characteristic Equation Roots: z

z

3

= 0 . 789

,

+ 2 . 3748

2

z

+

0 . 921

T = 0.001 sec. Forward-path Transfer Function Ω m ( z ) −1.43 × 10 =

−6

z

−6

z

j 0 .27 ,

+ 4. 98 × 10

−4

z

3

− 2 . 9635

2

z

−4

+ 4. 98 × 10

3

R (z )

− 2 . 963

3

z

z

+ 2 .9271 z

2

j 0 .27

− 2 . 384 × 10

2

z

−7

z

− 4. 89 × 10

z

− 4. 89 × 10

−4

− 0 .9636

z

− 2 . 384 × 10

+ 2 . 9271

2

z

0 . 921

3

E (z)

Closed-loop Transfer Function Ω m ( z ) −1.43 × 10 =

− 0 . 72695 = 0

z

−7

−4

− 0 . 9641

Characteristic Equation: z

3

− 2 .963

+ 2 . 9271

2

z

z

− 0 . 9641 = 0

Characteristic Equation Roots: z

= 0 . 99213

,

+

0 . 98543

j 0 . 02625 ,

0 . 98543

j 0 . 02625

(d) Error Constants: K

∗ p

= lim

1

Kv

z →1

=

Ka =

lim ( z z →1

T

1 T

G ho G p ( z )

2

− 1)

2

= lim

z →1

+ 1.7117

z

− 0 . 3064

− 1)( z − 0 .65)(

z

− 0 . 038

3 . 368 z (z

G ho G p ( z )

2

3. 368 z

= − lim

z →1

lim  ( z − 1) Gho Gp ( z)  = z →1

T

− 0 . 65)(

(z

2

z →1

Steady-state Errors: Step Input: Ramp Input:

e

e

∗ ss

∗ ss

= =

1 1+ K 1 K

+ 1. 7117

( 3.368 z lim

1

2

2

=0

p

=

v

366

T 14. 177

=∞ )

2

z

z

− 0 . 3064

− 0 . 038

=

14. 177

)

T

+ 1.7117 z − 0.3064 ( z − 1)( z − 0.038)

) ( z − 1) = 0

e

Parabolic Input:

I-27

∗ ss

=

1 K

=∞

∗ a

(a) Forward-path Transfer Function: (no zero-or der hold) T = 0.5 sec. G (z)

0 .1836 zK

= z

2

z

2

− 1. 0821

z

+ 0 . 0821

0 .1836 zK

= (z

− 1 )( z − 0 . 0821

T = 0.1 sec. G (z)

=

0 . 0787 zK

− 1. 6065

z

+ 0 . 6065

367

=

0 . 0787 zK (z

− 1)( z − 0 .6065)

)

(b)

Open-loop Transfer Function: (with zero-order hold) T = 0.5 sec. G (z)

=

K ( 0 . 06328 z

− 1. 0821

+ 0 . 02851

z

+ 0 . 08021

K ( 0 . 00426 z

+ 0 .003608

z

2

)

=

0 . 06328 K ( z (z

+ 0 .4505)

− 1 )( z − 0 . 0821

)

T = 0.1 sec. G (z)

=

z

I-28

2

− 1. 6065

Forward-path Transfer Function: G( z) =

z

+ 0 . 6065

(

)

=

0 . 00426 K ( z (z

− 1 )( z − 0 .6065)

) + 2.4644 z − 0.7408 )

0.0001546 K z + 3.7154 z + 0.8622

(

z z − 2.7236 z 3

2

2

368

+ 0 .8468

)

I-29 (a)

P (z )

=

z

3

−1

Q (z)

=

z

2

+ 1. 5 z − 1 = ( z − 0 . 5)(

The system is unstable for all values of K .

369

z

+ 2)

I-30

(a) Bode Plot:

The system is stable.

(b) Apply w-transformation, z

Then

G

=

ho

2

+ wT

2

− wT

G(w)

The Bode diagram of

G

ho

= G ho G ( z ) G (w )

z=

2 + wT

=

10( 1

2 − wT

− 0 .0025 w (1 + w )

is plotted as shown below.

The gain and phase margins are determined as follows: GM = 32 dB

PM = 17.7 deg.

370

2

w )

Bode plot of

I-31

G

ho

G (w ) :

2 16.67 N  1 − e −Ts   0.000295 ( z + 3.39 z + 0.714)  G hoG ( z ) = Z    =  s  s ( s + 1)( s + 12.5)  ( z − 1) ( z − 0.9486 )( z − 0.5354 )

The Bode plot of

G

ho

G(z)

is plotted as follows. The gain margin is 17.62 dB, or 7.6.

Thus selecting an integral value for N, the maximum number for N for a stable system is 7. Bode Plot of G ho G ( z )

371

I-32

(a)

G (s) c

=2+

200 s

Backward-rectangular Integration Rule: Gc ( z ) = 2 +

200T z −1

=

2 z − 2 + 200T z −1

=

2 + ( 200T − 2 ) z 1− z

−1

−1

Forward-rectangular Integration Rule: Gc ( z ) = 2 +

200Tz

=

z −1

( 2 + 200T ) z − 2 ( 2 + 200T ) − 2 z =

z −1

1−z

−1

−1

Trapazoidal Integration Rule: 200T ( z + 1)

Gc ( z ) = 2 +

(b)

G (s) c

2 ( z − 1)

=

( 4 + 200T ) z + 200T 2 ( z − 1)

−2

=

( 4 + 200T ) + ( 200T − 2 ) z

(

2 1−z

−1

)

= 10 + 0 .1 s

The controller transfer function does not have any integration term. The differentiator is realized by backward difference rule. G c ( z ) = 10 +

(c)

G (s) c

= 1 + 0 .2 s +

0.1 ( z − 1)

=

(10T + 0.1) z − 0.1

Tz

z

= (10T + 0.1) − 0.1z

5 s

Backward-rectangular Integration Rule: Gc ( z ) = 1 +

0.2 ( z − 1)

+

Tz

5T

( z − 1)

=

( T + 0.2 ) − 0.2 z

−1

+

T

5Tz

−1

1− z

−1

Forward-rectangular Integration Rule: Gc ( z ) = 1 +

0.2 ( z − 1)

+

Tz

5Tz z −1

=

( T + 0.2 ) − 0.2 z

−1

T

+

5T 1− z

−1

Trapezoidal Integration Rule:

Gc ( z ) = 1 +

0.2 ( z − 1) Tz

+

5T ( z + 1) 2 ( z − 1)

=

372

( T + 0.2 ) − 0.2z T

−1

+

( ) 2 (1 − z )

5T 1 + z

−1

−1

−1

−1

I-33

(a)

G (s) c

=

10 s

T = 0.1 sec

+ 12

(

Gc (z ) = 1 − z

−1

(

10 ( s + 1.5 ) s + 10

(

Gc (z ) = 1 − z

(

(c)

s

=

G (s) c

(

G (z) c

=

I-35

−1

)  z − 1 − z − e z

z

−1.2

 = 0.5825  z − 0.301

T = 1 sec

 10 ( s + 1.5 ) 

) Z  s ( s + 10)  = ( 1 − z ) Z    )  z − 1 + z − e 1.5 z

8.5

−1

−1

1.5 s

+

  s + 10  8.5

 = 10 ( z − 0.9052)   z − 0.368

−1

) Z  s + 1.55  = ( 1 − z )  z − e 1

z

−1

−0.155

 = z −1   z − 0.8564

+ 0 . 01 s

Gc (z ) = 1 − z

I-34

1

1 + 0 .4 s 1

(

1

−1

T = 0.1 sec

+ 1. 55

s

Gc (z ) = 1 − z

(d)

−1

−1

= 1− z

10

= 0.8333 1 − z

(b) G c ( s ) =

) Z  s ( s + 12)  = 0.8333( 1 − z ) Z  s − s + 12 

−1

) Z 

 −1  0.025 z + 0.975z  = 40  z − 0.975  = 40 ( 1 − z ) Z     z − 1 z − e−10   z − 0.0000454   s (1 + 0.01 s )  1 + 0.4 s

(a) Not physically realizable, since according to the form of Eq. (11-18), (b) Physically realizable. (c) Physically realizable. (d) Physically realizable. (e) Not physically realizable, since the leading term is 0.1z. (f) Physically realizable. (a)

G (s) c

= 1 + 10

s

K

P

=1

K

D

= 10

373

Thus

Gc ( z ) =

b

0

≠0

but a

0

= 0.

( T + 10) z − 10 Tz

(

) Z  s4  = 2T ( z + 1) 2

G ho Gp ( z ) = 1 − z

−1

G ( z ) = G c ( z ) Gho G p ( z ) =

( z − 1) 2T ( z + 1) [(T + 10 ) z − 10 ] z ( z − 1) 3

2

2

By trial and error, when T = 0.01 sec, the maximum overshoot of When T = 0.01 sec, G( z) =

0.02 ( z + 1) (10.01 z − 10 ) z ( z − 1)

Y ( z)

2

R (z )

=

y ( kT )

is less than 1 percent.

0.02 ( z + 1)( 10.01 z − 10 ) z − 1.7998 z + 1.0002 z − 0.2 3

2

When the input is a unit-step function, the output response y ( kT ) is computed and tabulated in the following for 40 sampling periods. The maximum overshoot is 0.68%, and the final value is 1. Sampling Periods k y ( kT ) ------------------------------

I-36

1  s3

(a) G ho Gp ( z ) = ( 1 − z −1 ) Z 

2  = 2T ( z + 1)   ( z − 1 )2

G ( z ) = G c ( z ) Gho G p ( z ) =

=

Characteristic Equation:

(

K PTz + KD ( z − 1 ) 2T

( z + 1) ( z − 1) 2

2

Tz

( K PT + KD ) z2 + K P Tz − K D 

2T

z ( z − 1)

)

2

(

)

z + 2 K PT + K DT − 1 z + 2K PT + 1 z − 2 K DT = 0 3

2

2

2

For two roots to be at z = 0.5 and 0.5, the characteristic equation should have z − z + 0 .25 as a 2 factor. Dividing the characteristic equation by z − z + 0 .25 and solving for zero remainder, we get 2

374

4K T

2

P

Solving for

K

P

and K

+ 2 K D T − 0 .25 = 0

D

K

The third root is at

z

− 0 . 5 K P T − 2 .5 K D T + 0 .25 = 0 2

a nd

from these two equations, we have P

=

0 . 0139 T

K

2

0 . 0972

=

D

T

= 1 − 2 K P T − 2 K D T = 0 . 7778 2

T

f or

= 0 . 01

sec

The forward-path transfer function is G( z) =

Y (z) R( z )

=

0.2222 ( z + 1) ( z − 0.8749) z ( z − 1)

z

3

+ 0 . 0278

2

0 .2222 z

− 1. 7778

2

z

2

z

+ 1. 0278

− 0 .1944 z

− 0 .1944

Unit-step Response:

(b)

(b) KP = 1, T = 0.01 sec G (z)

= G c ( z ) G ho G p ( z ) =

=

2T

δ+ ι + − β− γ δ + ι + β− γ T

K

z

D

z z

0 . 02

0 . 01

2

1

Tz

K

D

2

KD z

2

z z

1

0 . 01 z

− KD

2

The unit-step response of the system is computed for various values of

K

D

. The results are

tabulated below to show the values of the maximum overshoot. KD Max overshoot (%)

1.0 14

5.0 0.9

6.0 0.67

7.0 0.5

375

8.0 0.38

9.0 0.31

9.1 0.31

9.3 0.32

9.5 0.37

10.0 0.68

I-37

(

G ( z ) = G ho Gp ( z ) = 1 − z

−1

) Z  s

 = 0.02 ( z + 1)   z ( z − 1) 2

4

3

T = 0.1 sec

Closed-loop Transfer Function: Y (z ) R (z )

=

Gho Gp ( z)

0.02 ( z + 1)

=

1 + G ho Gp ( z )

2

=

z

With the w-transformation,

T

G ( jω

w

)

+w =

2 T

From the Bode plot of

The system is unstable.

z − 1.98 z + 1.02 2

20

+w

20

−w

G ( w) =

4 (1 − 0.05 w ) w

2

−w

the phase margin is found to be −5.73 degrees.

For a phase margin of 60 degrees, the phase-lead controller is G (w ) c

=

1 + aTw 1

+ Tw

=

1

+ 1.4286

w

1

+ 0 . 0197

w

The Bode plot is show below. The frequency-domain characteristics are: PM = 60 deg

M

GM = 10.76 dB

r

= 1.114

The transfer function of the controller in the z-domain is Gc ( z ) =

21.21 ( z − 0.9222 )

( z + 0.4344)

(b) Phase-lag Controller Design: Since the phase curve of the Bode plot of

G ( jω

w

)

is always below −180 degrees, we cannot

design a phase-lag controller for this system in the usual manner.

376

Bode Plots for Part (a):

G ho G p ( z )

=

Λ ε− ϕΜ Μ Ν β+ 1

z

−1

4500 K

Z

s

2

s

Ο β + Π = γΠ Θ β− γβ− K 0 . 002008

361 .2

377

z

1

z

z

0 . 001775

0 . 697

γ

γ

I-38

(a) Forward-path Transfer Function: ∗

Kv =

1 T

lim z →1

[ ( z − 1) G

]

G p ( z ) = lim

ho

K ( 2.008 z + 1.775)

= 1000

z − 0.697

z →1

Thus

K

∗ v

= 80 .1

(b) Unit-step Response:

Maximum overshoot = 60 percent.

(c) Deadbeat-response Controller Design: (K = 80.1) G ho Gp ( z ) =

G ho G p ( z

−1

)

=

Q(z P(z

−1 −1

)

=

0.16034 z + 0.14217

( z − 1) ( z − 0.697 )

0 . 16034 z

−1

1 − 1. 697 z

)

+ 0 .14217

−1

+ 0 . 697

z z

−2

Q (1 )

−2

= 0 . 3025

Digital Controller: −1

Gc ( z ) =

−1

P (z ) −1

Q (1) − Q (z )

= =

1 − 1.697 z + 0.697 z −1

0.3025 − 0.16034 z − 0.14217z z − 0.53 z − 0.47 2

M (z)

Closed-loop system transfer function: Unit-step response:

Y (z)

−2

3.3057 ( z − 1 ) ( z − 0.697 )

G ( z ) = G c ( z ) Gho G p ( z ) =

Forward-path transfer function:

−2

= 0 . 53

z

378

−1

+z

−2

+z

=

−3

3.3057 ( z − 1 ) ( z − 0.697 ) z − 0.53 z − 0.47 2

0 . 53 z

+L

z

+ 0 .47 2

I-39

G p (s)

2500

=

(

G hoG ( z ) = 1 − z

G ho

T = 0.05 sec

+ 25)

s(s

−1

) Z 

 2.146 z + 1.4215 =   s ( s + 25 )  ( z − 1 ) ( z − 0.2865 ) 2500

2

( ) = 2.146 z + 1.4215 z G( z ) = P ( z ) 1 − 1.2865 z + 0.2865z Q z

−1

−1

−1

−1

−2

−1

Q (1 )

−2

= 3 . 5675

( )=Q

Gc z

−1

( ) = 1 − 2.865z + 0.2865z (1) − Q ( z ) 3.5675 − 2.146 z − 1.4215 z P z

−1

−1

−2

−1

−1

−2

G (z) c

Forward-path Transfer Function: G (z)

= G c ( z ) G ho G p ( z ) =

2 .146 z 3 . 5675 z

+ 1.4215

− 2 . 146

z

0 . 6015 z

+ 0 . 3985

2

− 1.4215

Closed-loop System Transfer Function: M (z)

Unit-step response:

Y (z)

=

= 1 + 0 . 6015

z

379

−1

z

+z

2

−2

+z

−3

+L

=

(z

− 1 )( z − 0 .285)

3 . 5675 z

2

− 2 .146

z

− 1.4215

I-40

The characteristic equation is z + ( − 1.7788 + 0.1152 k1 + 22.12 k 2 ) z + 0.7788 + 4.8032 k1 − 22.12 k 2 = 0 2

For the characteristic equation roots to be at 0.5 and 0.5, the equation should be z

2

− z + 0 .25 = 0

Equating like coefficients in the last two equations, we have − 1.7788 + 0 .1152 0 . 7788

Solving for the value of

k and k 1

2

+ 4. 8032

k k

1

1

+ 22 .12 k 2 = − 1

− 22 . 12 k 2 = 0 .25

from the last two equations, we have

380

k

1

= 0 . 058

and k

2

= 0 . 035

.

#### E-Book Information

• Edition: 8

• Pages: 378

• Pages In File: 378

• Language: English

• Topic: 139

• Library: Ihtik

• Org File Size: 10,423,103

• Extension: pdf

• Toc: Kuo_AutomContrlSystems_8th_ISM
......Page 1
ch02
......Page 2
ch03
......Page 5
ch04
......Page 15
ch05
......Page 38
ch06
......Page 73
ch07
......Page 86
ch08
......Page 119
ch09
......Page 170
ch10
......Page 212
ch11
......Page 280
appendix H
......Page 331
appendix I
......Page 346

### Related Documents ##### Automatic Control Systems [DJVU]

Richard M. Phelan ##### Automatic Control Systems [DJVU]

Richard M. Phelan ##### Automatic Control Systems [PDF] ##### Automatic Control Systems [PDF]

Farid Golnaraghi, Benjamin C. Kuo ##### Automatic Control Systems [PDF]

Farid Golnaraghi, Benjamin C. Kuo ##### Automatic Control Systems [MOBI]

Farid Golnaraghi;Benjamin C. Kuo
{"ft_lang_label":"__label__en","ft_lang_prob":0.5996937,"math_prob":0.9998988,"size":254293,"snap":"2023-40-2023-50","text_gpt3_token_len":130906,"char_repetition_ratio":0.16024287,"word_repetition_ratio":0.20732713,"special_character_ratio":0.5965717,"punctuation_ratio":0.123093046,"nsfw_num_words":1,"has_unicode_error":false,"math_prob_llama3":0.99986935,"pos_list":[0,1,2,3,4,5,6,7,8,9,10,11,12],"im_url_duplicate_count":[null,2,null,2,null,2,null,2,null,2,null,2,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-12-06T00:28:11Z\",\"WARC-Record-ID\":\"<urn:uuid:74a3131a-e59c-4fdd-881b-4c1b5b027b02>\",\"Content-Length\":\"538708\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:91f3704c-7453-4551-9597-e44e49b5d3cf>\",\"WARC-Concurrent-To\":\"<urn:uuid:785f08be-a3d6-401d-9b28-eb7e62a72165>\",\"WARC-IP-Address\":\"172.67.200.59\",\"WARC-Target-URI\":\"https://vdoc.pub/documents/automatic-control-systems-10740ps2kjmo\",\"WARC-Payload-Digest\":\"sha1:GIE4DKT56IQKMNZOODO5HEK52TONLJK6\",\"WARC-Block-Digest\":\"sha1:QPUJER4CQBRG4CC7A7BH6DY6GPI56IQ7\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-50/CC-MAIN-2023-50_segments_1700679100575.30_warc_CC-MAIN-20231206000253-20231206030253-00172.warc.gz\"}"}