Linear Control System D'Azzo (5th Edition)

Problems 785

13.19. the set of closed-loop eigenvalues {À 0.5 Æ j0.6, À 60}. (c) Compute
the state and output responses for the unit-step input r(t) ¼ uÀ1(t)
and x(0) ¼ [1 0 1]T. Plot the responses.
For the plant represented by

" # "#
01 0
x_ ¼ x þ u
À2 À4 2

13.20. Design the feedback matrix K that assigns the set of closed-loop plant

eigenvalues as {À 3, À 6}. Determine the state responses with the
initial values x(0) ¼ [1 À1]T.

A SISO system represented by the state equation x_ p ¼ Apxp þ bpu has
the matrices

2 À1 0 0 3 213 213
Ap ¼ 64 0 À2 À2 75 bp1 ¼ 46 1 75 bp2 ¼ 64 0 75

À1 0 À3 1 0

13.21. For each bp: (a) Determine controllability. (b) Find the eigenvalues.
13.22. (c) Find the transformation matrix T that transforms the state equa-

tion to the control canonical (phase-variable) form. (d ) Determine
the state-feedback matrix kcT required to assign the eigenvalue spec-
trum sðAclÞ ¼ f À2ó À4ó À6g .
For the system of Prob. 13.20, determine the feedback matrix k

that assigns the closed-loop eigenvalue spectrum sðAclÞ ¼
fÀ1 þ j2ó À1 Àj2ó À2g.

Repeat Prob. 13.20 with

23 23
01 0 0

Ap ¼ 460 À2 À357 bp ¼ 46275 sfAclg ¼ fÀ1 þ j1 À 1 À j1ó À 4g

0 0 À6 1

13.23. x_ ¼ Ax þ Bu, where

2 3 23
À3 1 0 0 0

A ¼ 46666 À2 0 1 0 77577 B ¼ 66466 0 77577
0 1 1 0 1

000 1 1

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786 Linear Control System Analysis and Design

(a) Determine controllability. (b) Transform the state equation to
the control canonical form. Determine the matrix T. (c) Design the
feedback matrix K that assigns the closed-loop eigenvalue spectrum

sðAclÞ ¼ fÀ3 þ j2ó À 3 À j2ó À 5 À 10g

(d ) Obtain the time response of the closed-loop system with zero input
and the initial conditions x(0) ¼ [1 0 0 1]T.

CHAPTER 14

14.1. A unity-feedback control system has the following forward transfer
function:

GðsÞ ¼ smðs Kðs þ aÞ þ dÞ
þ bÞðs2 þ cs

14.2. The nominal values are K ¼ 10, a ¼ 4, b ¼ 2, c ¼ 2, and d ¼ 5. (a) With
m ¼1, determine the sensitivity with respect to (1) K, (2) a, (3) b, (4) c,
and (5) d. (b ) Repeat with m ¼ 0.
The range of values of a is 3 ! a !1, Mp 1.095 for a unit-step input,
and Ts 0.9 s. Design a state-variable feedback system that satisfies
the given specifications and is insensitive to variation of a. Assume

that all states xi are accessible and are fed back directly through the
gains ki. Obtain plots of y(t) for a ¼1, a ¼ 2, and a ¼ 3 for the final
design. Compare Mp, tp, ts, and K1. Determine the sensitivity functions
for variation of A and a (at a ¼1, 2, and 3).

14.3. Repeat Prob. 14.2 but with state X3 inaccessible.
14.4. Determine the sensitivity function for Example 1 in Sec.13.12 for varia-

14.5. tions in A. Insert the values of ki in Eq. (13.118) and then differentiate
14.6. with respect to A to obtain SAM .
For Example 2, Sec. 13.12, obtain Y(s)/R(s) as a function of KG [see Fig.
13.17 with a ¼1]. Determine the sensitivity function SKMG.
The desired control ratio for a tracking system is

MT ðsÞ ¼ ðs þ KGðs þ 1:5Þ À pÞ
1:1Þðs2 þ 2s þ 2Þðs

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Problems 787

This control ratio must satisfy the following specifications: e(t)ss ¼ 0
M
for r (t) ¼ R0uÀ1(t), and jS KG ð joÞj must be a minimum over the range

0 o ob. (a) Determine the values of KG and p that satisfy these

specifications. The transfer function of the basic plant is

Gx ðsÞ ¼ Kxðs þ 3Þ 5Þ ¼ Kxðs þ 3Þ
sðs þ 1Þðs þ s3 þ 6s2 þ 5s

14.7. A controllable and observable state-variable feedback-control system
is to be designed for this plant so that it produces the desired control
ratio above. (b) Is it possible to achieve MT (s) using only the plant in
the final design? If not,what simple unit(s) must be added to the system
to achieve the desired system performance? Specify the parameters of
these unit(s). Using phase-variable representation, determine kT for
the state-variable feedback-control system. Assume appropriate values
to perform the calculations.

For the state-feedback system shown, all the states are accessible, the
parameters a and b vary, and sensors are available that can sense all
state variables. The values of k1, k2, k3, and A are known for a desired
MT (s) based on nominal values for a and b. It is desired that the system
response y(t) be insensitive to parameter variations. (a) By means of a
block diagram indicate an implementation of this design to minimize
the effect of parameter variations on y(t) by use of physically realizable
networks. (b) Specify the required feedback transfer functions, ignor-
ing sensor dynamics.

14.8. For the state-feedback control system of Prob. 14.7 with a ¼ 4, b ¼ 3

and a and b do not vary, the specifications are as follows: e(t)ss ¼ 0 for
M
r(t) ¼ uÀ1(t), jS A ð joÞj must be a minimum over the range 0 o < ob,

and the dominant poles of MT (s) are given by s2 þ 4s þ13. Determine

(a) an MT(s) that achieves these specifications and (b) the values for KG,

k1, k2, and k3 that yield this MT (s).

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788 Linear Control System Analysis and Design

14.9. Use the Jacobian matrix to determine the stability in the vicinity of
the equilibrium points for the system described by:

ðaÞ x_1 ¼ 3x1 ¼ x12 À 2x1x2 þ u1 Â 1 ÃT

x_2 ¼ Àx1 þ x2 þ u1u2ó where u0 ¼ 0

ðbÞ x_1 ¼ 2x1 À 4x1x2

x_2 ¼ þx12 À 2x2

ðcÞ x_1 ¼ À4x12 þ x2

x_2 ¼ x1 À 2x2

14.10. For the systems described by the state equations

ð1Þ x_1 ¼ x1 À x12 À x1x2 À 2u1ó x_2 ¼ þx1 þ 2x2 À u1u2ó u0 ¼ ½1 0ŠT
ð2Þ x_1 ¼ x2ó x_2 ¼ 2x1 À 3x2 þ 0:25x32ó x_3 ¼ 2x1x3 þ x3
ð3Þ x_1 ¼ 4x2ó x_2 ¼ þ2x2 À sinx1

(a) Determine the equilibrium points for each system. (b) Obtain the
linearized equations about the equilibrium points. (c) Determine
whether the system is asymptotically stable or unstable in the neigh-
borhood of its equilibrium points.

CHAPTER 15

15.1. By use of Eqs. (15.13) and (15.14),verify the following Z transforms in
Table 15.1: (a) entry 4, (b) entry 7, (c) entry 12, (d ) entry 14.

15.2. Determine the inverse Z transform, in closed form, of

ðaÞ F ðzÞ ¼ ðz À zðz À 0:2Þ 0:41Þ ðbÞ F ðzÞ ¼ ðz À À4ðz2 þ zÞ À 0:5Þ
1Þðz2 À z þ 1Þ2ðz À 0:6Þðz

15.3. For the sampled-data control system of Fig. 15.14 the output
expression is:

C ðzÞ ¼ ðz 0:5Kðz2 À 0:48zÞ
À 1Þðz2 À z þ 0:26Þ

where T ¼ 1 s, r (t) ¼ uÀ1(t).
(a) Apply the partial fraction expansion to the expression C(z)/z
by the method of Sec. 15.7. (b) Using the resulting C(z) expression,
i.e., z[C(z)/z], of part (a) determine c(kT) by use of Table 15.1. For the
complex poles use entry 12 of Table 15.1. (c) Determine c(0) and c(1).

Copyright © 2003 Marcel Dekker, Inc.

Problems 789

15.4. Repeat Prob. 15.3 for

CðzÞ ¼ ðz zðz À 0:1Þðz þ 0:2Þ
À 1Þðz À 0:6Þðz À 0:8Þ

15.5. Repeat Prob.15.2 but use the power-series method to obtain the open
form for f (kT ).

15.6. Determine the value of c(2T ) for

CðzÞ ¼ z4 À 2:8z3 z4 þ z3 þ z2 þ 0:63
þ 3:4z2 À 2:24z

15.7. by use of the power series expansion method.
15.8. Determine the initial and final values of the Z transforms given in
15.9. Prob. 15.2, i.e., solve for c(0) and c(1),where R(z) ¼ z/(z À 1).
Let F(z) ¼ C(z)/E(z) in Prob. 15.2. Determine the difference equation
for c(kT ) by use of the translation theorem of Sec. 15.4.
Based upon the time domain to the s- to the z-plane correlation deter-
mine the anticipated values of tp and Ts for c(kT ) for the function

CðzÞ ¼ ðz À Kzðz þ 0:2Þ 0:74Þ
1Þðz2 À 1:4z þ

where T ¼ 0.1 s and r (t) ¼ uÀ1(t).
15.10. Determine the values of c(0) and c(1) for

CðzÞ ¼ ðz À Kzðz þ 0:1Þ þ 1Þ
1Þðz2 À 1:6z

15.11. where r (t) ¼ uÀ1(t).
For the sampled-data systems shown, determine C(z) and C(z)/R(z),
if possible, from the block diagram. Note: Write all basic equations
that relate all variables shown in the figure.

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790 Linear Control System Analysis and Design

15.12. (CAD Problem) For the sampled-data control system of Fig. 15.14,
15.13. Gx(s) is given by Eq. (15.15), where a ¼ 4 and T is unspecified.
(a) Determine G(z) and C(z)/R(z). (b) Assume a sufficient number of
values of T between 0.01and10 s, and for each value of Tdetermine the
maximum value that K can have for a stable response. (c) Plot Kmax vs.
T and indicate the stable region. (d ) Obtain the root locus for this
system with T ¼ 0.1 s. (e) Use Eq. (15.19) to locate the roots on the
root locus which have a z ¼ 0.65 and determine the corresponding
value of K. ( f ) For a unit step input determine c(kT ), Mp, tp, and ts.
For the sampled-data system of Fig.15.14 the plant transfer function is

Gx ðsÞ ¼ sðs þ Kx þ 10Þ
2Þðs

where T ¼ 0.05 s. The desired performance specifications are:
Mp ¼1.095 and ts ¼ 2 s. (a) Determine the value zD that the desired
dominant roots must have. (b) For this value of zD, by use of the root-
locus method, determine the value of Kx and the roots of the charac-
teristic equation. (c) For the value of Kx of part (b) determine the
resulting C(z)/R(z). (d ) For a unit step forcing function determine

the actual values of Mp and ts for the uncompensated system.

Copyright © 2003 Marcel Dekker, Inc.

Problems 791

CHAPTER 16
16.1. Given

ð1Þ GðsÞ ¼ sðs þ Kðs þ 3Þðs þ 20Þ Æ j25Þ
1:5Þðs þ 2 Æ j3Þðs þ 10

ð2Þ GðsÞ ¼ sðs þ K ðs þ 5Þðs þ 10Þ Æ j25Þ
1Þðs þ 2 Æ j2Þðs þ 10

16.2. and T ¼ 0.04 s. Map each pole and zero of G(s) by use of Eq. (16.8)
and z ¼ esT. Compare the results with respect to Fig. 16.5 and analyze

the warping effect.

Repeat Prob. 16.1, with T ¼ 0.01 s and 1 s, for:

ð1Þ s ¼ À1 ð2Þ s ¼ À500 ð3Þ s ¼ À1000 Æ j500

ð4Þ s ¼ À2 þ j0:5 ð5Þ s ¼ À0:1 þ j0:1 ð6Þ s ¼ À0:8 þ j0:6

16.3. The poles and zeros of the following desired control ratio of the PCT
system have been specified:

!!
CðsÞ ¼ Kðs þ 8Þ CðsÞ ¼ Kðs þ 8Þ
ð1Þ ð2Þ RðsÞ M ðs þ 4 Æ j8Þðs þ 6Þ
RðsÞ M ðs þ 5 Æ j7Þðs þ 10Þ

16.4. (a) What must be the value of K in order for y(t)ss ¼ r(t) for a step input?
(b) A value of T ¼ 0.1 s has been proposed for the sampled-data system
16.5. whose design is to be based upon the above control ratio. Do all the
16.6. poles and zeros lie in the good Tustin approximation region of Fig.
16.5? If the answer is no, what should be the value of T so that all the
poles and zeros do lie in this region?
(a) Obtain the PCT equivalent for the sampled-data system of Prob.
15.12 for T ¼ 0.1 s. (b) Repeat parts d through f of Prob. 15.12 for the
PCT equivalent. (c) Compare the results of part b with those of parts
d through f of Prob. 15.12. This problem illustrates the degradation in
system stability that results in converting a continuous-time system
into a sampled-data system.
(CAD Problem) The sampled-data control system of Prob.16.4 is to be
compensated in the manner shown in Fig.16.1. Determine a controller
by the DIR technique that reduces ts of Prob. 16.4, by one-half.
Determine Mp, tp, ts, and Km.
Repeat Prob. 16.5 by the PCT DIG technique. (a) By use of Eq. (16.8)
obtain [Dc(z)]TU and the control ratio [C(z)/R(z)]TU. (b) Obtain

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792 Linear Control System Analysis and Design

16.7. the values of Mp, tp, ts, and Km compare these values with those of
16.8. Prob. 16.5.
16.9. Repeat Prob. 16.4 for T ¼ 0.04.
16.10. Repeat Prob. 16.5 for T ¼ 0.04.
Repeat Prob. 16.6 for T ¼ 0.04 and compare the results with those of
16.11. Prob. 16.8.
16.12. (CAD Problem) For the basic system of Fig. 16.7, where Gx(s) ¼ Kx/
s(s þ 2): (a) Determine Mp, tp, ts, and K1 for z ¼ 0.65 and T ¼ 0.01 s
by the DIR method. (b) Obtain the PCT control system of part a
and the corresponding figures of merit. (c) Generate a controller
Dc(z) (with no ZOH) that reduces the value of ts of part a by approxi-
mately one-half for z ¼ 0.65 and T ¼ 0.01 s. Do first by the DIG
method (s plane) and then by the DIR method. Obtain Mp, tp, and ts
for c(t) of the PCT and c*(t) (for both the DIG and the DIR designs)
and Km. Draw the plots of c*(t) vs. *kT for both designs. (d ) Find a
controller Dc(z) (with no ZOH) that increases the ramp error coeffi-
cient of part a, with z ¼ 0.65 and T ¼ 0.01 s with a minimal degrada-
tion of the transient-response characteristics of part a. Do first by
the DIG method (s plane) and then by the DIR method. Obtain Mp,
tp, and ts for c(t) of the PCT and c*(t) for both the DIG and the DIR
designs and Km. Draw a plot of c*(t) vs. kT for both designs. (e)
Summarize the results of this problem in tabular form and analyze.
For part (b) use 4-decimal-digit accuracy; for parts (c) and (d ) 4-
decimal-digit accuracy; then repeat for 8-decimal-digit accuracy.
Compare results.

By use of the PCT DIG technique design a controller [Dc(z)]TU
[Eq. (16.46)] for Prob. 15.13 which will achieve the desired perfor-
mance specifications.

For the digital control system of Fig. 16.1 the plant and the ZOH
transfer function is:

Gz ðsÞ ¼ GzoðsÞGpðsÞ ¼ 2ð1 À eÀTsÞðs þ 3Þ
s2ðs þ 1Þðs þ 2Þ

The desired figures of merits (FOM) are Mp ¼ 1.043 and ts ¼ 2 s. (a) By
use of the PCT DIG technique, where T ¼ 0.02 s, determine Dc(s) that
will yield the desired dominant poles, based upon satisfying the spe-

cified FOM, for the system’s control ratio. (b) Determine the resulting

FOM for the compensated system of part a. (c) By use of Eq. (16.8)

obtain [Dc(z)]TU and the control ratio [C(z)/R(z)]TU. (d ) Obtain the
values of Mp, tp, ts, and Km for part (c) and compare these values with
those of part (b).

Copyright © 2003 Marcel Dekker, Inc.

Problems 793

16.13. (CAD Problem) For the basic system of Fig. 15.17, where Gx(s) ¼ Kx/
s(s þ 2): (a) Determine Mp, tp, ts, and K1 for z ¼ 0.65 and T ¼ 0.01 s by
the DIR method. (b) Obtain the PCT control system of part (a) and

the corresponding figures of merit. (c) Generate a digital controller

Dc(z) (with no ZOH) that reduces the value of ts of part (a) by approxi-
mately one-half for z ¼ 0.65 and T ¼ 0.01 s. Do first by the DIG
method (s plane) and then by the DIR method. Obtain Mp, tp, and ts
for c(t) of the PCT and c*(t) (for both the DIG and the DIR designs)

and Km. Draw the plot of c*(t) vs. kT for both designs. (d ) Find a
digital controller Dc(z) (with no ZOH) that increases the ramp error
coefficient of part (a), with z ¼ 0.65 and T ¼ 0.01 s with a minimal

degradation of the transient response characteristics of part (a). Do

first by the DIG method (s plane) and then by the DIR method.

Obtain Mp, tp, and ts for c(t) of the PCTand c*(t) for both the DIG and
the DIR designs) and Km. Draw the plots of c*(t) vs. kT for both
designs. (e) Summarize the results of this problem in tabular form

and analyze. Note: For part (b) use 4-decimal-digit accuracy; for

parts (c) and (d ) use 4-decimal-digit accuracy; then repeat using

8-digit accuracy. Compare the results.

Copyright © 2003 Marcel Dekker, Inc.

Answers to Selected Problems

CHAPTER 2

2.1. (a) Loop equations: Let R ¼ R2 þ R3 and R 0 ¼ R1/R

1 LD þ 1
eðtÞ ¼ R1 þ LD þ CD i1 À CD i2 ð1Þ
ð2Þ

11 ð3Þ
0¼À LD þ i1 þ R þ LD þ i2 ð4Þ
CD CD ð5Þ

(b) Node equations:

Node va:

11 1 11
þþ va À ðCDÞvb À R2 vc ¼ e
R1 R2 LD R1

Node vb:

1
ÀðCDÞva þ þ CD vb ¼ 0
LD

Node vc:
1 11
À R2 va þ þ vc ¼ 0
R2 R3

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796 Linear Control System Analysis and Design

(c) State equations: Let x1 ¼y1 ¼i/CD, x2 ¼ iL, x2 ¼ DiL, ic ¼ CDy1 ¼Cx1,

i1 Ài2 ¼ ic ¼ iL

2 13

A ¼ 646 0 C 775 ð6Þ
1 R0
À À ðR 0 þ 1ÞL
L

23 cT ¼ ½ 1 0Š ð7Þ
0

bT ¼ 4 1 5

ðR 0 þ 1ÞL

(d ) Note: Utilizing Eqs. (1), (2), (8), and (9), obtain

x2 ¼ Cx_1 ð8Þ

x_2 ¼ Cx€2 ð9Þ

GðDÞ ¼ y1 ðt Þ ¼ ½ðR 0 þ 1ÞLC ŠD2 1 0 C ŠD þ ½R 0 þ 1Š ð10Þ
uðtÞ þ ½R

2.2. (a) Loop equations:


1 1
E¼ R1 þ C1D i1 À C1D i2 ð1Þ


1 11
0¼À i1 þ R2 þ LD þ C1D þ C2D i2 ð2Þ
C1D

(b) Node 1 between R1 and R2, node 2 between R2 and L, and node 3

between L and C2


1 1 1 1
Node 1: R1 þ C1D þ R2 v1 À R2 v2 ¼ E

R1


1 11 1
Node 2: À v1 þ þ v2 À LD v3 ¼ 0
R2 R2 LD


1 1
Node 3: À LD v2 þ LD þ C2 D v3 ¼ 0

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Answers to Selected Problems 797

(c) Let x1 ¼v1 ¼vc1, x2 ¼ v3 ¼ vc2, x3 ¼ iL

2 À 1 0 À 1 3 213
6666664 R1C1 À 7757777
A ¼ 0 0 C1 b ¼ 6666664 R1C1 7757777
À1 1 0
1
C2
R2

L LL 0

2.3. Circuit 2: (b)

2 R1 þ R2 1 3 2 1 3
L L 577x 646 L 775u
466 À 1 À

x_ ¼ þ

C 00

2.6. (c)

GðDÞ ¼ D4 þ BD3 þ KD2 K2 þ ðKaKb À K 2Þ
þ BKaD

where Ka ¼ K1 þK2, Kb ¼ K2 þ K3, K ¼ Ka þ Kb
(d ) Let

x1 ¼ xa x2 ¼ x_1 x3 ¼ xb
x4 ¼ x_3 u ¼ f ðtÞ yðtÞ ¼ xb

2 0 1 0 03 203 203

A ¼ 6646 ÀK K2 0 0 7757 b ¼ 6466 1 5777 c ¼ 4666 0 7577
0 0 0 1 0 1

K2 0 Kb À1 0 0

2.10. (a) ðLD þ RÞi ¼ eó kii ¼ f ó xa ¼ xb
l1 l2

ðM2D2 þ B2D þ K2Þxb ¼ f2

ðM1D2 þ B1D þ K1Þxa þ l2 f2 ¼ f ¼ kii
l1

Let

l2 ¼ aó M1 þ aM2 ¼ b; B1 þ aB2 ¼ có and K1 þ aK2 ¼ d
l1 a a a

ðbD2 þ cD þ dÞxb ¼ kii

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798 Linear Control System Analysis and Design

(b) There are only three independent states:

x1 ¼ xbó x2 ¼ x_bó x3 ¼ ió and u ¼ e

x_1 ¼ x2

x_2 ¼ À d x1 À c x2 þ K1 x3
b b b

x_3 ¼ R x3 þ l u
L L

2.11. Traction force ¼ MD2x þ BDx

2.12. (b) T(t) ¼ (B1D)ya À (B1D)yb

0 ¼ À(B1D)ya þ [JD2 þ (B2 À B1)D þ K1]yb

(c) Let J ¼2J1 þJ2, y1 ¼x1 ¼ ya3, y2 ¼ x2 ¼ yb, x3 ¼ x2, and u ¼T
2 3
00 1 1

x_ ¼ 66646 0 77577x 66664 B1 57777u 1 0 0 !
0 0 0 1 0
0 1 þ 1 y¼ x
À
ÀK B1 B2
J
JJ

ðdÞ y1 ¼ JD2 þ B2D þK þ KŠ
u B1D½JD2 þ ðB2 À B1ÞD

y2 ¼ JD2 þ ðB2 1 B1ÞD þ K
u À

C1ad ! C1b
þ bÞðc þ aþb
2.18. ðaÞ ða d Þ þ D y1 ¼ x1

2.21. KT v1(t) ¼ JLaD3ym þ (JRa þ BLa)D2ym þ (BRa þ KTKb)ym

CHAPTER 3
3.2. ybðtÞ ¼ À2 þ 2t þ A1em1t þ A2em2t
where
A1 ¼ 2:01015 A2 ¼ À0:01015

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Answers to Selected Problems 799

Therefore

ybðtÞ ¼ À2 þ 2t þ 2:010105eÀ1:071t À 0:01015eÀ15:06t

Check at t ¼ 0: yb(0) ¼ À 2 þ 2.01015 À 0.01015 ¼ 0
3.5. (b) cðtÞ ¼ À0:36 þ 0:2t þ A1em1t þ A2em2t þ A3em3t

Utilizing the initial conditions yields

A1 ¼ 0:5 ffpffiffiffi À171:9 : ¼ A3Ã

A2 ¼ ð 2=10Þ

3.7. (a) Physical Àva1r0ia4bleÀs:1x016¼!vc and x2 ¼ iL. !
A¼ 0:1 À465 b¼
104 cT ¼ ½ 1 0Š
0

(b) vcðtÞ ¼ 48:95 þ A1em1t þ A2em2t where Dvc ¼ m1A1em1t þ m2A2em2t
Inserting the initial conditions yields

0 ¼ 48:95 þ A1 þ A2 0:5 Â 106 ¼ À475:5A1 À 9989:5A2ó
A1 ¼ 1:15 A2 ¼ À50:1

3.8. (b) x(t) ¼ 3 þ1.5eÀ3t À 4.5eÀt

ðcÞ xðtÞ ¼ 0:18 þ 0:2353t þ eÀ0:5t½À0:18 cos ð2tÞ À 0:16 sin ð2tފ

¼ 0:18 þ 0:2353t þ 0:242eÀ0:5t sin ð2t À 132Þ

3.13. "#
3. 16. ÀeÀ2t þ 2eÀ4t 2eÀ2t À 2eÀ4t
(ðtÞ ¼
ÀeÀ2t þ eÀ4t 2eÀ2t À eÀ4t
23
1 9eÀ4t
xðtÞ ¼ 646 2 À 3eÀ2t þ 2 577
3 À 3eÀ2t þ 9eÀ4t

44

yðtÞ ¼ x1 ¼ 1 À 3eÀ2t þ 9eÀ4t
2 2

ðaÞ xðtÞh ¼ (ðtÞxð0Þ ¼ eÀ2t !
À2eÀ2t

ðbÞ xðtÞ ¼ 1 À 2eÀt þ 2eÀ2t !
À2eÀt

3.17. (1) (a) 1 ¼ À1, 2,3 ¼ À1 Æj2
(ðtÞ ¼ Z1eÀt þ Z2eÀt e jt þ Z3eÀt eÀjt

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800 Linear Control System Analysis and Design

CHAPTER 4

4.1. (a) f (t) ¼ eÀt[t þ 0.333 sin (3t þ 180)]
(b) f (t) ¼ 1.3846 À eÀ5t þ 1.9612eÀ2t sin (3t þ 191.3)

4.2. (c) x(t) ¼ 1 À 0.9898eÀt þ 0.01701eÀ2t sin (3t þ 216.9)
4.4. (b) With zero initial conditions, the Laplace transform yields:

12 A B
XbðsÞ ¼ ¼ þ
sðs þ 10Þðs þ 1:2874Þðs þ 223:71Þ s s þ 10

þCþD
s þ 1:2874 s þ 223:71

where

A ¼ 4:167 Â 10À3 B ¼ 0:06445 Â 10À3
C ¼ À4:81 Â 10À3 D ¼ À1:128 Â 10À6

xbðtÞ ¼ 4:167 Â 10À3 þ 0:6445 Â 10À4eÀ10t À 4:81 Â 10À3eÀ1:2874t
À 1:128 Â 10À6eÀ223:7t

4.5. ðdÞ F ðsÞ ¼ 1 À 7 þ 4ðs 5 1Þ À 0:5 3Þ
s2 6s þ 6ðs þ

ðhÞ F ðsÞ ¼ 1 þ s 9 3 þ s þ 28 j1 þ s þ 30 j1
s þ 3À 3þ

4.6. ðbÞ iLðtÞ ¼ 0:4833eÀ3:684t À 0:5534 sin ðod t þ 61Þ

4.7. ðbÞ ðs2 þ 2:8s þ 4ÞX ðsÞ ¼ 2ðs2 þ 4:3s þ 6:13Þ
s

xðtÞ ¼ 3:0669 þ 1:5eÀ1:4t sin ð1:416t À 45:32Þ

ðd Þ ðs5 þ 4s4 þ 34s3 þ 110s2 þ 225s þ 250ÞX ðsÞ þ 4s2 À s À 56 ¼ s2 5 25
þ

xðtÞ ¼ 10:837eÀt sin ð2t À 85:233Þ
þ 0:99974 sin ð5t þ 180Þ þ 14:799eÀ2t

4.8. (a) Prob. 4.1(a): 0, Prob. 4.1(b): 18
13

(b) Prob. 4.2(a): 0, Prob. 4.2(c): 1

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Answers to Selected Problems 801

4. 11. ð1Þ ðbÞ xðtÞ ¼ 3 þ 3:8348eÀt sin ðt þ 237:53Þ þ 0:2353eÀ5t
ðcÞ xðtÞ ¼ 0:30t À 0:260 þ 0:2712eÀt sin ðt þ 102:53Þ À 0:0047eÀ5t

4.12. System (1)

ðaÞ (ðsÞ ¼ ½sIÀAŠÀ1 ¼ sþ5 s À1 !À1 1 sþ2 !
À2 þ2 ¼ Á 2 1
sþ5

Á ¼ s2 þ 7s þ 8 ¼ ðs þ 5:62Þðs þ 1:438Þ #
"
X1ðsÞ ! s2 þ 3s þ 1
X2ðsÞ 1 þ 12s2 þ 35s
XðsÞ ¼ ¼ Á s3 þ 1

XðsÞ ¼ ½sI À AŠÀ1xð0Þ þ ½sI À AŠÀ1bUðsÞ

(b) GðsÞ ¼ CÈðsÞB þ D

GðsÞ ¼ 1 ½01Š sþ2 1 ! 0!
Á 2 þ0

sþ5 1

¼ s2 sþ5 8 ¼ ðs þ sþ 5 1:438Þ
þ 7s þ 5:62Þðs þ

ðcÞ YðsÞ ¼ CXðsÞ ¼ ½ 0 1 ŠXðsÞ ¼ X2 ! yðtÞ ¼ x2ðtÞ
¼ 0:74836eÀ1:4384t À 0:37336eÀ5:5616t þ 0:6250

As a check: xi ð0Þ ¼ Lim ½sXi ðsފ ¼ 1 and x2ð0Þ ¼ Lim ½sX1ðsފ

s!1 s!1

¼ 1½sX2ðsފ ¼ 1 and yð1Þ ¼ 0:6250:

4.15. (1) s(A) ¼ {À 1 þj2, À1 Àj2, À1}

CHAPTER 5

Àð60s þ 3:8Þ
5. 5. GðsÞ ¼ ðs þ 10Þðs2 þ 0:15s þ 8:0054Þ

5.11. (2) ¼ À2, ¼ À3, ¼ À4, ¼ À5
(a) y ¼ [8 0 4 0]x
(b) y ¼ [4 À22 36 À18]x

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802 Linear Control System Analysis and Design
5. 12.
À10 2 À5ð1 À 3eÀ2t þ 2eÀ3t Þ 3
2Þðs þ 46 3 57
ðaÞ GðsÞ ¼ ðs þ 3Þ ðcÞ xðtÞ ¼ 10ð1 À eÀ3t Þ

3

CHAPTER 6
6.2. (a) 0 < K < 1280/9; s1,2 ¼ Æ j2.5827; for K ¼ 142.22

(c) 0 < K < 30; s1,2 ¼ Æ j2.2356; for K ¼ 30

6.3. (a) (s þ 2)(s þ 2.2 Æ j2.2) (c) (s þ 1 Æj1.414)(s À1 Æj 1.414)

( f ) (s À1)(s þ 2)(s À 3)(s þ 5)

6.4. (a) None (c) None ( f ) 2
6.10.
6.12. (a) For (1): Kp ¼ 1, Kv ¼ 153, Ka ¼ 0, (b) For r1(t): (1) e(t)ss ¼153 (2) e(t)ss ¼ 0
6.13. (a) 20, 0, 0 (d) 1, 6, 0

6.19. (a): (a) 5 (b) 1 (c) 1
2
6.25. 1 (c) 1
6.27. (d ): (a) 0 (b) 3

KGci ¼ ai þ KhKGci > 0 for a stable system and KGc0 ¼ a0 þ KhKGc0 for

e(t)ss ¼ 0
(b) K ¼ 4: (b) K ¼ 7: and K1 ¼6.666 sÀ1 e(t)ss ¼ 0.3;
(a) For K ¼ 100: K1 ¼68.97 sÀ1 and e(t)ss ¼ 0.0145; (b) for K ¼ 1000:

K1 ¼689.7 sÀ1 and e(t)ss ¼ 0.00145; (c) unstable for K ¼ 1000; stable

for K ¼ 100 and the corresponding FOM are Mp ¼ 1.345, tp ¼1.06 s,

and ts ¼ 3.286 s.

CHAPTER 7

7.2. (c) For K > 0, real-axis branches: none; four branches; g ¼ Æ 45,
Æ 135; angles of departure: Æ 45, Æ135; imaginary-axis crossing:
Æ j2 with K ¼ 0. Branches are straight lines and coincide with the

asymptotes.
(e) Real-axis branch: 0 to À1; five branches; g ¼ Æ 36, Æ108, 180;

s0 ¼ À 1.6; no breakaway or break-in points; angles of departure:
Æ 71.7, Æ 33.8; and the imaginary-axis crossing: Æ j1 with K ¼ 30.
Locus crosses the Æ 36 asymptotes.
7.4. (a) For K > 0, real-axis branches: 0 to À 10; three branches; g ¼ Æ 90;
s0 ¼ 0; break-in point: À3.9113 (K ¼ 3.33095); breakaway point:
À2.78698 (K ¼ 3.4388); and imaginary-axis crossing: none.

(b) K ¼ 0:05427

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Answers to Selected Problems 803

(c) eðtÞ ¼ 1:5951eÀ2:2164tsin ð2:1992t þ 207:274Þ
(d )

o M a, deg o M a, deg

01 0 1.4 0.9586 À44.7

0.2 0.9995 À6.1 1.8 0.9153 À58.5

0.4 0.998 À12.3 2.0 0.8854 À65.5

0.6 0.9951 À18.5 2.4 0.8103 À79.3

0.8 0.9903 À24.9 2.8 0.7202 À92.3

1.0 0.9832 À31.3 3.2 0.6252 À104.1

7.7. Poles: p1 ¼ À1ó p2 ¼ À7ó p3ó 4 ¼ 4 Æ j3
(a) For K > 0, real-axis branch: À1 to À 7; four branches; g ¼ Æ 45,
7.8.
7.11. Æ 135; angles of departure: 0, À180, Æ90; imaginary-axis cross-

ing Æ j4 with K ¼ 1105; break-in and breakaway pints coincide at

s ¼ À 4. For K < 0, real-axis branches: À1 to þ1 and À 7 to À1;
g ¼ Æ 90, 180; angles of departure; 0, Æ90, 180; imaginary-axis

crossing: s ¼ 0 with K ¼ À175.

ðbÞ À 175 < K < 1105 ðcÞ K ¼ 81
(1) (a) For K ¼ À 50K1: three branches; g ¼ Æ 90 for K1 <0; g ¼ 180
for K1 >0; s0 ¼ À 9.5; breakaway points: À 7.72 for K > 0, À 3.139 for
K < 0; break-in point: 5.362 for K < 0; imaginary-axis crossings:
at s ¼ 0 for K ¼ 50 and Æ j3.6993 for K ¼ À66.32.

(b) Stable for À66.32 < K < 50

(a) Real-axis branches: À3.2 to À1 (Note: The canceled pole and zero

at s ¼ À5 is a root of the characteristic equation and is not utilized

in determining the root locus for locating the remaining roots.)
6-branches: g ¼ Æ 36, Æ108, 180, s0 ¼ À 22.5; Break-in point:
s ¼ À16.5; Angle of departures (given in order of closeness to the
imaginary axis in the 2nd quadrant): 144.4, À94.4, 4.2;
Imaginary-axis crossing: Æ j17.20 for K ¼ 1.869 Â107.

(b) KGKH ¼ 1:0735; p1ó 2 ¼ Æj17:21ó p3ó 4 ¼ 33:28 Æ j28:05ó p5 ¼
À4:634ó p6 ¼ À44:46ó p7 ¼ À5

ðcÞ C ðsÞ ¼ ðs þ 4:634Þðs þ 5Þðs 1:869 Â 107ðs þ 3:2Þ þ 33:28 Æ j28:05Þ
RðsÞ þ 44:46Þðs Æ j17:21Þðs

7.12. ð1Þ ðaÞ A ¼ 1:44

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804 Linear Control System Analysis and Design

CHAPTER 8

8.1. (b) (1) Gð j0þÞ ¼ 1ff À90ó Gð j1Þ ¼ 0ff À 270ó ox ¼ oc ¼ 2:236,
Gð joxÞ ¼ À0:2223ó of ¼ 0:867
(2) Gð j0þÞ ¼ 1ff À90ó Gð j1Þ ¼ 0ff À360ó ox ¼ oc ¼ 9:75
Gð joxÞ ¼ À0:0952ó of ¼ 0:4977

8.4. (1) ðbÞ Gð j0þÞ ¼ 1ff À90ó Gð j1Þ ¼ 0ff À270ó ox ¼ 35:4ó
Gð joxÞ ¼ À0:05645

8.6. (a) It must have an initial slope of À 40 dB/decade
(b) The phase-angle curve approaches an angle of 180 as o ! 0

8.9. Case II:

GðsÞ ¼ sð1 þ 2337:5ð1 þ 0:25sÞ 0:05sÞ
5sÞð1 þ 0:125sÞð1 þ

8.10. ðaÞ K0 ¼ 31:7 ðbÞ K2 ¼ 2:56 ¼ ðoyÞ2
8.11. (a) Plant 1

GðsÞ ¼ sðs 0:4 j4Þ
þ1Æ

8.13. (a) N ¼ À 2, ZR ¼ 2, unstable (c) N ¼ 0, ZR ¼ 0, stable
8.14. (a) For stability K > 1.26759
8.15.
8.16. (a) Stable for K > À2 (e) Stable for 18 < K < 90.12

(1) ðaÞ K1 ¼ 5:5654ó of ¼ 4:845 ðbÞ K1 ¼ 3:345ó of ¼ 3:15
ðcÞ maximum K1 ¼ 1:9429 Â 105
(2) ðaÞ K0 ¼ 20:13ó of ¼ 8:3 ðbÞ K0 ¼ 8:58ó of ¼ 4:61
ðcÞ Stable for 0 < K0 < 1

CHAPTER 9

9.1. ð2Þ K1 ¼ 7:94ó Mm ¼ 1:326ó om ¼ 7:81ó of ¼ 6:94ó oc ¼ 15:81
(4) K1 ¼ 0:7075ó Mm ¼ 1:318ó om ¼ 0:658ó of ¼ 0:6054ó oc ¼ 1:508

9.3. Mm ¼ 1:30ó K1 ¼ 2:975ó om ¼ 2:75
9.5. (1) (a) Mm ¼ 1ó om ¼ 0 ðbÞ Mm ¼ 1:326ó om ¼ 5:4 ðcÞ Mm ¼ 1:016ó

om ¼ 2:24
9.8. (4) ðaÞ Mm ¼ 5:8904ó om ¼ 3:82 ðcÞ K0 ¼ 0:18 for Mm ¼ 1:12

ðdÞ om ¼ 0:39ó g ¼ 56:16
9.11. Using a CAD program: (a) Mm ¼ 1.11293, om ¼ 0.435 (b) K ¼ 1.338

(c) a ¼ 6

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Answers to Selected Problems 805
CHAPTER 10
10.3, 10.18.

System (1) Dominant Other K1ó sÀ1 Tp,s ts,s Mo
roots roots

Basic À1.25 Æ j 2.166 À6.501 1.72 1.67 3.36 0.14
Lag-compensated:
À1.03 Æ j 1.787 À5.267 16.16 1.8 5.37 0.41
a ¼ 10, T ¼ 2 À0.726

Lead-compensated: À2.23 Æ j 3.86 À3.123 3.13 0.916 1.78 0.10
a ¼ 0.1, T ¼ 0.25 À40.42

Lag-lead-compensated: À0.598 30.18 0.967 4.15 0.244
a ¼ 10, T1 ¼ 2, T2 ¼ 0.25 À2.08 Æ j 3.6 À2.897
À40.40

Tachometer-compensated: À1.2 1.02 — 3.39 0.0
À4.28 2.97 0.756 1.89 0.13
A ¼ 1.61, Kt ¼ 1, z ¼ 0.5 À3.40 Æ j 5.89
A ¼ 4.79, Kt ¼ 1, z ¼ 0.3 À1.86 Æ j 5.92

10.6. (a) a ¼ 4, b ¼ 2.6653 (b) a ¼ 1.126 (c) lag compensator
10.7.
(a) For the uncompensated system: from the performance specifica-
10.18.
tions, for z ¼ 0.5, are Mp ¼ 1.16, tp ¼ 2 s, ts ¼ 4.39 s, K ¼ 2.933,
K1 ¼1.76 sÀ1, and the control ratio poles are À 0.9235 Æ j 1.601
and À5.153.

(b) jsj ! 4=Ts ¼ 2; assuming a ¼ 2, z ¼ 0.5, and jsj ¼ 2 yields
b ¼ 4.726

(c) A ¼ 5.21 and a ¼ 0.4232

(d) A lead compensator; Mp ¼1.59, tp ¼ 0.917 s, ts ¼ 2.02 s, and
K1 ¼3.88 sÀ1

(a) For z ¼ 0.5 the uncompensated system yields the following data:
K ¼ 25.01, K1 ¼5 sÀ1, Mp ¼1.163, tp ¼ 0.725 s, ts ¼ 1.615 s.

(b) Based upon the desired values of z and s ¼ À7, the desired

dominant poles are P1,2D ¼ À 7 Æ j12.124. One possible solution
is: let a ¼ 40, b ¼ 0.01, c ¼ 0.1; by applying the angle condition

and the use of a CAD package the following results are obtained:
A ¼ 6.0464, Kh ¼11.18, K1 ¼126.1sÀ1, Mp ¼1.161, tp ¼ 0.27 s,
ts ¼ 0.426 s.

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806 Linear Control System Analysis and Design

10.20. See Prob.10.3 for basic system performance data. Based on the desired
performance specifications, the desired dominant poles are:
P1ó 2D ¼ À4 Æ j6:928. Let Kx ¼ 1250 K1 be the gain term of G(s). By
applying method 1, the following results are obtained: Kx ¼ 760.71,
Kt ¼ 0.5619, b ¼11.895, Mp ¼1.092, tp ¼ 0.725 s, ts ¼ 0.97 s, K1 ¼
2.369 sÀ1. The other poles are: À6.296 Æ j4.486 and À11. The plot of
c(t) has the desired simple second-order time response characteristics
due to the effect of the real root.

CHAPTER 11

11. 1. ðaÞ K ¼ 11935, of ¼ 3:10, g ¼ 51:16, Mp ¼ 1:161, tp ¼ 0:8654 só
ts ¼ 1:807 só K1 ¼ 3:6387 sÀ1:

(b) 1=T ¼ 0:302ó K ¼ 11935ó A ¼ 7:7084ó Mm ¼ 1:18ó
Mp ¼ 1:203ó tp ¼ 1:104 só ts ¼ 5:455 só K1 ¼ 28:05 sÀ1;
of ¼ 2:53ó g ¼ 51:1:
Results for an Mm ¼ 1:16 are not as good as for Mm ¼ 1:18:

(c) T ¼ 0:25, K ¼ 213200, Mm ¼ 1:564, Mp ¼ 1:186, tp ¼ 0:3125 só
ts ¼ 1:273 s, K1 ¼ 6:50 sÀ1; for Mm ¼ 1:16: Kx ¼ 175800, Mp ¼
1.076, tp ¼ 0:3372 só ts ¼ 1:105 só K1 ¼ 5:36 sÀ1ó of ¼ 7:10,
g ¼ 66:46:

(d) T1 ¼ 3:311, T2 ¼ 0:25, Kx ¼ 209000, Mm ¼ 1:564, Mp ¼ 1:212,
tp ¼ 0:3180 s, ts ¼ 2:857 s, K1 ¼ 63:72 sÀ1; for Mm ¼ 1:16: K ¼
175800, Mp ¼ 1:104, tp ¼ 0:3442 s, ts ¼ 3:526 só K1 ¼ 54:87 sÀ1ó
of ¼ 7:35ó g ¼ 62:71:
!
1 1 þ Ts B
11. 5. ðaÞ GðsÞ ¼ ó where T ¼ and ¼ 0:5
8K 1 þ Ts K

11. 12. ðaÞ At o ¼ 8 : Gð joÞ ¼ Cð joÞ=Eð joÞ % 6:3ff À119ó thus K1 ¼ 250ó
of ¼ 8ó and g ! 40:

ðbÞ Gc ¼ ðs þ 1Þð0:5s þ 1Þ
ð10s þ 1Þð0:05s þ 1Þ

11.17. (a) K2x ¼ 2.02, ofx ¼ 2.15, Mm ¼1.398, Mp ¼1.29, tp ¼1.39 s, ts ¼ 3.03 s
(c) For g ¼ 50: u ¼ ofT ¼ 8:6T ¼ 1:19 yields T ¼ 0.1384. From the
graphical construction, with o1 of=20 and o2 ¼ 20 of ¼ 172,
obtain Kt ¼ 24.28 and A ¼ 180, thus

C ðsÞ ¼ 2907ðs þ 1Þðs þ 7:225Þ
EðsÞ s2ðs þ 1:108Þðs þ 406:3Þ

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Answers to Selected Problems 807

The 0-dB crossing slope is À12 dB/oct at o ¼ 6, but the zero at
À7.225 will have a moderating effect on the degree of stability.
(d ) Mm ¼1.47, Mp ¼1.30, tp ¼ 0.34 s, ts ¼1.03 s, K2 ¼ 46.65; since the
values of Mp are essentially the same, a good improvement in
system performance has been achieved. g ¼ 51 and of ¼ 9:12.

CHAPTER 12

12. 2. GcðsÞ 2:31ðs þ 20Þ
sþ 21

12.4. With this cascade compensator Mp ¼1.12, tp ¼1.493 s, ts ¼ 2.941 s,
K1 ¼1.53. The specified MT (s) yields Mp ¼1.123, tp ¼1.493, ts ¼ 2.948,
K1 ¼1.424.
As a first trial, use

Gc % 2:994ðs þ 2:1Þ
s þ 2:223

12.8. With this cascade compensator Mp ¼1.03, tp ¼ 2.37 s, ts ¼ 2.94 s,
Kt ¼ 0.75. The specified MT (s) yields Mp ¼1.06, tp ¼ 2.2 s, ts ¼ 3.04 s,
K1 ¼0.79.
(a) The pole at the origin of H(s) ensures c(t)ss ¼ 0; make pole Àb

nondominant: b ¼ 100

ðbÞ KH ¼ 2500
(c) Chxaracteristic equation: s4 þ104s3 þ10,400s2 þ120,000s þ

320,000 ¼ 0. Routh stability criterion reveals that all roots lie

in the left-half s plane. A plot of Lm GxH reveals that its 0-dB
axis crossing slope is À6 dB/oct, indicating a stable system.

The system poles are p1 ¼ À4, p2 ¼ À 8.06, p3,4 ¼ À 45.97 Æ
j311.7.jy(tp)j ¼ 0.0025, tp ¼ 0.163 s. For jy(tp)j ¼ 0.01, KH ¼ 630,
which produces very little change in the value of tp.

CHAPTER 13

13.3. (1) (a) Not observable, (b) controllable, (c)GðsÞ ¼ 2=ðs À 2Þ, (d) one
13.5. observable and two controllable states, (e) unstable
(4) (a) Observable, (b) controllable, (c) GðsÞ ¼ ðs À 1Þ=ðs þ 1Þ ðs þ 2Þ,
(d) Two observable and two controllable states, and (e) stable
(a) Add a cascade compensator GcðsÞ ¼ 1=ðs þ 1Þ
ðbÞ HeqðsÞ ¼ k3s2 þ ð2k3 þ k2Þs þ k1

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808 Linear Control System Analysis and Design

(d) A desired control ratio is

MT ðsÞ ¼ Y ðsÞ ¼ ðs2 þ 4s A þ 25Þ
RðsÞ þ 13Þðs

ðeÞ With A ¼ 162:5ó kT ¼ ½1 0:16923 0:08Š
ðgÞ GeqðsÞ ¼ 325=½sðs2 þ 29s þ 113ފó K1 ¼ 2:876

ðhÞ Mp ¼ 1:12ó tp ¼ 1:09 só ts ¼ 1:66 s

13.8. For system (1) (a)

M ðsÞ ¼ ðs þ 12 11ó700 þ 20Þ
Æ j21Þðs

k1 ¼ 1 k2 ¼ 0:091026 k3 ¼ 0:003077

(b) yðtÞ ¼ 1 þ 1:025eÀ12t sin ð21t þ 171:1Þ À 1:1584eÀ20t
Mp ¼ 1:057ó tp ¼ 0:215 só ts ¼ 0:273 só K1 ¼ 11 sÀ1

13.12. (a)

MT ðsÞ ¼ Y ðsÞ ¼ ðs2 1:467ðs þ 1:5Þðs þ 2Þ 2Þ
RðsÞ þ 2s þ 2Þðs þ 1:1Þðs þ

where Gc ðsÞ s þ 1:5 ó A ¼ 1:467
sþ1

Y ðsÞ=RðsÞ

¼ s4 þ ½7 þ Að0:5k4 þ k3 þ Aðs þ 1:5Þðs þ 2Þ þ 2:5k3 þ 3:5k2 þ k1ފs2
k2ފs3 þ ½11 þ Að3k4

þ ½5 þ Að2:5k4 þ 1:5k3 þ 3k2 þ 3:5k1ފs þ 3Ak1

where k1 ¼ 1ó k2 ¼ 0:8635ó k3 ¼ À2:43541ó k4 ¼ 0:5521
ðcÞ Mp ¼ 1:015ó tp ¼ 3:75 só ts ¼ 2:79 só K1 ¼ 0:524 sÀ1
13.16. (a)

195ðs þ 2Þ
MT ðsÞ ¼ ðs2 þ 2s þ 2Þðs þ 1:95Þðs þ 100Þ

(b) By applying the Guillemin-Truxal method, a reduced-order

compensator is obtained by dividing the denominator of Gc(s) by
its numerator to obtain Gc(s) % 1/(s þ 99.9). The figures of merit
for MT(s) are Mp ¼1.042, tp ¼ 3.18 s, and ts ¼ 4.23 s, where for the
system implemented with the reduced-order compensator

Mp ¼1.056, tp ¼ 3.12 s, and ts ¼ 4.44 s.
 Ã
13.19. kT ¼ À8 À2:5

5 1 1 !
3 3 À1
x1ðt Þ ¼ eÀ3t À eÀ6t x2ðtÞ ¼ À5eÀ3t þ 4eÀ6t xð0Þ ¼

13.20. (1) (a) Completely controllable, (b) s ( Ap) ¼ {À1, À2, À3},

Copyright © 2003 Marcel Dekker, Inc.

Answers to Selected Problems 809

23
6 5 1
ðcÞ T ¼ 64 3 2 1 75ó ðd Þ kT ¼ Â À33 Ã
À42 À6

021

CHAPTER 14

14.1. For m ¼1: (a)

SkT ðsÞjk¼15 ¼ s4 s4 þ 7s3 þ 25s2 þ 39s
þ 7s3 þ 25s2 þ 54s þ 75

14.6. (a) p ¼ À21, KG ¼ 30.8 (b) No; need a compensator of the form

Gc ¼ ðs Aðs þ aÞ cÞ
þ bÞðs þ

14.10. For system (2), (a) Three equilibrium points are

23 23 23
0 À21 À12
xb ¼ 64 0 75 xc ¼ 46 0 57
xa ¼ 64 0 75

0 2 À2

(b) For xb,

2 3
01 0

Jxb ¼ 64 2 À3 0:5x3 75

2 0 2x1 þ 1 xo
3

CHAPTER 15

15.5. ðaÞ zÀ1 þ 1:8zÀ2 þ 2:19zÀ3 þ 2:252zÀ4 þ Á Á Á
15.7. ðaÞ f ð0Þ ¼ 0; system is stableó thus f ð1Þ ¼ 1:95122
15.8. ðaÞ ð1 À 2zÀ1 þ 1:41zÀ2 À 0:41zÀ3ÞCðzÞ ¼ ðzÀ1 À 0:2zÀ2ÞEðzÞ
By use of Theorem 1:

cðkT Þ ¼ 2c½ðk À 1ÞT Š À 1:41c½ðk À 2ÞT Š þ 0:41c½ðk À 3ÞT Š

þ e½ðk À 1ÞT Š À 0:2e½ðk À 2ÞT Š

15.11. System 2
CÃðsÞ ¼ G1G2ÃðsÞM ÃðsÞ

Copyright © 2003 Marcel Dekker, Inc.

810 Linear Control System Analysis and Design

C Ã ðsÞ ¼ 1 þ G1G2ÃðsÞ
RÃðsÞ G1H ÃðsÞ þ G1G2ÃðsÞ

CðzÞ ¼ G1G2ðzÞ

RðzÞ 1 þ G1H ðzÞ þ G1G2ðzÞ

CHAPTER 16 z-domain value
16.1. (1)
Eq. (16.8) Z ¼ esT
s Plane
À 1.5 0.94176 0.94176
À3 0.8868 0.88692
À 20 0.4286 0.44933
À2Æj3 0.9167 Æ j 0.1106 0.91671 Æ j 0.11054
10 Æ j 25 0.4201 Æ j 0.5917 0.39208 Æ j 0616063

16.2. (2) zTU ¼ À0.2857, z ¼ 0.006738ödifference due to warping
16.3. (3) zTU ¼ À0.71598 Æ j0.11834, z ¼ 0.20586 Æ j0.69589ödifference due
to warping
(1)

!
CðsÞ Ks þ 8K
RðsÞ T ¼ ðs2 þ 10s þ 174Þðs þ 10Þ ! 8K

¼ ð10Þð174Þ ! K ¼ 217:5

(b) For T ¼ 0.1! Æj(0.6114/T ) ¼ Æ j6.114 and (À0.1)/T ¼ À 1!
thus, the zero and poles of the desired control ratio do not lie in the
good Tustin region of Fig. 16.5. Thus:

0:6114 ! 7 ! T 0:08734ó and 0:1 ! 10 ! T 0:1 0:01;
T T 10

therefore,T 0.01.

16.5.

Dc ðz Þ ¼ Kcðz À 0:6703Þ where
ðz À 0:06703Þ

K ¼ 3:795 Â 10À2K1 À 0:2566 ! Kc ¼ 6:7615

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Answers to Selected Problems 811

Mp ¼ 1:062ó tp ¼ 0:4 só ts ¼ 0:6þsó and K1 ¼ 5:1577 sÀ1

CðsÞ ¼ ðs þ 0:4061 Æ j 251:2ðs þ 0:005Þ þ 3:183Þðs þ 200þÞ
RðsÞ 0:4747Þðs þ 0:0005Þðs

Mp ¼ 1:078ó tp ¼ 6:98 só ts ¼ 10:7 só K1 ¼ 4:187ó and A ¼ 10

The remaining solutions are left to the instructor to obtain.
16.10. (a)

Gz ðz Þ ¼ 4:967 Â 10À5Kxðz þ 0:9934Þ ! Kx ¼ 2:34105
ðz À 1Þðz À 0:9802Þ

CðzÞ ¼ 1:163 Â 10À4ðz þ 0:9934Þ
RðzÞ ðz À 0:9900 Æ j0:01152Þ

(b) The PCT block diagram is similar to that of Fig. 16.14 d with
DcðsÞ ¼ 1 and GpaðsÞ=T ¼ 200=ðs þ 200Þ ¼ GAðsÞ:

GPC ðsÞ ¼ GA Gx ðsÞ ¼ sðs 200Kx 200Þ ó Kx ¼ 2:34105
þ 2Þðs þ

CðsÞ ¼ GPC ðsÞ ¼ 466:21

RðsÞ 1 þ GPC ðsÞ ðs þ 0:99408 Æ j1:159Þðs þ 200:012Þ

The continuous-time system control ratio, for z ¼ 0.65, is:

CðsÞ ¼ GxðsÞ ¼ 2:368
RðsÞ 1 þ GxðsÞ s þ 1 Æ j1:170

(c) DIGöTo reduce ts by about 1/2, select DcðsÞ ¼ Aðs þ 2Þ=ðs þ 5Þ;
GðsÞ ¼ DcðsÞGpðsÞ ¼ À1 for z ¼ 0.65: K ¼ 200AKx ¼ 2874.6;
Kx ¼ 2.368, and A ¼ 6.0705.

CðsÞ ¼ ðs þ 2:463 Æ K þ 200:1Þ ;
RðsÞ j2:881Þðs

½Dc ðz ފTU ¼ 5:982ðz À 0:9802Þ
z À 0:9512

CðzÞ ¼ ½DcðzފTU GzðzÞ ¼ 7:0359  10À4ðz þ 0:9934Þ
RðzÞ 1 þ ½DcðzފTU GzðzÞ z À 0:9752 Æ j0:02810

DIRöFor Kx ¼ 2.368 and z ¼ 0.65, Dc(z) = A(z À 0.9802)/
ðz À 0:09802Þ where A ¼ 1865:3.

CðzÞ ¼ DcðzÞGzðzÞ ¼ 0:2194ðz þ 0:9934Þ
RðzÞ 1 þ DcðzÞGzðzÞ z À 0:4393 Æ j0:3507

Copyright © 2003 Marcel Dekker, Inc.

812 Linear Control System Analysis and Design

(d ) DIGöFor Kx ¼ 2.368 and z ¼ 0.65, Dc(s) = 0.1A(s + 0.01)/
ðs þ 0:001Þ and A ¼ 9:384893:

CðsÞ ¼ DcðsÞGPC ðsÞ
RðsÞ 1 þ DcðsÞGPC ðsÞ

CðsÞ ¼ ðs þ 0:98955 Æ 468:4ðs þ 0:01Þ þ 200:01Þ
RðsÞ j1:1595Þðs þ 0:01008Þðs

½Dc ðz ފTU ¼ 0:9385315ðz À 0:9999Þ
z À 0:99999

CðzÞ ¼ ½DcðzފTU GzðzÞ
RðzÞ 1 þ ½DcðzފTU GzðzÞ

¼ 1:1633 Â 10À4ðz þ 0:9934Þðz À 0:9999Þ
ðz À 0:990087 Æ j0:01148126Þðz À 0:99989923Þ

DIRöFor Kx ¼ 2.368, z ¼ 0.65, and

½Dc ðz ފTU ¼ Kzcðz À 0:9999Þ ! Kzc ¼ 0:986259 obtain:
z À 0:99999

CðzÞ ¼ DcðzÞGzðzÞ ¼ 1:1163Â10À4ðz þ0:9934Þðz À0:9999Þ
RðzÞ 1þDcðzÞGzðzÞ ðz À0:990087Æj0:01145146Þðz À0:99989923Þ

(e) The tables summarize the results of the various designs presented
on the following page. The desired improvements using s-domain
compensation design rules (for both s- and z-domain analysis)
are obtained. Since the dominant poles and zeros lie in the
good Tustin region for the lag and lead units, there is a good
correlation between the DIG and DIR results.

System Zeros Poles of C/R
Domain of C/R

Basic (a) Continuous s — À1.000 Æ j 1.170

and (b)

DIG s — À0.9941 Æ j 1.159, À200.012

DIR z À0.9934 0.9900 Æ j 0.01152

(c) Lead DIG s — À2.463 Æ j 2.881, À200.1

z À0.9934 0.9752 Æ j 0.02810

DIR z À0.9934 0.4393 Æ j 0.3507

(d) Lag DIG s À0.01 À 0.9895 Æ j 1.160, À0.01008, À200.01

z À0.9934 Æ0.990087 Æ j 0.011481, 0.999899

DIR z 0.9999 0.990087 Æ j 0.001145146, 0.999899

Copyright © 2003 Marcel Dekker, Inc.

Answers to Selected Problems 813

System Domain Kx Mp tp s ts s K1 sÀ1

Basic (a) and (b) Continuous s 2.368 1.0682 2.69 3.90 1.184

DIG s 2.341 1.0675 2.72 3.94 1.165

DIR z 2.341 1.068 2.70 3.93 1.171

(c) Lead DIG s 2.368 1.068 1.096 1.59 2.875

z 1.068 1.09 1.58 2.874

DIR z 1.067 0.05 0.07 2.84

(d) Lag DIG s 1.0767 2.71 4.15 11.71

z 1.077 2.71 4.16 11.71

DIR z 1.076 2.72 4.17 11.68

16.13. (a)

Gz ðz Þ ¼ GzoGx ðzÞ ¼ 2:073 Â 10À7ðz þ 3:695Þðz þ 0:2653Þ
ðz À 1Þðz À 0:9900Þðz À 0:9704Þ

CðzÞ ¼ 2:073 Â 10À7ðz þ 3:695Þðz þ 0:2653Þ
RðzÞ ðz À 0:9959 Æ j0:0047756Þðz À 0:9687Þ

Mp ¼ 1:067ó tp ¼ 6:94 só ts ¼ 9:92 só and K1 ¼ 0:41604 sÀ1

(b)

GðsÞ ¼ GA ðsÞGx ðsÞ ¼ sðs þ 200Kx þ 200Þ
1Þðs þ 3Þðs

C ðsÞ ¼ ðs þ 0:4082 Æ 251:2 3:184Þðs þ 200þÞ
RðsÞ j0:4773Þðs þ

Mp ¼ 1:067ó tp ¼ 6:94 só ts ¼ 9:90 só and K1 ¼ 0:4187 sÀ1

(c) PCT lead designöDcðsÞ ¼ KscðsÞðs þ 1Þ=ðs þ 3Þ

CðsÞ ¼ ðs þ 0:8439 Æ 1454 þ 4:312Þðs þ 200þÞ
RðsÞ j0:9866Þðs

Mp ¼ 1:064ó tp ¼ 3:47 só ts ¼ 4:88 só K1 ¼ 0:808 sÀ1ó and
A = 5.7882

(d ) PCT lag designöDcðsÞ ¼ ðA=10Þ½ðs þ 0:005Þ=ðs þ 0:0005ފ

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