This document provides solutions to problems from various chapters related to control systems. The problems cover topics such as transfer functions, block diagrams, time response analysis, stability analysis, root locus analysis and frequency response analysis. Specific solutions include closed-loop transfer functions, state space representations, time domain specifications, pole-zero locations, stability criteria, compensator design and performance analysis.

1. BANSW 05/18/2011 14:14:20 Page 897
Answers
to Selected Problems
Chapter 1
19. c. xðtÞ ¼
2
5
À eÀ4t 2
5
cos 3t þ
8
15
sin 3t
20. b. xðtÞ ¼ ÀeÀt
þ 9teÀt
þ 5eÀ2t
þ t À 2
Chapter 2
3. b. xðtÞ ¼
15
26
eÀ2t
À
3
10
eÀ4t
À
18
65
cos 3t À
1
65
sin 3t
7.
YðsÞ
XðsÞ
¼
s3
þ 6s2
þ 2s þ 8
s3 þ 5s2 þ 7s þ 1
8. c.
d3
x
dt3
þ 10
d2
x
dt2
þ 11
dx
dt
þ 18x ¼
dfðtÞ
dt
þ 2fðtÞ
16. a.
VoðsÞ
ViðsÞ
¼
1
2
s þ 1
18. b.
VoðsÞ
ViðsÞ
¼
s2
þ 2s þ 2
s4 þ 2s3 þ 3s2 þ 3s þ 2
33.
u2ðsÞ
TðsÞ
¼
3
20s2 þ 13s þ 4
34.
u2ðsÞ
TðsÞ
¼
10
800s2 þ 4000s þ 550
43.
u2ðsÞ
EaðsÞ
¼
0:0833
sðs þ 0:75Þ
897

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Chapter 3
1. _x ¼
À
2
3
À
1
3
1
3
À
1
3
À
2
3
2
3
À
1
6
À
1
3
À
1
6
2
6
6
6
6
6
6
4
3
7
7
7
7
7
7
5
iL1
iL2
no
2
4
3
5 þ
2
3
1
3
1
6
2
6
6
6
6
6
6
4
3
7
7
7
7
7
7
5
ni
y ¼ 0 0 1½ Š
iL1
iL2
no
2
4
3
5
Note: L1 is left-most inductor in Figure P3.1 in the text.
11. a. _x ¼
0 1 0 0
0 0 1 0
0 0 0 1
À17 À5 À1 À5
2
6
6
6
4
3
7
7
7
5
x þ
0
0
0
1
2
6
6
6
4
3
7
7
7
5
rðtÞ
cðtÞ ¼ 10 7 0 0½ Šx
14. a.
YðsÞ
RðsÞ
¼
10
s3 þ 5s2 þ 2s þ 3
19. _x ¼
À
Deq
Jeq
0
Kt
Jeq
N1
N2
1 0 0
À
Kb
La
N2
N1
0 À
Ra
La
2
6
6
6
6
6
4
3
7
7
7
7
7
5
vL
uL
ia
2
6
4
3
7
5 þ
0
0
1
La
2
6
6
4
3
7
7
5ea
y ¼ 0
N2
N1
0
vL
uL
ia
2
4
3
5
Chapter 4
14. xðtÞ ¼
1
5
1 À
ffiffiffi
5
p
2
eÀt
cosð2t þ 26:6
Þ
20. a. z ¼ 0:375; vn ¼ 4 rad/s; Ts ¼ 2:67 s; Tp ¼ 0:847 s; %OS ¼ 28:06
23. a. s ¼ À20þÀ j29:59
35. s ¼ À6:3723 and À 0:6277
36. a. s3
À 8s2
À 11s þ 8 ¼ 0 b. s ¼ 9:11; 0:534; À1:64
40. yðtÞ ¼
2
5
À
2
5
eÀ5t
43.
FðtÞ ¼
1:0455eÀ0:20871t
À 0:045545eÀ4:7913t
0:21822eÀ0:20871t
À 0:21822eÀ4:7913t
À0:21822eÀ0:20871t
þ 0:21822eÀ4:7913t
À0:045545eÀ0:20871t
þ 1:0455eÀ4:7913t
#
898 Answers to Selected Problems

3. BANSW 05/18/2011 14:14:20 Page 899
xðtÞ ¼
1:0455eÀ0:20871t
À 0:045545eÀ4:7913t
À0:21822eÀ0:20871t
þ 0:21822eÀ4:7913t
#
yðtÞ ¼ 0:60911eÀ0:20871t
þ 0:39089eÀ4:7913t
73. D ¼ 0:1824 N-m-s=rad
76. R ¼ 456 V
Chapter 5
2.
CðsÞ
RðsÞ
¼
G3ðG1G2 þ 1Þ
1 þ G1H1
4.
CðsÞ
RðsÞ
¼
G1G2 þ G3
1 þ G3H þ G1G2H þ G2G4
6.
CðsÞ
RðsÞ
¼
G1G5
1 þ G1G2 þ G1G3G4G5 þ G1G3G5G6G7 þ G1G5G8
9.
CðsÞ
RðsÞ
¼
G4G6 þ G2G5G6 þ G3G5G6
1 þ G6 þ G1G2 þ G1G3 þ G1G2G6þ
G1G3G6 þ G4G6G7 þ G2G5G6G7 þ G3G5G6G7
26.
CðsÞ
RðsÞ
¼
G1G2G3G4
3 þ G2G3G4 þ 2G3G4 þ 2G4
27.
CðsÞ
RðsÞ
¼
G1G6G7ðG2 þ G3ÞðG4 þ G5Þ
1 À G6G7H3ðG2 þ G3ÞðG4 þ G5Þ À G6H1 À G7H2 þ G6G7H1H2
28.
CðsÞ
RðsÞ
¼
s3
þ 1
2s4 þ s2 þ 2s
30. b. _x ¼
À5 1 0 0
0 À5 0 0
0 0 À7 1
0 0 0 À7
2
6
6
6
4
3
7
7
7
5
x þ
0
1
0
1
2
6
6
6
4
3
7
7
7
5
rðtÞ
y ¼ À
3
4
1 À
5
4
À1
x
37. _x ¼
0 1 0 0
À2 0 1 0
0 0 0 1
À1 À1 0 0
2
6
6
6
4
3
7
7
7
5
x þ
0
0
0
1
2
6
6
6
4
3
7
7
7
5
r
y ¼ c ¼ 1 1 0 0½ Šx
74. DL ¼ 4192:11 N-m-s/rad
Chapter 6
1. 2 rhp, 3 lhp
3. 3 rhp, 2 lhp
Answers to Selected Problems 899

4. BANSW 05/18/2011 14:14:20 Page 900
4. 1 rhp, 0 lhp, 4 jv
5. 0 rhp, 2 lhp, 2 jv
9. Stable (0 rhp, 4 lhp)
17. 1 rhp, 2 lhp, 4 jv
23. K 1
39. a. À3 K 14:22; b. 1:291 rad=s
41. a. 0 K 14:69; b. K ¼ 14:69; c. s ¼ þÀ j1:069 and s ¼ À3:5 þÀ j1:558
42. À
2
3
K 0
Chapter 7
4. estepð1Þ ¼ 0; erampð1Þ ¼ 105; eparabolað1Þ ¼ 1
7. _eð1Þ ¼ 0:9
10. a. %OS ¼ 16:3%; b. Ts ¼ 0:1143 sec; c. estepð1Þ ¼ 0;
d. erampð1Þ ¼ 0:0714; e. eparabolað1Þ ¼ 1
13. a. Kp ¼ 0:25; Kn ¼ 0; Ka ¼ 0; b. eð1Þ ¼ 40; 1; 1; respectively;
c. Type 0
20. K ¼ 110; 000
27. b ¼ 1; K ¼ 1:16; a ¼ 7:76; or b ¼ À1; K ¼ 5:16; a ¼ 1:74
31. a. K ¼ 207; 936; a ¼ 415:872; b: K ¼ 3; 492:8; a ¼ 69:86
35. K1 ¼ 125; 000; K2 ¼ 0:016
41. a. Step: eð1Þ ¼ 1:098; ramp: eð1Þ ¼ 1
Chapter 8
16. Breakaway point ¼ À1:92; asymptotes: sa ¼ À4; jv-axis crossing ¼ þÀ j 5:91
19. b. Asymptotes: sa ¼ À3; c. K ¼ 194; d. K ¼ 16:9
20. K ¼ 9997; a ¼ 7
23. a. sa ¼ À
5
2
; b. s ¼ À1:38; À3:62; c. 0 K 126; d. K ¼ 10:3
26. b. K ¼ 14:0; c. Ts ¼ 4:23 s; Tp ¼ 1:698 s; d. s ¼ À7:0; e. 0 K 89:7
30. a ¼ 9
39. a. 0 K 5745; b. K ¼ 1090; c. K ¼ 689
42. a. K ¼ 170:1; b. K ¼ 16:95
Chapter 9
1. GcðsÞ ¼
s þ 0:1
s
; K ¼ 92:7 for the uncompensated, and 91.8 for the compen-
sated system; KpO
¼ 2:575; KpN
¼ 1; %OSO ¼ %OSN ¼ 16:3; TsO
¼ 2:53 sec;
TsN
¼ 2:58 sec
900 Answers to Selected Problems

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9. a. s ¼ À2:5 þÀ j5:67; b. Angle ¼ À59:27
; c. s ¼ À5:87
d. K ¼ 225:79; e. s ¼ À11:59; À1:36
10. a. s ¼ À2:2 þÀ j3:81; b. s ¼ À5:694; c. K ¼ 25:1;
d. s ¼ À1:29; f. Ka ¼ 4:4
14. a. GcðsÞ ¼
s þ 7
s þ 37:42
; K ¼ 5452; dominant poles ¼ À4:13 þÀ j10:78
24. a. Kuc ¼ 10; Kc ¼ 9:95; b. Kpuc
¼ 1:25; Kpc
¼ 6:22;
c. %OSuc ¼ %OSc ¼ 4:32;
d. Uncompensated: exact second-order system, approximation OK;
compensated: closed-loop pole at À0:3, closed-loop zero at À0:5, simulate
e. Approach to final value longer than settling time of uncompensated system
f. GLLCðsÞ ¼
404:1ðs þ 0:5Þðs þ 4Þ
ðs þ 2Þðs þ 4Þðs þ 0:1Þðs þ 28:36Þ
yields approximately a 5 times
improvement in speed.
25. GcðsÞ ¼
ðs þ 9:81Þðs þ 0:1Þ
s
; K ¼ 2:077
28. Poles ¼ À0:734 þÀ j1:21; À2:48; zeros—none
Chapter 10
10. System 1: 0 K 490:2; System 2: 0 K 1:4; System 3: 1 K 1
(Answers are from exact frequency response)
11. a. System 1: GM ¼ À6:38 dB; FM ¼ À20:3
(Answers are from exact frequency
response)
15. c. vBW ¼ 2:29 rad=s
23. System 2: Ts ¼ 2:23 sec, Tp ¼ 0:476 s, %OS ¼ 42:62 (Answers are from exact
frequency response)
44. GM ¼ 1:17 dB, FM ¼ 6:01
(Answers are from exact frequency response)
Chapter 11
1. a. K ¼ 2113 (Answer is from exact frequency response)
2. a. K ¼ 2365 (Answer is from exact frequency response)
3. a. K ¼ 575 (Answer is from exact frequency response)
12. GcðsÞ ¼
s þ 2:701
s þ 5:954
; K ¼ 7936
(Answer is from exact frequency response)
21. GcðsÞ ¼
ðs þ 0:097Þðs þ 2:16Þ
s
; K ¼ 25:189
(Answer is from exact frequency response)
Chapter 12
1. d. For function i: TðsÞ ¼
s þ 3
s2 þ ðk2 þ 8Þs þ ðk1 þ 16Þ
Answers to Selected Problems 901

6. BANSW 05/18/2011 14:14:20 Page 902
3. b. For function i: GðsÞ ¼
6:25
s
À
27:5
s þ 10
þ
71:25
s þ 20
; TðsÞ ¼
200ðs2
þ 7s þ 25Þ
4s3 þ as2 þ bs þ c
where a ¼ ð25k3 À 110k2 þ 285k1 þ 120Þ
b ¼ ð750k3 À 2200k2 þ 2850k1 þ 800Þ
c ¼ 5000k3
and C ¼ ½ 1 1 1 Š; B ¼ ½ 71:25 À27:5 6:25ŠT
was used
11. a. Uncontrollable; b. Controllable; c. Controllable
14. K ¼ ½ 92:35 36:78 À7 Š for a characteristic polynomial of
ðs þ 6Þðs2
þ 8s þ 45:78Þ ¼ s3
þ 14s2
þ 93:78s þ 274:7
23. L ¼ ½ À671:19 1472:4 ŠT
for a characteristic polynomial of s2
þ 144s þ 14; 400
Chapter 13
3. a. fðkTÞ ¼ 229:5ð0:4Þk
À 504ð0:6Þk
þ 275:5ð0:8Þk
6. c. GðzÞ ¼ 0:395
ðz þ 0:2231Þ
ðz À 0:2231Þðz2 þ 0:1857z þ 0:04979Þ
8. b. GðzÞ ¼ 0:0517
z2
þ 2:2699z þ 0:2995
ðz À 1Þðz À 0:2231Þðz À 0:4065Þ
9. a. TðzÞ ¼
G1ðzÞG2ðzÞ
1 þ G1ðzÞG2HðzÞ
14. 0 K 15:76
15. a. Kp ¼
1
2
; eÃ
ð1Þ ¼
2
3
; Kn ¼ 0; eÃ
ð1Þ ¼ 1; Ka ¼ 0; eÃ
ð1Þ ¼ 1
17. K ¼ 14:18 for 16:3% of overshoot; 0 K 109:28 for stability
902 Answers to Selected Problems