Norman Nise Control Systems Engineering 6th Solultions.PDF
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Norman Nise Control Systems Engineering 6th Solultions.PDF


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c
v
f \u2212=\u2202
\u2202
0
3 
00
01
1 vT
u
f \u3b2=\u2202
\u2202
 
0
02
1 =\u2202
\u2202
u
f
 
00
01
2 vT
u
f \u3b2\u2212=\u2202
\u2202
 
0
02
2 =\u2202
\u2202
u
f
 
0
01
3 =\u2202
\u2202
u
f
 
*
0
02
3 kT
u
f \u2212=\u2202
\u2202
 
 
Then just by direct substitution. 
 
3-39 Chapter 3: Modeling in the Time Domain 
Copyright © 2011 by John Wiley & Sons, Inc. 
 
b. 
Substituting values one gets: 
 
\u23a5\u23a6
\u23a4\u23a2\u23a3
\u23a1
\u23a5\u23a5
\u23a5
\u23a6
\u23a4
\u23a2\u23a2
\u23a2
\u23a3
\u23a1
\u2212
\u2212+
\u23a5\u23a5
\u23a5
\u23a6
\u23a4
\u23a2\u23a2
\u23a2
\u23a3
\u23a1
\u23a5\u23a5
\u23a5
\u23a6
\u23a4
\u23a2\u23a2
\u23a2
\u23a3
\u23a1
\u2212
\u2212
\u2212+\u2212
=
\u23a5\u23a5
\u23a5
\u23a6
\u23a4
\u23a2\u23a2
\u23a2
\u23a3
\u23a1
2
1
*
0
00
00
*
00
00
*
0
0
0
0
0)(
u
u
kT
vT
vT
v
T
T
ck
Tv
Tvd
v
T
T
\u3b2
\u3b2
\u3b2\u3bc\u3b2
\u3b2\u3b2
&
&
&
 
 
[ ]
\u23a5\u23a5
\u23a5
\u23a6
\u23a4
\u23a2\u23a2
\u23a2
\u23a3
\u23a1
=
v
T
T
y *100 
32. 
a. The following basic equations characterize the relationships between the state, input, and output 
variables for the HEV common forward path of the figure: 
)()( tuKtu cAa \u22c5= 
)()()()()( tktuKtetutIRIL bcAbaaaaA \u3c9\u22c5\u2212=\u2212=\u22c5+\u22c5 & (1) 
)()()( tTtTtTJ cftot \u2212\u2212=\u22c5\u3c9& , where tot m veh wJ J J J= + + , 
)()( tIktT at \u22c5= , )()( tktT ff \u3c9\u22c5= 
b. Given that the state variables are the motor armature current, Ia(t), and angular speed, \u3c9 (t), we re-
write the above equations as: 
 
)()()( tu
L
Kt
L
ktI
L
RI c
a
A
a
b
a
a
a
a +\u22c5\u2212\u22c5\u2212= \u3c9& (2) 
 
)(1)()( tT
J
t
J
k
tI
J
k
c
tottot
f
a
tot
t \u2212\u2212= \u3c9\u3c9& (3) 
In matrix form, the resulting state-space equations are: 
3-40 Chapter 3: Modeling in the Time Domain 
Copyright © 2011 by John Wiley & Sons, Inc. 
 \u23a5\u23a6
\u23a4\u23a2\u23a3
\u23a1
\u23a5\u23a5
\u23a5\u23a5
\u23a6
\u23a4
\u23a2\u23a2
\u23a2\u23a2
\u23a3
\u23a1
\u2212
+\u23a5\u23a6
\u23a4\u23a2\u23a3
\u23a1
\u23a5\u23a5
\u23a5\u23a5
\u23a6
\u23a4
\u23a2\u23a2
\u23a2\u23a2
\u23a3
\u23a1
\u2212
\u2212\u2212
=\u23a5\u23a6
\u23a4\u23a2\u23a3
\u23a1
c
c
tot
a
A
a
tot
f
tot
t
a
b
a
a
a
T
u
J
L
K
I
J
k
J
k
L
k
L
R
I
10
0
\u3c9\u3c9&
& (4) 
 \u23a5\u23a6
\u23a4\u23a2\u23a3
\u23a1\u23a5\u23a6
\u23a4\u23a2\u23a3
\u23a1=\u23a5\u23a6
\u23a4\u23a2\u23a3
\u23a1
\u3c9\u3c9
aa II
10
01
 (5) 
 
 
 
 
Copyright © 2011 by John Wiley & Sons, Inc. 
F O U R 
 
 Time Response 
 
SOLUTIONS TO CASE STUDIES CHALLENGES 
 
Antenna Control: Open-Loop Response 
The forward transfer function for angular velocity is, 
 
G(s) = 
\u3c90(s)
VP(s) = 
24
(s+150)(s+1.32) 
a. \u3c90(t) = A + Be-150t + Ce-1.32t 
b. G(s) = 
24
s2+151.32s+198
 . Therefore, 2\u3b6\u3c9n =151.32, \u3c9n = 14.07, and \u3b6 = 5.38. 
c. \u3c90(s) = 24s(s2+151.32s+198) = 
 
Therefore, \u3c90(t) = 0.12121 + .0010761 e-150t - 0.12229e-1.32t. 
d. Using G(s), 
\u3c90
\u2022\u2022 +151.32\u3c90
\u2022 +198\u3c9 0 = 24vp (t) 
Defining, 
x1 = \u3c90
x2 = \u3c90\u2022
 
Thus, the state equations are, 
 
x1
\u2022 = x2
x2
\u2022 = \u2212198x1 \u2212151.32x2 + 24vp(t)
y = x1
 
In vector-matrix form, 
 
x
\u2022 = 0 1\u2212198 \u2212151.32
\u23a1 
\u23a3 \u23a2 
\u23a4 
\u23a6 \u23a5 x +
0
24
\u23a1 
\u23a3 \u23a2 
\u23a4 
\u23a6 \u23a5 vp (t); y = 1 0[ ]x 
4-2 Chapter 4: Time Response 
 
Copyright © 2011 by John Wiley & Sons, Inc. 
 
e. 
Program: 
'Case Study 1 Challenge (e)' 
num=24; 
den=poly([-150 -1.32]); 
G=tf(num,den) 
step(G) 
 
Computer response: 
ans = 
 
Case Study 1 Challenge (e) 
 
 
Transfer function: 
 24 
------------------- 
s^2 + 151.3 s + 198 
 
 
 
 
Ship at Sea: Open-Loop Response 
a. Assuming a second-order approximation: \u3c9n2 = 2.25, 2\u3b6\u3c9n = 0.5. Therefore \u3b6 = 0.167, \u3c9n = 1.5. 
Ts = 
4
\u3b6\u3c9n = 16; TP = 
\u3c0
\u3c9n 1-\u3b62 
 = 2.12 ; 
%OS = e-\u3b6\u3c0 / 1 - \u3b62 x 100 = 58.8%; \u3c9nTr = 1.169 therefore, Tr = 0.77. 
b. \u3b8 s 2.25
s s 2 0.5 s 2.25+ +
= = 1
s
s 0.5+
s 2 0.5 s 2.25+ +
\u2212 
 = 1
s
s 0.25+ 0.25
2.1875
2.1875+
s 0.25+ 2 2.1875+
\u2212 
 Answers to Review Questions 4-3 
Copyright © 2011 by John Wiley & Sons, Inc. 
 = 1
s
s 0.25+ 0.16903 1.479+
s 0.25+ 2 2.1875+
\u2212 
Taking the inverse Laplace transform, 
\u3b8(t) = 1 - e-0.25t ( cos1.479t +0.16903 sin1.479t) 
c. 
Program: 
'Case Study 2 Challenge (C)' 
'(a)' 
numg=2.25; 
deng=[1 0.5 2.25]; 
G=tf(numg,deng) 
omegan=sqrt(deng(3)) 
zeta=deng(2)/(2*omegan) 
Ts=4/(zeta*omegan) 
Tp=pi/(omegan*sqrt(1-zeta^2)) 
pos=exp(-zeta*pi/sqrt(1-zeta^2))*100 
t=0:.1:2; 
[y,t]=step(G,t); 
Tlow=interp1(y,t,.1); 
Thi=interp1(y,t,.9); 
Tr=Thi-Tlow 
'(b)' 
numc=2.25*[1 2]; 
denc=conv(poly([0 -3.57]),[1 2 2.25]); 
[K,p,k]=residue(numc,denc) 
'(c)' 
[y,t]=step(G); 
plot(t,y) 
title('Roll Angle Response') 
xlabel('Time(seconds)') 
ylabel('Roll Angle(radians)') 
 
Computer response: 
ans = 
 
Case Study 2 Challenge (C) 
 
 
ans = 
 
(a) 
 
 
Transfer function: 
 2.25 
------------------ 
s^2 + 0.5 s + 2.25 
 
 
omegan = 
 
 1.5000 
 
 
zeta = 
 
 0.1667 
 
 
Ts = 
 
 16 
4-4 Chapter 4: Time Response 
 
Copyright © 2011 by John Wiley & Sons, Inc. 
 
 
Tp = 
 
 2.1241 
 
 
pos = 
 
 58.8001 
 
 
Tr = 
 
 0.7801 
 
 
ans = 
 
(b) 
 
 
K = 
 
 0.1260 
 -0.3431 + 0.1058i 
 -0.3431 - 0.1058i 
 0.5602 
 
 
p = 
 
 -3.5700 
 -1.0000 + 1.1180i 
 -1.0000 - 1.1180i 
 0 
 
 
k = 
 
 [] 
 
 
ans = 
 
(c) 
 
 
 
 Answers to Review Questions 4-5 
Copyright © 2011 by John Wiley & Sons, Inc. 
ANSWERS TO REVIEW QUESTIONS 
 
1.Time constant 
2. The time for the step response to reach 63% of its final value 
3. The input pole 
4. The system poles 
5. The radian frequency of a sinusoidal response 
6. The time constant of an exponential response 
7. Natural frequency is the frequency of the system with all damping removed; the damped frequency of 
oscillation is the frequency of oscillation with damping in the system. 
8. Their damped frequency of oscillation will be the same. 
9. They will all exist under the same exponential decay envelop. 
10. They will all have the same percent overshoot and the same shape although differently scaled in time. 
11. \u3b6, \u3c9n, TP, %OS, Ts 
12. Only two since a second-order system is completely defined by two component parameters 
13. (1) Complex, (2) Real, (3) Multiple real 
14. Pole's real part is large compared to the dominant poles, (2) Pole is near a zero 
15. If the residue at that pole is much smaller than the residues at other poles 
16. No; one must then use the output equation 
17. The Laplace transform of the state transition matrix is (sI -A)-1 
18. Computer simulation 
19. Pole-zero concepts give one an intuitive feel for the problem. 
20. State equations, output equations, and initial value for the state-vector 
21. Det(sI-A) = 0 
 
SOLUTIONS TO PROBLEMS 
1. 
a. Overdamped Case: 
C(s) = 
9
s(s2 + 9s + 9)
 
 
Expanding into partial fractions, 
 
9 1 0.171 1.171(s) -
s(s 7.854)(s 1.146) s (s 7.854) (s 1.146)
C = = ++ + + + 
Taking the inverse Laplace transform, 
c(t) = 1 + 0.171 e-7.854t - 1.171 e-1.146t 
4-6 Chapter 4: Time Response 
 
Copyright © 2011 by John Wiley & Sons, Inc. 
b. Underdamped Case: 
 
 
 K2 and K3 can be found by clearing fractions with K1 replaced by its value. Thus, 
 9 = (s2 + 3s + 9) + (K2s + K3)s 
 or 
 9 = s2 + 3s +9 + K2s2 + K3s 
 Hence K2 = -1 and K3 = -3. Thus, 
 
 
 
 
 
 
 
 
c(t) = 1 - 
2
3
 e-3t/2 cos(
27
4 t - \u3c6) 
 = 1 - 1.155 e -1.5t cos (2.598t - \u3c6) 
 where 
\u3c6 = arctan ( 3
27
 ) = 30o 
 
 Solutions to Problems 4-7 
Copyright © 2011 by John Wiley & Sons, Inc. 
c. Oscillatory Case: 
 
 The evaluation of the constants in the numerator are found the same way as they were for