Control Systems Engineering By norman-s-nise 6th Edition -Test Bank

Control Systems Engineering By norman-s-nise 6th Edition -Test Bank

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Solutions to Design Problems 4-73
Copyright © 2011 by John Wiley & Sons, Inc.
8
2 0.5333
(5 4 )
n M
ζω = =
+ . Solving for
ωn yields
ωn = 0.4510. But,
0.4510.
(5 4 ) 15
n
K K
M
ω = = =
+ Thus, K = 3.051.
76.
The transfer function for the capacitor voltage is V C(s)
V(s) =
1
Cs
R+Ls+ 1
Cs
= 10 6
s2+Rs+106 .
For 20% overshoot, ζ =
– ln (%OS
100 )
π2 + ln 2 (%OS
100 )
= 0.456. Therefore, 2ζωn = R = 2(0.456)(10 3) =
912Ω.
77.
Solving for the capacitor voltage using voltage division, VC (s) = Vi (s) 1/(CS)
R + LS + 1
CS
. Thus, the
transfer function is VC (s)
Vi (s) = 1/(LC)
s2 + R
L s + 1
LC
. Since Ts = 4
Re = 10−3 , Re = R
2L = 4000 . Thus
R = 8 KΩ . Also, since 20% overshoot implies a damping ratio of 0.46 and
2
ζωn = 8000,
ωn = 8695.65 = 1
LC . Hence, C = 0.013
μF .
78.
Using voltage division the transfer function is,
VC (s)
Vi (s) =
1
Cs
R + Ls + 1
Cs
=
1
LC
s2 + R
L s + 1
LC
Also, 3 4 4 8
7 10 Re
2
s
L
T x R R
L
−
= = = = . Thus, 1143
R
L = . Using Eq. (4.39) with 15% overshoot,
ζ
= 0.5169. But, 2
ζωn = R/L. Thus, 5
1 1
1105.63 (10 )
n LC L
ω −
= = = . Therefore, L = 81.8 mH
and R = 98.5 Ω.

4-74 Chapter 4: Time Response
Copyright © 2011 by John Wiley & Sons, Inc.
79.
For the circuit shown below
R1 =
L =
i1(t) i2(t) o
write the loop equations as
R 1 L s+ I 1 s R 1 I 2 s− V i s=
R 1 I 1 s− R 1 R 2 1
C s
+ + I 2 s+ 0=
Solving for I2(s)
I 2 s
R 1 L s+ V i s
R 1− 0
R 1 L s+ R 1−
R 1− R 1 R 2 1
C s
+ +
=
( )
But, V o s 1
C s I 2 s= . Thus,
V o s
V i s
R 1
R 2 R 1+ C L s 2 C R 2 R 1 L+ s R 1+ +
=
Substituting component values,
Vo (s)
Vi (s) = 1000000
1
(R2 + 1000000)C
s2 + (1000000CR2 + 1)
(R2 + 1000000)C s + 1000000 1
(R2 + 1000000)C
For 8% overshoot, ζ = 0.6266. For Ts = 0.001, ζωn = 4
0.001 = 4000. Hence, ωn = 6383.66. Thus,
2
2
1
1000000 6383.66
( 1000000)R C =
+
or,
2
1
0.0245 1000000
C R
= + (1)
Also,

Solutions to Design Problems 4-75
Copyright © 2011 by John Wiley & Sons, Inc.
1000000 C R 2 1+
R 2 1000000+ C 8000= (2)
Solving (1) and (2) simultaneously, 2 8023R = Ω, and C = 2.4305 x 10 -2 μF.
80.
sI − A = s 0
0 s
⎡
⎣⎢
⎤
⎦⎥ − (3.45 − 14000K c ) −0.255x10−9
0.499x1011 −3.68
⎡
⎣⎢
⎤
⎦⎥
= s − (3.45 − 14000Kc ) 0.255x10−9
−0.499x1011 s + 3.68
⎡
⎣⎢
⎤
⎦⎥
sI − A = s2 + (0.23 + 0.14x10 5 Kc )s + (51520Kc + 0.0285)
(2
ζωn )2 = [2* 0.9] 2 *(51520K c + 0.0285) = (0.23 + 0.14x105 K c ) 2
or
K c
2 − 8.187x10−4 Kc − 2.0122x10−10 = 0
Solving for Kc,
K c = 8.189x10−4
81.
a. The transfer function from Chapter 2 is,
Yh(s) − Ycat (s)
Fup (s) = 0.7883(s + 53.85)
(s2 + 15.47s + 9283)(s2 + 8.119s + 376.3)
The dominant poles come from s 2 + 8.119s + 376.3 . Using this polynomial,
2
ζωn = 8.119, and
ωn
2 = 376.3 . Thus,
ωn = 19.4 and
ζ = 0.209 . Using Eq. (4.38), %OS =
51.05%. Also,Ts = 4
ζωn
= 0.985 s, and Tp =
π
ωn 1−
ζ2 = 0.166 s . To find rise time, use
Figure 4.16. Thus,
ωn Tr = 1.2136 or Tr = 0.0626 s.
b. The other poles have a real part of 15.47/2 = 7.735. Dominant poles have a real part of 8.119/2 =
4.06. Thus, 7.735/4.06 = 1.91. This is not at least 5 times.
c.
Program:
syms s
numg=0.7883*(s+53.85);
deng=(s^2+15.47*s+9283)*(s^2+8.119*s+376.3);
‘G(s) transfer function’
G=vpa(numg/deng,3);
pretty(G)
numg=sym2poly(numg);
deng=sym2poly(deng);
G=tf(numg,deng)
step(G)

4-76 Chapter 4: Time Response
Copyright © 2011 by John Wiley & Sons, Inc.
Computer response:
ans =
G(s) transfer function
.788 s + 42.4
——————————————
2 2
(s + 15.5 s + 9280.) (s + 8.12 s + 376.)
Transfer function:
0.7883 s + 42.45
—————————————————-
s^4 + 23.59 s^3 + 9785 s^2 + 8.119e004 s + 3.493e006
The time response shows 58 percent overshoot, Ts = 0.86 s, Tp = 0.13 s, Tr = 0.05 s.
82.
a. In Problem 3.30 we had
⎥
⎦
⎤
⎢
⎣
⎡
⎥
⎥
⎥
⎦
⎤
⎢
⎢
⎢
⎣
⎡
−
−+
⎥
⎥
⎥
⎦
⎤
⎢
⎢
⎢
⎣
⎡
⎥
⎥
⎥
⎦
⎤
⎢
⎢
⎢
⎣
⎡
−
−
−+−
=
⎥
⎥
⎥
⎦
⎤
⎢
⎢
⎢
⎣
⎡
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
β
β
β
μ
β
β
β
&
&
&
[ ]
⎥
⎥
⎥
⎦
⎤
⎢
⎢
⎢
⎣
⎡
=
v
T
T
y *
100
When 02 =u the equations are equivalent to

Solutions to Design Problems 4-77
Copyright © 2011 by John Wiley & Sons, Inc.
100
00
*
00
00
*
00
0)(
uvT
vT
v
T
T
ck
Tv
Tvd
v
T
T
⎥
⎥
⎥
⎦
⎤
⎢
⎢
⎢
⎣
⎡
−+
⎥
⎥
⎥
⎦
⎤
⎢
⎢
⎢
⎣
⎡
⎥
⎥
⎥
⎦
⎤
⎢
⎢
⎢
⎣
⎡
−
−
−+−
=
⎥
⎥
⎥
⎦
⎤
⎢
⎢
⎢
⎣
⎡
β
β
β
μ
β
β
β
&
&
&
[ ]
⎥
⎥
⎥
⎦
⎤
⎢
⎢
⎢
⎣
⎡
=
v
T
T
y *
100
Substituting parameter values one gets
1
**
0
2.5
2.5
4.21000
0058.024.00217.0
0058.0004167.0
u
v
T
T
v
T
T
⎥
⎥
⎥
⎦
⎤
⎢
⎢
⎢
⎣
⎡
−+
⎥
⎥
⎥
⎦
⎤
⎢
⎢
⎢
⎣
⎡
⎥
⎥
⎥
⎦
⎤
⎢
⎢
⎢
⎣
⎡
−
−
−−
=
⎥
⎥
⎥
⎦
⎤
⎢
⎢
⎢
⎣
⎡
&
&
&
[ ]
⎥
⎥
⎥
⎦
⎤
⎢
⎢
⎢
⎣
⎡
=
v
T
T
y *
100
b.
)det(
)(
4.21000
0058.024.00217.0
0058.0004167.0
)(
1
1
AI
AI
AI −
−
=
⎥
⎥
⎥
⎦
⎤
⎢
⎢
⎢
⎣
⎡
+−
−+−
+
=−
−
−
s
sAdj
s
s
s
s
[ ]
)0048.00398.0)(6419.2(
0126.011.06817.2
)17.2)(0058.0(58.0)4.2)(24.0()04167.0(
1000
24.00217.0
0058.0
4.2100
0058.024.0
)04167.0()det(
2
23
+++=
+++=
+−+++=
−
+−
+
+−
−+
+=−
sss
sss
sss
s
s
s
ss AI
To obtain the adjoint matrix we calculate the cofactors:
)64.2(
4.2100
0058.024.0
11 +=
+−
−+
= ss
s
s
C
)4.2(0217.0
4.20
0058.00217.0
12 +−=
+
−−
= s
s
C
17.2
1000
24.00217.0
13 =
−
+−
= s
C
58.0
4.2100
0058.00
21 =
+−
= s
C

4-78 Chapter 4: Time Response
Copyright © 2011 by John Wiley & Sons, Inc.
)4.2)(04167.0(
4.20
0058.004167.0
22 ++=
+
+
= ss
s
s
C
)04167.0(100
1000
004167.0
23 +−=
−
+
= s
s
C
)24.0(0058.0
0058.024.0
0058.00
31 +−=
−+
= s
s
C
1001.04117.2
4.20217.0
0058.004167.0 2
32 ++=
+−
+
= ss
s
s
C
01.02817.0
24.00217.0
004167.0 2
33 ++=
+−
+
= ss
s
s
C
Then we have
⎥
⎥
⎥
⎦
⎤
⎢
⎢
⎢
⎣
⎡
+++
++−+++
+−−+
=−
01.02817.0)04167.0(10017.2
)1101.04417.2()4.2)(04167.0()4.2(0217.0
)24.0(0058.058.0)64.2(
)(
2
2
sss
sssss
sss
sAdj AI
Finally
[ ]
)0048.00398.0)(6419.2(
02.0
520
)0048.00398.0)(6419.2(
3844.10520
0
2.5
2.5
)0048.00398.0)(6419.2(
01.028171.0)04167.0(10017.2
)()(
22
2
2
1
1
+++
+
−=
+++
−−
=
⎥
⎥
⎥
⎦
⎤
⎢
⎢
⎢
⎣
⎡
−
+++
+++
=−= −
sss
s
sss
s
sss
sss
ss
U
Y BAIC
c. 100% effectiveness means that 11 =u or s
sU 1
)(1 = , so by the final value theorem
820.1168
1
)0048.00398.0)(6419.2(
)02.0(520
)()( 200
−=
+++
+
−==∞ →→ ssss
s
sLimssYLimy ss
(virus copies per mL of plasma)
The closest poles to the imaginary axis are 0661.00199.0 j±− so the approximate settling time
will be 210
0199.0
4 =≈sT days.
83.
a.
Substituting s
2650
F(s) =Δ into the transfer function and solving for ΔV(s) gives:

Solutions to Design Problems 4-79
Copyright © 2011 by John Wiley & Sons, Inc.
)101908()101908(
2650
1908
F(s)
V(s) +⋅
+=
+⋅
=
⋅
Δ
=Δ s
B
s
A
sss
Here: 265
0)101908(
2650 =
=+⋅
=
ss
A and 2650 505, 620
1190.8
B s s
= = −
= −
Substituting we have:
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛
×+
−=
+⋅
−=Δ − )1024.5(
11
265
)101908(
505620265
V(s) 3
ssss
Taking the inverse Laplace transform, we have:
m/sin),()1(265)( 3
1024.5 tutetv ⋅−=Δ −
×−
b.
>> s=tf(‘s’);
>> G=1/(1908*s+10);
>> t=0:0.1:1000;
>> y1=2650*step(G,t);
>> y2=265*(1-exp(-5.24e-3.*t));
>> plot(t,y1,t,y2)
>> xlabel(‘sec’)
>> ylabel(‘m/s’)

Copyright © 2011 by John Wiley & Sons, Inc.
F I V E
Reduction of Multiple
Subsystems
SOLUTIONS TO CASE STUDIES CHALLENGES
Antenna Control: Designing a Closed-Loop Response
a. Drawing the block diagram of the system:
+
–
10
Π
iu K 150
s+150
u o0.16
s (s+1.32)

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