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Norman Nise Control Systems Engineering 6th Solultions.PDF

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0 5 10 15
-0.2
0
0.2
0.4
0.6
0.8
1
1.2
1.4
1.6

SOLUTIONS TO DESIGN PROBLEMS
69.
Writing the equation of motion, ( fvs + 2)X(s) = F(s). Thus, the transfer function is
X(s)
F(s)
= 1/ fv
s + 2
fv
. Hence, Ts = 4a =
4
2
fv
= 2 fv , or fv = Ts2 .
70.
The transfer function is, F(s) = 1/ M
s2 + 1
M
s + K
M
. Now,
4 44 81Re
2
sT M
M
= = = = . Thus,
1
2
M = . Substituting the value of M in the denominator of the transfer function yields,
2 2 2s s K+ + . Identify the roots 1,2 1 2 1s j K= \u2212 ± \u2212 . Using the imaginary part and substituting
into the peak time equation yields 1
Im 2 1p
T
K
\u3c0 \u3c0= = = \u2212 , from which 5.43K = .

Solutions to Design Problems 4-71

Copyright © 2011 by John Wiley & Sons, Inc.
71.
Writing the equation of motion, (Ms2 + fvs +1)X(s) = F(s) . Thus, the transfer function is

X(s)
F(s)
= 1/ M
s2 + fv
M
s + 1
M
. Since Ts =10 = 4\u3b6\u3c9n , \u3b6\u3c9n = 0.4 . But,
fv
M
= 2\u3b6\u3c9n = 0.8.Also,
from Eq. (4.39) 17% overshoot implies \u3b6 = 0.491. Hence, \u3c9n = 0.815. Now, 1/M = \u3c9n2 = 0.664.
Therefore, M 1.51. Since 2 0.8, 1.21v n v
f f
M
\u3b6\u3c9= = = .
72.
Writing the equation of motion: (Js2+s+K)\u3b8(s) = T(s). Therefore the transfer function is

\u3b8(s)
T(s) =
1
J
s2+
1
Js+
K
J
.
\u3b6 =
- ln (
%OS
100 )
\u3c02 + ln2 (%OS100 )
= 0.358.

Ts =
4
\u3b6\u3c9n =
4
1
2J
= 8J = 3.

Therefore J =
3
8
. Also, Ts = 3 =
4
\u3b6\u3c9n =
4
(0.358)\u3c9n . Hence, \u3c9n = 3.724. Now,
K
J = \u3c9n2 = 13.868.
Finally, K = 5.2.

73.
Writing the equation of motion
[s2+D(5)2s+
1
4 (10)
2]\u3b8(s) = T(s)
The transfer function is
\u3b8 (s)
T(s) =
1
s2+25Ds+25

Also,
\u3b6 =
- ln (
%OS
100 )
\u3c02 + ln2 (%OS100 )
= 0.358
and
2\u3b6\u3c9n = 2(0.358)(5) = 25D
Therefore D = 0.14.

4-72 Chapter 4: Time Response

Copyright © 2011 by John Wiley & Sons, Inc.
74.
The equivalent circuit is:

where Jeq = 1+(
N1
N2 )
2 ; Deq = (
N1
N2 )
2; Keq = (
N1
N2 )
2. Thus,

\u3b81(s)
T(s) =
1
Jeqs2+Deqs+Keq
. Letting
N1
N2 = n and substituting the above values into the transfer
function,

\u3b81(s)
T(s) =
1
1+n2
s2 +
n2
1+n2
s +
n2
1+n2
. Therefore, \u3b6\u3c9n = n
2
2(1+n2)
. Finally, Ts =
4
\u3b6\u3c9n =
8(1+n2)
n2
= 16. Thus
n = 1.
75.
Let the rotation of the shaft with gear N2 be \u3b8L(s). Assuming that all rotating load has been reflected
to the N2 shaft, JeqLs
2 + DeqLs + K( )\u3b8 L(s) + F(s)r = Teq (s) , where F(s) is the force from the
translational system, r = 2 is the radius of the rotational member, JeqL is the equivalent inertia at the
N2 shaft, and DeqL is the equivalent damping at the N2 shaft. Since JeqL = 1(2)2 + 1 = 5 and DeqL =
1(2)2 = 4, the equation of motion becomes, 5s2 + 4s + K( )\u3b8L(s) + 2F(s) = Teq (s). For the
translational system (Ms2 + s)X(s) = F(s) . Substituting F(s) into the rotational equation of
motion, 5s2 + 4s + K( )\u3b8L(s) + Ms2 + s( )2X(s) = Teq (s).
But,\u3b8 L(s) = X(s)r =
X(s)
2
and Teq (s) = 2T (s). Substituting these quantities in the equation
above yields (5 + 4M)s 2 + 8s + K( )X(s)
4
= T (s). Thus, the transfer function is
X(s)
T(s)
= 4 /(5 + 4M)
s2 + 8
(5 + 4M) s +
K
(5 + 4M)
. Now,
4 415 (5 4 )8Re
2(5 4 )
sT M
M
= = = = +
+
.
Hence, M = 5/2. For 10% overshoot, \u3b6 = 0.5912 from Eq. (4.39). Hence,
Solutions to Design Problems 4-73

Copyright © 2011 by John Wiley & Sons, Inc.
82 0.5333
(5 4 )n M
\u3b6\u3c9 = =+ . Solving for \u3c9n yields \u3c9n = 0.4510. But,
0.4510.
(5 4 ) 15n
K K
M
\u3c9 = = =+ Thus, K = 3.051.
76.
The transfer function for the capacitor voltage is
VC(s)
V(s) =
1
Cs
R+Ls+
1
Cs
=
106
s2+Rs+106
.
For 20% overshoot, \u3b6 =
- ln (
%OS
100 )
\u3c02 + ln2 (%OS100 )
= 0.456. Therefore, 2\u3b6\u3c9n = R = 2(0.456)(103) =
912\u3a9.
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 = 4Re =10
\u22123, Re = R
2L
= 4000. Thus
R = 8 K\u3a9 . Also, since 20% overshoot implies a damping ratio of 0.46 and
2\u3b6\u3c9n = 8000, \u3c9n = 8695.65 = 1LC . Hence, C = 0.013 \u3bcF.
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 87 10
Re
2
s
LT x R R
L
\u2212= = = = . Thus, 1143R
L
= . Using Eq. (4.39) with 15% overshoot, \u3b6
= 0.5169. But, 2\u3b6\u3c9n = R/L. Thus, 5
1 11105.63
(10 )n LC L
\u3c9 \u2212= = = . Therefore, L = 81.8 mH
and R = 98.5 \u3a9.

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\u2212 V i s=
R 1 I 1 s\u2212 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\u2212 0
R 1 L s+ R 1\u2212
R 1\u2212 R 1 R 2 1C 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, \u3b6 = 0.6266. For Ts = 0.001, \u3b6\u3c9n = 40.001 = 4000. Hence, \u3c9n = 6383.66. Thus,

2
2
11000000 6383.66
( 1000000)R C
=+
or,
2
10.0245
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 = \u3a9, and C = 2.4305 x 10-2 \u3bcF.

80.
sI \u2212 A = s 0
0 s
\u23a1
\u23a3 \u23a2
\u23a4
\u23a6 \u23a5 \u2212
(3.45 \u221214000Kc) \u22120.255x10\u22129
0.499x1011 \u22123.68
\u23a1
\u23a3 \u23a2
\u23a4
\u23a6 \u23a5
= s \u2212 (3.45 \u221214000Kc) 0.255x10
\u22129
\u22120.499x1011 s + 3.68
\u23a1
\u23a3 \u23a2
\u23a4
\u23a6 \u23a5

sI \u2212 A = s2 + (0.23 + 0.14x105 Kc )s + (51520Kc + 0.0285)
(2\u3b6\u3c9n )2 = [2* 0.9]2 *(51520Kc + 0.0285) = (0.23 + 0.14x105 Kc)2

or
Kc
2 \u2212 8.187x10\u22124 Kc \u2212 2.0122x10\u221210 = 0
Solving for Kc,
Kc = 8.189x10\u22124

81.
a. The transfer function from Chapter 2 is,

Yh(s) \u2212 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\u3b6\u3c9n = 8.119, and \u3c9n2 = 376.3. Thus, \u3c9n = 19.4 and \u3b6 = 0.209. Using Eq. (4.38), %OS =
51.05%. Also,Ts = 4\u3b6\u3c9n = 0.985 s, and Tp =
\u3c0
\u3c9n 1\u2212 \u3b6 2
= 0.166 s . To find rise time, use
Figure 4.16. Thus,\u3c9nTr = 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: