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Norman Nise Control Systems Engineering 6th Solultions.PDF

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loop, G1e(s) =
K
(s+0.01)
s2
1+K
(s+0.01)
s2
=
K (s+0.01)
s2+Ks+0.01K
, where K =
Kc
J .
Form Ge(s) = G1e(s)
(s+0.01)
s2
= K
(s+0.01)2
s2(s2+Ks+0.01K)
.
System is Type 2. Therefore, estep = 0,
b. eramp = 0,
c. eparabola =
1
Ka
=
1
0.01 = 100
d. T(s) =
Ge(s)
1+Ge(s)
=
K(s+0.01)2
s4+Ks3+1.01Ks2+0.02Ks+10-4K

s4 1 1.01K 10-4K
s3 K 0.02K 0 0 < K
s2 1.01K-0.02 10-4K 0 0.0198 < K
s1
0.0201 K2 0.0004 K\u2212
1.01 K 0.02\u2212 0 0 0.0199 < K
s0 10-4K 0 < K

Thus, for stability K =
Kc
J > 0.0199
59.
a. Following Figure P7.29, the transfer function from f\u3b4 to e is given by:
*
1
f
f r
r Ge
KG\u3b4
\u2212= +
For
1
f s
\u3b4 = we have that in steady state
7-36 Chapter 7: Steady-State Errors
Copyright © 2011 by John Wiley & Sons, Inc.
2
2
4
2
0.8
( )
1 (0)
b
ae bK
a
\u2212
\u221e =
+

It can be seen from this expression that if ( )K s is type 1 or larger ( ) 0e \u221e = .
b. From Figure P7.29:
*
1
f r
f r
G r KGr
KG\u3b4
+= +

The error is now defined as
* *1
1 1
1 1
f r r f r
f r r
G r KG KG G r KGr
KG KG\u3b4
+ + \u2212 +\u2212 = \u2212 =+ +

In steady state this expression becomes:
4 2 4 4 2
2 2 2 2 2
4 4
2 2
1 (0) 0.8 (0) 1 0.2 (0)
1 (0)
1 (0) 1 (0)f
b b b b bK K K
r a a a a a
b bK K
a a
\u3b4
+ \u2212 \u2212 + \u2212
\u2212 = =
+ +

It can be seen in this equation that the steady state error cannot be made zero.

SOLUTIONS TO DESIGN PROBLEMS

60.
Pot gains: K1 =
3\u3c0
\u3c0 = 3; Amplifier gain: K2 ; Motor transfer function: Since time constant = 0.5, \u3b1

= 2. Also,
K
\u3b1 =
100
10 = 10. Hence, K = 20. The motor transfer function is now computed as
C(s)
Ea(s)
=
20
s(s+2) . The following block diagram results after pushing the potentiometers to the right past the
summing junction:

Solutions to Problems 7-37

Copyright © 2011 by John Wiley & Sons, Inc.

Finally, since Kv = 10 =
60K2
2 , from which K2 =
1
3 .
61.
First find Kv: Circumference = 2\u3c0 nautical miles. Therefore, boat makes 1 revolution
in
2\u3c0
20 = 0.314 hr.
Angular velocity is thus,
1
0.314
rev
hr =
2\u3c0
3600 x 0.314
sec = 5.56 x 10
-3
sec .
For 0.1o error, e(\u221e) = 1/10
o
360o
x 2\u3c0 rad = 5.56 x 10-3Kv . Thus Kv = 3.19 =
K
4 ; from which,
K = 12.76.

62.
a. Performing block diagram reduction:

K1 \u2212 s \u221213+ 13
100
s2 +14 +100
2
(s+0.5)(s2+ 9.5s +78)
3s
s+ 0.2
+ + +
---
-s
R(s) C(s)
s s

7-38 Chapter 7: Steady-State Errors
Copyright © 2011 by John Wiley & Sons, Inc.
-
+R(s) C(s)
K1
2
(s+ 0.5)(s 2+ 9. 5 + 78)
+
-
100 (s+ 0. 2)
s 3 + 14. 2 s2 + 402. 8 s + 20
\u2212 s
\u2212 s \u2212 13
+ 13s s

K1
+ +
--
Ge(s)
\u2212 s
R(s) C(s)

200\u2212 s 2 12.8 s\u2212 2.6\u2212
s 7 37.2 s 6 942.15 s 5 13420 s 4 1.0249×105 s 3 4.6048×105 s 2 2.2651×105 s 10140+ + + + + + +
Ge(s) =

System is unity feedback with a forward transfer function, Gt(s), where
Gt(s) = 200 K 1
s 2 12.8 s\u2212 2.6\u2212
s 7 37.2 s 6 942.15 s 5 13420 s 4 1.0269×105 s 3 4.5792×105 s 2 2.2599×105 s 10140+ + + + + + +
\u2212

Thus, system is Type 0.

b. From Gt(s), Kp =
520K1
10140 = 700. Thus, K1 = 13650.
c. T s
G t
1 G t+
=
For K1 = 13650,
T s 2730000 s
2 12.8 s\u2212 2.6\u2212
s 7 37.2 s 6 942.15 s 5 13420 s 4 1.0269×105 s 3 2.2721×106 s 2\u2212 3.517×107 s 7108140+ + + + + +
\u2212=

Because of the negative coefficient in the denominator the system is unstable and the pilot would not
be hired.

Solutions to Problems 7-39

Copyright © 2011 by John Wiley & Sons, Inc.
63.
The force error is the actuating signal. The equivalent forward-path transfer function is
Ge (s) = K1s(s + K1K2 ) . The feedback is H(s) = Des + Ke . Using Eq. (7.72)
Ea (s) = R(s)1 + Ge(s)H (s) . Applying the final value theorem,
ea _ ramp (\u221e) = lims \u21920
s 1
s 2
\u239b
\u239d \u239e \u23a0
1 + K1(Des + Ke )
s(s + K1K2 )
= K2
Ke
< 0.1. Thus, K2 < 0.1Ke. Since the closed-loop system
is second-order with positive coefficients, the system is always stable.
64.
a. The minimum steady-state error occurs for a maximum setting of gain, K. The maximum K possible
is determined by the maximum gain for stability. The block diagram for the system is shown below.
-
+ K
(s +10)(s 2 + 4s +10)3
3
\u3c9i _desired ( s) \u3c9o (s)Vi(s)

Pushing the input transducer to the right past the summing junction and finding the closed-loop
transfer function, we get

T (s) =
3K
(s + 10)(s2 + 4s +10)
1 + 3K
(s +10)(s 2 + 4s +10)
= 3K
s3 +14s2 + 50s + (3K +100)
Forming a Routh table,

s3 1 50
s2 14 3K+100
s1 \u22123K + 600
14

0
s0 3K+100 0

7-40 Chapter 7: Steady-State Errors
Copyright © 2011 by John Wiley & Sons, Inc.
The s1 row says -\u221e < K < 200. The s0 row says \u2212 100
3
< K. Thus for stability,
\u2212 100
3
< K < 200. Hence, the maximum value of K is 200.
b. Kp = 3K100 = 6 . Hence, estep (\u221e) =
1
1 + Kp =
1
7
.
c. Step input.

65.
Substituting values we have
)10)(67.2(
140625)(
1.0
++=
\u2212
ss
esP
s
,
005.0
005.0)( += s
LsG
The proportional error constant
L
ss
e
s
LLimsPsGLimK
s
ssP
44.5273
)10)(67.2(
140625
005.0
005.0)()(
1.0
00
=+++==
\u2212
\u2192\u2192

1.0
44.52731
1
1
1 =+=+= LKe Pss
which gives 31071.1 \u2212×=L .
66.
a. The open loop transmission is
89.2
48500)( 2 +
+=
ss
KsKsGP IP . The system is type 2.
b. The Transfer function from disturbance to error signal is
)(4850089.2
48500
89.2
485001
89.2
48500
)(
)(
23
2
2
IPIP KsKss
s
s
KsK
ss
ss
sD
sE
+++\u2212=+
++
+\u2212=
Using the Final value theorem
01
)(4850089.2
48500)( 2300 =\u239f\u239f\u23a0
\u239e
\u239c\u239c\u239d
\u239b
+++\u2212== \u2192\u2192 sKsKss
ssLimssELime
IP
ssss

c. We calculate
89.2
48500
)()(2
0
I
sa
KsPsGsLimK ==
\u2192
so 05.0
48500
89.21 ===
Ia
SS KK
e . So
we get 0120.0=IK
d. The system\u2019s characteristic equation is 0
012.0
89.2
485001 2 =
+
++ s
sK
ss
P or
0021.5844850089.2 23 =+++ sKss P . The Routh array is:

Solutions to Problems 7-41

Copyright © 2011 by John Wiley & Sons, Inc.
3s 1 PK48500
2s 89.2 021.584
s
89.2
21.584140165 \u2212PK
1 PK48500

The dominant requirement is given by the third row 00417.0>PK

67.
a.
Sensor
+
-
Input
transducer
Desired
force
Input
voltage
Controller Actuator Pantograph
dynamics
Spring
Fup
Yh-Ycat
Spring
displacement
Fout1
100
K 1
1000
0.7883( s + 53.85)
( s2 + 15.47 s + 9283 )( s2 + 8.119 s + 376 .3) 82300
1
100

+
-
Desired
force
Controller Actuator Pantograph
dynamics
Spring
Fup
Yh-Ycat
Spring
displacement
Fout1
1000
0.7883( s + 53.85)
(s2 + 15.47s + 9283 )(s 2 + 8.119 s + 376 .3) 82300
K
100

b.
G(s) =
Yh(s) \u2212 Ycat (s)
Fup(s)
= 0.7883(s + 53.85)
(s2 + 15.47s + 9283)(s2 +8.119s + 376.3)

Ge(s) = (K/100)*(1/1000)*G(s)*82.3e3

0.6488K (s+53.85)
Ge(s) =
(s2 + 8.119s + 376.3) (s2 + 15.47s + 9283)
Kp = 0.6488K*53.85/[(376.3)(9283)] = K*1.0002E-5
7-42 Chapter 7: Steady-State Errors
Copyright © 2011 by John Wiley & Sons, Inc.

Maximum K minimizes the steady-state error. Maximum K possible is that which yields stability.
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