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


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G3(s)=G2(s)/(1+G2(s)H1(s) 
Transfer function: 
6 s^2 + 156 s + 918 
------------------------------ 
s^3 + 190 s^2 + 5404 s + 38046 
Solutions to Problems 7-13 
ans = 
G4(s)=G1(s)G3(s) 
Transfer function: 
6 s^3 + 210 s^2 + 2322 s + 8262 
------------------------------------------------------ 
s^7 + 222 s^6 + 11808 s^5 + 273542 s^4 + 3.16e006 s^3 
+ 1.777e007 s^2 + 3.835e007 s 
ans = 
Ge(s)=G4(s)/(1+G4(s)H2(s)) 
Transfer function: 
6 s^4 + 252 s^3 + 3792 s^2 + 24516 s + 57834 
------------------------------------------------------- 
s^8 + 229 s^7 + 13362 s^6 + 356198 s^5 + 5.075e006 s^4 
+ 3.989e007 s^3 + 1.628e008 s^2 + 2.685e008 s 
+ 8262 
ans = 
T(s)=Ge(s)/(1+Ge(s)) 
Transfer function: 
6 s^4 + 252 s^3 + 3792 s^2 + 24516 s + 57834 
------------------------------------------------------- 
s^8 + 229 s^7 + 13362 s^6 + 356198 s^5 + 5.075e006 s^4 
+ 3.989e007 s^3 + 1.628e008 s^2 + 2.685e008 s 
+ 66096 
ans = 
Poles of T(s) 
ans = 
-157.1538 
-21.6791 
Solutions to Problems 7-13 
Copyright © 2011 by John Wiley & Sons, Inc. 
-14.0006 
-11.9987 
-11.1678 
-7.0001 
-5.9997 
-0.0002 
Kp = 
7 
ans = 
sGe(s)= 
ans = 
sGe(s) 
Transfer function: 
6 s^5 + 252 s^4 + 3792 s^3 + 2.452e004 s^2 
+ 5.783e004 s 
-------------------------------------------------------- 
s^8 + 229 s^7 + 1.336e004 s^6 + 3.562e005 s^5 
+ 5.075e006 s^4 + 3.989e007 s^3 + 1.628e008 s^2 
+ 2.685e008 s + 8262 
Kv = 
0 
ans = 
s^2Ge(s)= 
ans = 
s^2Ge(s) 
Transfer function: 
6 s^6 + 252 s^5 + 3792 s^4 + 2.452e004 s^3 
+ 5.783e004 s^2 
-------------------------------------------------------- 
s^8 + 229 s^7 + 1.336e004 s^6 + 3.562e005 s^5 
+ 5.075e006 s^4 + 3.989e007 s^3 + 1.628e008 s^2 
+ 2.685e008 s + 8262 
Solutions to Problems 7-15 
Ka = 
0 
essstep = 
12.5000 
essramp = 
Inf 
essparabola = 
Inf 
 
35. 
 The equivalent forward transfer function is, G(s) = 
10K1
s(s+1+10Kf)
 . 
Also, T(s) = 
G(s)
1 + G(s) = 
10K1
s2+(10Kf+1)s+10K1
 . From the problem statement, Kv = 
10K1
1+10Kf
 = 10. 
Also, 2\u3b6\u3c9n = 10Kf+1 = 2(0.5) 10K1 = 10K1 . Solving for K1 and Kf simultaneously, K1 = 10 and 
Kf = 0.9. 
 
 
36. 
 
We calculate the Velocity Error Constant, 
7-14 Chapter 7: Steady-State Errors 
Copyright © 2011 by John Wiley & Sons, Inc. 
)842)(07895.0(
)3393.0108.02)(895.7(00842.0
94.31261.14393.95418.135
)2.557704724.144316.34(
00
)()(
+++
+++
++++
++\u2212\u2212
\u2192\u2192
==
sss
sss
sssss
ssss
ssv
LimsPssGLimK
 
623.00357.0
94.31
2.557 == 
For a unit ramp input the steady state error is 605.11 ==
v
ss K
e . The input slope is 6.15
605.1
25 = 
37. 
a. 
 
 
From the point of view of e(t) the above block diagram is equivalent to the original. In this unity 
feedback block diagram the open loop transmission is \u3b2
\u3b1
\u3b2\u3b1 +=+
=
s
s
ssG
1
1
)( , the system is 
type 0. 
b. The position error constant is \u3b2
\u3b1== )0(GK P . The steady state error is 
\u3b2\u3b1
\u3b2
\u3b2
\u3b1 +=+
=+= 1
1
1
1
P
ss K
e . 
 
 
 
Solutions to Problems 7-15 
Copyright © 2011 by John Wiley & Sons, Inc. 
38. 
 2
0
1 2
sR(s) sD(s)G (s)e( ) lim
1 G (s)G (s)s\u2192
\u2212\u221e = + , where G1(s) = 
1
s+5 and G2 = 
100
s+2 . From the problem statement, 
 R(s) = D(s) = 
1
s . Hence, 0
1001 49s 2e( ) lim 1 100 111
s 5 s 2
s\u2192
\u2212 +\u221e = = \u2212
+ + +
. 
39. 
 Error due only to disturbance: Rearranging the block diagram to show D(s) as the input, 
 
 
 Therefore, 
 -E(s) = D(s) 
K2
s(s+4)
1 + 
K1K2(s+2)
s(s+3)(s+4)
 = D(s) 
K2(s+3)
s(s+3)(s+4) + K1K2(s+2)
 
 
 For D(s) = 
1
s , eD(\u221e) = 0lims\u2192 sE(s) = - 
3
2K1
 . 
 Error due only to input: eR(\u221e) = 
1
Kv
 = 
1
K1K2
6
 = 
6
K1K2
 . 
 Design: 
 eD(\u221e) = - 0.000012 = - 
3
2K1
 , or K1 = 125,000. 
 eR(\u221e) = 0.003 = 
6
K1K2
 , or K2 = 0.016 
 
 
 
 
 
7-16 Chapter 7: Steady-State Errors 
Copyright © 2011 by John Wiley & Sons, Inc. 
40. 
a. The open loop transmission is 
2
35)()( += ssPsG , so 2
35)()(
0
==
\u2192
sPsGLimK
sP
. For a unit 
step input 0541.0
1
1 =+= Pssr K
e . Since the input is threefold that we have that 
1622.0)0541.0(3 ==ssre 
b. 
 
Solutions to Problems 7-17 
Copyright © 2011 by John Wiley & Sons, Inc. 
 
 
c. The transfer function from disturbance to error signal is 
37
7
2
751
2
7
)(
)(
+\u2212=
++
+\u2212=
s
s
s
sD
sE
 
Using the final value theorem 1892.0
37
71
37
7)(
00
==\u239f\u23a0
\u239e\u239c\u239d
\u239b\u2212\u239f\u23a0
\u239e\u239c\u239d
\u239b
+\u2212== \u2192\u2192 sssLimssELime ssssd 
d. 
 
7-18 Chapter 7: Steady-State Errors 
Copyright © 2011 by John Wiley & Sons, Inc. 
 
 
e. 351.01892.01622.0 =+=+= ssdssrtot eee 
f. 
 
Solutions to Problems 7-19 
Copyright © 2011 by John Wiley & Sons, Inc. 
 
 
 
 
 
 
 
 
41. 
C(s)
R(s)
= G1(s)G2(s)
1 + G2 (s)H1(s) ; \u2234
Ea1 (s)
R(s)
= G1(s)
1 + G2 (s)H1(s)
ea1(\u221e) = lims \u21920
sR(s)G1(s)
1 + G2 (s)H1(s)
 
 42. 
System 1: 
Forming a unity-feedback path, the equivalent unity feedback system has a forward transfer function of 
 
Ge (s) =
10(s + 10)
s(s + 2)
1 + 10(s +10)(s + 3)
s(s + 2)
= 10(s + 10)
11s 2 + 132s + 300 
a. Type 0 System; b. Kp = Kp = lims\u2192 0 Ge (s) = 1/ 3 ; c. step input; d. e(\u221e) = 
1
1 + Kp = 3/4; 
 e. ea \u2212step (\u221e) = lims\u2192 0
sR(s)
1+ G(s)H(s) = lims\u2192 0
s 1
s
\u239b 
\u239d \u239e \u23a0 
1+ 10(s + 10)(s + 4)
s(s + 2)
= 0 . 
System 2: 
Forming a unity-feedback path, the equivalent unity feedback system has a forward transfer function of 
 
7-20 Chapter 7: Steady-State Errors 
Copyright © 2011 by John Wiley & Sons, Inc. 
Ge (s) =
10(s + 10)
s(s + 2)
1 + 10(s + 10)s
s(s + 2)
= 10(s + 10)
s(11s +102) 
a. Type 1 System; b. Kv = lims\u2192 0 sGe (s) = 0.98 ; c. ramp input; d. e(\u221e) =
1
Kv
= 1.02; 
e. ea \u2212ramp (\u221e) = lims \u21920
sR(s)
1 + G(s)H(s) = lims\u21920
s 1
s2
\u239b 
\u239d \u239e \u23a0 
1 + 10(s +10)(s +1)
s(s + 2)
= 1
50
. 
43. 
System 1. Push 5 to the right past the summing junction: 
 
(s+ 5)(s+ 8)
R(s) C(s)+
-
5(s+4)
2
 
Produce a unity-feedback system: 
R(s) C(s)+
1
-
(s+ 5)(s+ 8)
5(s+4)
-
 
 
Thus, 2
5( 4)
5( 4)( 3)( 7)( ) 5( 4) 15 411
( 3)( 7)
e
s
ss sG s s s s
s s
+
++ += =+ + ++ + +
. Kp = 
20
41
. estep = 
1
1+Kp
 = 0.67, eramp = \u221e, 
eparabola = \u221e. 
Checking for stability, from first block diagram above, T(s) = 2
5( 4)
20 61
s
s s
+
+ + . The system is stable. 
System 2. Push 20 to the right past the summing junction and push 10 to the left past the pickoff point: 
5( 4)
( 3)( 7)
s
s s
+
+ +
 
5( 4)
( 3)( 7)
s
s s
+
+ +
 
Solutions to Problems 7-21 
Copyright © 2011 by John Wiley & Sons, Inc. 
 
R(s) C(s)+
-
200(s+4)
(s+5)(s+8)
1
40
 
Produce a unity-feedback system: 
 
R(s) C(s)+
-
200(s+4)
(s+5)(s+8)
-39
40
-
 
 
Thus, 2
200( 4)
200( 4)( 3)( 7)( )
200( 4) 39 185 7591
( 3)( 7) 40
e
s
ss sG s
s s s
s s
+
++ += =+ \u2212 \u2212\u239b \u239e\u2212 \u239c \u239f+ + \u239d \u23a0
. Kp = 
200(4) 1.05
759
= \u2212\u2212 . 
estep = 
1
1+Kp
 = -20, eramp = \u221e, eparabola = \u221e. 
Checking for stability, from first block diagram above, 2
( ) 200( 4)( )
1 ( ) 15 41
e
e
G s sT s
G s s s
+= =+ + + . 
Therefore, system is stable and steady-state error calculations are valid. 
 
44. 
a. Produce a unity-feedback system: 
 
200( 4)
( 3)( 7)
s
s s
+
+ + 
200( 4)
( 3)( 7)
s
s s
+
+ + 
7-22 Chapter 7: Steady-State Errors 
Copyright © 2011 by John Wiley & Sons, Inc. 
R(s) C(s)+