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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)+