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


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fout
1
s
1
s
1
s
3x4x x 12x
 
 
Writing the state and output equations 
x1
\u2022 = x2
x2
\u2022 = x3
x3
\u2022 = x4
x4
\u2022 = \u221223.59x4 \u2212 9785x3 \u2212 81190x2 \u2212 3493000x1 + 0.01 fdesired \u2212 0.01 fout
 
But, 
fout = 42.45* 82300 x1 + 0.7883 *82300 x2 
Substituting fout into the state equations yields 
x4
\u2022 = \u22123527936.35x1 \u221281838.7709x2 \u2212 9785x3 \u2212 23.59x4 + 0.01fdesired 
Putting the state and output equations into vector-matrix form. 
x
\u2022 =
0 1 0 0
0 0 1 0
0 0 0 1
\u22123.528x106 \u221281840 \u22129785 \u221223.59
\u23a1 
\u23a3 
\u23a2 
\u23a2 
\u23a2 
\u23a4 
\u23a6 
\u23a5 
\u23a5 
\u23a5 
+
0
0
0
0.01
\u23a1 
\u23a3 
\u23a2 
\u23a2 
\u23a2 
\u23a4 
\u23a6 
\u23a5 
\u23a5 
\u23a5 
fdesired
y = fout = 3494000 64880 0 0[ ]x
 
 
 
 
 
 
 
 
 
 
 
Solutions to Problems 5-101 
 
Copyright © 2011 by John Wiley & Sons, Inc. 
80. 
 
a. The transfer function obtained in Problem 3.30 is 
 
0126.011.06817.2
3844.10520
23
1 +++
\u2212\u2212=
sss
s
U
Y
 by inspection we write the phase-variable form 
 
1
3
2
1
3
2
1
1
0
0
6817.211.00126.0
100
010
u
x
x
x
x
x
x
\u23a5\u23a5
\u23a5
\u23a6
\u23a4
\u23a2\u23a2
\u23a2
\u23a3
\u23a1
+
\u23a5\u23a5
\u23a5
\u23a6
\u23a4
\u23a2\u23a2
\u23a2
\u23a3
\u23a1
\u23a5\u23a5
\u23a5
\u23a6
\u23a4
\u23a2\u23a2
\u23a2
\u23a3
\u23a1
\u2212\u2212\u2212
=
\u23a5\u23a5
\u23a5
\u23a6
\u23a4
\u23a2\u23a2
\u23a2
\u23a3
\u23a1
&
&
&
 
[ ]
\u23a5\u23a5
\u23a5
\u23a6
\u23a4
\u23a2\u23a2
\u23a2
\u23a3
\u23a1
\u2212\u2212=
3
2
1
05203844.10
x
x
x
y 
b. We renumber the phase-variable form state variables in reverse order 
 
1
1
2
3
1
2
3
1
0
0
6817.211.00126.0
100
010
u
x
x
x
x
x
x
\u23a5\u23a5
\u23a5
\u23a6
\u23a4
\u23a2\u23a2
\u23a2
\u23a3
\u23a1
+
\u23a5\u23a5
\u23a5
\u23a6
\u23a4
\u23a2\u23a2
\u23a2
\u23a3
\u23a1
\u23a5\u23a5
\u23a5
\u23a6
\u23a4
\u23a2\u23a2
\u23a2
\u23a3
\u23a1
\u2212\u2212\u2212
=
\u23a5\u23a5
\u23a5
\u23a6
\u23a4
\u23a2\u23a2
\u23a2
\u23a3
\u23a1
&
&
&
 
[ ]
\u23a5\u23a5
\u23a5
\u23a6
\u23a4
\u23a2\u23a2
\u23a2
\u23a3
\u23a1
\u2212\u2212=
1
2
3
05203844.10
x
x
x
y 
And we rearrange in ascending numerical order to obtain the controller canonical form: 
 
1
3
2
1
3
2
1
0
0
1
100
001
0126.011.06817.2
u
x
x
x
x
x
x
\u23a5\u23a5
\u23a5
\u23a6
\u23a4
\u23a2\u23a2
\u23a2
\u23a3
\u23a1
+
\u23a5\u23a5
\u23a5
\u23a6
\u23a4
\u23a2\u23a2
\u23a2
\u23a3
\u23a1
\u23a5\u23a5
\u23a5
\u23a6
\u23a4
\u23a2\u23a2
\u23a2
\u23a3
\u23a1 \u2212\u2212\u2212
=
\u23a5\u23a5
\u23a5
\u23a6
\u23a4
\u23a2\u23a2
\u23a2
\u23a3
\u23a1
&
&
&
 
[ ]
\u23a5\u23a5
\u23a5
\u23a6
\u23a4
\u23a2\u23a2
\u23a2
\u23a3
\u23a1
\u2212\u2212=
3
2
1
3844.105200
x
x
x
y 
c. To obtain the observer canonical form we rewrite the system\u2019s transfer function as: 
 
32
32
1 0126.011.06817.21
3844.10520
sss
ss
U
Y
+++
\u2212\u2212
= 
We cross-multiply to obtain 
5-102 Chapter 5: Reduction of Multiple Subsystems 
 
Copyright © 2011 by John Wiley & Sons, Inc. 
 
\u23a5\u23a6
\u23a4\u23a2\u23a3
\u23a1 +++=\u23a5\u23a6
\u23a4\u23a2\u23a3
\u23a1 \u2212\u2212 32132 0126.011.06817.213844.10520 sssYUss 
Combining terms with like powers of integration: 
 
[ ] [ ] [ ]
[ ] [ ] \u23a5\u23a6
\u23a4\u23a2\u23a3
\u23a1 \u239f\u23a0
\u239e\u239c\u239d
\u239b \u2212\u2212+\u2212\u2212+\u2212=
\u2212\u2212+\u2212\u2212+\u2212=
YR
s
YR
s
Y
s
YR
s
YR
s
Y
s
Y
0126.03844.10111.052016817.21
0126.03844.10111.052016817.21 32
 
We draw the signal flow graph: 
 
 
The following equations follow: 
211 6817.2 xxx +\u2212=& 
rxxx 52011.0 312 \u2212+\u2212=& 
rxx 38.100126.0 13 \u2212\u2212=& 
1xy = 
Which lead to observer canonical form: 
 
Solutions to Problems 5-103 
 
Copyright © 2011 by John Wiley & Sons, Inc. 
1
3
2
1
3
2
1
38.10
520
1
000126.0
1011.0
016817.2
u
x
x
x
x
x
x
\u23a5\u23a5
\u23a5
\u23a6
\u23a4
\u23a2\u23a2
\u23a2
\u23a3
\u23a1
\u2212
\u2212+
\u23a5\u23a5
\u23a5
\u23a6
\u23a4
\u23a2\u23a2
\u23a2
\u23a3
\u23a1
\u23a5\u23a5
\u23a5
\u23a6
\u23a4
\u23a2\u23a2
\u23a2
\u23a3
\u23a1
\u2212
\u2212
\u2212
=
\u23a5\u23a5
\u23a5
\u23a6
\u23a4
\u23a2\u23a2
\u23a2
\u23a3
\u23a1
&
&
&
; [ ]
\u23a5\u23a5
\u23a5
\u23a6
\u23a4
\u23a2\u23a2
\u23a2
\u23a3
\u23a1
=
3
2
1
001
x
x
x
y 
 
d. 
>> A=[-0.04167 0 -0.0058; 0.0217 -0.24 0.0058; 0 100 -2.4]; 
>> B=[5.2;-5.2;0]; 
>> C=[0 0 1]; 
>> [V,D]=eig(A); 
>> Bd=inv(V)*B 
 
Bd = 
 
 1.0e+002 * 
 
 -0.9936 + 0.0371i 
 -0.9936 - 0.0371i 
 1.9797 
 
>> Cd = C*V 
 
Cd = 
 
 0.9963 0.9963 1.0000 
 
 
 
5-104 Chapter 5: Reduction of Multiple Subsystems 
 
Copyright © 2011 by John Wiley & Sons, Inc. 
>> D 
 
D = 
 
 -0.0192 + 0.0658i 0 0 
 0 -0.0192 - 0.0658i 0 
 0 0 -2.6433 
So a diagonalized version of the system is 
 
1
3
2
1
3
2
1
97.197
71.336.99
71.336.99
6433.200
00658.00192.00
000658.00192.0
uj
j
x
x
x
j
j
x
x
x
\u23a5\u23a5
\u23a5
\u23a6
\u23a4
\u23a2\u23a2
\u23a2
\u23a3
\u23a1
\u2212\u2212
+\u2212
+
\u23a5\u23a5
\u23a5
\u23a6
\u23a4
\u23a2\u23a2
\u23a2
\u23a3
\u23a1
\u23a5\u23a5
\u23a5
\u23a6
\u23a4
\u23a2\u23a2
\u23a2
\u23a3
\u23a1
\u2212
\u2212\u2212
+\u2212
=
\u23a5\u23a5
\u23a5
\u23a6
\u23a4
\u23a2\u23a2
\u23a2
\u23a3
\u23a1
&
&
&
 
[ ]
\u23a5\u23a5
\u23a5
\u23a6
\u23a4
\u23a2\u23a2
\u23a2
\u23a3
\u23a1
=
3
2
1
19963.09963.0
x
x
x
y 
 
 
81. 
a. Substituting all values and transfer functions into the respective blocks of the system 
(Figure 4), we get: 
 
Solutions to Problems 5-105 
 
Copyright © 2011 by John Wiley & Sons, Inc. 
 
Moving the last pick-off point to the left past the 06154.0
875.4
3.0 ==
toti
r
 block and changing the position of the 
back-emf feedback pick-off point, so that it becomes an outer loop, we obtain the block-diagram shown below. In 
that diagram the 
s\u22c5226.7
1
 block (representing the total inertia) has two parallel feedback blocks. Reducing these two 
blocks into one, we have the following equivalent feedback transfer function: 
01908.0
1384.0
226.7
13787.01
226.7
1
)(
)()( +=+
=\u3a9=
s
s
s
sT
ssGeq 
Vehicle 
Speed 
V(s) 
Torque 
Controller Ref. 
Signal
 
0.6154 
Motor 
Angular 
Speed 
\u3a9 (s) _ 
_ 
0.1 
2 
Friction 
Torque 
Tf (s) 
UC(s) 
1.8 
Eb (s) 
Back emf 
875.4
3.0
Speed
Error
0.5 
s
s 40100 +
 
s
s 610 +
Feedback 
Speed Signal 
s\u22c5226.7
1
 
TL (s)
+
Feedback 
Current Signal 
KCS Ia(s) 
Speed 
Current Sensor 
Sensitivity 
Speed Sensor 
Sensitivity 
 
1
Ua(s) 
Armature 
Res. 
Ra(s) 
Armature 
Current 
Ia(s) 
Motive
Torque
T(s)
0.0443 
5-106 Chapter 5: Reduction of Multiple Subsystems 
 
Copyright © 2011 by John Wiley & Sons, Inc. 
\u3a9 (s) 
_
_
0.1 
2 
Tf (s)
UC(s) 
1.8 
Ia(s) V(s)Ev(s)
0.5
0.0443
s
s 40100 +
 
s
s 610 +
+  _ _ 
Rv(s) 
KSS\u3a9(s) 
s\u22c5226.7
1
 
TL (s)
KCS Ia(s) 
+
+
T(s)
0.06154
UA
0.03787 
Eb(s)
1 
Replacing that feedback loop with its equivalent transfer function, Geq(s), we have: 
 
\u3a9 (s) 
2 
UC(s) 
1.8 
Ia(s) V(s)Ev(s)
0.5
0.0443 
s
s 40100 +
 
s
s 610 +
+  _ _ 
Rv(s) 
KSS\u3a9(s) 
01908.0
1384.0
+s
KCS Ia(s) 
+
T(s)
0.06154 
UA (s) 
Eb(s)
1 
Geq(s)
 
Moving the armature current pick-off point to the right past the 
)(
)(
sI
sT
a
 and Geq(s) blocks, the above block-diagram 
becomes as shown below. 
Solutions to Problems 5-107 
 
Copyright © 2011 by John Wiley & Sons, Inc. 
\u3a9 (s) 
2 
UC(s) 
1.8 
Ia(s) V(s)Ev(s)
0.5
0.0443 
s
s 40100 +
 
s
s 610 +
+  _ _
Rv(s) 
KSS\u3a9(s) 
01908.0
1384.0
+s
KCS Ia(s) 
+
T(s)
0.06154
U
A
 (s)
Eb(s)
1 
2491.0
01908.0+s
 
The latter, in turn, can be reduced to that shown next as the cascaded blocks in the feedback to the torque controller 
are replaced by the single block: 
4982.0
01908.0
)(
)( +=\u3a9
s
s
sIK aCS and the inner feedback loop is replaced by its 
equivalent transfer function: 
 
 
 
 0.5173
2491.0
2
01908.0
2491.01
01908.0
2491.0
)(
)(
+=×++
+=\u3a9
s
s
s
sU
s
A
 
 
\u3a9 (s) UC(s)  V(s)Ev(s)
0.0443 
s
s 40100 +
 
s
s 610 +
+  _ _ 
Rv(s) 
KSS\u3a9(s) 
5173.0
2491.0
+s
KCS Ia(s)
0.06154
U
A
 (s)
4982.0
01908.0+s
+ 
 
5-108 Chapter