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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

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

\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

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

>> 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

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

\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

\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