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compensated system is given by
k(s + 15)(s + 19.9014)
G(s) = ~s +1)(s + 3) s + 9)
Next, we add the PI controller. Select a pole at s = O and a zero at s - -0.5 so
that the transfer function of the PID controller is grven by
G(s) = k(s + 0.5)(s + 15)(s + 19.9014)
s~s+ ) s+3~s+9)
504 PERFORMANCE, STABILITY, DYNAMICS, AND CONTROL
Fig.538 PID controller for Example 5.11.
Now we draw the root-locus of the PID system as shown in Fig. 5.38b and pick
the point corresponding to < :0.3. We get k = 6.1112 and closed-loop poles at
-3.6846 + j12.2343, -11.2460, and -0.4969.
The unit-step responses of the basjc, PD-, and PID-compensated systems are
shown in Fig. 5.38c. It may be observed that the PID-compensated system meets
the design requirements.
Example 5.12
For the system of Example 5.10, design a major-loop feedback to achieve the
sameperformance. .
So/ution. We have found in Example 5.10 that Ozc = 86.2498. With this, we
obtain the equrvalent pole location, zc - 2.9261, and kh = 1]zc = 0.3418. We then
LINEAR SYSTEMS, THEORY, AND DESIGN:A BRIEF REVIEW 505
plot the root-locus using MATLAB and obtain the value of the gain as 40.2971,
which is equal to kkh so that k - 117.8967. The reader may verify that this response
is identical to that of the PD controller of Example 5.10.
Example 5.13
For the control system shown in Fig. 5.39a, determine the gain kh so that the
minor loop operates with a damping ratio of 0.707 and the complete system has a
damping ratio of 0.4.
So/ution. Consider the minorloop.We draw the root-locus using MATLAB4
as shown in Fig. 5.39b and pick the point on the root-locus corresponding to < -
0.707. Then, we get kh = 14.1018 and p -.5.8471, -1.0765 :1: j1.1194. Having
designed the minor loop and knowing tI:e value of kh, we can simplify the system
a) Controlsystem
Fig.539 Minor-Ioop design for Example 5.13.
7
i
,ii
506 PERFORMANCE, STABILITY, DYNAMICS, AND CONTROL
en
x
<
o
tio
E
c) Outer-Ioop block diagram
Reaf Axis
d) Root-Iocus of outer loop
Fig.539 Minor-Ioop design for Example 5.13,continued.
LINEAR SYSTEMS, THEORY, AND DESIGN: A BRIEF REVIEW 507
block diagram as shown in Fig. 5.39c. We have
Gi(s) :
s3 + 8S2 + (kh + 15)s
For this system, we draw the root-locus as shown in Fig. 5.39d and obtain k -
69.5452 and p = -4.9495, -1.5202 1: j3.4222 for operating at <:0.4. This
completes the design.
5.10 State-Space Analysis and Design
The classical method of analyses discussed in the previous sections is called
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