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method of compensation is usually used in aircraft control systems to improve the
response to individual degrees of freedom like pitch, roll, or yaw before closing
the outer loop as we will discuss in Chapter 6.
Example 5.9
For the system shown in Fig. 5:33, 1) design a PI compensator to reduce the
steady-state error to zero and 2) a lag compensator to reduce the steady-state error
by a factor of 10 for a step input without affecting the transient response. Assume
that the system is required to operate with a damping ratio of < = 0.2.
LINEAR SYSTEMS, THEORY, AND DESIGN: A BRIEF REVIEW 495
Fig.533 Controlsystem ofExample 5.9.
Solution. Wehave
k
G(s) - ~s +2 (s +5)(s+12)
The first step is to draw the root-locus of the basic (uncompensated) system and
deternune the value of the gain for operation at < :0.2. Using MATLAB,4 the
root-locus is drawn as shown in Fig. 5.34a. For operation with < - 0.2, the values
of the gain and closed-loop pole locations are k - 679.086 and p = -16.2416.
-1.3792 + j6.8773.
The position constant K p and the steady-state error e(oo) are given by
k 679.086
Ki, = P~P2P3 = 2~5 v 12 - 5.6590
1 1
e(oo) = ~+kp = 1+5 6590 - 0.1502
With this, we get the steady-state value of the out)ut (for a unit-step input),
y(oo) = 1 - e(oo):0.8498.
Design of a Plcompensator. The PI compensator is characterized by a pole at
the origin and a zero close to it. Let us choose the zero at s - -0.05. With this, the
open-loop transfer function of the PI-compensated system is
Gc(s):- k(s+0.05)
= s~s+2 (s+5) s+~2~
Now let us determine the value of the gain k so that the PI-compensated sys-
tem operates at a damping ratio of 0.2, while the steady-state error is driven to
zero. Using MATLAB,4 we draw the root-locus for the PI-compensated system as
shown in Fig. 5.34b. For operating with g = 0.2, we obtain k - 673.175 and p =
-16.2118, -1.3728 j:: j6.8408, and -0.0426.
Comparing these results with those obtained earlier for the basic system, we
observe that the dominant second-order complex poles that determine the transient
response are virtually unchanged because the pole at s - -0.0426 almost cancels
with the zero at s - -0.05. The pole at s = -16.2118 is so far away on theleft-hand
side of the s-plane that its influence is negligible. In view of this, the system will
essentially behave like a second-order system with dominant poles at -1.3728 :1
j6.8408.
[
496 PERFORMANCE, STABILITY, DYNAMICS, AND CONTROL
a)
Fig. 5.34 Root-Iocii for the control system of Example 5.9.
Design of the lag compensr;ttor. We have to design the lag compensator to
achieve a reduction in the steady-state error by a factor of 10, i.e.,
e(oo) : 0/~02
- 0.01502
Then,
1 - e(oo) 1 - 0.01502
Kp= e(oo)= 00~502 -64.7895
For the lag-compensated system,
zck
Kp = p *2*5*12
LINEAR SYSTEMS, THEORY, AND DESIGN: A BRIEF REVIEW 497
c)
Fig. 5 34 Root-Iocii for the controlsystem of Example 5.9, continued.
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