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   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

　
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