曝光臺 注意防騙 網曝天貓店富美金盛家居專營店坑蒙拐騙欺詐消費者

Here, we have three unknowns k, zc, and Pc and one relation as above. To begin
with, let us assume k ~ 679.086 (uncompensated gain). Then, we can choose one
of the two remauung unknowns arbitrarily. Let us choose zc = 0.05 so that we get
Pc = 0.0044. The open-loop transfer function of the Iag-compensated system is
given by
                                     k(s + 0.05)
G(s) = ~s +0 0044 (s +2)(s +5)(s +12)
Now we can draw the root-locus as shown in Fig. 5:34c and determine the value
of the gain and closed-loop poles for operating with < = 0.2. We get k - 670.3530
and p = -16.2019, -1.3796 -1- j6.8304, and -0.0433.
     Thus, the pole locations are similar to those observed for the PI compensator.
     With this new value of the gain k = 670.3530, the steady-state error is slightly
changed. We have
                  e
Kp =  p-, * 2z~k5 ,
                         5 *12
               0.05 * 670.3530
      = 0 0044 *2*5 *12
    - 63.4804
which is close to the target value of 0.01502. Hence, we need not repeat the design
process.
7
  I
  i
 {
                1
e(oo) -  ~+ K p
                    1
. =~+63480~
       - 0.0155
498           PERFORMANCE, STABILITY, DYNAMICS, AND CONTROL
Fig.535   Unit-step responses ofExample 5.9.
     Now to test the designs of PI- and lag-compensated systems, we have obtained
unit-step responses of the basic, PI-, and Iag-compensated systems as shown in
Fig. 5.35. We observe that the design objectives are met. The transient response
of the PI- and lag-compensated systems are almost identical to that of the basic
system. Furthermore, as t assumes large values, the steady-state error for the PI
compensator approaches zero and that for the lag compensator approaches the
target value of 0.01502.
                                         Example 5.10
     For the following system, design 1) a PD compensator and 2) alead-lag compen-
sator so that the peak time is reduced by a factor of 3, while the percent overshoot
remains unchanged at 25.  8%.
                                   k
                                             G(s) = s~s + 3~s + 5~
Solution.
   PD compensator. The first step is to draw the root-locus of the basic system
as shown in Fig. 5.36a. Using Eq. (5.75), we find that the damping ratio Lhat
corresponds to 25.38qro overshoot is equal to 0.4. For the basic (uncompensated)
system, the value of the gain that corresponds to < = 0.4 is equal to 28.3825, and
LINEAR SYSTEMS, THEORY, AND DESIGN:A BRIEF REVIEW      499
Root~Locus of Basic System
Real Axis
a)
   T ------ -~'- - -- -                                                                 , jco      
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┃                                                                                    ┃~el  _    ┃
　
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