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

We have three poles at s - 0, -3, -5 and no finite zeros. Therefore, the missing
 zeros arelocated at s -
    As s  + co, G(s) ~ 1~3 = O,i.e., G(s) has three zeros at s - c:o.
Consider
                                                                          G(s) : s                                                     (5.134)
 This system has a zero ats  =  0 and a pole atinfinity because,as s ~ oo,  G(s) +  oo.
 Similarly, G(s) -.  l]s has a zero at infinity because G(s) -y  0 as s -+  oo.
        5) Asymptotes. The asymptotes give the behavior of the root-locus as the param-
 eter k approaches infinity. The point, ofintersection of the asymptotes with the real
 axis uo (see Fig.5:15a) and the slopes ofthe asymptotes M  at this point are given by
cro = Ppoles - E zeros
       (5,135)
                        Fr P - r.z
                                  M = tan np U. ~-                         (5.136)
where np and nz are the number of open-loop poles and zeros, respectively, and
n  -. 0, :1:1, +2,.... The running index n gives the slopes of the asymptotes that
form the branches:of the root-locus as k -* oo.
      /maginary axis crossing.     Another characteristic feature that is ofinterest in
the root-locus method is the point where the root-locus crosses the imaginary axis
because the system stability changes at this point. If the imaginary axis crossing
is from right to left of the s-plane, the closed-loop becomes stable as the gain is
increased. Ifit is from left to right, then the closed-loop system becomes unstable
on increasing the gain.
      The point(s) where the root-locus crosses the imaginar)r axis can be determined
by 1) using the Routh's criterion and finding the values of the gain k that give all
the zeros in any one row of the Routh's ta6ble-or 2) substituting s =  jco in the
characteristic equation, setting both real and imaginary parts to zero and solving
for the gain k and frequency a>. We will illustrate this second procedure in the
following example.
                                          Example 5.5           .
Sketch the root-locus for the unity feedback system with
       G(s)=- k(s+4)
                                                                   ,  s~s +    )(s + 2)(s + 5)
    Solution.   We have four poles at s = 0, -1, -2, -5 and only one finite zero
at s - -4. Therefore, the other three missing zeros are at inftnity. We have four
branches of the root-locus. Furthermore, the root-locus will be symmetrical. with
respect to the real axis. That segment of the real axis forms a part of the root-locus,
which lies to the left of the odd number of poles ancVor zeros. Thus, the real axis
segment between the poles at 0 and -1 and between the pole at -2 and zero at
-4 and all the real axis that is to the left of the pole at -5 forms the branches of
the root-locus.
LINEAR SYSTEMS, THEORY, AND DESIGN: A BRIEF REVIEW       471
　
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