Solution 4.6.1.6 The characteristic equation is 1+ K (s + 1)(s+10) =0;; or s 2 +11s +(10+K) (s +1)(s+10) =0;; or equivalently s 2 +11s +(10+K)=0: The MATLAB program K=1 p=[1 11 10+K] roots(p) K=5 p=[1 11 10+K] roots(p) K=10 p=[1 11 10+K] roots(p) K=20.25 p=[1 11 10+K] roots(p) K=50 p=[1 11 10+K] roots(p) K=100 p=[1 11 10+K] roots(p) K=[1 5 10 20.25 50 100] gh = zpk([],[-1 -10],1) [R,K] = rlocus(gh,K) plot(R,'kd') print -deps rl4616.eps generates the following output 1 EDU>sm4616 K= 1 p= 1 11 11 ans = -9.8875 -1.1125 K= 5 p= 1 11 15 ans = -9.4051 -1.5949 K= 10 p= 2 1 11 20 ans = -8.7016 -2.2984 K= 20.2500 p= 1.0000 11.0000 30.2500 ans = -5.5000 -5.5000 K= 50 p= 1 11 60 ans = -5.5000+ 5.4544i -5.5000- 5.4544i 3 K= 100 p= 1 11 110 ans = -5.5000+ 8.9303i -5.5000- 8.9303i K= 1.0000 5.0000 10.0000 20.2500 50.0000 100.0000 Zero/pole/gain: 1 ------------ (s+1) (s+10) R= Columns 1 through 4 -9.8875 -9.4051 -8.7016 -5.5000 -1.1125 -1.5949 -2.2984 -5.5000 Columns 5 through 6 -5.5000+ 5.4544i -5.5000+ 8.9303i -5.5000- 5.4544i -5.5000- 8.9303i 4 K= 1.0000 5.0000 10.0000 20.2500 50.0000 100.0000 EDU> The plot of the points is shown in Figure ?? -10 -9 -8 -7 -6 -5 -4 -3 -2 -1 -10 -8 -6 -4 -2 0 2 4 6 8 10 Figure 1: Plot of solutions 5