Solution 4.6.1.2 The characteristic equation is 1+ K (s +4)(s +8) =0;; or s 2 +12s +(32+K) (s +4)(s+8) =0;; or equivalently s 2 +12s +(32+K)=0: The MATLAB program K=0.1 p=[1 12 K+32] roots(p) K=0.51 p=[1 12 K+32] roots(p) K=4 p=[1 12 K+32] roots(p) K=20 p=[1 12 K+32] roots(p) K=50 p=[1 12 K+32] roots(p) K=[0.1 0.51 4 20 50] gh = zpk([],[-4 -8],1) [R,K] = rlocus(gh,K);; plot(R,'kd') print -deps rl4612.eps generates the following output K= 0.1000 1 p= 1.0000 12.0000 32.1000 ans = -7.9748 -4.0252 K= 0.5100 p= 1.0000 12.0000 32.5100 ans = -7.8682 -4.1318 K= 4 p= 1 12 36 ans = 2 -6 -6 K= 20 p= 1 12 52 ans = -6.0000+ 4.0000i -6.0000- 4.0000i K= 50 p= 1 12 82 ans = -6.0000+ 6.7823i -6.0000- 6.7823i K= 0.1000 0.5100 4.0000 20.0000 50.0000 3 Zero/pole/gain: 1 ----------- (s+4) (s+8) EDU> The plot of the points is shown in Figure 1 -8 -7.5 -7 -6.5 -6 -5.5 -5 -4.5 -4 -8 -6 -4 -2 0 2 4 6 8 Figure 1: Plot of solutions 4