Solution 4.6.1.3 The characteristic equation is 1+ K s(s + 4)(s+8) =0;; or s 3 +12s 2 +32s+ K s(s +4)(s+8) =0;; or equivalently s 3 +12s 2 +32s+ K =0: The MATLAB program K=0.1 p=[1 12 32 K] roots(p) K=0.51 p=[1 12 32 K] roots(p) K=3.8 p=[1 12 32 K] roots(p) K=4 p=[1 12 32 K] roots(p) K=4.2 p=[1 12 32 K] roots(p) K=20 p=[1 12 32 K] roots(p) K=50 p=[1 12 32 K] roots(p) K=[0.1 0.513.844.22050] gh = zpk([],[0 -4 -8],1) [R,K] = rlocus(gh,K);; plot(R,'kd') print -deps rl4613.eps 1 generates the following output K= 0.1000 p= 1.0000 12.0000 32.0000 0.1000 ans = -8.0031 -3.9937 -0.0031 K= 0.5100 p= 1.0000 12.0000 32.0000 0.5100 ans = -8.0158 -3.9681 -0.0160 K= 3.8000 2 p= 1.0000 12.0000 32.0000 3.8000 ans = -8.1138 -3.7617 -0.1245 K= 4 p= 1 12 32 4 ans = -8.1196 -3.7490 -0.1314 K= 4.2000 p= 1.0000 12.0000 32.0000 4.2000 3 ans = -8.1253 -3.7364 -0.1383 K= 20 p= 1 12 32 20 ans = -8.5194 -2.5655 -0.9151 K= 50 p= 1 12 32 50 ans = -9.0830 -1.4585+ 1.8378i -1.4585- 1.8378i 4 K= 0.1000 0.5100 3.8000 4.0000 4.2000 20.0000 50.0000 Zero/pole/gain: 1 ------------- s(s+4) (s+8) EDU> The plot of the points is shown in Figure ?? -10 -9 -8 -7 -6 -5 -4 -3 -2 -1 0 -2 -1.5 -1 -0.5 0 0.5 1 1.5 2 Figure 1: Plot of solutions 5