Solution 4.6.1.8 The characteristic equation is 1+ K s(s + 2)(s +40) =0;; or s 3 +42s 2 +80s+ K s(s +2)(s +40) =0;; or equivalently s 3 +42s 2 +80s+ K =0: The MATLAB program K=1 p=[1 42 80 K] roots(p) K=5 p=[1 42 80 K] roots(p) K=50 p=[1 42 80 K] roots(p) K=100 p=[1 42 80 K] roots(p) K=[1 5 50100] gh = zpk([],[0 -2 -40],1) [R,K] = rlocus(gh,K) plot(R,'kd') print -deps rl4618.eps generates the following output EDU>sm4618 K= 1 1 p= 1 42 80 1 ans = -40.0007 -1.9868 -0.0126 K= 5 p= 1 42 80 5 ans = -40.0033 -1.9320 -0.0647 K= 50 p= 1 42 80 50 ans = 2 -40.0328 -0.9836+ 0.5306i -0.9836- 0.5306i K= 100 p= 1 42 80 100 ans = -40.0656 -0.9672+ 1.2492i -0.9672- 1.2492i K= 1 5 50 100 Zero/pole/gain: 1 -------------- s(s+2) (s+40) R= -40.0007 -40.0033 -40.0328 -40.0656 -1.9868 -1.9320 -0.9836- 0.5306i -0.9672- 1.2492i -0.0126 -0.0647 -0.9836+ 0.5306i -0.9672+ 1.2492i 3 K= 1 5 50 100 EDU> The plot of the points is shown in Figure 1 -45 -40 -35 -30 -25 -20 -15 -10 -5 0 -1.5 -1 -0.5 0 0.5 1 1.5 Figure 1: Plot of solutions 4