Solution 4.6.1.11 The characteristic equation is 1+ K(s +4) s 2 (s +2)(s +50) =0;; or s 4 +52s 3 +100s 2 + Ks+4K s 2 (s +2)(s +50) =0;; or equivalently s 4 +52s 3 +100s 2 + Ks+4K =0: The MATLAB program K=1 p=[1 52 100 K 4*K] roots(p) K=2 p=[1 52 100 K 4*K] roots(p) K=5 p=[1 52 100 K 4*K] roots(p) K=50 p=[1 52 100 K 4*K] roots(p) K=500 p=[1 52 100 K 4*K] roots(p) K=[1 2 5 50 500] gh = zpk([-4],[0 0 -2 -50],1) [R,K] = rlocus(gh,K) plot(R,'kd') print -deps rl46111.eps generates the following output EDU>sm46111 K= 1 1 p= 1 52 100 1 4 ans = -50.0004 -2.0103 0.0053+ 0.1994i 0.0053- 0.1994i K= 2 p= 1 52 100 2 8 ans = -50.0008 -2.0202 0.0105+ 0.2812i 0.0105- 0.2812i K= 5 p= 2 1 52 100 5 20 ans = -50.0019 -2.0485 0.0252+ 0.4412i 0.0252- 0.4412i K= 50 p= 1 52 100 50 200 ans = -50.0192 -2.3250 0.1721+ 1.3001i 0.1721- 1.3001i K= 500 p= 1 52 100 500 2000 3 ans = -50.1902 0.6268+ 3.5518i 0.6268- 3.5518i -3.0633 K= 1 2 5 50 500 Zero/pole/gain: (s+4) ---------------- s^2 (s+2) (s+50) R= Columns 1 through 4 -50.0004 -50.0008 -50.0019 -50.0192 -2.0103 -2.0202 -2.0485 -2.3250 0.0053+ 0.1994i 0.0105+ 0.2812i 0.0252+ 0.4412i 0.1721+ 1.3001i 0.0053- 0.1994i 0.0105- 0.2812i 0.0252- 0.4412i 0.1721- 1.3001i Column 5 -50.1902 -3.0633 0.6268+ 3.5518i 0.6268- 3.5518i K= 1 2 4 5 50 500 EDU> The plot of the points is shown in Figure 1 -60 -50 -40 -30 -20 -10 0 10 -4 -3 -2 -1 0 1 2 3 4 Figure 1: Plot of solutions 5