Solution 6.8.3.4 The root locus is shown in Figure 1. Eventually two limbs of the root locus Im(s) Re(s) -10 -2 -1 Figure 1: Root locus will cross into the right half plane. The MATLAB program alpha = atan(0.75/19)*180/pi p1 =0 p2 = 1 p3 = 10 zeta = 0.2 omegan =linspace(2.5,3,200);; s=-zeta*omegan + j* omegan *sqrt(1-zeta^2);; ang = (angle(s + p1)+angle(s+p2)+angle(s+p3))*180/pi;; plot(omegan,ang) print -deps 6834a.eps 1 2.5 2.55 2.6 2.65 2.7 2.75 2.8 2.85 2.9 2.95 3 174 175 176 177 178 179 180 181 182 183 184 Figure 2: Angle versus real part of s creates the plot shown in Figure 2. Wesee that the root locus crosses the rayofconstantdamping ratio  =0:8ats = ;0:5598+ j2:7424 Indeed, at s = ;0:5598+ j2:7424, where the net angle is 180:03  . Then the gain to place closed loop poles at s = ;0:5598+ j2:7424 is K = jsjjs +1jjs +10jj s=;0:5598+j2:7424 =85:2322: The natural frequency of the closed loop poles is ! n = p 0:5598 2 +2:7424 2 =2:799: The MATLAB program alpha = atan(0.75/19)*180/pi p1 =0 p2 = 1 p3 = 10 zeta = 0.2 omegan =linspace(2.5,3,200);; s=-zeta*omegan + j* omegan *sqrt(1-zeta^2);; ang = (angle(s + p1)+angle(s+p2)+angle(s+p3))*180/pi;; plot(omegan,ang) 2 0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5 0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 Figure 3: Comparison of step responses print -deps 6834a.eps s=-zeta*omegan(120) + j*omegan(120)*sqrt(1-zeta^2) K=abs(s + p1)*abs(s+p2)*abs(s+p3) omegana = omegan(120) sc = conj(s) tn2 = zpk([],[s sc],omegana^2) g=zpk([],[-p1 -p2 -p3],K) tc = feedback(g,1) T=linspace(0,5,200);; [Yn2,T] = step(tn2,T);; [Ytc,T] = step(tc,T);; plot(T,Yn2,'k-',T,Ytc,'k--') print -deps sr6834.eps Prints and plots the step responses of T c (t) and T N2 (t) shown in Figure 3 3