Solution 6.8.1.15 The MATLAB dialogue EDU>g = zpk([-0.5],[0 -5,-20],5) Zero/pole/gain: 5 (s+0.5) -------------- s(s+5) (s+20) EDU>rlocus(g) EDU>print -deps rl68115a.eps EDU> draws and saves the root locus is shown in Figure 1. -20 -15 -10 -5 0 5 -2.5 -2 -1.5 -1 -0.5 0 0.5 1 1.5 2 2.5 Real Axis I mag A x i s Figure 1: Root locus Then the MATLAB program z=0.5 p1 =0 p2 = 5 1 15.3 15.32 15.34 15.36 15.38 15.4 15.42 15.44 15.46 15.48 15.5 -180.4 -180.2 -180 -179.8 -179.6 -179.4 -179.2 Figure 2: Angle versus ! n p3 = 20 zeta = 0.8 omegan =linspace(15.3,15.5,400);; s=-zeta*omegan + j* omegan *sqrt(1-zeta^2);; ang = (angle(s + z)-angle(s + p1)-angle(s+p2)-angle(s+p3))*180/pi;; plot(omegan,ang) print -deps 68115a.eps creates the plot shown in Figure 2. Wesee that the root locus crosses the line of constantdamping at about s = ;12:3471 + j9:2603. Indeed, at s = ;12:3471+ j9:2603 the net angle is 180:00  . Then the gain to place closed loop poles at s = ;12:3471+ j9:2603 is K = jsjjs+5js+20j] js +0:5j s=;12:3471+j9:2603 = 145:7574: The natural frequency of the closed loop poles is ! n = p 12:3471 2 +9:2603 2 =15:4338: The MATLAB program z=0.5 2 p1 =0 p2 = 5 p3 = 20 zeta = 0.8 omegan =linspace(15.3,15.5,400);; s=-zeta*omegan + j* omegan *sqrt(1-zeta^2);; ang = (angle(s + z)-angle(s + p1)-angle(s+p2)-angle(s+p3))*180/pi;; plot(omegan,ang) print -deps 68115a.eps s=-zeta*omegan(268) + j*omegan(268)*sqrt(1-zeta^2) K=(abs(s + p1)*abs(s+p2)*abs(s+p3))/abs(s+z) omegn=omegan(268) sc = conj(s) tn2 = zpk([],[s sc],omegan(268)^2) g=zpk([-z],[-p1 -p2 -p3],K) tc = feedback(g,1) T=linspace(0,6,200);; [Yn2,T] = step(tn2,T);; [Ytc,T] = step(tc,T);; plot(T,Yn2,'k-',T,Ytc,'k--') print -deps sr68115.eps Prints and plots the step responses of T c (t) and T N2 (t) shown in Figure 3 3 0 1 2 3 4 5 6 0 0.2 0.4 0.6 0.8 1 1.2 1.4 Figure 3: Comparison of step responses of T c and T N2 for  =0:4 4