Solution 6.8.1.17 The MATLAB dialogue EDU>g = zpk([-1],[0 -4 -20],10) Zero/pole/gain: 10 (s+1) -------------- s(s+4) (s+20) EDU>rlocus(g) EDU>print -deps rl68117.eps EDU> draws and saves the root locus is shown in Figure 1. -20 -15 -10 -5 0 5 -8 -6 -4 -2 0 2 4 6 8 Real Axis I mag A x i s Figure 1: Root locus Then the MATLAB program z=1 p1 =0 p2 = 4 1 23 23.02 23.04 23.06 23.08 23.1 23.12 23.14 23.16 23.18 23.2 -180.3 -180.2 -180.1 -180 -179.9 -179.8 -179.7 Figure 2: Angle versus ! n p3 = 20 zeta = 0.5 omegan =linspace(23,23.2,400);; 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 68117a.eps creates the plot shown in Figure 2. Wesee that the root locus crosses the line of constant damping ratio at about s = ;11:5559+ j20:0154. Indeed, at s = ;11:5559+ j20:0154 the net angle is 180:00  . Then the gain to place closed loop poles at ss = ;11:5559+j20:0154 is K = jsjjs+4js +20j] js +1j s=;11:5559+j20:0154 =474:6876: The natural frequency of the closed loop poles is ! n = p 11:5559 2 +20:0154 2 =23:1118: The MATLAB program z=1 2 p1 =0 p2 = 4 p3 = 20 zeta = 0.5 omegan =linspace(23,23.2,400);; 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 68117a.eps pause s=-zeta*omegan(224) + j*omegan(224)*sqrt(1-zeta^2) K=(abs(s + p1)*abs(s+p2)*abs(s+p3))/abs(s+z) omegn=omegan(224) sc = conj(s) tn2 = zpk([],[s sc],omegan(224)^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 sr68117.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