Solution 6.8.1.14 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 rl68114a.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 20.35 20.4 20.45 20.5 20.55 20.6 -180.4 -180.3 -180.2 -180.1 -180 -179.9 -179.8 -179.7 -179.6 -179.5 Figure 2: Angle versus ! n p3 = 20 zeta = 0.6 omegan =linspace(20.4,20.6,400);; 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 68114a.eps creates the plot shown in Figure 2. Wesee that the root locus crosses the line of constant damping at about s = ;12:3038 + j16:4050. Indeed, at s = ;12:3038+ j16:4050 the net angle is 180:000  . Then the gain to place closed loop poles at s = ;12:3038+ j16:4050 is K = jsjjs +5js +20j] js +0:5j s=;12:3038+j16:4050 =330:1661: The natural frequency of the closed loop poles is ! n = p 12:3038 2 +16:4050 2 =20:5063: The MATLAB program 2 z=0.5 p1 =0 p2 = 5 p3 = 20 zeta = 0.6 omegan =linspace(20.4,20.6,400);; 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 68114a.eps s=-zeta*omegan(213) + j*omegan(213)*sqrt(1-zeta^2) K=(abs(s + p1)*abs(s+p2)*abs(s+p3))/abs(s+z) omegn=omegan(213) sc = conj(s) tn2 = zpk([],[s sc],omegan(213)^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 sr68114.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