Solution 6.8.1.13 The root locus is shown in Figure 1. Eventually two limbs of the root locus -35 -30 -25 -20 -15 -10 -5 0 5 -20 -15 -10 -5 0 5 10 15 20 Real Axis I mag A x i s Figure 1: Root locus will cross into the righthalf plane. Indeed, for ! d =15twopoles are in the right half plane. The MATLAB program p1 =0 p2 = 1 p3 = 30 omegad = 15 x=linspace(2.6,2.8,400);; s=x+j*omegad;; ang = (angle(s + p1)+angle(s+p2)+angle(s+p3))*180/pi;; plot(x,ang) print -deps 68113a.eps creates the plot shown in Figure 2. Then s =2:7719 + j15. Indeed, at s =2:7719+ j15 the net angle is 180:00  . Then the gain to place closed loop poles at s =2:7719+ j15 is K = jsjjs+1js +30jj s=2:7719+j15 =8503:4: 1 2.6 2.65 2.7 2.75 2.8 2.85 179.6 179.8 180 180.2 180.4 180.6 180.8 181 181.2 181.4 181.6 Figure 2: Angle versus real part of s The natural frequency of the closed loop poles is ! n = p 2:7719 2 +15 2 =15:254: The MATLAB program p1 =0 p2 = 1 p3 = 30 omegad = 15 x=linspace(2.6,2.8,400);; s=x+j*omegad;; ang = (angle(s + p1)+angle(s+p2)+angle(s+p3))*180/pi;; plot(x,ang) print -deps 68113a.eps s=x(344) + j*omegad K=abs(s + p1)*abs(s+p2)*abs(s+p3) omegan = abs(s) tn2 = zpk([],[s conj(s)],omegan^2) g=zpk([],[-p1 -p2 -p3],K) tc = feedback(g,1) T=linspace(0,1,100);; [Yn2,T] = step(tn2,T);; 2 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 -10 -5 0 5 10 15 20 Figure 3: Comparison of step responses of T c and T N2 for  =0:4 [Ytc,T] = step(tc,T);; plot(T,Yn2,'k-',T,Ytc,'k--') print -deps sr68113.eps Prints and plots the step responses of T c (t) and T N2 (t) shown in Figure 3 3