Solution 6.8.1.18 The MATLAB dialogue EDU>g = zpk([-1],[0 -4 -30],10) Zero/pole/gain: 10 (s+1) -------------- s(s+4) (s+30) EDU>rlocus(g) EDU>print -deps rl68118.eps EDU> draws and saves the root locus shown in Figure 1. Then the MATLAB -30 -25 -20 -15 -10 -5 0 5 -10 -8 -6 -4 -2 0 2 4 6 8 10 Real Axis I mag A x i s Figure 1: Root locus program z=1 p1 =0 p2 = 4 1 -16.7 -16.68 -16.66 -16.64 -16.62 -16.6 -16.58 -16.56 -16.54 -16.52 -16.5 -180.4 -180.3 -180.2 -180.1 -180 -179.9 -179.8 -179.7 -179.6 -179.5 -179.4 Figure 2: Angle versus real part of s p3 = 30 omegad = 10 x=linspace(-16.7,-16.5,400);; s=x+j*omegad;; ang = (angle(s + z)- angle(s + p1)-angle(s+p2)-angle(s+p3))*180/pi;; plot(x,ang) print -deps 68118a.eps creates the plot shown in Figure 2. Then s = ;16:6268 + j10. Indeed, at s = ;16:6268+ j10 the net angle is 180:00  . Then the gain to place closed loop poles at s = ;16:6268+ j10 is K = jsjjs +1js +30j js +1j s=;16:6268+j10 =281:2841: The natural frequency of the closed loop poles is ! n = p 16:6268 2 +10 2 =19:4023: The MATLAB program z=1 p1 =0 2 p2 = 4 p3 = 30 omegad = 10 x=linspace(-16.7,-16.5,400);; s=x+j*omegad;; ang = (angle(s + z)- angle(s + p1)-angle(s+p2)-angle(s+p3))*180/pi;; plot(x,ang) print -deps 68118a.eps pause s=x(147) + j*omegad K=(abs(s + p1)*abs(s+p2)*abs(s+p3))/abs(s+z) omegan = abs(s) tn2 = zpk([],[s conj(s)],omegan^2) g=zpk([-z],[-p1 -p2 -p3],K) tc = feedback(g,1) T=linspace(0,5,100);; [Yn2,T] = step(tn2,T);; [Ytc,T] = step(tc,T);; plot(T,Yn2,'k-',T,Ytc,'k--') print -deps sr68118.eps Prints and plots the step responses of T c (t) and T N2 (t) shown in Figure 3 3 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 Figure 3: Comparison of step responses of T c and T N2 for  =0:4 4