Solution 6.8.3.3 T c (s) = K(s +20) s(s +2)(s+20)+K = K(s +20) s 3 +22s 2 +(40+K) : The root locus, shown in Figure 1, is Eventually twolimbs of the root locus Im(s) Re(s) -30 -2 -1 Figure 1: Root locus will cross into the righthalfplane, but near the origin the pole at s = ;20 has very little in uence on the root locus. Near the origin the root locus looks very like that of the simpler transfer function G(s)= K s(s +2) : 1 1 1.05 1.1 1.15 1.2 1.25 1.3 1.35 1.4 1.45 1.5 170 175 180 185 190 195 Figure 2: Angle versus real part of s So a good place to start looking for the roots is s = ;1j!, where ! = 1 "p 1; 2  # = 0:75: Indeed, the contribution of the pole at s = ;30 at s = ;1+j0:75 is 2:2605  . The MATLAB program p1 =0 p2 = 2 p3 = 20 zeta = 0.8 omegan =linspace(1,1.5,200);; s=-zeta*omegan + j* omegan *sqrt(1-zeta^2);; ang = (angle(s + p1)+angle(s+p2)+angle(s+p3))*180/pi;; plot(omegan,ang) print -deps 6833a.eps creates the plot shown in Figure 2. Wesee that the root locus crosses the rayofconstant damping ratio  =0:8ats = ;0:82. Indeed, at s = ;0:9608+ j7206, where the net angle is 180:03  . 2 Then the gain to place closed loop poles at s = ;0:83 + j3is K = jsjjs +2jjs +20jj s=;0:9608+j0:7206 =28:9371: The natural frequency of the closed loop poles is ! n = p 0:9608 2 +0:7206 2 =1:2010: The MATLAB program p1 =0 p2 = 2 p3 = 20 zeta = 0.8 omegan =linspace(1,1.5,200);; s=-zeta*omegan + j* omegan *sqrt(1-zeta^2);; ang = (angle(s + p1)+angle(s+p2)+angle(s+p3))*180/pi;; plot(omegan,ang) print -deps 6833a.eps s=-zeta*omegan(81) + j*omegan(81)*sqrt(1-zeta^2) K=abs(s + p1)*abs(s+p2)*abs(s+p3) omegana = omegan(81) zeta = 0.8 sc = conj(s) tn2 = zpk([],[s sc],omegana^2) g=zpk([],[0 -2],K) h=zpk([],[-20],1) tc = feedback(g,h) T=linspace(0,5,200);; [Yn2,T] = step(tn2,T);; [Ytc,T] = step(tc,T);; plot(T,Yn2,'k-',T,Ytc,'k--') print -deps sr6833.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 5 10 15 20 25 T N2 T c Figure 3: Comparison of step responses 4