Solution 6.8.1.16 The MATLAB dialogue EDU>g = zpk([-1],[0 -4 -10],5) Zero/pole/gain: 5 (s+1) -------------- s(s+4) (s+10) EDU>rlocus(g) EDU>print -deps rl68116.eps EDU> draws and saves the root locus is shown in Figure 1. -10 -8 -6 -4 -2 0 2 4 -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=1 p1 =0 p2 = 4 1 8.4 8.41 8.42 8.43 8.44 8.45 8.46 8.47 8.48 8.49 8.5 -181.2 -181 -180.8 -180.6 -180.4 -180.2 -180 -179.8 -179.6 Figure 2: Angle versus ! n p3 = 10 zeta = 0.8 omegan =linspace(8.4,8.5,400);; 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 68116a.eps creates the plot shown in Figure 2. Wesee that the root locus crosses the line of constant damping ratio at about s = ;6:7314+ j5:0486. Indeed, at s = ;6:7314+ j5:0486 the net angle is 180:00  . Then the gain to place closed loop poles at s = ;12:3471+ j9:2603 is K = jsjjs+4js +10j] js +1j s=;6:7314+j5:0486 =38:0319: The natural frequency of the closed loop poles is ! n = p 6:7314 2 +5:0486 2 =8:4143: The MATLAB program z=1 2 p1 =0 p2 = 4 p3 = 10 zeta = 0.8 omegan =linspace(8.4,8.5,400);; 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 68116a.eps pause s=-zeta*omegan(58) + j*omegan(58)*sqrt(1-zeta^2) K=(abs(s + p1)*abs(s+p2)*abs(s+p3))/abs(s+z) omegn=omegan(58) sc = conj(s) tn2 = zpk([],[s sc],omegan(58)^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 sr68116.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