Solution 10.8.8.12 The MATLAB program w=logspace(-2,3,200);; s=j*w;; K=100 z1 = 5 p1 = 1+j*1 p2 = 1-j*1 p3 = 20 p4 = 40 hold off mag = 20*log10( abs((K.*(s+z1)) ./((s+p1).*(s+p2).*(s+p3).*(s+p4))));; phase = (angle(s+z1)-angle(s+p1)-angle(s+p2) -angle(s+p3)-angle(s+p4))*180/pi;; plot(phase,mag);; grid on axis([-260,-60,-80 40]) print -deps 108812logmag.eps g=zpk([-z1],[-p1 -p2 -p3 -p4],100) K=linspace(0,1600,500);; [P,K] = rlocus(g,K);; plot(real(P),imag(P),'k.') grid on print -deps 108812rl.eps pause subplot(2,1,1),semilogx(w,mag) grid on axis([0.1 1000 -80 20]) subplot(2,1,2), semilogx(w,phase) grid on axis([0.1 1000 -280 0]) print -deps 108812bodema.eps draws the log magnitude plot shown in Figure 1 With K =100thegain margin is about 44 dB. Converting to a real number wehave 10 52=20 =398:107: Thus, the system is stable for 0 <K<39;;810:7: The same information is available from the Bode magnitude and phase plots as shown in Figure 2 By tracing the arrows from the phase scale at ;180  1 -260 -240 -220 -200 -180 -160 -140 -120 -100 -80 -60 -80 -60 -40 -20 0 20 40 The point (-1,0) Figure 1: Log magnitude chart for K = 100 2 10 -1 10 0 10 1 10 2 10 3 -80 -60 -40 -20 0 20 10 -1 10 0 10 1 10 2 10 3 -250 -200 -150 -100 -50 0 Figure 2: Bode phase and magnitude plots for K =100 3 -80 -70 -60 -50 -40 -30 -20 -10 0 10 -50 -40 -30 -20 -10 0 10 20 30 40 50 Figure 3: Root locus to where it intercepts the phase plot and then going up to the magnitude plot and thence to the magnitude scale we nd, again, that the gain margin is about 52 dB. Wealsoseethat the ! =25when the phase is 180  .This means that the root locus will cross the imaginary axis for ! = 25, The root locus is shown in Figure 3. As can be seen, twolimbs of the root locus cross into the righthalf place at ! 25. 4