Solution 9.10.2.4 The MATLAB program load bodeid4 topm = size(bodeid4) top = topm(1,1) w=bodeid4(1:top,1);; mag = bodeid4(1:top,2);; phase = bodeid4(1:top,3);; semilogx(w,mag);; grid on print -deps 91024mag.eps semilogx(w,phase);; grid on print -deps 91024phase.eps can be used to load and plot the data. You will havetoedit this data le. You only wantthe rst 138 lines. There is some other data mistakenly appended. That should be deleted. Figure 1 shows the magnitude data with some asymptotes added. There is a pole at ts = ;1and what appears to be a double pole near s = ;5. The phase plot, shown in Figure 2 also inidcates the presence of three poles. The gains is K =10 4=20 =1:5849 Then, our rst guess at the transfer function is G(s) = 1:5849 (1 + s)(1 + s=5) 2 = 39:622 (1 + s)(1 + s=5) 2 The comparison of the Bode magnitude plot and the proposed magnitude plot is shown in Figure 3 A similar comparison of the phase plots is shown in Figure 4 The gain is a little low, the magnitude ts almost perfectly,and the phase does not t perfectly.Thefact that the phase t is worse than the magnitude t suggests wehavecomplex poles with a large damping ratio. So we next try G(s) = 2 (1+ s)[(s= p 50) 2 +0:2s +1] = 100 (s +1)(s +5;j5)(s+5+j5) 1 10 -2 10 -1 10 0 10 1 10 2 10 3 -160 -140 -120 -100 -80 -60 -40 -20 0 20 Figure 1: Bode magnitude plot 2 10 -2 10 -1 10 0 10 1 10 2 10 3 -300 -250 -200 -150 -100 -50 0 Figure 2: Bode phase plot 3 10 -2 10 -1 10 0 10 1 10 2 10 3 -160 -140 -120 -100 -80 -60 -40 -20 0 20 Figure 3: Comparison of acutal and propose Bode magnitude plots 4 10 -2 10 -1 10 0 10 1 10 2 10 3 -300 -250 -200 -150 -100 -50 0 Figure 4: Comparison of acutal and propose Bode phase plots 5 10 -2 10 -1 10 0 10 1 10 2 10 3 -160 -140 -120 -100 -80 -60 -40 -20 0 20 Figure 5: Comparison of acutal and propose Bode magnitude plots 6 10 -2 10 -1 10 0 10 1 10 2 10 3 -300 -250 -200 -150 -100 -50 0 Figure 6: Comparison of acutal and propose Bode phase plots 7 10 -2 10 -1 10 0 10 1 10 2 10 3 -160 -140 -120 -100 -80 -60 -40 -20 0 20 Figure 7: Comparison of acutal and propose Bode magnitude plots The results are shown in Figures 5 and 6. The matchisbetterinsome places but worse elsewhere. So we next try reducing the natural frequency of the dampe poles to obtain G(s) = 2 (s + 1)[(s= p 18) 2 +(1=3)s+1) = 36 (s + 1)(s 2 +6s +18) The comparisons of magnitude and angle shown in Figures 7 and 7 show that weare close. Wehaveagoodmatchthus 36 (s + 1)(s 2 +6s +18) The MATLAB program that does this analysis is load bodeid4 topm = size(bodeid4) 8 10 -2 10 -1 10 0 10 1 10 2 10 3 -300 -250 -200 -150 -100 -50 0 Figure 8: Comparison of acutal and propose Bode phase plots 9 top = topm(1,1) w=bodeid4(1:top,1);; mag = bodeid4(1:top,2);; phase = bodeid4(1:top,3);; semilogx(w,mag);; grid on print -deps 91024mag.eps semilogx(w,phase);; grid on print -deps 91024phase.eps p1 = 1 p2 = 5 p3 = 5 Ktc = 1.5849 K=Ktc *abs(p1)*abs( p2) * abs(p3) wp = logspace(-2,3,20);; jw = j*wp;; mag1 = 20*log10( (K)./ ( abs(jw +p1).* abs(jw + p2).*abs(jw+p3)));; semilogx(w,mag,'k-',wp,mag1,'kd') grid on print -deps 91024mag1.eps phase1 = ( -angle(jw +p1) -angle(jw + p2) - angle(jw+p3) )*180/pi;; semilogx(w,phase,'k-',wp,phase1,'kd') grid on print -deps 91024phase1.eps pause p1 = 1 p2 = 5 - j*5 p3 = 5 + j*5 Ktc = 2 K=(Ktc *p1* p2 * p3) mag2 = 20*log10( (K )./ ( abs(jw +p1).* abs(jw + p2).*abs(jw+p3)));; semilogx(w,mag,'k-',wp,mag2,'kd') print -deps 91024mag2.eps phase2 = ( -angle(jw +p1) -angle(jw + p2) - angle(jw+p3) )*180/pi;; semilogx(w,phase,'k-',wp,phase2,'kd') grid on print -deps 91024phase2.eps p1 = 1 p2 = 3 -j*3 p3 = 3 + j*3 Ktc = 2 K=Ktc*p1*p2*p3 mag3 = 20*log10( K./ ( abs(jw +p1).* abs(jw + p2).*abs(jw+p3) ) );; semilogx(w,mag,'k-',wp,mag3,'kd') 10 grid on print -deps 91024mag3.eps phase3 = ( -angle(jw +p1) -angle(jw + p2) - angle(jw+p3) )*180/pi;; semilogx(w,phase,'k-',wp,phase3,'kd') grid on print -deps 91024phase3.eps 11