Solution 9.10.2.9 The MATLAB statements load bodeid9 topm = size(bodeid9) top = topm(1,1) w=bodeid9(1:top,1);; mag = bodeid9(1:top,2);; phase = bodeid9(1:top,3);; semilogx(w,mag);; grid on axis([0.1,100,-100,40]) print -deps 91029mag.eps semilogx(w,phase);; grid on axis([0.01,100,-330,-180]) print -deps 91029phase.eps can be used to load the data and plot the magnitude response. Figure 1 shows the magnitude data with some asymptotes added. The transfer func- tion clearly has twopoles at the origin and twocomplex poles. The high frequency asymptote has a slope of ;60 dB/dec, so there is also a zero present. The phase plot, shown in Figure 2, is interesting because even at lowfrequencies, the phase is;220  .This seems to indicate that the complex poles are in uencing the phase as muchatlowfrequency at the zero. The \hump" in the magnitude plot is over 10 dB, so weexpect the damping ratio to be small. The resonant frequency is about 10 rad/sec. The gain term can be approximated by remembering that at lowfrequency only the gain term and the term 1=s 2 will be present. Thus our rst guess is The term 1=s 2 by itself will cross the vertical line through ! =1rad/sec at 0 db. The plot actually crosses at ;6 dB, so wecan nd the gain to be K =10 ;6=20 =0:5012: Then our rst guess at the transfer function is G(s) = 0:5(1+ s=10) s 2 ((s= p 104) 2 +(1=52)s+1) = 2:2(s+10) s 2 (s +2;j10)(s+2+j10) : Wecheckthe accuracy of the model bycomparing the actual magnitude and phase to the phase of the derived transfer function, as shown in Figures 3 1 10 -1 10 0 10 1 10 2 -100 -80 -60 -40 -20 0 20 40 Figure 1: ABode magnitude plot 2 10 -2 10 -1 10 0 10 1 10 2 -300 -280 -260 -240 -220 -200 -180 -160 -140 Figure 2: ABode phase plot 3 10 -2 10 -1 10 0 10 1 10 2 10 3 -100 -80 -60 -40 -20 0 20 40 Figure 3: Comparison of actual and derived Bode magnitude plots 4 10 -2 10 -1 10 0 10 1 10 2 10 3 -300 -250 -200 -150 -100 Figure 4: Comparison of actual and derived Bode phase plots 5 10 -2 10 -1 10 0 10 1 10 2 10 3 -100 -80 -60 -40 -20 0 20 40 Figure 5: Comparison of actual and derived Bode magnitude plots and 4 The magnitude t is very good, but the phase t is not as good. It looks likewe need to decrease the damping ratio so wetry G(s) = 0:5(1+ s=10) s 2 ((s= p 101) 2 +(1=101)s+1) = 5:05(s+10) s 2 (s +1;j10)(s+1+j10) : As shown in Figures 5 and 6, the t is very good. The MATLAB program that does this analysis is load bodeid9 topm = size(bodeid9) top = topm(1,1) w=bodeid9(1:top,1);; mag = bodeid9(1:top,2);; 6 10 -2 10 -1 10 0 10 1 10 2 10 3 -320 -300 -280 -260 -240 -220 -200 -180 -160 Figure 6: Comparison of actual and derived Bode phase plots 7 phase = bodeid9(1:top,3);; semilogx(w,mag);; grid on axis([0.1,100,-100,40]) print -deps 91029mag.eps semilogx(w,phase);; grid on axis([0.01,100,-300,-140]) print -deps 91029phase.eps z1 = 10 p1 = 0 p2 = 0 p3 = 2 -j*10 p4 = 2 + j*10 Ktc = 0.5 K=(Ktc * p3 * p4)/(z1) wp = logspace(-2,3,20);; jw = j*wp;; mag1 = 20*log10( (K*abs(jw + z1) ) ./ ( abs(jw +p1).* abs(jw + p2).*abs(jw+p3).*abs(jw + p4) ) );; semilogx(w,mag,'k-',wp,mag1,'kd') grid on axis([0.01,1000,-100,40]) print -deps 91029mag1.eps phase1 = (angle(jw + z1) -angle(jw +p1) -angle(jw + p2) - angle(jw+p3) -angle(jw + p4) )*180/pi;; semilogx(w,phase,'k-',wp,phase1,'kd') grid on axis([0.01,1000,-330,-90]) print -deps 91029phase1.eps z1 = 10 p1 = 0 p2 = 0 p3 = 1-j*10 p4 = 1 + j*10 Ktc = 0.5 K=(Ktc * p3 * p4)/(z1) mag2 = 20*log10( (K*abs(jw + z1) ) ./ ( abs(jw +p1).* abs(jw + p2).*abs(jw+p3).*abs(jw + p4) ) );; semilogx(w,mag,'k-',wp,mag2,'kd') grid on axis([0.01,1000,-100,40]) print -deps 91029mag2.eps phase2 = (angle(jw + z1) -angle(jw +p1) -angle(jw + p2) - angle(jw+p3)-angle(jw+p4) )*180/pi;; 8 semilogx(w,phase,'k-',wp,phase2,'kd') grid on axis([0.01,1000,-330,-160]) print -deps 91029phase2.eps 9