Solution: 7.9.2.9 R + C G c G p Σ - Figure 1: Cascade Compensation with UnityFeedback For the system of Figure 1 wehave G p = 1 (s +3)(s +7) The PI compensator can be put in the form: G c = k pi + k i s = k pi s + k i s = k pi (s + k i =k pi ) s : The constant k pi serves as the gain of the compensator and the ratio k i =k pi determines the zero of the compensator. In this problem we rstelect to make k i k pi =3;; and the zero of the PI compensator cancels the pole of the plantats = ;3 As as consequence G c (s)G p (s)= K pi s(s +7) : The root locus is shown in Figure 2 Toachieve t p < 0:5s;; wemust have ! d >  1 =6:4rad/s: The gain to place the closed loop poles at s = ;3:5 j! d is K pi = jsjjs +7j j s=;3:5+j! d = ! 2 n =53:21: The damping ratio is  =cos[tan ;1 (6:4=3:5)]= 0:48: 1 Im(s) Re(s) -3.5 ω s + 7 s -7 d Figure 2: Root Locus of Gain Compensated System Then K v =lim s!0 s 53:21 s(s +7) =7:6: The step response shown in Figure 3 is generated bytheMATLAB dialogue EDU>g = zpk([],[0 -7],53.21) Zero/pole/gain: 53.21 ------- s(s+7) EDU>tc = feedback(g,1) Zero/pole/gain: 53.21 ------------------ (s^2 + 7s+53.21) EDU>step(tc) EDU>print -deps sr7929aa.eps 2 Time (sec.) Amplitude Step Response 0 0.4 0.8 1.2 1.6 0 0.2 0.4 0.6 0.8 1 1.2 1.4 Figure 3: Step response of compensated system Thus, we meet the spec cations. We can try to improveonthis result bymoving the zero of the PI com- pensator around, but as the following analysis shows it is hard to get much improvementonthe pole/zero cancellation approach. Figure 4 shows the angles involved in making the calculation ; 1 ; 2 ;  3 = ;180  : The following Matlab program nds a,andK v over a range of acceptable values of ! d . zeta = cos(atan(6.4/3.5)) omegan = sqrt( 3.5^2 + 6.4^2) K=omegan^2 kv = K/7 p1 = 3 p2 = 7 3 θ 1 θ 3 α Im(s) Re(s) ω -7 -3 -x θ 2 V 2 = s + 2 V 1 = s V 3 = s + a V 4 = s + 6 d -a Figure 4: Satisfying angle condtion along line of constantdamping ratio kplant=1 zeta = 0.48 omegan = linspace(6.4,12,200);; sigma = zeta*omegan omegad = omegan*sqrt(1-zeta^2) s=-sigma + j*omegad alpha = angle(s)+angle(s + p1)+angle(s+p2)-pi z=sigma + ( omegad./tan(alpha)) K=(abs(s).*abs(s+p1).*abs(s+p2))./abs(s +z) Kv = (K.*z)/(p1*p2) plot(z,Kv) print -deps 7929.eps Figure 5 shows K v as a function of the zero location. Weseethat for a damping ratio of  =0:48, the cancellation of the pole at s = ;3provides very nearly the maximum K v . If welet =0:45 and repeat the analysis, we get the plot shown in Figure 6. The step response is shown in Figure 7, for the zero at s = ;3. The overshoot is 20.5%. 4 -3 -2 -1 0 1 2 3 4 -12 -10 -8 -6 -4 -2 0 2 4 6 8 Figure 5: K v as a function of zero location 5 -2 -1 0 1 2 3 4 -6 -4 -2 0 2 4 6 8 10 Figure 6: K v as a function of zero location 6 Time (sec.) A mp li tu d e Step Response 0 0.4 0.8 1.2 1.6 0 0.2 0.4 0.6 0.8 1 1.2 1.4 Figure 7: Step response for  =0:45 and zero at s = ;3 7