Solution: 7.9.9.11 R + C G c G p Σ - Figure 1: Cascade Compensation with UnityFeedback For the system of Figure 1 wehave G p = 4000(s+6)(s+15) s(s +8)(s+3;j12)(s+3+j12) : The system is unusual in that it has twozeros, and these zeros greatly reduce the complexityofthe design. The designer of the ap control has also done us a big favor by providing an enormous amountofgain. We rst consider a compensator with tworealzeros and tworeal poles, G c (s)= 1:6(s+1:2)(s+3:3) s(s +140) : Wehavechosen to add another integrator. Since the requirementonK v is so large, wemightaswell go ahead and make the system type 2. For this gain, the closed loop system is T c (s)= 6400(s+1:2)(s+3:3)(s+6)(s+15) (s +40:01)(s+6:974)(s 2 +3:273s+2:808)(s 2 +103:7s+ 2911) : The tworeal poles are not dominant. Wehavetwosetsofcomplex poles s 2 +3:273s+2:808 = (s +1:6365;j:3603)(s+1:6365+ j1:3503);; and s 2 + 103:7s+ 2911 = (s +51:85;j14:91)(s+51:85+ j14:91): The root locus is shown in Figure 2 The rst set of complex poles is almost critically damped. The second set of complex poles are marked by the boxes in Figure 2 The step response of this system is shown in Figure 3. If we 1 -140 -120 -100 -80 -60 -40 -20 0 20 -50 -40 -30 -20 -10 0 10 20 30 40 50 Real Axis Imag Axis Figure 2: Root locus for Proposed Compensator 2 Time (sec.) Amplitude Step Response 0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2 0 0.2 0.4 0.6 0.8 1 1.2 1.4 Figure 3: Step Response for K c =1:6 3 Time (sec.) Amplitude Step Response 0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2 0 0.2 0.4 0.6 0.8 1 1.2 1.4 Figure 4: Step Response for K c =2:5 4 -20 -18 -16 -14 -12 -10 -8 -6 -4 -2 0 -6 -4 -2 0 2 4 6 Real Axis Imag Axis Figure 5: Root Locus 0 < 4000K c < 75) increase the gain of the compensator to 2.5, then T c (s)= 10;;000(s+1:2)(s+3:3)(s+6)(s +15) (s +21:12)(s+6:756)(s+2:152)(s+1:44)(s 2 +122s +8059) : The step response is shown in Figure 4. As can be seen, weget a little more overshoot and a little less undershoot. Wecertainly meet the speci cations on percentovershoot and rise time with either of these systems. As an alternativeapprochwetry cancelling the complex poles. Wecould build suchacompensator using the Tow-Thomas biquadratic opamp circuit from Chapter 3. Consider the compensator G c (s)= K c (s +3;j12)(s+3+j12) (s +7)(s +20) : Then Figure 5 shows the rlocus for 0 < 4000K c < 75. If wechoose K c = 75 4000 =0:01875 5 Time (sec.) Amplitude Step Response 0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2 0 0.2 0.4 0.6 0.8 1 1.2 1.4 Figure 6: Step Response for 4000K c =75 6 Time (sec.) Amplitude Step Response 0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2 0 0.2 0.4 0.6 0.8 1 1.2 1.4 Figure 7: Step Response for Near Cancellation Then the step response is shown in Figure 6. Since wenever achieveperfect cancellation weconsider G c (s)= K c (s +3:2;j13)(s+3:2+j13) (s +7)(s +20) : The step response is shown in Figure 7 Wenow just makesome minor adjustments. The following dialogue yields a compensator that gives the step response in Figure 8. EDU>g = zpk([-6 -15 -3.2+j*13 -3.2-j*13],[0 -6.2 -8 -20 -3+j*12 -3-j*12],60) Zero/pole/gain: 60 (s+6) (s+15) (s^2 + 6.4s + 179.2) --------------------------------------- s (s+6.2) (s+8) (s+20) (s^2 + 6s + 153) EDU>tc = feedback(g,h) Zero/pole/gain: 7 Time (sec.) Amplitude Step Response 0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2 0 0.2 0.4 0.6 0.8 1 1.2 1.4 Figure 8: Step Response for Near Cancellation for Final Compensator 60 (s+15) (s+6) (s^2 + 6.4s + 179.2) ---------------------------------------------------------------- (s+18.8) (s+5.924) (s^2 + 10.12s + 60.82) (s^2 + 5.349s + 142.8) EDU>step(tc,2) EDU>print -deps pitchstepe.eps EDU> The compensator is G c (s)= 0:015(s+3:2;j13)(s+3:2+j13) (s +6:2)(s+20) : There are, of course, better solutions whichtheinterested student can nd with a little e ort. 8