Solution: 7.9.2.3 Σ G C G P + R C Figure 1: Cascade Compensation with UnityFeedback For the system of Figure 1 wehave G p = 1 (s +2)(s +4) Part (a) 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 k i k pi =2;; and the zero of the PI compensator cancels the pole of the plantats = ;2 As as consequence G c (s)G p (s)= K pi s(s +4) : The root locus is shown in Figure 2 The gain to place the closed loop poles at s = ;2j3is K pi = js +2jjs +4jj s=;2+j3 =2 2 +3 2 =13: The compensator is then G c (s)= 13(s +2) s : The following MATLAB dialogue generates the step response shown in Fig- ure 3. 1 Im(s) Re(s) -4 -2 3 s + 4 s Figure 2: Root Locus of Compensated System EDU>clear all EDU>g = zpk([],[0 -4],13) Zero/pole/gain: 13 ------- s (s+4) EDU>h= 1 h= 1 EDU>tc = feedback(g,h) Zero/pole/gain: 13 --------------- (s^2 + 4s + 13) EDU>step(tc,3) EDU>print -deps pistep.eps EDU> 2 Time (sec.) Amplitude Step Response 0 0.5 1 1.5 2 2.5 3 0 0.2 0.4 0.6 0.8 1 1.2 1.4 Figure 3: Step Response of Compensated System In this case the step response can be predicted exactly using the formula for percentovershoot in Chapter 6, because the pole cancellation has resulted in T c (s)beingT N2 (s). 3