Solution: 7.9.2.11 R + C G c G p Σ - Figure 1: Cascade Compensation with UnityFeedback For the system of Figure 1 wehave G p = 1 (s +1)(s +5) 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 =0:1: The gain to place the dominantclosedlooppoles at s = ;3 j4is K pi = js +1jjs+5jj s=;3+j4 =20:24: The following MATLAB dialogue generates the step response shown in Fig- ure 2. 1 EDU>g = zpk([-0.1],[0 -3 -5],20.24) Zero/pole/gain: 20.24 (s+0.1) ------------- s(s+3) (s+5) h= 1 Zero/pole/gain: 20.24 (s+0.1) --------------------------------- (s+0.0582) (s^2 + 7.942s + 34.78) EDU>step(tc,20) EDU>print -deps pi11stepa.eps The creep to steady state is still very slow. If weplace the poles at s = ;3j12, then K =148.Thestep response is then that shown in Figure 3. Thus, if the physical system can stand the single quickoscillation, then wecanreduce the e ect of the creep towards steady state somewhat. 2 Time (sec.) Amplitude Step Response 0 2 4 6 8 10 12 14 16 18 20 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 Figure 2: Step Response of Compensated System K = 20.24 3 Time (sec.) Amplitude Step Response 0 2 4 6 8 10 12 14 16 18 20 0 0.2 0.4 0.6 0.8 1 1.2 1.4 Figure 3: Step Response of Compensated System K = 148 4