Solution: 7.9.2.10 R + C G c G p Σ - Figure 1: Cascade Compensation with UnityFeedback For the system shown in Figure 1 wehave G p (s)= 10 (s +4)(s+20) Two compensators are being considered G c1 (s)= K c (s +0:5) s and G c2 (s)= K c (s +4) s Using G c1 (s)wehave G c1 (s)G p (s) = K c (s +0:5) s 10 (s + 4)(s+20) = 10K c (s +0:5) s(s + 4)(s+20) : Since the pole and zero of the PI compensator are very close together and distant from the desired closed loop pole locations, wecanassume that the closed loop poles will be at s = ;12j9. Note with  =0:8wehavea3-4-5 righttriangle. The situation is shown in Figure 2. That is weignore the net angle contribution of the PI compensator and compute the gain required to place the poles at s = ;12j9as K = jV 1 jjV 2 j 10 = p 8 2 +9 2  p 8 2 +9 2 10 =14:5 For the second compensator wehave G c1 (s)G p (s) = K c (s +4) s 10 (s +4)(s +20) = 10K c s(s +20) : 1 Re(s) Im(s) -12 9 j 9 j 6 Closed Loop Poles for G c1 (s) -4-20 V 1 V 2 V 3 V 4 Figure 2: Closed Loop System Re(s) j Im(s) Closed Loop Poles for G c1 (s) V 1 V 2 7.5 -20 -10 j 7.5 Figure 3: Closed Loop Poles Using G c2 (s) This is exactly T N2 (s)asshown in Figure 3. K = jV 1 jjV 2 j 10 = p 10 2 +7:5 2  p 10 2 +7:5 2 10 =15:625 2 (a) For G c1 (s)wehave K v = lim s!0 sG c (s)G p (s) = lim s!0 s 10K c (s +0:5) s(s +4)(s +20) = lim s!0 145(s+0:5) (s +4)(s +20) = 0:90625 Thus e ss = 1 K v =1:103: For G c2 (s)wehave K v = lim s!0 sG c (s)G p (s) = lim s!0 s 10K c s(s +20) = lim s!0 156:25 s +20 = 7:8125 Thus e ss = 1 K v =0:128 Thus the second compensator has muchbetter steady state error character- istics. 3 t c(t) Figure 4: Closed Loop System t c(t) Figure 5: Step Response Using G c2 (s) (b) For G c1 (s), the response to a step will be c(t)= h 1+be ; t +2jMje ;12t cos(9t+ ) i : It is the term be ; t that will delay the settling time. As discussed in the text, the response is likely to look like that shown in Figure 4. For G c2 (s) the response can be determined exactly from the formulas for T N2 (s) and will look likeFigure 5. Thus G c2 will yield both a shorter settling time and less steady state error to a ramp input. 4