solution 7.9.1.3 (a) Wehave G p = 10 (s+3)(s+30) and G c = K c (s+0:5) s+ b Figure 1(a) shows the root locus without the lag compensator, while g- ure 1(b) shows the root locus with the lag compensator, assuming we place the pole of the lag comensator near the origin. At the pointwherethetwo Re(s) Im(s) Re(s) Im(s) (a) (b) -30 -3 -30 -3 Figure 1: Cascade Compensation with UnityFeedback root loci cross the line of constant damping  =0:8, the loci are very simi- lar. For the locus of part(a), wecan nd the intersection analytically. The break-out pointisats = ;16:5. To nd the imaginary part of the proposed closed loop poles, we note that  =cos ;1 (0:8) = 36:87  : Then tan = x 16:5 ;; or x =16:5tan =16:5tan36:87  =12:375: 1 Thus the closed loop poles that meet the speci cations are s = ;16:5j12:375 The gain required to place the closed loop poles at the desired locations is 10K c = p 13:5 2 +12:375 2  p 13:5 2 +12:375 2 =335:4 Then the closed loop transfer function is T c (s)= 335:4 (s +16:5;j12:375)(s+16:5+j12:375) ;; and the step response is C(s) = 335:4 s(s+16:5;j12:375)(s+16:5+j12:375) = A s + M s+16:5;j12:375 + M  s+16:5+j12:375 : A = sC(s) j s=0 = T c (s) j s=0 = 335:4 (16:5;j12:375)(16:5+j12:375) = 335:4 16:5 2 +12:375 2 = 0:7884 M = (s+16:5;j12:375)C(s)j s=;16:5+j12:375 = T c (s) j s=;16:5+j12:375 = 335:4(s+16:5;j12:375) s(16:5;j12:375)(16:5+j12:375) s=;16:5+j12:375 = 335:4 (16:5+12:375)(j24:75) = ;0:3942+j0:526 = 0:6570 6 ;2:2143: The time response is then c(t)= h 0:7884+ 1:3140e ;16:5t cos(12:375t;2:2143) i 1(t): We see that the steady state error to a step input is e ss = 1;0:7884 = 0:2116;; which exceeds our speci cation. 2 (b) We knowthat e ss = 1 1+K p ;; or K p = 1;e ss e ss : Then for e ss  0:1;; wehave K p  9: Next note that K p = lim s!0 335:4(s+0:5) (s+ b)(s+3)(s+30) = 1:8633 b : Thus for K>9, wemust have b < 1:8633 9 ;; or b<0:207: Wechoose b =0:1. Wenow nd the angle of G c G p (s)= 335:4(s+0:5) (s+0:1)(s+3)(s+30) at s = ;16:5+j12:375. 6 G c G p (s = ;16:5+j12:375) = [180  ;tan ;1 (12:375=16)];[180  ;tan ;1 (12:375=16:4)] ;[180;tan ;1 (12:375=13:5)];tan ;1 (12:375=13:5) = tan ;1 (12:375)=16:4);tan ;1 (12:375=16) = ;179:3  : Thus, wesee that the root locus with the lag compensator added passes very close to s = ;16:5+j12:375, so close, in fact, that wecanassume it does pass through this point, and choose G c (s)= 335:4(s+0:5) (s+0:1) : 3 Then T c (s)= 335:4(s+0:5) (s+0:426)(s+16:34;j12:16)(s+16:34+j12:16) ;; and the step response is C(s) = 335:4(s+0:5) s(s+0:426)(s+16:34;j12:16)(s+16:34+ j12:16) = A s + B s+0:426 + M s+16:34;j12:16 + M  s+16:34+ j12:16 : A = sC(s) j s=0 = T c (s) j s=0 = 335:4(0:5) (0:426)(16:34;j12:16)(16:34+ j12:16) = 0:949 B = (s+0:426)C(s)j s=;0:426 = 335:4(s+0:5) s(s+16:34;j12:16)(s+16:34+ j12:16) s=;0:426 = ;0:1452 M = (s+16:34;j12:16)C(s)j s=;16:34+j12:16 = 335:4(s+0:5)(s+16:34;j12:16) s(s+0:426)(s+16:34;j12:16)(s+16:34+j12:16) s=;16:34+j12:16 = ;0:401905+j0:542472 = 0:6751 6 ;2:2083 The closed loop response is c(t)= h 0:949;0:1452e ;0:426t +1:3502e ;16:34t cos(12:16t;2:2083) i : The steady state error to a step input is e ss = 1;0:949 = 0:051;; whichmore than meets the speci cation. 4 0 0.2 0.4 0.6 0.8 1 0 0.5 1 1.5 2 2.5 Time in Seconds Response with Lag Compensation Respone with Gain Compensation Figure 2: Time Response with Gain and Lag Compensation The time responses of the two systems are shown in Figure 2. Note that the response with gain compensation reaches steady state rapidly, but unfortunately the steady state is o bytwentypercentfrom the desired value. The lag compensated system comes within vepercentofthe desired steady state but exhibits the long slowcreep to nal value characteristic of this kind of compensator. 5