Solution: 7.9.2.1 For the system of Figure 1 wehave G p = 1 (s+2)(s+4) The PI compensator can be put in the form: R + C G c G p Σ - Figure 1: Cascade Compensation with UnityFeedback 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 the ratio k i =k pi has been speci ed. As a consequence, all that is required is to nd where the root locus crosses a parallel line through s = j3. This determination is shown in Figure 2. Eachofthe vectors in Figure 2 is the polar representation of one of the factors in G c G p evaluated at s = ;x + j3. That is the vector V 1 is the polar representation of the factor s in the denominator of G c G p , V 2 the polar representation of the factor s+0:2, V 3 the polar representation of s +2,and V 4 the polar representation of s +4. Thus, wearetrying to nd the solution to  180  ;tan ;1  3 x;0:2  ;  180  ;tan ;1  3 x  ;  180  ;tan ;1  3 x;2  ;tan ;1  3 4;x  = ;180  : The assumption is that the root locus will cross the line s = j3foravalue of x close to 3. This assumption is based on the fact that near s = ;3+j3 1 θ 1 θ 3 α j Im(s) Re(s) j 3 -4 -2 -0.2 -x θ 2 V 2 = s + 0.2 V 1 = s V 3 = s + 2 V 4 = s + 4 Figure 2: Satisfaction of Angle Condition For Desired Closed Loop Poles   1 .For instance,if weletx=2.9, then ; 1 ; 2 ; 3 =  180  ;tan ;1  3 2:7  ;  180  ;tan ;1  3 2:9  ;  180  ;tan ;1  3 0:9  ;tan ;1  3 1:1  = 131:99  ;134:03  ;106:7  ;69:86  = 178:6  : Thus, wesee that, indeed, the root locus will pass very close to s = ;3+j3. Letting x =2:95 yields ; 1 ; 2 ; 3 = 132:51  ;134:52  ;107:57  ;70:71  = ;180:29  : We see that in twotrieswearevery close. Letting x =2:945 yields ; 1 ; 2 ; 3 = 132:46  ;134:47  ;107:48  ;70:62  = ;180:1  ;; whichisclose enough. The gain that places a closed loop pole at s = ;2:945+j3, and another at s = ;2:945;j3isobtained bysolving K c jV 3 j jV 1 jjV 2 jjV + 4 j =1;; (1) 2 or K c = jV 1 jjV 2 jjV 4 j jV 3 j : (2) The gain to place the poles at this location is then K c = K pi = p 17:67 p 9:893 p 10:11 p 16:54 =10:34 The PI compensator is then G c = 10:34(s+0:2) s : Since k pi = K c =10:34, and k i k pi =0:2;; wehave k i =0:2k pi =0:210:34 = 2:07: Part (b) K v = lim s!0 sG c (s)G p (s)=lim s!0 s  10:34(s+0:2) s(s +2)(s+4)  = 10:340:2 8 =0:2585 Then the steady state error to a ramp input is e ss = 1 K v = 1 0:2585 =3:87;; or 387%. Part (c) To nd the time response we rst nd the closed loop system in factored form. The system has three closed loop poles, and one zero. The location of twoofthe closed loop poles is known, namely s = ;2:945j3. The third closed loop pole is near the zero at s = ;0:2, as shown in Figure 3. Tolocate this zero, wemerely searchalongthereal axis looking for a gain of 10.34. That is wecompute K c = jsjjs+2jjs+4j js+0:2j 3 j Im(s) Desired Closed Loop Poles j 3 -j 3 -4 -2 -0.2 Third Closed Loop Pole Figure 3: Root Locus of Compensated System 4 for ;0:2 <s<0. For example s = ;0:1yields K c = 0:11:9:9 0:1 =7:41: Since we need K c ;10:34 wemovetotheleft and try s = ;0:15, yielding K c = 0:151:853:85 0:05 = 21:4: This is too large, so wemovebacktotherighttos = ;0:12, to obtain K c = 0:121:883:88 0:08 =10:94: Wearenowvery close. If wetrys = ;0:118, we obtain K c = 0:1181:8823:882 0:082 =10:5: Trying s = ;0:1171 nally yields K c = 0:11711:88293:8829 0:0829 =10:33: Then T c (s)= 10:34(s+0:2) (s+0:1171)(s+2:945;j3)(s+2:945+ j3) Part (d) The repsonse to a step will haveatime to peak of roughly t p   ! d =  3 =1s. The output C(s)forastep is C(s) = 1 s T c (s) = 10:34(s+0:2) s(s +0:1171)(s+2:945;j3)(s+2:945+j3) At this pointweknowthegeneral form of c(t). That is, c(t)= h 1+Be ;0:1171t +2jMje ;3t cos(3t+) i 1(t);; 5 where  = 6 M.Soweknow quite a bit. What remains to be done is to nd the constants B and M. B = (s+0:1171)C(s) j s=;0:1171 = 10:34(s+0:2) s(s +2:945;j3)(s+2:945+ j3) j s=;0:1171 = 10:34(0:0829) (;0:1171)(2:8279;j3)(2:8279+ j3) = ;0:4307 M = (s+2:945;j3)C c (s) j s=;2:945+j3 = (s+2:945;j3)  10:34(s+0:2) s(s +0:1171)(s+2:945;j3)(s+2:945+j3)  j s=;2:945+j3 = 10:34(;2:745+j3) (;2:945+ j3)(;2:8279+j3)(j6) = 0:4043 6 2:3506 rad: Thus, nally wehave c(t)= h 1;0:4307e ;0:1171t +0:80864e ;2:945t cos(3t2:3506) i 1(t) The time response is shown in Figure 4. The response is over damped. By looking at the individual components of the response wecanseewhy. The coecientofthe term e ;0:1171t is very large and the time constantalso very large. As a result, this term decays very slowly.Notethat the response of the cosine term is over very rapidly,leaving the decayoftheterm e ;0:1171t to dominate the behavior. 6 -1 -0.5 0 0.5 1 051015 20 25 30 35 40 Time in Seconds Closed Loop Response 0.80864exp(-2.945t)cos(3t+2.3506) -0.4307exp(-0.1171t) Figure 4: Step Response of Compenstated System 7