Solution: 7.9.2.2 Σ G C G P + R C Figure 1: Cascade Compensation with UnityFeedback For the system of Figure 1 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 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+0:1, V 3 the polar representation of s+2,and V 4 the polar representation of s+4. Thus, we are trying to nd the solution to ;180  =  180  ;tan ;1  3 x;0:1  ;  180  ;tan ;1  3 x  ;  180  ;tan ;1  3 x;2  ;tan ;1  3 4;x  : 1 θ 1 θ 3 α Im(s) Re(s) 3 -4 -2 -0.1 -x θ 2 V 2 = s + 0.1 V 1 = s V 3 = s + 2 V 4 = s + 4 Figure 2: Satisfaction of Angle Condition For Desired Closed Loop Poles 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 .For instance,if weletx=2.9, then ; 1 ; 2 ; 3 =  180  ;tan ;1  3 2:8  ;  180  ;tan ;1  3 2:9  ;  180  ;tan ;1  3 0:9  ;tan ;1  3 1:1  = 133:03  ;134:03  ;106:7  ;69:86  = 177:6  : Thus, wesee that, indeed, the root locus will pass very close to s = ;3+j3. Letting x =2:95 yields ; 1 ; 2 ; 3 = 133:53  ;134:52  ;107:57  ;70:71  = ;179:3  : We see that in twotrieswearevery close. Letting x =2:96 yields ; 1 ; 2 ; 3 = 133:63  ;134:61  ;107:7  ;70:9  = ;179:6  : Letting x =2:97 yields ; 1 ; 2 ; 3 = 133:73  ;134:7  ;107:9  ;71:05  = ;179:95  : 2 If we let x =2:97 we will be close enough. The gain that places a closed loop pole at s = ;2:97+j3, and another at s = ;2:97;j3isobtained bysolving K c jV 3 j jV 1 jjV 2 jjV 4 j =1;; (1) 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 17:821 p 9:941 p 10:06 p 17:24 =10:17 The PI compensator is then G c = 10:17(s+0:1) s : Since k pi = K c =10:17 and k i k pi =0:1;; wehave k i = 0:1k pi = 0:110:16 = 1:017: Part (b) K v = lim s!0 sG c (s)G p (s)=lim s!0 s  10:17(s+0:1) s(s +2)(s+4)  = 10:170:1 8 =0:1271 Then the steady state error to a ramp input is e ss = 1 K v = 1 0:1271 =7:867;; or 787%. Part (c) 3 Im(s) Desired Closed Loop Poles 3 - 3 -4 -2 -0.1 Third Closed Loop Pole Re(s) Figure 3: Root Locus of Compensated System 4 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 = ;2:97j3. The third closed loop pole is near the zero at s = ;0:1, as shown in Figure 3. Tolocatethis pole, wemerely searchalongthereal axis looking for a gain of 10.17. That is wecompute K c = jsjjs+2jjs+4j js+0:1j for ;0:1 <s<0. For example s = ;0:07 yields K c = 0:071:93:93 0:03 = 17:7: Since we need K c =10:17 wemovetotherightandtry s = ;0:05, yielding K c = 0:051:953:95 0:05 = 7:7: This is too small, so wemovebacktothe righttos = ;0:06, to obtain K c = 0:061:963:96 0:04 = 11:47: Wearenowvery close. If wetrys = ;0:058, we obtain K c = 0:0581:9423:942 0:042 = 10:57: Trying s = ;0:057 yields K c = 0:0571:9433:943 0:043 = 10:16: Then T c (s)= 10:16(s+0:1) (s +0:057)(s+2:97;j3)(s+2:97+ j3) Part (d) 5 The output C(s) for a step is C(s) = 1 s T c (s) = 10:16(s+0:1) s(s+0:057)(s+2:971;j3)(s+2:971+ j3) At this pointweknowthegeneral form of c(t). That is, c(t)= h 1+Be ;0:057t +2jMje ;2:97t cos(3t+) i 1(t);; where  = 6 M.Soweknow quite a bit. What remains to be done is to nd the constants B and M. B = (s+0:057)C(s) j s=;0:057 = 10:16(s+0:1) s(s +2:971;j3)(s+2:971+ j3) s=;0:057 = 10:16(0:043) (;0:057)(2:913;j3)(2:913+ j3) = ;0:4382 M = (s+2:97;j3)C c (s) j s=;2:97+j3 = (s+2:971;j3)  10:16(s+0:1) s(s +0:057)(s+2:971;j3)(s+2:971+j3)  s=;2:971+j3 = 10:16(;2:87+j3) (;2:97+j3)(;2:913+ j3)(j6) = 0:3982 6 2:3536 rad: Thus, nally wehave c(t)= h 1;0:4382e ;0:057t +0:7964e ;2:971t cos(3t+2:3536) 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 ;0:057t 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 decayofthe term e ;0:057t to dominate the behavior. 6 -1 -0.5 0 0.5 1 0102030405060 Time in Seconds Closed Loop Response 0.80864exp(-2.97t)cos(3t+2.3538) -0.438exp(-0.057t) Figure 4: Step Response of Compensated System 7