solution: problem 7.9.1.1 Σ G C G P + R C Figure 1: Cascade Compensation with UnityFeedback For the system of Figure 1 wehave G p = 10 (s + 1)(s+3)(s+30) and G c = K(s +0:1) s + b Figure 2 shows the root locus for the plant G p .Because the pole at s = ;30 is so far from the twopoles near the origin wewould expect the root locus near the origin to be very similar to that for G p = K (s + 1)(s+3) : That is, wewould expect the root locus to pass very near s = ;2+j2. Figure 3 gives the vector representation of G p necessary to compute the sum of angles along a horizontal line that passes through s = j2. 6 G p (x + j2) = ; 1 ; 2 ; 2 : Table 1 summarizes the search for the actual crossing point. As can be seen x -1.85 -1.9 -1.93 -1.92 6 GH ;176:6  ;179:1  ;180:7  ;180:2  Table 1: from the table we make only a slight error in assuming that the root locus passes through s = ;2+j2. Therefore, wewill assume it does pass through s = ;2+j2. The gain required to put poles at s = ;2j2can be obtained from the vector representation of G p in Figure 3. K = jV 1 jjV 2 jjV 3 j 10 = p 5 p 5 p 2 2 +28 2 10 =14 1 Re(s) -30 -3 -1 Im(s) Figure 2: Root Locus for Gain Compensation of G p Re(s) Im(s) -30 -3 -2 -1 2 -x θ 1 θ 2 θ 3 s + 3 = V 2 s + 30 = V 3 s + 1 = V 1 Figure 3: Angle Calculation for G p at s = ;x + j2 2 -30 -3 -2 -1 Desired Closed Lop Pole Im(s) Re(s) 2 V 3 V 2 V 1 V 5 V 4 Figure 4: Gain Computation with Lag Present The steady state error to a step input for gain compensation is e ss = 1 1+K p ;; where K p = lim s!0 140 (s + 1)(s+3)(s+30) =1:559 Hence e ss = 1 1+1:559 =0:391: Thus, the steady state error is 39%. Toreduce this error weintroduce the lag compensator G c = s +0:1 s + b Wesetthegainto1.0for the reasons shown in Figure 4. The gain calculation to place closed loop poles at s = ;2j2 with the lag compensator present is K = jV 1 jjV 2 jjV 3 jjV 4 j 10jV 5 j  jV 1 jjV 2 jjV 3 j 10 because jV 4 j jV 5 j  1: Further, the angle contributions of the pole and zero of the lag compensator cancel almost exactly since the twovectors V 4 and V 5 are nearly colinear. Thus the addition of the lag compensator does not change the root locus 3 muchinthevicinityofs = ;2+j2. What does change is the steady state error, because K p changes. The new K p is K p = lim s!0 140(s+0:1) (s + b)(s+1)(s+ 3)(s+30) = 1:559 0:1 b Thus, K p changes by the factor 0:1=b.Ifwe put ;b close to s = ;0:1 but closer to the origin, then wecan dramatically increase K p .Inthe present case wewant 1 1+K p =0:1;; which implies K p =9: Then wemust have 140(0:1) 3(30)b =9;; or b = 14 990 =0:0173 Thus, the overall lag compensator is G c = 14(s +0:1) s +0:0173 : The step response is shown in Figure 5 As can be seen, and expected, the response has a long slow creep to steady state. 4 Time (sec.) Amplitude Step Response 0 10 20 30 40 50 60 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 Figure 5: Unit Step Response of Compensated System 5