Solution: 7.9.6.8 R + C G c G p Σ - For the system shown abovewehave G p (s)= 10 (s +1)(s +2) and G c (s)= K c (s + z 1 )(s + z 2 ) (s + p 1 )(s +15) : To meet the accuracy requirement one of the poles of the compensator must be at s =0.Tohaveaclosed loop transfer function with three poles, one of the poles of the plantmust be cancelled. There are thus twochoices for the compensator, either (a) G c (s)= K c (s + 1)(s+ a) s(s + 15) ;; or (b) G c (s)= K c (s + 2)(s+ a) s(s +15) : In either case, the location of the zero can be found bysatisfying the angle condition at s = ;4 j2, as shown in Figure 1. For case (a), satisfaction of the angle condition of the root locus at s = ;4 j2 requires that = [180  ;tan ;1 (2=4)]+ [180  ;tan ;1 (2=2)]+ tan ;1 (2=11);180  = 153:43  +135  +10:3  ;180  = 118:73: -1 or -2 -a -15 -4 j 2 θ 1 θ 2 θ 3 α Re(s) j Im(s) Figure 1: Satisfaction of Angle Condition at s = ;4+j2 1 Hence a = 4; 2 tan(180  ;118:73  ) = 4; 2 tan(61:27  ) = 4;1:096 = 2:903: The gain to place the poles at s = ;4j2isthen K c = jsjjs+2jjs+15j 10js+2:903j s=;4+j2 = p 2 2 +4 2 p 2 2 +2 2 p 2 2 +11 2 10 p 2 2 +1:1 2 = 6:2 The following MATLAB dialogue veri es these calculations and nds the unit step response shown in Figure 2. EDU>gcgp = zpk([-2.903],[0 -2 -15],62) Zero/pole/gain: 62 (s+2.903) -------------- s(s+2) (s+15) EDU>tc = feedback(gcgp,1) Zero/pole/gain: 62 (s+2.903) --------------------- (s+9) (s^2 + 8s + 20) EDU>step(tc,4) EDU>print -deps sr7968a.eps EDU> The unit step response is c(t)=[1+1:4483e ;9t +2:9361e ;4t cos(2t; 2:5568)]1(t): 2 Time (sec.) Amplitude Step Response 0 0.5 1 1.5 2 2.5 3 3.5 4 0 0.2 0.4 0.6 0.8 1 1.2 1.4 Figure 2: Unit step response for case (a) For case (b), the analysis is identical, with = [180  ;tan ;1 (2=4)]+ [180  ;tan ;1 (2=3)]+ tan ;1 (2=11);180  = 153:43  +146:31  +10:3  ;180  = 130:05: Hence a = 4; 2 tan(180  ;130:05  ) = 4; 2 tan(49:95  ) = 4;1:68 = 2:319: The gain to place the poles at s = ;4j2isthen K c = jsjjs+1jjs+15j 10js+2:319j s=;4+j2 = p 2 2 +4 2 p 2 2 +3 2 p 2 2 +11 2 10 p 2 2 +1:68 2 = 6:9 3 Time (sec.) Amplitude Step Response 0 0.5 1 1.5 2 2.5 3 3.5 4 0 0.2 0.4 0.6 0.8 1 1.2 1.4 Figure 3: Unit step response for case (b) The following MATLAB dialogue veri es these calculations and plots the unit step response shown in Figure 3. EDU>gcgp = zpk([-2.319],[0 -1 -15],69) Zero/pole/gain: 69 (s+2.319) -------------- s(s+1) (s+15) EDU>tc = feedback(gcgp,1) Zero/pole/gain: 69 (s+2.319) ----------------------------- (s+8.001) (s^2 + 7.999s + 20) EDU>step(tc,4) EDU>print -deps sr7968b.eps EDU> 4 The unit step response is c(t)=[1+2:45e ;8t +4:507e ;4t cos(2t; 2:4426)]1(t): 5