Solution 7.9.6.4 For the system of Figure 1 wehave Σ G C G P + R C Figure 1: Cascade Compensation with UnityFeedback G p = 1 (s +0:5) 2 : Part a. Wehave G c G p (s)= K c (s +0:5) 2 s(s + a) 1 (s +0:5) 2 = K c s(s + a) The root locus is shown in Figure 2. For the poles to be at s = ;4 j2we must have a =8.Then, as shown in the gure, K c = jsjjs +8jj s=;4+j2 = p 2 2 +4 2  p 2 2 +4 2 =20: Thus T c (s)= 20 s 2 +4s +20 This is T N2 (s)for! n = p 20 and  =cos  tan ;1 (0:5)  =cos(26:57  )=0:89 As a result the system will be essentially critically damped. The peak occurs at t p =  2 =1:57 s. but will be barely discernible since the overshoot is less than one per cent. K v = lim s!0 s 20 s(s +8) = 20 8 The steady state error to a ramp input is e ss = 1 K v = 8 20 =0:4;; and the steady state error to a ramp input is 40% 1 Im(s) -a = -8 2 - 2 s + 8 s -4 Re(s) Figure 2: Root Locus for First Compensator Part b. G c (s)G p (s)= K c (s +0:5)(s+0:1) s(s + a) 1 (s +0:5) 2 = K c (s +0:1) s(s + a)(s +0:5) : The root locus and angle calculation to place closed loop poles at s = ;4j2 is shown in Figure 3. The angle equation that must be solved at s = ;4+j2 is ;  1 ; 2 ;  3 = ;180  Im(s) 2 s + a s -0.1 -0.5 -a s + 0.1 s + 0.5 θ 1 θ 2 θ 3 α Re(s) Figure 3: Root locus and angle calculation for second compensator 2 or  3 = +180  ;  1 ; 2 = [180  ;tan ;1 (2=3:9)]+ 180  ;[180  ;tan ;1 (2=4)];[180  ;tan 1; (2=3:5)] = 180  + 152:85  ;153:43  ;150:255  = 29:17  : Then a =4+ 2 tan(29:17  ) = ;7:584 Then, asshown in the gure, K c = jsjjs+7:584jjs+0:5j js +0:1j s=;4+j2 = p 20 p 16:25 p 16:85 p 19:21 = 16:88 Thus T c (s)= 16:88(s+0:1) (s + )(s +2; j4)(s+2+j4) : The third closed loop pole can be found bysearching the interval [0;;;0:5] for a closed loop pole location that will yield a gain of 16.88. Wedonot need the third pole of the closed loop compensator to compute the steady state error to a ramp input, because with unityfeedbackwecanuse the forward loop transfer function to compute K v =lim s!0 s 16:88(s+0:1) s(s +0:5)(s+7:854) =0:445: However, wecan nd the the third pole witht the MATLAB statements gcgp = zpk([-0.1],[0 -0.5 -7.584],16.88) tc = feedback(gcgp,1) whose output is Zero/pole/gain: 16.88 (s+0.1) ------------------- s (s+0.5) (s+7.584) 3 Zero/pole/gain: 16.88 (s+0.1) --------------------------- (s+0.08441) (s^2 + 8s + 20) Thus, the steady state error to a ramp input is e ss = 1 K v = 1 0:43 =2:25: Thus the steady state error to a ramp input is 225%, far worse than the rst design, primarily because of the zero at s = ;0:1whichcauses T c to havea pole close to the origin. The following MATLAB dialogue generates the closed loop transfer func- tion and the step response shown in Figure 4 EDU>gcgp = zpk([-0.1],[0 -7.584],16.88) Zero/pole/gain: 16.88 (s+0.1) ------------- s (s+7.584) EDU>tc = feedback(gcgp,1) Zero/pole/gain: 16.88 (s+0.1) -------------------- (s+24.39) (s+0.0692) EDU>gcgp = zpk([-0.1],[0 -0.5 -7.584],16.88) Zero/pole/gain: 16.88 (s+0.1) ------------------- s (s+0.5) (s+7.584) 4 EDU>tc = feedback(gcgp,1) Zero/pole/gain: 16.88 (s+0.1) --------------------------- (s+0.08441) (s^2 + 8s + 20) EDU>step(tc,30) EDU>print -deps sr7964b.eps EDU> Time (sec.) Step Response 5 10 15 20 25 30 0.2 0.4 0.6 0.8 1 1.2 1.4 Figure 4: Step response of compensated syteme, part (b) By partial fractions we nd the step response to be c(t)=[1;0:1612e ;0:0844t +1:8816e ;4t cos(2t;2:0328)]1(t) 5