Solution 4.6.5.4 (a) The closed loop system is T c (s) = K c s(s +2) 1+ K c s(s +2) = K c s 2 +2s + K c : The characteristic equation, in terms of K c ,is p(s)=s 2 +2s + K c : The desired characteristic equation is p(s)=(s +1;j)(s+1+j)=s 2 +2s +2: Thus to obtain the desired poles wemust have K c =2: The closed loop transfer function is thus T c (s)= 2 s 2 +2s +2 : The MATLAB dialogue EDU>tc = zpk([],[-1-j*1 -1+j*1],2) Zero/pole/gain: 2 -------------- (s^2 + 2s+2) EDU>step(tc) EDU>print -deps sr4654a.eps EDU> creates the step response shown if Figure 1. 1 Time (sec.) A mp li tu d e Step Response 0 1 2 3 4 5 6 0 0.2 0.4 0.6 0.8 1 1.2 1.4 Figure 1: Step response of compensated system 2 (b) The closed loop transfer function is T c (s) = K c (s +3) s(s +2)(s + p) 1+ K c (s +3) s(s +2)(s+ p) = K c (s +3) s 3 +(p +2)s 2 +(2p+ K c )s +3K p : The characteristic equation, in terms of K c ,is p(s)=s 3 +(p+2)s 2 +(2p+ K c )s +3K c : The desired characteristic equation is p(s)=(s +4) 2 (s + q)=s 3 +(8+q)s 2 + (16+ 8q)s +16q: Equating coecients in likepowers of s weobtainthe equations 8+q = p +2 16+ 8q = 2p + K c 3K c = 16q These equations can be rearranged as: q ;p = ;6 8q ;2p;K c = ;16 16q;3K c = 0 In matrix form wehave 2 6 4 1 ;1 0 8 ;2 ;1 16 0 ;3 3 7 5 2 6 4 q p K c 3 7 5 = 2 6 4 ;6 ;16 0 3 7 5: The MATLAB program, a=4 z=3 A=[1-1 0;;2*a -2 -1;;a^2 0 -z] b=[2-2*a;; -a^2;; 0] x=inv(A)*b 3 ageneralization of the speci c problem weare trying to solve, yields the output a= 4 z= 3 A= 1 -1 0 8 -2 -1 16 0 -3 b= -6 -16 0 x= -6 0 -32 EDU> yields a solution, but it is not a satisfactory solution. However, if wemove the zero slightly,froms = ;3tos = ;2:5, weget a= 4 4 z= 2.5000 A= 1.0000 -1.0000 0 8.0000 -2.0000 -1.0000 16.0000 0 -2.5000 b= -6 -16 0 x= 10 16 64 EDU> whichisaviable solution, namely G c (s)= 64(s +2:5) s +16 : As another example wecould leavethezero at s = ;3and let the double pole of the characteristic equation be a s = ;5. In that case weget a= 5 5 z= 3 A= 1 -1 0 10 -2 -1 25 0 -3 b= -8 -25 0 x= 27.0000 35.0000 225.0000 EDU> whichisalsosatisfactory,yielding G c (s)= 225(s+3) s +35 : Chapter 5 will shed a good deal of lightonwhysome solutions are good and some bad, and will also showushowtochoose viable solutions. For G c (s)= 64(s +2:5) s +16 ;; the MATLAB dialogue 6 EDU>gcgp = zpk([-2.5],[0 -2 -16],64) Zero/pole/gain: 64 (s+2.5) -------------- s(s+2) (s+16) EDU>tc = feedback(gcgp,1) Zero/pole/gain: 64 (s+2.5) -------------- (s+10) (s+4)^2 EDU>step(tc) EDU>print -deps sr4654b.eps EDU> creates the step response shown if Figure 2. 7 Time (sec.) A mp li tu d e Step Response 0 0.5 1 1.5 0 0.2 0.4 0.6 0.8 1 1.2 1.4 Figure 2: Step response for z =2:5 8