Solution: 7.9.5.3 R + C G c G p Σ - Figure 1: Cascade Compensation with UnityFeedback For the system of Figure 1 wehave G p = 10 s(s + 1)(s +10) and G c (s)= K c (s +1)(s + z) (s +12)(s+ p) : Figure 2 shows the evaluation of G c G p (s)ats = ;3+j2incomponent form. That is, eachofthe vectors in Figure 2 is the polar representation of one of the factors in G c G p evaluated at s = ;3+j2. The vector V 1 , for instance, 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 + p,andsoon. Notice that the pole at s = ;1hasbeen cancelled by the zero at s = ;1 and so those twotermsdo not enter into the angle and magnitude calculations to be made next. Thus, G c G p j s=;3+j2 = K c jV 3 j 6 (jV 1 j 6  1 )(jV 2 j 6  2 )(jV 3 j 6  3 )(jV 4 j 6  4 ) = K c jV 3 j jV 1 jjV 2 jjV 3 jjV 4 j 6 ( ;  1 ;  2 ;  3 ;  4 ) The evaluation of G c G p at s = ;3+j2hasnowbeen broken down into acomposite magnitude and a composite angle. For the root locus to pass through this point, wemust have ;  1 ;  2 ;  3 ;  4 = ;180  : our tactic will be to, iteratively choose p and then compute =  1 +  2 +  3  4 ; 180  = h 180  ; tan ;1 (2=3) i + h 180  ; tan ;1 (2=3; p) i +tan ;1 (2=7) +tan ;1 (2=9) = ;180  ;; 1 θ 1 θ 2 α -3 -z Im(s) Re(s) 2 V 5 = s + z V 2 = s + p V 4 = s + 12 V 1 = s θ 3 V 3 = s + 10 -10 -p XXX X -12 Figure 2: Satisfaction of Angle Condition For Desired Closed Loop Poles moving p closer and closer to the origin. The Matllab program shown below does this. s=-3+j*2 p1 = 12 p2 = 10 pp = 0.1 p(1) = 0.1 dp = 0.001 i=1 while pp>0.001 p(i) = pp;; alpha(i) = angle(s + p(i)) +angle(s) + angle(s + p1) + angle(s + p2) -pi;; alphad(i) = alpha(i)*180/pi;; z(i) = 3 + 2/tan(alpha(i));; k(i) = (abs(s)*abs(s + p1)*abs(s + p2)*abs(s + p(i) ) )/(10 *abs(s + z(i) ));; kv(i) = (10 * k(i) * z(i))/(p1*p2*p(i));; pp = pp -dp;; i=i+1;; end Figure 3 shows K v as a function of the pole being added near the origin. The following MATLAB dialogue identi es p, z,andK. 2 0 0.01 0.02 0.03 0.04 0.05 0.06 0.07 0.08 0.09 0.1 0 100 200 300 400 500 600 Figure 3: K v as a function of p 3 EDU>p(89) ans = 0.01200000000000 EDU>k(89) ans = 27.38679999999999 EDU>z(89) ans = 0.53126323630362 EDU> Then the compensator is G c (s)= 27:4(s+0:531) (s +0:012)(s+12) The following MATLAB dialogue generates the step response for the com- pensated system, shown in Figure 4 EDU>g = zpk([-0.531],[0 -0.012 -10 -13],274) Zero/pole/gain: 274 (s+0.531) ------------------------- s(s+0.012) (s+10) (s+13) EDU>h = 1 h= 1 4 Time (sec.) Amplitude Step Response 0 1 2 3 4 5 6 0 0.2 0.4 0.6 0.8 1 1.2 1.4 Figure 4: Step Response of Compensated System EDU>tc = feedback(g,h) Zero/pole/gain: 274 (s+0.531) ------------------------------------------- (s+15.85) (s+0.7735) (s^2 + 6.384s + 11.86) EDU>step(tc,6) EDU>print -deps lagld3step.eps EDU> The following dialogue generates the ramp response shown in Figure 5. EDU>t = 0:0.01:6;; EDU>u = t;; EDU>lsim(tc,u,t) EDU>print -deps lagld3ramp.eps EDU> 5 Time (sec.) Amplitude Linear Simulation Results 0 1 2 3 4 5 6 0 1 2 3 4 5 6 Figure 5: Ramp Response of Compensated System 6