Solution: 7.9.5.5 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 +0:5)(s+10) and G c (s)= K c (s + z 1 )(s + z 2 ) (s +20)(s+ p) : Figure 2 shows the evaluation of G c G p (s)ats = ;4+j3incomponent 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 = ;4+j3. 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 = ;0:5has been cancelled by the zero at s = ;0:5and so those twoterms do not enter into the angle and magnitude calculations to be made next. Thus, G c G p j s=;4+j3 = 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 = ;4+j3hasnowbeen 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 (3=4) i + h 180  ; tan ;1 [3=(4; p)] i + tan ;1 (3=6) +tan ;1 (3=16); 180  ;; 1 θ 1 θ 2 α -4 -z Im(s) Re(s) 3 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 -20 Figure 2: Satisfaction of Angle Condition For Desired Closed Loop Poles moving p closer and closer to the origin, evaluating K v and K c as wego. At each iteration we will havetorecompute the location of the zero. The MATLAB program shown belowdoesthis. s=-4+j*3 p1 = 20 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, that meet the spec- i cations 2 0 0.01 0.02 0.03 0.04 0.05 0.06 0.07 0.08 0.09 0.1 0 50 100 150 200 250 300 350 400 450 Figure 3: K v as a function of p 3 EDU>kv(93) ans = 1.103282131050450e+02 EDU>k(93) ans = 5.731420341078142e+02 EDU>z(93) ans = 0.30799545394164 EDU>p(93) ans = 0.00800000000000 EDU> Then the compensator is G c (s)= 573(s +0:308)(s+0:5) (s +0:008)(s+20) The following MATLAB dialogue generates the step response for the com- pensated system, shown in Figure 4 EDU>g = zpk([-.308],[0 -0.008 -10 -20],573) Zero/pole/gain: 573 (s+0.308) ------------------------- s(s+0.008) (s+10) (s+20) 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>h = 1 h= 1 EDU>tc = feedback(g,h) Zero/pole/gain: 573 (s+0.308) ------------------------------------------ (s+22.11) (s+0.3469) (s^2 + 7.55s + 23.01) EDU>step(tc,6) EDU>print -deps lagld5step.eps EDU> The following dialogue generates the ramp response shown in Figure 5. 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 EDU>step(tc,6) EDU>print -deps lagld5step.eps EDU>t = 0:0.1:6;; EDU>u = t;; EDU>lsim(tc,u,t) EDU>print -deps lagld5ramp.eps EDU> 6