Solution: 7.9.3.2 For the system of Figure 1 wehave R + C G c G p Σ - Figure 1: Cascade Compensation with UnityFeedback G p = 10 s(s +0:5)(s+20) ;; and G c = K c (s + b) s +10 : Figure 2 shows the vector evaluation of G c G p at s = ;3+j3. Togetthe root locus to pass through this pointwemust have 6 G c G p (s) j s=;3+j3 = ;180  : Eachofthe vectors, as labeled in Figure 2, is the polar representation of one of the factors in G c G p evaluated at s = ;3+j3. Thus G c G p j s=;3+j3 = Kjs + bj 6 (jsj 6  1 )(js +0:5j 6  2 )(js +10j 6  3 )(js+20j 6  4 ) = Kjs + bj jsjjs +0:5jjs+10jjs+20j 6 ( ; 1 ; 2 ; 3 ; 4 ) The evaluation of G c G p at s = ;3+j3hasnowbeen broken down into acomposite magnitude and a composite angle. For the root locus to pass through s = ;3+j3wemust have ; 1 ; 2 ; 3 ; 4 = ;180  : The gain that places a closed loop pole at s = ;3+j3, and another at s = ;3;j3isobtained by solving 10Kjs+ bj jsjjs +0:5jjs+10jjs+20j =1;; (1) or K c = jsjjs +0:5jjs+10jjs+20j 10js+ bj : (2) 1 -20 -10 -b -0.5 Im(s) Re(s) s s + 0.5s + b s + 10 s + 20 -3 j 3 θ 1 θ 2 θ 3 θ 4 α Figure 2: Cascade Compensation with UnityFeedback It should be clear that it is the angle condition that drives this whole busi- ness. The angle condition will be used to nd b. Once b is determined then K can easily be calculated. All of the angles in equation(1) are known except .Sowecanwrite =  1 +  2 +  3 +  4 ;180  = [180  ;tan ;1 (3=3)]+ [180  ;tan ;1 (3=2:5)][tan ;1 (3=17)]+ [tan ;1 (3=7)];180  = 135  +129:81  +10:01  +23:2  ;180  = 118  Wenow use simple trigonometry to nd ;b = ;3+ 3 tan(180  ;118  ) = ;3+ 3 tan(62  ) = ;3+1:6 = ;1:4 Wecannow nd the gain using equation(2). K c = jsjjs+0:5jjs+10jjs+20j js +1:4j = 4:2433:9057:61617:263 103:4 = 49:2! 49 2 Thus the complete compensator is G c = 49(s +1:4) s +10 : The following Matlab dialogue leads to the step response shown in Figure 3 EDU>g = zpk([-1.4],[0 -0.5 -10 -20],492) Zero/pole/gain: 492 (s+1.4) ----------------------- s(s+0.5) (s+10) (s+20) EDU>h = 1 h= 1 EDU>tc = feedback(g,h) Zero/pole/gain: 492 (s+1.4) ------------------------------------------ (s+21.83) (s+4.562) (s^2 + 4.112s + 6.918) EDU>step(tc,4) EDU>print -deps lead1astep.eps EDU> AMATLAB program to compute all the quantities in this problem is s=-3+j*3 theta1 = angle(s) theta1d = theta1*180/pi theta2 = angle(s + 0.5) theta2d = theta2*180/pi theta3 = angle(s + 10) theta3d = theta3*180/pi theta4 = angle(s + 20) theta4d = theta4*180/pi alpha = theta1 + theta2 + theta3 + theta4 - pi 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: Step Response of Compensated System alphad = alpha*180/pi a=3+(3 / tan(alpha) ) abs1=abs(s) abs2 = abs(s + 0.5) abs3 =abs(s +10) abs4 =abs(s + 20) abs5 = abs(s + a) Kc = (abs(s)*abs(s + 3)*abs(s + 10)*abs(s + 20) )/abs(s + a) 4