Σ G C G P + R C Figure 1: Cascade Compensation with UnityFeedback Solution 7.9.3.3 For the system of Figure 1 wehave G p = 5 s(s +1) ;; and G c = K c (s +4) s + b : Toachieve t p =0:785 s. wesolve t p =  ! d for ! d =  t p = 3:4142 0:785 =4rad/s. Since ! d = ! n p 1;  2 , and  =0:8, wehave ! n = ! d p 1; 2 = 4:0 p 0:36 =6:667 Then the real part of the complex roots is ;! n = ;0:86:667 = ;5:34: Figure 2 shows the vector evaluation of G c G p at s = ;5:34+ j4. Toget the root locus to pass through this pointwemust have 6 G c G p (s) j s=;5:34+j4 = ;180  : Eachofthevectors in Figure 2 is the polar representation of one of the factors in G c G p evaluated at s = ;5:34 + j4. That is the vector V 1 is the polar representation of the factor s in the denominator of G c G p , the vector V 2 the polar representation of the factor s+1,the vector V 3 the polar representation of the factor s+b,andthe vector V 4 is the polar representation of the factor s +4. Thus 1 θ 1 θ 2β α Re(s) Im(s) -1 -4-b -5.33 4.0 V 3 = s + b V 1 = s V 2 = s + 1 V 4 = s + 4 Figure 2: Satisfaction of Angle Condition at s = ;5:55+ j4:0 G c G p 9s) j s=;5:34+j4 = 5K c jV 4 j 6 (jV 1 j 6  1 )(jV 2 j 6  2 )(jV 3 j 6 ) = 5K c jV 4 j jV 1 jjV 2 jjV 3 j 6 ( ; ; 1 ;  2 ) The evaluation of G c G p at s = ;5:34 + j4 has nowbeenbroken down into acomposite magnitude and a composite angle. For the root locus to pass through s = ;5:34+ j4wemust have ; ;  1 ; 2 = ;180  : The gain that places a closed loop pole at s = ;5:34 + j4, and another at s = ;5:34; j4isobtained bysolving 5K c jV 4 j jV 1 jjV 2 jjV 3 j =1;; (1) or K c = jV 1 jjV 2 jjV 3 j 5jV 4 j : (2) 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 c can easily be calculated. All of the angles in equation(1) are known except .Sowecanwrite = ; 1 ; 2 + 180  = 180  +[180  ;tan ;1 (4=1:33)];[180  ;tan ;1 (4=5:34)];[180  ;tan ;1 (4=4:33)] = 180  +108:46  ;143:13  ;137:29  = 8:04  2 Wenow use simple trigonometry to nd b =5:34+ 4 tan(8:04  ) =5:34+ 4 0:1412 =5:34+ 28:35 = 33:66 Wecannow nd the gain using equation(2). K c = jV 1 jjV 2 jjV 3 j 5jV 5 j = 6:675:928:61 54:219 =53:38 Thus the complete compensator is G c = 53:38(s+4) s +33:66 : To nd the third closed loop pole weuse the fact that for a system where G c G p has a pole zero excess of twoorgreater, the sum of the closed loop poles is equal to the sum of the poles of G c G p .Inthis case wehave (;5:34+ j4)+ (;5:34; j4)+ = ;33:66;1;; This yields = ;34:66+ 10:68 = ;23:98;24: The closed loop transfer function is T c (s)= 266:9(s+4) (s +5:34+ j4)(s+5:34;j4)(s+24) Wecancheck this result with the following MATLAB dialogue EDU>g = zpk([-4],[0 -1 -33.66],266.9) Zero/pole/gain: 266.9 (s+4) ----------------- s (s+1) (s+33.66) EDU>h = 1 h= 1 EDU>tc = feedback(g,h) 3 Time (sec.) Step Response 0.5 1 1.5 2 2.5 3 0.2 0.4 0.6 0.8 1 1.2 1.4 Figure 3: Step Response of Compensated System Zero/pole/gain: 266.9 (s+4) -------------------------------- (s+23.98) (s^2 + 10.68s + 44.51) EDU>step(tc,3) EDU>print -deps sr7933.eps EDU> The step response is shown in Figure 3 4