Problem 7.9.5.6 R + C G c G p Σ - Figure 1: Cascade Compensation with UnityFeedback For the system of Figure 1 G p = 1 s(s +0:2)(s+0:5) and G c = K c (s + z 1 )(s+ z 2 ) (s + p 1 )(s +30) Part (a) Toget exactly three poles in the closed loop transfer function wemust cancel twoofthe poles of the plant. Thus G c = K c (s +0:2)(s+0:5) (s + p 1 )(s +30) and G c G p (s)= K c s(s + p 1 )(s +30) The calculation of p 1 and K c is based on Figure 2. Eachofthe vectors in Figure 2 is the polar representation of one of the factorsin G c G p evaluated at s = ;4+j3. That is the vector V 1 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 1 , and V 3 the polar representation of s +30.Thus, G c G p j s=;4+j3 = K c (jV 1 j 6  1 )(jV 2 j 6  2 )(jV 3 j 6  3 ) = K c jV 1 jjV 2 jjV 3 j 6 ( 1 ; 2 ;  3 ) 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 = ;180  : 1 Re(s) Im(s) -4 3 s + p = V 2 θ 1 -30 θ 2 θ 3 s + 30 = V 3 s = V 1 -p 1 Figure 2: Satisfying Angle Condition for Desired Closed Loop Poles Since all the angles except  2 are known, wecanwrite  2 = 180  ;  1 ;  2 = 180  ; h 180  ;tan ;1 (3=4) i ; tan ;1 (3=26) = 180  ;143:1  ;6:6  = 30:3  Then p 1 = 4+ 3 tan(30:3  ) = 9:136 The gain that places a closed loop pole at s = ;4+j3, and another at s = ;4; j3isobtained by solving K c j jV 1 jjV 2 jjV 3 j =1;; (1) or K c = jV 1 jjV 2 jjV 3 j: (2) The gain to place the poles at this location is then K c = p 25 p 35:4 p 685 = 778:4 2 The lead compensator is then G c = 778:4(s+0:2)(s+0:5) (s +9:136)(s+30) : Wenextcheck the steady state accuracy. K v = lim s!0 sG c (s)G p (s) = lim s!0 s  779 s(s +9:136)(s+30)  = 7790:20:5 9:13630 = 2:84 Thus, the steady state accuracy to a ramp input e ss = 1 K v = 1 2:84 =0:35;; or roughly 35%. Not very good. The following MATLAB dialogue creates and saves the step response of the compensated system. EDU>gcgp = zpk([],[0 -9.136 -30],778.4) Zero/pole/gain: 778.4 ------------------ s(s+9.136) (s+30) EDU>tc = feedback(gcgp,1) Zero/pole/gain: 778.4 ------------------------- (s+31.14) (s^2 + 8s + 25) EDU>step(tc,4) EDU>print -deps sr7956.eps EDU> 3 Time (sec.) Amplitude Step Response 0 0.5 1 1.5 2 2.5 3 0 0.2 0.4 0.6 0.8 1 1.2 1.4 Figure 3: Step response of compensated system 4