Problem 7.9.5.7 R + C G c G p Σ - Figure 1: Cascade Compensation with UnityFeedback For the system of Figure 1 G p = 1 s(s +0:1)(s+5) and G c = K c (s + z 1 )(s +0:2) (s + p 1 )(s+30) Togetexactly four poles in the closed loop transfer function wemust cancel one of the poles of the plant. Thus G c = K c (s +0:2)(s+5) (s +0:1)(s+ p 2 ) and G c G p (s)= K c (s +0:2) s(s +0:1)(s+ p 2 )(s +30) The calculation of p 2 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+0:1, V 3 the polar representation of s +0:2, etc. 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 4 j 6  3 )jV 5 j 6  4 = K c jV 3 j jV 1 jjV 2 jjV 3 jjV 5 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  : 1 Re(s) Im(s) -4 3 θ 1 -30 θ 2θ 3 V 1 = s V 2 = s + 0.1 V 3 = s + 0.2 V 4 = s + p 2 θ 4 α V5 = s + 30 -p 2 -0.2 -0.1 Figure 2: Satisfying Angle Condition for Desired Closed Loop Poles Since all the angles except  3 are known, wecanwrite  3 = +180  ; 1 ;  2 ; 4 = h 180  ;tan ;1 (3=3:8) i +180  ; h 180  ;tan ;1 (3=4) i ; h 180  ;tan ;1 (3=3:9) i ;tan ;1 (3=26) = 141:7  + 180  ;143:1  ;142:4  ;6:6 = 29:6  Then p 2 = 4+ 3 tan(29:6  ) = 9:29 The gain that places a closed loop pole at s = ;4+j3, and another at s = ;4; j3isobtained by solving K c jjV 3 j jV 1 jjV 2 jjV 4 jjV 5 j =1;; (1) or K c = jV 1 jjV 2 jjV 4 jjV 5 j jV 3 j (2) The gain to place the poles at this location is then K c = p 25 p 24:2 p 36:96 p 685 p 23:44 = 808:5 2 The lead compensator is then G c = 808:5(s+0:2)(s+5 (s +9:29)(s+30) : K v = lim s!0 sG c (s)G p (s) = lim s!0 s  808:5(s+0:2) s(s +0:1)(s+9:29)(s+30)  = 5:8 Thus, the steady state accuracy to a ramp input e ss = 1 K v =0:173: or 17%. The following MATLAB dialogue generates the step response shown in Figure 3. EDU>gcgp = zpk([-0.2],[0 -0.1 -9.29 -30],808.8) Zero/pole/gain: 808.8 (s+0.2) ------------------------- s(s+0.1) (s+9.29) (s+30) EDU>tc = feedback(gcgp,1) Zero/pole/gain: 808.8 (s+0.2) ------------------------------------------- (s+31.18) (s+0.2075) (s^2 + 8.001s + 25.01) EDU>step(tc,4) EDU>print -deps sr7957.eps EDU> The root locus is shown in Figure 4 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: Root locus of compensated system -35 -30 -25 -20 -15 -10 -5 0 5 -20 -15 -10 -5 0 5 10 15 20 Real Axis Imag Axis Figure 4: Root locus of compensated system 4