Problem 7.9.5.8 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 +1) (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 + 1)(s+5) (s + p 1 )(s +30) and G c G p (s)= K c (s +1) s(s +0:1)(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 factors in 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 +1,V 4 the polar representation of s + p 1 and V 5 the polar representation of s +30Thus, 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 j 3 θ 1 -30 θ 2θ 3 V 1 = s V 2 = s + 0.1 V 4 = s + p 2 θ 4 α V 5 = s + 30 -p 1 -0.1 -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) i +180  ; h 180  ; tan ;1 (3=4) i ; h 180  ;tan ;1 (3=3:9) i ;tan ;1 (3=26) = 135  +180  ; 143:1  ; 142:4  ;6:6 = 22:86  Then b = 4+ 3 tan(22:86  ) = 11:12 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 24:21 p 59:65 p 685 p 18 = 1172:2 2 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 The lead compensator is then G c = 1172:2(s+1)(s +5 (s+11:12)(s+30) : K v = lim s!0 sG c (s)G p (s) = lim s!0 s  1172:2(s+1) s(s +0:1)(s+11:12)(s+30)  = 35:15 Thus, the steady state accuracy to a ramp input e ss = 1 K v =0:0285: or 2.85%. Prettygood.The following MATLAB dialogue generates the step response and the root locus for the compensated system, shown in Figures 3 and 4 respectively. The time response is c(t)=[1+0:8144e ;1:477t ; 0:04816e ;31:74t +3:03e ;4t cos(3t+2:1928)]1(t) 3 -35 -30 -25 -20 -15 -10 -5 0 5 -20 -15 -10 -5 0 5 10 15 20 Real Axis Imag Axis Figure 4: Step response of compensated system 4