Solution: 7.9.9.12 R + C G c G p Σ - Figure 1: Cascade Compensation with UnityFeedback For the system of Figure 1 wehave G p = 1 s(s +3) Toachieve t p < 0:4sec;; wemust have ! d >  0:4 =7:85! 8rad/sec: Tokeep PO under 20% wechoose  =1= p 2. The wewishtoplace the poles at s = ;8j8: Wechoose a compensator of the form G c (s)= K c (s +3)(s+ z) (s + p 1 ) 2 Wehavethree degrees of freedom here, so we xa=0.15. Toachieve our requirements on K v our strategy will be to move p towards the origin in small steps, recalculating p 1 , K c and K v at eachstep. Welocate the zero using the angle condition 6 p 1 = 6 (s + z)+pi; 6 (s); 6 (s + p) 2 ;; with s = ;8+j8. The following Matlab program implements this stategy. p(1)= 0.1 a=0.15 x=0.1 1 0 0.01 0.02 0.03 0.04 0.05 0.06 0.07 0.08 0.09 0.1 0 50 100 150 200 250 Figure 2: Plot of K v versus p kplant=1 dx = 0.005 i=1 s=-8+j*8 while x > 0.001 p(i) = x alpha(i) = ( angle(s + a) + pi - angle(s) - angle(s + p(i) ) )/ 2;; p1(i) = 8 + ( 8 / tan(alpha(i) ) );; k(i) = (abs(s)* abs(s + p(i))*abs(s + p1(i))*abs(s + p1(i)) )/ (kplant*abs(s + a) );; kv(i) = (kplant*k(i)*a)/((p(i)*p1(i)^2) );; ess(i) = 1 / kv(i);; i=i+1 x=x-dx end Figure 2 show K v as a function of p, the magnitude of the pole introduced near the origin. The following Matlab dialogue generates the compensator G c (s)= 5003(s+0:15)(s+3) (s +27:41) 2 ;; 2 and the step and ramp responses shown in Figures 3 and 4. EDU>kv(2) ans = 1.051560162494277e+01 EDU>p(2) ans = 9.500000000000000e-02 EDU>p1(2) ans = 2.740945177471859e+01 EDU>k(2) ans = 5.003422410091059e+03 EDU>g = zpk([-0.15],[0 -0.095 -27.41 -27.41], 5003) Zero/pole/gain: 5003 (s+0.15) ----------------------- s(s+0.095) (s+27.41)^2 EDU>h = 1 h= 1 EDU>tc = feedback(g,h) 3 Time (sec.) Amplitude Step Response 0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2 0 0.2 0.4 0.6 0.8 1 1.2 1.4 Figure 3: Step Response of Compensated System Zero/pole/gain: 5003 (s+0.15) -------------------------------------- (s+38.76) (s+0.1513) (s^2 + 16s + 128) EDU>step(tc,2) EDU>print -deps missrollstep.eps EDU>t = 0:0.01:4;; EDU>u = t;; EDU>lsim(tc,u,t) EDU>print -deps missrollramp.eps The root locus is shown in Figure 5 This solution is more than adequate, but it requires a large gain. 4 Time (sec.) Amplitude Linear Simulation Results 0 0.5 1 1.5 2 2.5 3 3.5 4 0 0.5 1 1.5 2 2.5 3 3.5 4 Figure 4: Ramp Response of Compensated System 5 -35 -30 -25 -20 -15 -10 -5 0 -8 -6 -4 -2 0 2 4 6 8 Real Axis Imag Axis Figure 5: Root locus of Compenstated System 6