Solution: 7.9.5.2 For the system of Figure 1 wehave R + C G c G p Σ - Figure 1: Cascade Compensation with UnityFeedback G p = 1 (s+1)(s+2) and and G c (s)= K c (s+1)(s+ z 2 ) (s+0:2)(s+20) : Figure 2 shows the evaluation of G c G p (s)ats = ;3+j2incomponent form. That is, eachofthe vectors in Figure 2 is the polar representation of one of the factors in G c G p evaluated at s = ;3+j2. The vector V 1 , for instance, 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 +2,andsoon. Notice that the pole at s = ;1hasbeen cancelled by the zero at s = ;1 and so those twotermsdo not enter into the angle and magnitude calculations to be made next. Thus, G c G p (s) j s=;3+j2 = K c jV 3 j 6 (jV 1 j 6  1 )(jV 2 j 6  2 )(jV 4 j 6  3 ) = K c jV 3 j jV 1 jjV 2 jjV 4 j 6 ( ; 1 ; 2 ; 3 ) The evaluation of G c G p at s = ;3+j2hasnowbeen broken down into acomposite magnitude and a composite angle. For the root locus to pass through this point, wemust have ; 1 ; 2 ; 3 = ;180  : Since all the angles except are known, wecanwrite =  1 + 2 +  3 ;180  = h 180  ;tan ;1 (2=2:8) i + h 180  ;tan ;1 (2=1) i +tan ;1 (2=17);180  = 144:46  +116:57  +6:71  ;180  =87:74  1 θ 1 θ 2 -2 α -3 -a Im(s) Re(s) 2 V 3 = s + a V 2 = s + 2 V 1 = s + 0.2 θ 3 V 4 = s + 10 -20 -0.2 Figure 2: Satisfaction of Angle Condition For Desired Closed Loop Poles Then a = 3+ 2 tan(87:74  ) =3:08 The gain that places a closed loop pole at s = ;3+j2, and another at s = ;3;j2isobtained by solving K c jV 3 j jV 1 jjV 2 jjV 4 j =1;; (1) or K = jV 1 jjV 2 jjV 4 j jV 3 j : (2) The gain to place the poles at this location is then K = p 11:84 p 5 p 293 p 4:0062 =65:8 The lead/lag compensator is then G c (s)= 65:8(s+1)(s+3:08) (s+0:2)(s+20) 2 -20 -3 -3.08 -2 2 -2 Im(s) Re(s) -0.2 Figure 3: Root Locus of Compensated System Part b. K p = lim s!0 G c (s)G p (s)=lim s!0  65:8(s+3:08) (s+0:2)(s+2)((s+20)  = 65:83:08 0:2220 =25:325 Then the steady state error to a step input is e ss = 1 1+K p = 1 26:325 =0:0395;; or about 4%. Thus, the system will trackastep quite well. Part c. The system has three closed loop poles, under the assumption of perfect cancellation. Weknow where twoofthethree closed loop poles are, namely at s = ;3  j2. The next task is to nd the third closed loop pole. The root locus in Figure 3 shows where that third closed loop pole must be. One wayto nd the pole is to searchalongthe real axis between s = ;20 and 3 V 4 = s + 20 V 3 = s + 3.08 V 2 = s + 2 V 1 = s = 0.2 Re(s) Im(s) -20 -3.08 -2 -0.2 Figure 4: Calculation of Gain Along Real Axis s -17 -16.5 -16.3 16.2 K 54.31 61.64 64.44 65.8 Table 1: SearchforClosed Loop Pole s = ;3:08 until we nd a value of s that yields a gain of 65.8. Figure 4 shows how this calculation is made. That is K c = jV 1 jV 2 jjV 4 j jV 3 j Table 1 summarizes the searchfor the third closed loop pole. Thus, the third closed loop pole is at approximately s = ;16:2. the closed loop transfer function is T c (s)= 65:8(s+3:08) (s+16:2)(s+3;j2)(s+3+j2) The following MATLAB program checks this result and generates the step response shown in Figure 5 gcgp = zpk([-3.08], [-0.2 -2 -20],Kc) tc = feedback(gcgp,1) step(tc,3) print -deps sr7952.eps The damped frequency is 2 r./s. and the damping ratio is  = cos[tan ;1 (2=3)]= cos(33:69  )=0:83 Thus,if we had guessed the time response it would be as shown in Figure 6. The formula for overshoot predicts less than 2% overshoot, but weknow 4 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 5: Step Response of Compensated System 1.5 s. Over shoot = 10% t c(t) Figure 6: Estimated Time Response 5 from experience that the formula can be too optimistic, so a good guess is 10%. The time to peak should be t p =  ! d  1:5s.: Wecan see that the actual response is actually better than our guess. d C(s) = 65:8(s+3:08) s(s+16:2)(s+3;j2)(s+3+j2) = 1 s + B s+16:2 + M s +3;j2 + M  s+3+j2 : A = sC(s) j s=0 = (s+16:2)  65:8(s+3:08) (s+3;j2)(s+3+j2)(s+16:2)  s=0 = 0:962 B = (s+16:2)C(s)j s=;16:2 = (s+16:2)  65:8(s+3:08) s(s+3;j2)(s+3+j2)(s+16:2)  s=;16:2 = 65:8(;13:2) (;16:2)(;13:2;j2)(;13:2+j2) = ;868:56 ;2887:488 = 0:299 M = (s+3;j2)C c (s) j s=;3+j2 = (s+3;j2)  65:8(s+3:08) s(s+3;j2)(s+3+j2)(s+16:2)  s=;3+j2 = 0:6305 6 ;2:743 rad: Thus, nally wehave c(t)= h 1+0:299e ;16:2t +1:368e ;3t cos(2t;2:743) i 1(t) The time response is shown in Figure 7. The actual per centovershoot around twopercent, and the time to peak is 0.8 s. 6 -1.5 -1.0 -0.5 0.0 0.5 1.0 1.5 0.00 1.00 2.00 3.00 Time in Seconds Desired Steady State Total Response 1.368exp(-3t)cos(2t-2.744) 0.3exp(-16.2t) Figure 7: Step Response of Compensated System 7