Solution: 7.9.6.1 Σ G C G P + R C Figure 1: Cascade Compensation with UnityFeedback For the system of Figure 1 wehave G p = 1 (s+1)(s+2) and G c (s)= K c (s+ 1)(s+ z 2 ) s(s+10) so that the closed loop system has poles at s = ;3j2. 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 bythezero at s = ;1, under perfect cancellation, and so those twoterms do not enter into the angle and magnitude calculations to be made next. Thus, G c G p 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=3) i + h 180  ;tan ;1 (2=1) i +tan ;1 (2=7);180  = 146:31  +116:57  +15:95  ;180  =98:82  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 θ 3 V 4 = s + 10 -10 Figure 2: Satisfaction of Angle Condition For Desired Closed Loop Poles Then a =3; 2 tan(180  ;98:82  ) =3; 2 tan(81:18  ) =2:69 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 13 p 5 p 53 p 4:096 =29: The lead/pi compensator is then G c (s)= 29(s+1)(s+2:69) s(s+10) Part b. K v = lim s!0 sG c (s)G p (s)=lim s!0 s  29(s+2:69) s(s+2)((s+10)  = 292:69 210 =3:9 2 -10 -3 -2.69 -2 2 -2 Im(s) Re(s) Figure 3: Root Locus of Compensated System Then the steady state error to a ramp input is e ss = 1 K v = 1 3:9 =0:256;; or 26%. Thus, the system will not trackaramp very 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 = ;10 and s = ;2:69 until we nd a value of s that yields a gain of 29. Figure 4 shows how this calculation is made. That is K c = jV 1 jjV 2 jjV 4 j jV 3 j Table 1 summarizes the searchforthe third closed loop pole. Thus, the closed loop transfer function is T c (s)= 29(s+2:69) (s+6)(s+3;j2)(s+3+j2) 3 V 4 = s + 10 V 3 = s + 2.69 V 2 = s + 2 V 1 = s Re(s) Im(s) -10 -2.69 -2 Figure 4: Calculation of Gain Along Real Axis s -8 -7 -6 -5 K 18.08 24.36 29 32.45 Table 1: SearchforClosed Loop Pole The damped frequency is 2 rad/s, and the damping ratio is  = cos[tan ;1 (2=3)] = cos(33:69  )=0:83 Thus, an estimate of the time response can be made, as shown in Figure 5. The formula for overshoot predicts less than 2% overshoot, but weknow from experience that the formula is often too optimistic, so a good guess is 10%. The time to peak should be t p =  ! d  1:5s.: Wecanverify our estimate by nding the time response by partial frac- 1.5 s. Over shoot = 10% t c(t) Figure 5: Estimated Time Response 4 tion expansion. C(s) = 29(s+2:69) s(s+6)(s+3;j2)(s+3+j2) = 1 s + B s+6 + M s +3;j2 + M  s+3+j2 : B = (s+6)C(s) j s=;6 = (s+6)  29(s+2:69) s(s +3;j2)(s+3+j2)(s+6)  s=;6 = 29(;3:31) (;6)(;3;j2)(;3+j2) = ;99:99 78 = 1:2307 M = (s+3;j2)C c (s) j s=;3+j2 = (s+3;j2)  29(s+2:69) s(s+3;j2)(s+3+j2)(s+6)  s=;3+j2 = 1:1287 6 ;2:988 rad: Thus, nally wehave c(t)= h 1+1:2307e ;6t +2:2574e ;3t cos(2t;2:988) i 1(t) The time response is shown in Figure 6. The actual per centovershoot is six per cent, and the time to peak is 0.8 s. 5 -2.5 -2.0 -1.5 -1.0 -0.5 0 0.5 1.0 1.5 1 2 3 Time in Seconds Total Response 2.2574exp(-3t)cos(2t-2.988) 1.2306exp(-6t) steady state step response Figure 6: Step Response of Compensated System 6