Solution: 7.9.4.1 Σ G C G P + R C Figure 1: Cascade Compensation with UnityFeedback For the system of Figure 1 wehave G p = 1 s(s +1) and G c = k pd + k d s: Part (a) The PD compensator can be put in the form: G c = k d s+ k pd = k d (s+k pd =k d )=k d (s+a);; whichisapurezero. The calculation of k d and the ratio k pd =k d 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 = ;3+j2. 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+0:5, and V 3 the polar representation of s +a.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 ) = K c jV 3 j jV 1 jjV 2 j 6 ( ; 1 ; 2 ) 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 = ;180  : Since all the angles except are known, wecanwrite =  1 + 2 ;180  = h 180  ;tan ;1 (2=3) i + h 180  ;tan ;1 (2=2) i ;180  = 146:31  +135  ;180  =101:31  1 θ 1 θ 2 α -3 -a Im(s) Re(s) 2 V 3 = s + a V 2 = s + 1 V 1 = s -1 Figure 2: Satisfaction of Angle Condition For Desired Closed Loop Poles Then a = 3; 2 tan(180  ;101:31  ) = 3; 2 tan(78:69  ) =2:6: 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 j =1;; (1) or K = jV 1 jjV 2 j jV 3 j ;; (2) for s = ;3+j2. The gain to place the poles at this location is then K = p 13 p 8 p 4:16 =5:0: The PD compensator is G c =5(s+2:6): Since k d =5:0, and k pd k d =2:6;; 2 wehave k pd = 2:6k d =2:65=13: Part (b) K v = lim s!0 sG c (s)G p (s)=lim s!0 s  5(s+2:6 s(s+1)  = 52:6 1 =13 Then the steady state error to a ramp input is e ss = 1 K v = 1 13 =0:077;; or less than eightpercent. Part (c) To nd the time response we rst nd the closed loop system in factored form. The system has two closed loop poles, and one zero. The location of the twoclosed loop poles is known, namely s = ;3j2. The root locus of the compensated system is shown in Figure 3. Thus, the closed loop system in factored form is: T c (s)= 5(s+2:6) (s+3;j2)(s+3+j2) The output C(s) for a step is C(s) = 1 s T c (s) = 5(s+2:6) s(s+3;j2)(s+3+j2) At this pointweknowthegeneral form of c(t). That is, c(t)= h 1+2jMje ;3t cos(2t+ ) i 1(t);; where  = 6 M.Soweknow quite a bit. What remains to be done is to nd the constants B and M. M = (s+3;j2)C c (s) j s=;3+j2 = (s+3;j2)  5(s+2:6) s(s+3;j2)(s+3+j2)  j s=;3+j2 = 1= p 2 6 ;2:356 rad: Thus, nally wehave c(t)= h 1+2= p 2e ;3t cos(2t;2:356) i 1(t) The time response is shown in Figure 4. 3 Im(s) 2.586 2 -3 Desired Closed Loop Poles Re(s) Figure 3: Root Locus of Compensated System 4 0 0.2 0.4 0.6 0.8 1 1.2 Closed Loop Response 0123 Time in Seconds Figure 4: Step Response of Compensated System 5