Solution 7.9.5.1 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 +0:5) and G c = K c (s +1:5) s + p 1 : Part a. The calculation of p 1 and K c is based on Figure 2, and the fact that every pointonthe root locus of the compensated system satis es 6 G c G p = ;180  : 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, V 3 the polar representation of s+b,and V 4 the polar representation of s +1:5. Thus, G c G p (s) j s=;3+j2 = K c jV 4 j 6 (jV 1 j 6  1 )(jV 2 j 6  2 )(jV 3 j 6 ) = K c jV 4 j jV 1 jjV 2 jjV 3 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 = +180  ; 1 ; 2 = h 180  ;tan ;1 (2=1:5) i +180  ; h 180  ;tan ;1 (2=3) ih 180  ;tan ;1 (2=2:5) i = 126:87  +180  ;146:31  ;141:34  = 19:22  1 Re(s) Im(s) -b -1.5 -0.5 α β -3 2 s + b = V 3 s + 1.5 = V 4 s + 0.5 = V 2 θ 2 θ 1 s = V 1 Figure 2: Satisfying Angle Condition for Desired Closed Loop Poles Then b =3+ 2 tan(19:22  ) =3+5:74 = 8:74: The gain that places a closed loop pole at s = ;3+j2, and another at s = ;3;j2isobtained by solving K c jV 4 j jV 1 jjV 2 jjV 3 j =1;; (1) or K = jV 1 jjV 2 jjV 3 j jV 4 j : (2) The gain to place the poles at this location is then K c = p 13 p 10:25 p 36:91 p 6:25 =28:1: The lead compensator is then G c = 28(s +1:5) s +8:74 : Part b. K v =lim s!0 sG c (s)G p (s)=lim s!0 s  28(s+1:5 s(s +0:5)(s+8:74)  = 281:5 0:58:74 =9:63: Then the steady state error to a ramp input is e ss = 1 K v = 1 9:63 =0:104;; or roughly ten percent. 2 Part c. Toget the steady state error to twopercentmeansincreasing K v by roughly afactorof ve. That is wemust have K v =50.If weaddthelag compen- sator then K v = lim s!0 s  28(s+1:5)(s+ z 2 ) s(s +0:5)(s+8:74)(s+ p 2 )  = 28(1:5)z 2 (0:5)(8:74)p 2 = 9:61z 2 p 2 : Equating this to 50 yields z 2 p 2 = 50 9:63 =5:2: If we let z 2 =0:2, then p 2 = z 2 5:2 = 0:2 5:2 =0:0385: Then the nal compensator is: G c = 28(s+1:5)(s+0:2) (s +8:74)(s+0:0385) : Note that wehaveassumed that the addition of the lag compensator will not change the previously calculated value of gain byanappreciable amount. Part d. The closed loop system has four closed loop poles, twoofwhichweknow, namely the closed loop poles at approximately s = ;3j2. The closed loop poles locations are approximate because of the angle di erence between the pole and zero of the lag compensator, and because wechoose not to adjust the gain. The root locus of the compensated system is shown in Figure 3. We see that there must be a closed loop pole between the loop transfer pole at s = ;8:74 and the loop transfer zero at s = ;1:5andaclosedlooppole between the loop transfer pole at s = ;0:5and the loop transfer zero at s = ;0:2. The easiest wayto ndthem isbytrial and error along the real axis. Tables 1 and 2 summarize the search. These values are approximate because wehavenot readjusted the gain K c for the angle contributions of the pole and zero of the lag. The following MATLAB program veri es these calculations 3 Re(s) -0.5-1.5 -8.74 -0.2 Im(s) Figure 3: Root Locus of Compensated System s -8 -7 -6 -4 -3.0 -3.86 K 6.894 14.71 20.6 27.65 30.34 27.98 Table 1: Searchfor Closed Loop Pole s -0.4 -0.3 -0.205 -0.201 -0.202 0.2023 K 0.548 1.18 13.27 64 32 28.2 Table 2: Searchfor Closed Loop Pole 4 gp = zpk([],[0 -0.5], 1) gc = zpk([-0.2 -1.5],[-8.74 -0.0385], 28) gcgp = series(gc,gp) tc = feedback(gcgp,1) The actual closed loop transfer function is T c (s)= 28(s+1:5)(s+0:2) (s +2:62;j2)(s+2:62+ j2)(s+3:83)(s+0:2023) e. The output C(s)for a unit step input is C(s) = 1 s T c (s) = 28(s+1:5)(s+0:2) s(s +3;j2)(s+3+j2)(s+3:86)(s+0:0203)0:202) At this pointweknowthegeneral form of c(t). That is, c(t)= h 1+Be ;3:86t + Ce ;0:2023t +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, C, and M. B = (s +3:83)C(s)j s=;3:83 = (s +3:83)  28(s+1:5)(s+0:2) s(s +2:62;j2)(s+2:62+ j2)(s+3:83)(s+0:2023)  s=;3:83 = 3:13 C = (s +0:2023)C(s)j s=;0:2023 = (s +0:2023)  28(s +1:5)(s+0:2) s(s +2:62;j2)(s+2:62+ j2)(s+3:83)(s+0:2023)  s=;0:2023 = ;0:01166 M = (s +2:62;j2)C c (s) j s=;2:62+j2 = (s +2:62;j2)  28(s+1:5)(s+0:2) s(s +2:62;j2)(s+2:62+ j2)(s+3:83(s+0:2023)  s=;2:62+j2 = 2:2092 6 3:005rad: 5 Time (sec.) Step Response 0.5 1 1.5 2 2.5 3 0.2 0.4 0.6 0.8 1 1.2 1.4 Figure 4: Step Response of Compensated System Thus, nally wehave c(t)= h 1+3:13e ;3:83t ;0:01166e ;0:2023t +4:1836e ;2:62t cos(2t;3:005) i 1(t) The time response is shown in Figure 4. The time response of the pole at s = ;0:2023 is not shown because it is almost instantly zero. The following MATAB program generates the time response shown in Figure 4. gp = zpk([],[0 -0.5], 1) gc = zpk([-0.2 -1.5],[-8.74 -0.0385], 28) gcgp = series(gc,gp) tc = feedback(gcgp,1) step(tc,3) print -deps sr7951.eps 6