Re(s) j Im(s) -p -0.5 -0.1 α β -3 j 2 s + p = V 3 s + 0.5 = V 4 s + 0.1 = V 2 θ 2 θ 1 s = V 1 Figure 1: Satisfying Angle Condition for Desired Closed Loop Poles Problem 7.9.5.4 For the system of Figure ?? wehave G p = 1 s(s +0:5) and G c = K c (s +0:5)(s+ z) (s +0:1)(s+ p) As a rst pass welet G c = K c (s +0:5)(s+0:5) (s +0:1)(s+ p) so that G c G p (s)= K c (s +0:5) s(s +0:1)(s+ p) The calculation of p and K c is based on Figure 1. Eachofthevectors in Figure 1 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+0:1, V 3 the polar representation of s+p,andV 4 the polar representation of s +0:5. Thus, G c G p 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 ) 1 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=2:5) i +180  ; h 180  ;tan ;1 (2=3) ih 180  ;tan ;1 (2=2:9) i = 141:34  +180  ;146:31  ;145:41  = 29:62  Then b = 3+ 2 tan(29:62 circ ) = 3+3:517 = 6:517 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 c = 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 12:41 p 16:37 p 10:25 = 16:05 The lead compensator is then G c = 16:05(s+0:5) s +6:517 : Wenextchecktoseeif the requirementonsteady state accuracy is met. K v = lim s!0 sG c (s)G p (s) 2 = lim s!0 s  16:06(s+0:5 s(s +0:1)(s+6:517)  = 16:050:5 0:16:517 = 12:31 Thus, the steady state accuracy to a ramp input e ss = 1 K v = 1 12:31 =0:081;; or roughly eightpercent. This does not quite meet the speci cation, so we redesign the compensator bymoving the zero to s = ;0:8. That is G c = K c (s +0:5)(s+0:8) (s +0:1)(s+ p) so that G c G p (s)= K c (s +0:8) s(s +0:1)(s+ p) : If wenowrepeate the analysis for this new compensator weobtain: G c = 19:48(s+0:5)(s+0:8) (s +0:1)(s+7:1) : Then K v = lim s!0 s  19:48(s+0:8) s(s +0:1)(s+7:1)  = 19:48(:8) (0:1)(7:1) = 21:96 .Thus e ss = 1 K v = 1 21:96 =0:046;; and the speci cation on steady state accuracy is met. Part (b) To nd the time response we rst nd the closed loop system in factored form. The system has three closed loop poles, twoofwhichweknow, namely the closed loop poles at approximatley s = ;3 j2. The root locus of the 3 Re(s) -0.8-7.1 -0.1 Im(s) Figure 2: Root Locus of Compensated System s -6 -5 -4 -3 -2 -1.5 -1.2 K 7.48 12.24 15.1 16.21 16.5 16.8 -19.47 Table 1: Searchfor Closed Loop Pole 4 compensated system is shown in Figure 2. Wesee that there must be a closed loop pole between the loop transfer pole at s = ;7:1 and the loop transfer zero at s = ;0:8andTheeasiest wayto nd the third pole is by trial and error along the real axis. Tables 1 summarizes the search. Thus the closed loop system in factored form is: T c (s)= 19:49(s+0:8) (s +3;j2)(s+3+j2)(s+1:2) Part (c) The output C(s)forastep is C(s) = 1 s T c (s) = 19:48(s+0:8) s(s +3;j2)(s+3+j2)(s+1:2) The following MATLAB program checks this result, nds the step response and saves it as an encapsulated postscript le. gcgp = zpk([-a], [-p1 -p2 -p3],Kc) tc = feedback(gcgp,1) step(tc,3) print -deps sr7954.eps Atthispointweknow the general form of c(t). That is, c(t)= h 1+Be ;1:2t +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 +1:2)C(s)j s=;1:2 = (s +1:2)  19:49(s+0:8) s(s +3;j2)(s+3+j2)(s+1:2)  s=;1:2 = 0:895 M = (s +3;j2)C c (s) j s=;3+j2 = (s +3;j2)  19:49(s+0:8) s(s +3;j2)(s+3+j2)(s+1:2)  s=;3+j2 = 1:4925 6 2:2587 5 -2 -1 0 1 2 012345 Time in Seconds Total Response 2.9852exp(-3t)cos(2t + 2.2586) 0.895exp(-1.2t) Figure 3: Step Response of Compensated System Thus, nally wehave c(t)= h 1+0:895e ;1:2t +2:9849e ;3t cos(2t +2:2587) i 1(t) The time response is shown in Figure 3. 6