Solution 5.8.3.3 For the system of Figure 1 let G H +R C Figure 1: Standard Closed Loop Con guration GH = K(s +2) (s +1;j3)(s+1+j3)(s+15)(s+20) = The rst step in drawing the root locus is to plot the poles and zeros of GH.Thepoles of GH are not the closed loop pole locations, but they can be used to nd the closed loop poles. The closed loop zeros can be found immediately: they are the zeros of G and the poles of H.Thepoles and zeros of GH serves as landmarks that help in nding the poles of the closed loop system. The portion of the root locus on the real axis is the shaded regions shown in Figure 2. These regions are determined byinvoking the rule that states that root locus on the real axis is found to the left of an odd countofpoles and zeros of GH. The root locus has four poles and one nite zero. One of the limbs of the root locus will end at the nite zero for K = 1.Theother three limbs Im(s) Re(s) X X XX -3-15-20 Figure 2: Root Locus on Real Axis 1 will end at so-called `zeros at in nity.' These zeros are at the `end' of each of the three asymptotes. The asymptotes are at  =60  ;;180  ;; and 300  . The asymptotes intersect at  i = [0 + (;1) + (;1)+(;15) + (;20)];[;3] 3 = ;12:3: The twopotential root loci are shown in Figure 3. If a break-in and break-out pointexistbetween s = ;3ands = ;15 then the root locus will look likepart (a) of the gure. If there are no break-in and break-out points the root locus will look likepart(b) of the gure. Tocheckforthe possibilityofbreak-in and break-out points the gain can be calculated at some representativevaules along the real axis. The plot will look likepart (a) of Figure 4 if there are break-in and break-out points, and like part(b) if there are no break-in or break-out points. Figure 4. The gain can be plotted using Figure 5. A representativepointonthe line segmentisshown in the gure. For this choice of s we know that jGH(s)j= Kjs +3j js +1;j3jjs+1+3jjs +15jjs+20j =1 Thus K = js +1;j3jjs+1+j3jjs+15jjs+40j js +3j In this case we are not really sure that a break-in and break-out pointare present. When we use MATLAB to drawtheroot locus at the end of the problem we will nd out for sure. To nd angle of departure, whichmayormaynot help us decide if break-in and break-out points are present, werefer to Figure 6 and invoke the angle condition ; 1 ; 2 ; 3 ; 4 = ;180  : As the circle shrinks in radius all the angles except  1 can be computed. That is,  1 = tan ;1 (3=2)+ 180  ;tan ;1 (3=14);tan ;1 (3=19);90  = 56:31  + 180  ;90  ;12:09  ;8:9726  = 125:24  The angle of departure is really a neutral indicator in this case. The MATLAB dialogue 2 Im(s) Re(s) -20 Im(s) Re(s) -3-20 (a) (b) -15 -15 -3 Figure 3: TwoPotential Root Loci 3 Im(s) Re(s)-3 Re(s)-3 K break-out break-in Im(s) Re(s)-15 Re(s) -15 K (a) (b) X X -15 -15 -20 -20 -3 -3 X X X X -20 -20 Figure 4: Gain Versus Position for s 2 [;15;;;3] 4 Im(s) Re(s) -15 s -20 s + 20 s + 25 s+3 s + 1 + j3 s +1 - j3 -3 Figure 5: Computation of Gain Along Real Axis Im(s) Re(s) -15 3 -20 θ 1 θ 2 α θ 3 θ 4 ?3 ?3 Figure 6: Calculation of angle of Departure 5 -25 -20 -15 -10 -5 0 5 -10 -8 -6 -4 -2 0 2 4 6 8 10 Real Axis I mag A x i s Figure 7: MATLAB generated root locus EDU>gh = zpk([-3],[-15 -20 -1+j*3 -1-j*3],10) Zero/pole/gain: 10 (s+3) ----------------------------- (s+15) (s+20) (s^2 + 2s + 10) EDU>rlocus(gh) EDU>print -deps rl5833f.eps EDU> draws the root locus shown in Figure 7. In this case, wesee that the poles and GH are clusted tightly enough that there are no break-in and break-out points on the interval [;15;;;3]. 6