Solution 5.8.1.16 The rst step 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.The poles 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 1. 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. j Im(s) Re(s) -1-2-10-20 Figure 1: Root Locus on Real Axis 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 will end at so-called `zeros at in nity.' These zeros are at the `end' of eachofthe three asymptotes. The asymptotes are at  0 =60  ,  1 =180  and  2 = 300  . The asymptotes intersect at  i = [0 + (;1) + (;10) + (;20)];[;2] 4;1 = ; 29 3 = ;9:67 There are twopotential root loci as shown in Figure 2 (a) and (b). To decide whichlocus is the correct one, weneed to look for potential break-in and break-out points along the real axis between s = ;10 and s = ;2. We also need to nd the break-out pointbetween s +;1ands =0. Consider rst the determination of break-out and break-in points on the segment[;10;;;2]. If there are no break-in or break out points the plot of K versus position along the real axis will look likeFigure 3 (b). If a break-in and a break-out pointdoexist the plot will look likeFigure 3 (a). 1 j Im(s) Re(s) -1 -2-10 (a) j Im(s) Re(s) -1 -2-10 (b) Figure 2: TwoPotential Root Loci 2 j Im(s) Re(s)-1 -2-10-20 (a) Re(s)-1 -2-10-20 K (b) j Im(s) Re(s)-1 -2-10-20 (b) Re(s)-1 -2-10-20 K break-out break-in Figure 3: TwoPotential Gain Patterns 3 j Im(s) Re(s) -1 -2 -10 s s + 10 s + 2 s + 1 -20 s + 20 s Figure 4: Computation of Gain Along Real Axis s ;9 ;8 ;7 ;6 ;5 ;4 ;3:5 K 113.14 224 327.6 420 500 576 625.6 Table 1: Gain Values for Selected Points in [;10;;;3] The gain is plotted using Figure 4. Arepresentativepointontheline segmentisshown in the Figure. For this choice of s we know that jGH(s)j = Kjs +2j jsjjs +1jjs +10jjs +20j =1 Thus K = jsjjs +1jjs +10jjs +20j js +2j For instance, for s = ;9, K = (9)(8)(1)(11) 7 = 113:14 Table 1 summarizes the calculation of the gain on the interval [;10;;;3]. Since the gain increases constantly across the interval the correct root locus is that shown in Figure 2 (b). The break-out pointbetween s = ;1and s =0.Inthis case weare certain that there is a break-out point. Table 2 summarizes the searchforthemaximum gain. 4 s ;0:9 ;0:8 ;0:7 ;0:6 ;0:5 ;0:4 ;0:3 -0.2 -0.1 K 14.72 23.55 28.99 31.26 30.88 28.22 23.6 17.24 9.33 Table 2: Gain Values for Selected Points in [;1;;;0] The breakout pointissomewhere around s = ;0:6because this is about where the maximum gain along the real axis occurs. 5