Solution 5.8.1.18 For the system of Figure 1 let For the system of Figure 1 G H + R CΣ Figure 1: Standard Closed Loop Con guration GH = K(s +1)(s+2) s 3 (s +30)(s+40) 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 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 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  . Im(s) Re(s) -2-4 Figure 2: Root Locus on Real Axis 1 s ;1:9 ;1:8 ;1:6 ;1:4 ;1:2 ;1 ;0:8 ;0:6 ;0:4 ;0:2 ;0:1 K 1.04 2.16 4.66 7.54 10.9 15 20.35 28.24 42.62 83.45 163.76 Table 1: Gain Values for Selected Points in [;2;;0] The asymptotes intersect at  i = [0 + (1)+ (1)+ (;2) + (;4)];[0] 4;1 = ;2 The three potential root loci are shown in Figure 3. If a break-in and break-out point exist between s =0ands = ;2 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 likeeither part (b) or part (c) of the gure. Tocheckfor the possibilityofbreak-in and break-out points the gain is 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 is plotted using Figure 5. A representativepointonthe line segmentisshown in the gure. For this choice of s weknowthat jGH(s)j= Kjsj js;1;jjjs;1+jjjs+2jjs+4j =1 Thus K = js;1;jjjs;1+jjjs+2jjs+4j jsj For instance, for s = ;1:9, K = ( p 1 2 +2:9 2 )( p 1 2 +2:9 2 )(0:1)(2:1) 1:9 = 1:04 Table 1 summarizes the calculation of the gain on the interval [;2;;0]. As can be seen, the gain increases monotonically so that the correct root locus is that shown in Figure 3 (b). Figure 6 shows how the angle of departure is calculated. ; 1 ; 2 ; 3 ; 4 = ;180  : 2 Im(s) Re(s) -2 -4 Im(s) Re(s) -2 -4 (a) (b) Im(s) Re(s) -2 -4 (c) Figure 3: Three Potential Root Loci 3 Im(s) Re(s) -2 Re(s) -2 K break-out break-in Im(s) Re(s) -2 Re(s) -2 K (a) (b) Figure 4: Gain Versus Position for s 2 [;2;;0] Im(s) Re(s) -2 s 1 j -j -4 s + 4 s + 2 s s - 1 + j s - 1 - j Figure 5: Computation of Gain Along Real Axis 4 Im(s) Re(s) -2 1 j -j -4 θ 1 θ 2 α θ 3 θ 4 Figure 6: Calculation of angle of Departure As the circle shrinks in radius all the angles except  1 can be computed. That is,  1 = tan ;1 (1=1)+ 180  ;tan ;1 (1=3);tan ;1 (1=5);90  = 45  +180  ;18:43  ;11:31  = 105:3  Todetermine if the root locus crosses the j!-axis, wechecktosee if there is any ! for which 6 GH(j!)=;180  .Figure 7 shows how this calculation is made. 6 GH(j!) = ; 1 ; 2 ; 3 ; 4 = 90  ;(180  ;tan ;1 (! ;1));(180  ;tan ;1 (! + 1));tan ;1 (!=2);tan ;1 (!=4) The value of 6 GH(j!) for some selected values of ! is shown in Table 2 From the table it is clear that the root locus never crosses the imaginary axis and hence the correct root locus is that of Figure 3 (c). 5 ! 0:1 0:3 0:5 0:7 1:0 1:2 6 GH(j!) ;178:6  ;175:4  ;171:4  ;166:4  ;157:2  ;150:8  Table 2: Angle of Loop Transfer Function Along Imaginary Axis Im(s) Re(s) -2 1 -4 θ 1 θ 2 α θ 3 θ 4 ω Figure 7: Angle Calculation Along Imaginary Axis 6