Solution 5.8.1.27 G H C R + - Figure 1: For the system shown abovewehave GH(s)= K s(s +6)(s +20) : The rst step is to plot the poles and zeros of GH in the s-plane and then nd the root locus on the real axis. The shaded regions of the real axis in Figure 2 showwhere the root locus occurs. The rule is that root locus occurs on the real axis to the left of an odd countofpoles and zeros. That is, if you stand on the real axis and look to your right you must countan odd numberofpoles and zeros. The next step is to compute the asymptotes. The numberofasymptotes is p ex =Number of poles of GH ;Numberof nite zeros of GH: Im(s) Re(s) -6 -20 Figure 2: The asymptotes occur at the angles  ` =  1+2` p ex  180  ` =0;;1;;:::p ex ;1: In the present case there are three poles and no nite zeros so p ex =3.Thus, there are three asymptotes at  0 =  1+(2)(0) 3  180  =60   1 =  1+(2)(1) 3  180  =180   1 =  1+(2)(2) 3  180  =300  The asymptotes intersect the real axis at  i = Sum of poles of GH ;Sum of zeros of GH p ex = [(;6)+ (;20)];[0] 3 = ;8:67 Since there are no nite zeros and three poles, three closed loop poles will migrate to zeros at in nity,onesuchzero at the `end' of eachasymptote. The root locus is shown in Figure 3 To nd where the root locus crosses the raycorresponding to  =1= p 2, remember that 6 GH(s)=;180  at all points on the root locus, including the point where it crosses the ray.Figure 4 shows the angles made byeach of the poles and zeros along this ray.Thatis, ; 1 ; 2 ; 3 = ;180  : Since  1 = ;135  the equation reduces to  2 +  3 =45  : Nowitismerely a matter picking some points along the rayandevaluating the angles. For starters, weknowthatthe pointmust lie to the rightof s = ;3+j3since at this point  2 =45  .Sowechoose as a starting point s = ;2+j2. Then  2 +  3 = tan ;1 (2=4)+ tan ;1 (2=18) = 32:91  Im(s) Re(s) -20 -6 Figure 3: Completed Root Locus Im(s) Re(s) θ 1 θ 3 -6 -20 θ 2 Figure 4: Satisfaction of Angle Condition Along RayofConstantDamping  =0:707 s ;2:6+j2:6 ;2:57 + j2:57 ;2:56+ j2:56  2 +  3 45:9  45:23  45:01  Table 1: SearchforMaximum in Gain s -22 -21 -20.5 -20.7 -20.8 -20.9 K 704 315 148 213 246 280 Table 2: Searchforclosedlooppole This means that wemust makeour next guess to the left. Table 1 sumarizes the searchfor the solution to this equation. The gain to place the poles on imaginary axis is then K = jsjjs +6jjs+20j = p 2:56 2 +2:56 2 q (6;2:56) 2 +2:56 2 q (20;2:56) 2 +2:56 2 ) = 273:6 The third closed loop pole can be found byavarietyofmethods. One is to search along the negativerealaxistotheleft of s = ;20 until a gain of 275 is found. The Table 2 summarizes that search. Thus, the third closed loop pole is around s = ;20:9. It is clear that the closed loop pole associated with the loop transfer function pole at s = ;20 migrates very slowly towards ;1.