Solution 5.8.1.26 G H C R + - Figure 1: For the system shown abovewehave GH(s)= K(s +1) s 2 (s +4)(s +40) : 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 on the real axis. 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 countanodd number of poles 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) -40 -1-4 2 Figure 2: 1 The asymptotes occur at the angles  ` =  1+2` p ex   180  ` =0;;1;;:::p ex ; 1: In the presentcasethere are four poles and one nite zero 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 = [(;4)+ (;40)]; [(;1)] 3 = ;14:33 Since there is one nite zero and four poles, three closed loop poles will migrate to zeros at in nity,onesuchzeroatthe`end' of eachasymptote. The root locus is shown in Figure 3 To nd where the root locus crosses the imaginary axis, remember that 6 GH(s)=;180  at all points on the root locus, including the point where is crosses the imaginary axis. Figure 4 shows the angles made byeachof the poles and zeros. The angle equation is then ; 2 1 ; 2 ; 3 = ;180  : Since 2 1 = ;180  the equation reduces to ;  2 ;  3 =0: That is tan ;1 (!); tan ;1 (!=4)tan ;1 (!=40) = 0: Table 1 sumarizes the searchfor the solution to this equation. The gain to 2 Im(s) Re(s) -40 -1-4 2 Figure 3: Completed Root Locus Im(s) Re(s) -40 -1-4 ω θ 1 αθ 3 θ 4 Figure 4: Satisfaction of Angle Condition on Imaginary Axis ! 9 10 10.5 10.8 10.75 10.78 ; 2 ;  3 4:94  2:05  0:706  ;0:08  0:0525  ;0:025  Table 1: SearchforMaximum in Gain 3 s ;1:1 ;1:05 ;1:02 ;1:024 ;1:0237 -1.02375 K 1365 2533 6442 5067 5129 5119 Table 2: SearchforClosed Loop Pole near s = ;1 s ;41:8 ;42:5 ;43 42.98 K 2914 4189 5151 5112 Table 3: Search for Closed Loop Pole Left of s = ;40 place the poles on imaginary axis is then K = jsj 2 js +4jjs +40j js +1j k s=;j10:78 = 5113 To nd the other two closed loop poles that result from K =5132,we need to search along the real axis between s = ;4 and s = ;1foroneof them, and to the left of s = ;40 for the other one. Tables 2 and 3 summarize the search. Thus, the other twoclosed loop poles are near s = ;1:02375 and s = ;42:98. 4