Solution 5.8.1.28 G H C R + - Figure 1: For the system shown above GH(s)= K(s +1) s(s +5)(s +15)(s+30) : 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) -1-5 -15-30 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 = [(;5)+ (;15)+ (;30)]; [;1] 3 = ;16:333 Since there are 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 a 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 ;  4 = ;180  : Since  1 = ;135  the equation reduces to  2 +  3 +  4 =45  : For s = ;2+j2  2 +  3 +  4 = tan ;1 (2=3)+ tan ;1 (2=13)+ tan ;1 (2=28) = 46:5  2 Im(s) Re(s) -15 -5 -1 Figure 3: Completed Root Locus Im(s) Re(s) -15 -5 -1 θ 1 θ θ 2 θ α 3 4 Figure 4: Satisfaction of Angle Condition Along RayofConstantDamping  =0:707 3 s ;1:9+j1:9 ;1:95+ j1:95 ;1:94 + j1:94 -1.945+j1.945  2 +  3 43:63  45:06  44:8  44:9  Table 1: SearchforMaximum in Gain 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 the line of constant damping ratio is K = jsjjs +5jjs +15jjs+30j js +1j j s=;1:95+j1:95 = 1707:6: 4