Solution 5.8.1.32 G H C R + - Figure 1: For the system shown above GH(s)= K s(s +1)(s+100) : 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 number of poles and zeros for root locus to occur on that portion of the real axis. The next step is to compute the asymptotes. The numberofasymptotes is p ex = Number of poles of GH ;Number of zeros of GH: Im(s) Re(s) -100 -1 Figure 2: 1 The asymptotes occur at the angles  ` =  1+2` p ex   180  ` =0;;1;;:::p ex ;1: In the presentcase 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   2 =  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 = [(;100)+ (;1) + (0)];[0] 3 = ;33:67 Since there are no nite zeros all three closed loop poles will migrate to zeros at in nity,onesuchzero at the `end' of each asymptote. The resulting root locus is shown in Figure 3. Note that there is a break-out pointbetween s = ;1ands =0.We knowthatfor lowgaintheclosed loop poles occur near the poles of GH.Asthe gain increases the closed loop poles migrate away from the poles of GH towards the zeros of GH.Inthis case all three zeros are zeros at in nity.Thetwoclosed loop poles migrating towards each other on the interval [;1;;0] eventually meet and break-out of the real axis to migrate towards the asymptotes. To nd where the root locus crosses the imaginary axis, weinvoke the angle requirement, namely that at eachpointonthe root loucs 6 GH(s)= ;180  . The angle calculation along the imaginary axis is shown in Figure 4 The gure shows the three vector components of GH for evaluation ats = j! The root locus crosses the j!-axis where ;( 1 +  2 +  3 )=;180  : Since  1 =90  ,we only need to nd the value of ! for which  2 +  3 =90  : 2 Im(s) Re(s) -100 -1 Figure 3: Completed Root Locus Im(s) Re(s) -100 -1 ω s s + 1 s + 10 θ 1 θ 2θ 3 Figure 4: Computation of Angle of GH on Imaginary Axis 3 ! 5 10 15 20  1 +  2 81:55  90  94:7  98:4  Table 1: SearchforCrossing PointonImaginary Axis That is tan ;1 (!)+tan ;1 (!=10) = 90  : Table 1 summarized the searchforthe crossing point. The gain required to place twoofthe three closed loop poles on the imaginary axis is K = [jsjjs+1jjs+100j] j s=j10 = 1421:3 4