s -2 -2.5 -2.7 -3 -3.1 -3.3 K 8.4 10.31 10.67 10.8 10.73 10.44 Table 1: Selected Gains Along Real Axis ;5 <s<;1 Solution 5.8.1.11 For the system of Figure 1 let G H +R C Figure 1: Standard Closed Loop Con guration GH = 10K (s + 1)(s +5)(s+30) 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.Thezeros of GH also help in nding the poles of the closed loop system. The root locus is shown in Figure 2. The root locus on the real axis occurs between the poles at s = ;1 and s = ;5 and to the left of the pole at s = ;30 The break-out pointbetween s = ;5ands = ;1willoccurat s ;3. The pole at s = ;30 is far enough awaythat its in uence on the break-out point will be minimal. It is as if wehad GH = K (s +1)(s+5) We can nd the break-out pointbycomputing the gain along the real axis for ;5 <s<;1. That calculation is shown in Figure 3. Table 1. Summarizes these calculations. From the table wesee that the in uence of the third pole has been to movethe break-out point back slightly towards s = ;1. That is the break-out pointisats = ;3. 1 Im(s) Re(s) -30 -5 -1 Figure 2: Root Locus 2 Re(s) j Im(s) V 3 s Gain at point s = x 10 V 1 V 2 V 3x -5 -30 -1 V 2 V 1 Figure 3: Gain Calculation for ;5 <s<;1 The pole zero excess is p =3;0=3: The number of asymptotes is equal to p and hence there will be three asymp- totes at  ` =  (1+ 2`) p  180  ` =0;;1;;2 = 60  ;; 180  ;; 300  The three asymptotes intersect the real axis at:  i = P polesofGH ; P zeros of GH p = ;30;5;1;0 3 = ;12 Wecould also nd the break-out pointby nding the critical points of K(s), namely the values of s for which dK=ds =0. dK ds = d ds h ;(s 3 +36s 2 +185s+ 150) i = ;(3s 2 +72s +185) 3 The solutions are s = ;129:07: The value of interest is s = ;2:93. We can nd the gain that places closed loop pole on the j!-axis in a number of ways. Analytically,wecould write out the denominator of the closed loop transfer function in unfactored form to obtain: 1+GH = 1+ 10K (s +1)(s+5)(s+30) = s 3 +36s 2 +185s+150+10K (s +1)(s + 4)(s+20) Toobtain 1 + GH =0requires that s 3 +36s 2 +185s +150+10K =0+j0 Setting s = j! in this last equation yields ;j! 3 ;36! 2 + j185! +150+10K = (80+ 10K ;25! 2 )+j(;! 3 +104!) If this expression is to equal 0+ j0wemust have 150+ 10K ;36! 2 =0 and ;! 3 +185! =0 The second equation is easily solved, yielding ! =  p 185 = 13:6 Substituting into the rst equation then yields K = 36! 2 ;150 10 = 651 These calculations are satisfying in the sense that they are precise. As a general rule, even for more complicated transfer functions, the equations are usually reasonably easy to solve. Consider the following alternativeanalysis which students usually nd unappealing because of its imprecise nature. 4 θ 1 θ 2 θ 3 Re(s) Im(s) -30 -5 -1 ω Figure 4: Angle Satisfaction On Real Axis We know that at every pointonthe root locus wehave GH =1 6 ;180  Thus, at those points where the root locus crosses the j!-axis, the angle condition must be satis ed. As shown in Figure 4, wesimplydrawavector from eachpoleand zero of GH to the pointonthej!-axis where wewishtoevaluate 6 GH.Each vector is the evaluation of the corresponding term in the factored form of GH.For the current problem 6 GH = ; 1 ; 2 ; 3 Table 2 summarizes the search. In this case weknowabout where to start because wehavealready solved the problem byothermeans. Even if we did not haveagoodstarting pointitwould not takemuchlonger to nd the crossing point, because if 6 GH was muchgreater than or muchlessthan ;180  wewould simply makealarger jump along the j!-axis to make our next guess. The resulting gain to place poles at s = j13:6, based on Figure 5 is K = jV 1 jjV 2 jjV 3 j 10 = 13:6414:4932:94 10 = 650:86 5 ! 13 13.4 13.6 13.7 6 GH ;178  ;179:3  ;180  ;180:3  Table 2: Angle GH Calculations Along j!-axis Im(s) -30 Gain at point s = 13.6 Re(s) V3 = s + 30 V 2 = s + 5 V 1 = s + 1 θ 3 = s + 30 θ 2 = s + 5 θ 1 = s + 1 V 3 V 1 V 2xx | s + 1 | | s + 5| | s + 30 |xx 10 = 10 = 651 GH = a + 1- - s + 5 - s + 30 = θ 1 θ 2 - - θ 3 - = -180 0 Figure 5: Gain Calculation on Imaginary Axis at ! =13:6 This is very close to the precise answer of 651 obtained earlier, and certainly good enough. 6