1 Solution 5.8.1.6 For the system of Figure 1 wehave G H +R C Figure 1: Standard Closed Loop Con guration GH = 10K(s +8) s(s +4)(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 =0ands = ;4 and between the pole at s = ;30 and the zero at s = ;8. There is no break-out pointbetween the pole at s = ;30 and the zero at s = ;8. This can be seen bycalculating the gain along the real axis in this region. The gain at selected points in this region is shown in Table 1. The computation of eachpointisbased on Figure 3. The gain rises constantly aswemovefroms = ;28 to s = ;10. There are no maxima or minima. s -28 -26 -24 -22 -20 -18 -16 -14 -12 -10 K 6.72 12.7 18 22.6 26.7 30.2 33.6 37.3 43.2 60 Table 1: Selected Gains Along Real Axis ;20 <s<;10 The pole zero excess is p =3;1=2: 2 Re(s) Im(s) -4 -8 Figure 2: Root Locus 3 Re(s) Im(s) V 3 V 2 V 1 Gain at point s = x -4 -8-30 V 4 s V 3 V 2 xV 1 V 4 | | |||| | | Figure 3: Computation of Gain Along Real Axis The numberofasymptotes is equal to p and hence there will be two asymp- totes at  ` =  (1 + 2`) p  180  ` =0;;1 = 90  ;; 270  The twoasymptotes intersect the real axis at:  i = P poles ofGH ; P zeros ofGH p = ;30;4;0;(;8) 2 = ;13 The break-out pointbetween s =0ands = ;4can be found either by computing some gains along this stretchofthe real axis, or by nding the critical points of dK=ds.Finding this break-out pointisadetail that is usually not required. We know the break-out must occur, and that the break-out pointwillbenears = ;2 and that is usually good enough. 4 By contrast, the analytical method yields, dK ds = d ds " ;(s 3 +34s 2 +120s) 10(s +8) # = ;(2s 3 +58s 2 +644s +960) 10(s +8) 2 To nd the critical pointofinterest wemust solve 2s 3 +58s 2 +644s + 960 = 0: This is a factorization problem of the same order of dicultyasthe orginal root locus problem, just the kind of problem wetryto avoid by using the root locus method.So the searchalong the real axis for potential break-in and break-out points is in most cases easier to do and also less fraught with error.