Solution 5.8.1.1 G H + R CΣ Figure 1: Standard Closed Loop Con guration For the system of Figure 1 wehave GH = K(s+2) s(s +1) 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 rootlocus is shown in Figure 2. Note that there is root locus between the poles at s =0ands = ;1andtothe left of the zero at s = ;2. All locations between the poles and to the left of the zero are closed loop poles for some gain. As can be seen, root locus also appears o the real axis for the following reasons. Weknow that there is a closed loop pole associated with eachofthepoles of GH. Heuristically,asthe the gain is increased from zero through small values one can think of a pole emanating from eachof the poles of GH.Asthegain is increased, these twopoles migrate towards each other. For a certain gain, the largest gain that can be found along the real axis between s =0and s = ;1, they `collide,' that is for some gain there is a double real pole in the closed loop system. As the gain is increased further, the twopoles break out of the real axis. They migrate to the left because weknowfrom the `rules' that one of these poles must eventually go to the nite zero at s = ;2 and the other to the zero at s = ;1. Since there is root locus everywhere to the left of the zero at s = ;2the logical conclusion is that the twopolesbreak in somewhere to the left of s = ;2. That is, for some gain the closed loop system has adouble pole to the left of the zero at s = ;2. For higher gains one pole 1 XX -2 -1 Re(s) Im(s) Figure 2: Root Locus migrates righttowards s = ;2and the other left towards s = ;1.Thus, insofar as the real axis is concerned, the value of gain at the break-in point represents the minimum gain. That is, if wecompute the gain required to place closed loop poles anywhere to the left of s = ;2onthe real axis, the minimum value will occur at the break-in point. To nd the break out and break in points along the real axis, wecan do one of twothings: build a table of speci c values of K versus s along the real axis, or nd the critical points of K as a function of s.Thesecond method has analytical appeal, and in this case is not dicult to use: d ds K(s) = d ds ;s(s +1) s +2 = ;(s 2 +4s+2) (s+2) 2 The critical points ( maximums and minimums ) occur where the deriva- tiveisequal to zero or where ;(s 2 +4s+2)=0;; namely s = ;2 p 2 2 -2 -1 Re(s) Im(s) V 3 V 2 V 1 s Gain at point s = V 2 V 1 x V 3 Figure 3: Gain Calculation to Find Break-in/Break-out Points s -0.8 -0.7 -0.6 -0.5 -0.4 K 0.1333 0.1615 0.1714 0.1667 0.15 Table 1: K versus s for s Between Poles of GH Note that the break in at s = ;2; p 2andthebreak-out at s = ;2+ p 2 are equal distance from the zero of GH.Actually, the root locus o the real axis is a circle centered at the zero at s = ;2. This is always the case if GH has exactly one pole and twozeros. We could also nd the approximate break-in and break-out points by computing the gain at speci c points along the real axis. The calculation is based upon Figure 3. Weknow that GH = K(s+2) s(s+1) ;; and that for any s on the root locus, including those along the real axis, K = jsjjs+1j js+2j and that the magnitudes in this equation are simply the lengths of the vec- tors drawn from the poles and zero ofGH to the chosen s. These magnitudes are shown in Figure 3. Tables 1 and 2 can be used to nd the approximate break out and break in points. One might ask whyanapproximate calculation would be preferred over an exact calculation. An answer to that is the following. In this problem, 3 s -4 -3.5 -3.4 -3.2 -3 K 6.0 5.883 0.5.826 5.8667 6.0 Table 2: K versus s for s to the Left of s = ;2 GH has twopoles and the equation we needed to solveto nd the critical points was second order. If GH had four poles the equation for the critical points of K(s)would havebeen fourth order. With a modern calculator, and supposing we made no error in calculating dK=ds, this is not a big problem. However, nding dK=ds if K(s)isfourth order is tedious and one can easily makeamistake, a mistake that can be hard to spot. On the other hand, the tables can be generated with a simple calculator and they are self correcting. That is, if wemakeamistakeincomputing a particular value of gain, it should stickoutlikeasore thumb. Further, wegenerally are not interested in nding break-in and break-out points exactly,wejustwantto knowtheir approximate locations. 4