Solution 5.8.1.5 G H + R CΣ Figure 1: Standard Closed Loop Con guration For the system of Figure 1 wehave GH = 10K(s +10) s(s +4)(s +20) 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. The zeros 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 only between the poles at s =0ands = ;4andbetween the pole at s = ;20 and the zero at s = ;10. There is no break out pointbetween the pole at s = ;20 and the zero at s = ;10. 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 gain increases monotonically.There is no s -19 -18 -17 -16 -15 -14 -13 -12 -11 K 3.167 6.3 9.47 12.8 16.5 21 27.3 38.4 69.3 Table 1: Selected Gains Along Real Axis ;20 <s<;10 minimum or maximum hence no break in or break out points. The pole zero excess is p =3;1=2: 1 Re(s) Im(s) -4-10 -20 Figure 2: Root Locus The number of asymptotes is equal to p and hence there will be twoasymp- totes at  ` = " (1 + 2`) p # 180  ` =0;;1 = 90  ;; 270  The twoasymptotes intersect the real axis at:  i = P real part of poles of GH ; P real part of zeros of GH p = ;20 ;4;0;(;10) 2 = ;7 The break out pointbetween s =0ands = ;4canbefound either by computing some gains along this stretchoftherealaxis, or by nding the critical points of dK=ds. The gains can be computed using Figure 3. Eachvector in the gure represents one of the terms in GH in polar form. Table 2 summarizes these gain calculations for selected points along the real axis. This table can be constructed with relatively little e ort. The search converges, because as we compute eachgainwebegin to isolate the break-out pointandknowwhether to moveleftorrighttomakethenextcomputation. By contrast, the analytical method yields, dK ds = d ds " ;(s 3 +14s 2 +40) 10(s +10) # 2 Im(s) V 3 V 2 V 1 Gain at point s = x -4-10-20 V 4 s V 3 V 2 x V 1 V 4 Re(s) Figure 3: Finding break-out point s -3.5 -3 -2.5 -2.2 -2 -1 K 0.444 0.729 0.875 0.904 0.9 0.633 Table 2: Gain K for ;4 <s<0 = ;(2s 3 +126s 2 +14s +100) 10(s +10) 2 To ndthecritical pointofinterest wemust solve 2s 3 +126s 2 +14s +100=0: This is a factorization problem of the same order as the root locus. Hence, in this case the searchinTable 2 is more ecient. To ndthe gain that yields closed loop poles with damping ratio  = 1= p 2, wemust locate the pointwheretherootlocus crosses the rayshown in Figure 4. This rayisdrawn at an angle of 45  from the negativereal axis because the angle of the rayis  = cos ;1 (1= p 2) = 45  This comes from the basic de nition of the damping ratio as the cosine of the angle the raymakes with the negative real axis. Note that any pole that lies on this ray has the same damping ratio. Wenow use the angle condition to nd where the root locus crosses this ray.We know that at every pointontheroot locus wehave GH =1 6 ;180  3 θ 1 θ 2θ 3 Re(s) Im(s) -20 -10 -4 Line of constant damping -a a Figure 4: Pointwhere root locus crosses ray Thus, at those points wherethe root locus crosses the ray,the angle condition must be satis ed. Wemerely havetosearch along the rayuntil 6 GH = ;180  as shown in Figure 4. The root locus only crosses this rayatonepointso that there is only one point where ; 1 ; 2 ; 2 = ;180  : The table below shows the searchforthis point. The searchwas started at s = ;2:5+j2:5for the following reason. The composite angle of GH at s = ;2+j2is ; 3 ;45  ;135  = ; 3 ;180  > ;180  ;; since > 3 all along the ray. Hence, if we start looking further out, we may actually start close to the rightanswer. The table belowsummarizes the search. along the real axis. We see that the rst guess was almost on the money.Then due to the nonlinearityofthe arctangent function, wehave to makeabout vemore attempts to get close to the exact answer. In a practical problem wedon't need an exact answer just a good approximation. So if wetake s +;2:51 +j2:51, wehaveasmuch accuracy as weneed.The 4 Re( s) -2.5 -2.4 -2.3 -2.35 -2.34 6 GH ;183:7  ;181:5  ;179:3  ;180:4  ;180:2  Table 3: Gain K for ;4 <s<0 Gain to place poles at this location is K = jV 1 jjV 2 jjV 3 j 10jV 4 j = 2:305 Another big advantage to the latter approachisthat it ties in nicely with the Nyquist Analysis discussed in Chapter 10. 5