Solution 5.8.1.23 G H C R + - Figure 1: For the system shown above GH(s)= K(s +3) s(s + 2)(s +50) The rst step in drawing the root locus 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 show where the root locus occurs. The rule is that root locus occurs on the real axis to the left of an odd countof poles and zeros. That is, if you stand on the real axis and look to your right you must countanodd number of poles and zeros. The next step is to compute the asymptotes. The numberofasymptotes is p ex =Number of poles of GH ;Numberof nite zeros of GH: Im(s) Re(s) -50 -2-3 Figure 2: The asymptotes occur at the angles  ` =  1+2` p ex   180  ` =0;;1;;:::p ex ;1: In the presentcase there are three poles and one nite zero so p ex =2. Thus, there are two asymptotes at  0 =  1+(2)(0) 2  180  =90   1 =  1+(2)(1) 2  180  =270  The asymptotes intersect the real axis at  i = Sum of poles of GH ; Sum of zeros of GH p ex = [(;2)+ (;50)]; [(;3)] 2 = ;24:5 Since there is one nite zero and three poles, twoclosed loop poles will migrate to zeros at in nity,onesuchzeroatthe`end' of eachasymptote. Twopossible root loci are shown in Figure 3. Of the two, that shown in Figure 3 (a) is by far the more probable. Torule out case (b) requires computing the gain along the real axis between s = ;50 and s = ;3Before doing that, wenotethatthere is also a break-out pointbetween s = ;2and s =0.Weknow that for lowgain the closed loop poles occur near the poles of GH.Asthegain increases the closed loop poles migrate awayfrom the poles of GH towards the zeros of GH.Inthis case all twozeros are zeros at in nity. The twoclosed loop poles migrating towards eachotheron the interval [;2;;0] eventually meet and break-out of the real axis. The question is where do they go after they break-out? To answer that question wecompute the gain between s = ;50 and s = ;3. The gain is calculated by solving the equation jGH(s)j =1 for points along the real axis between s = ;50 and s = ;3. That is K = jsjjs +2jjs +50j js +3j Im(s) Re(s) -50 -2-3 Im(s) Re(s) -50 -2 (a) (b) -3 Figure 3: Completed Root Locus S -28 -26 -25.5 -25 -24.5 -24 -23.5 K 641 651 652 653 653.8 653.7 653.1 Table 1: SearchforMaximum in Gain S -5.3 -5.2 -5.1 -5.0 -4.9 -4.8 -4.7 K 339.9 338.9 338 337.5 337.3 337.5 338.2 Table 2: SearchforMaximum in Gain Table 1 summarizes the searchforpeaks and valleys. From the table wesee that there is a maximum around s = ;24. Having found the maximum, weknow the minimum will exist and so our earlier supposition is veri ed, The correct root locus is that shown in Figure 3 (a). The actual location of the break-in can be found from Table 2. In both cases the searchwas started the vicinityofwhere the break-out or break-in point was expected to be. In the case of the break-out, the approximate break-out can be determined byconsidering G(s)= K s(s +50) ;; whichhas a break-out of s = ;25. The approximate break-in pointcanbe found by considering G(s)= K(s +3) s(s +2) ;; which has a break-in near s = ;4. The actual break-in and break-out points should be close to these approximations.