Solution 5.8.1.13 For the system of Figure 1 let For the system of Figure 1 G H + R CΣ Figure 1: Standard Closed Loop Con guration GH = K(s +1)(s+2) s 3 (s +30)(s+40) 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. In the presentcasesince we are just given GH the zeros of G and poles of H cannot be distinguished. This distinction is not necessary,however, to drawtheroot locus. In this particular problem the exact form of the root locus is not easy to draw without some work. The locus maylookeither like Figure 2 or like Figure 3. One mightbeinclined to lean towards the latter locus for the following reason. The poles of GH at s = ;30 and s = ;40 are quite distant from the other poles and zeros. Thus, near the origin these poles do not contribute muchtothe total angle of GH.Near the origin the shape of the root locus is probably close to that of GH = K(s +1)(s +2) s 3 : The root locus for this GH is shown in Figure 4. If one plots the root locus accurately, one obtains the locus shown in Figure 5. This can be seen to be a combination of all three loci wespeculated about. That is why for systems with veormore poles, the details of the root locus are often not easy to determine without some calculation. In this case wewould havetosearch for break-in and break-out points. The root locus on the real axis occurs: 1 -1-2 -30-40 -22.3 Re(s) Im(s) 3 Figure 2: One Potential Root Locus 2 -1-2-30-40 -22.3 Re(s) Im(s) 3 Figure 3: AlternativeRootLocus 3 -1 -2 Re(s) Im(s) X 3 Figure 4: Root Locus 1. Between the three poles at the origin and the zero at ;1. 2. Between the zero at ;2andthepoleat;30 3. To the left of the pole at ;40. In eachofthese segments of the real axis, the total countofpoles and zeros to rightisodd. The pole zero excess is p =5;2=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 poles of GH ; P zeros of GH p = ;40;30;0;(;1;2) 3 = ;22:3 4 -1-2 Re(s) 30 -30 -40 Im(s) 3 Figure 5: Root Locus 5 θ 1 α β θ 2 θ 3 Re(s) Im(s) -2-3-30-40 3 ω Figure 6: Computation of Angle of GH on Imaginary Axis Re( s) j29 j30 j32 j31.5 6 GH ;175:89  ;177:6  ;180:87  ;180:06  Table 1: SearchforImaginary Axis Crossing To nd where the root locus crosses the imaginary axis, weconsult Fig- ure 6. The root locus only crosses the j!-axis at one pointsothat there is only one pointwhere ; ;3 1 ; 2 ; 2 = ;180  : The table below shows the searchfor this point. The searchwas started at s = j29 for the following reason. Since the twozeros are close to the three poles at the origin, the net angle contribution from the twozeros and three poles is roughly 180  ;270  = ;90  : Thus to get a net angle of ;180  we need about another ;90  .Ifwe pick s  j30 weget;45  from the pole at s = ;30 and a little less than ;45  from the pole at s = ;40. Thus, the total should be close to ;180  .Table 1 summarizes the search. We see that the rst guess was close. Three more iterations yield the crossing point. Gain to place poles at this location is K = Product of vectors from poles Product of vectors from zeros 6 = 31:5 3 43:550:9 31:5231:56 = 69;;590: 7