Solution 5.8.1.4 For the system of Figure 1 let wehave G H +R C Figure 1: Standard Closed Loop Con guration GH = K (s +3) 4 : 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. There is no root locus on the real axis. p =4;0=4: The number of asymptotes is equal to p and hence there will be four asymp- totes at  ` =  (1 + 2`) p  180  ` =0;;1;;2:3 The four asymptotes intersect the real axis at:  i = P poles ofGH ; P zeros ofGH p = ;3;3;3;3;0 4 = ;3 Since the asymptotes meet at the quadruple pole, we can see that along the entire length of eachofthe asymptotes the sum of the angles of the three poles must equal ;180  .Thus, the root locus is just the four asymptotes. 1 Re(s) Im(s) 4 -3 3 Figure 2: Rootlocus We can verify this bycomputing dK ds = d ds h ;(s +3) 4 i = ;4(s +3) 4 Hence, the break-out occurs at s = ;3. We can nd the gain that places closed loop pole on the j!-axis in a number of ways. Analytically,wecould write out the denominator of the closed loop transfer function in unfactored form to obtain: 1+GH = 1+ K (s +3) 4 = s 3 +9s 2 +9s +27 (s +3) 3 Toobtain 1 + GH =0requires that s 4 +12s 3 +54s 2 +108s +81+K =0 Setting s = j! in this last equation yields = ! 4 ;12j! 3 +;54! 2 ++108j!+81+K = (81+ K + ! 4 ;54! 2 )+j(;12! 3 +108!) 2 If this expression is to equal 0+ j0wemust have (81+ K + ! 4 ;54! 2 )=0 and j(;12! 3 +108!)=0 The second equation is easily solved, yielding ! =  q (9) = 3 Substituting into the rst equation then yields K = 54(3 2 );3 4 ;81 = 324 These calculations are satisfying in the sense that they are precise. As a general rule, even for more complicated transfer functions, the equations are usually reasonably easy to solve. However, they obscure the connection with Nyquist theory.For that reason, weo erthe following alternativeanalysis. We know that at every pointonthe root locus wehave GH =1 6 ;180  Thus, at those points where the root locus crosses the j!-axis, the angle condition must be satis ed. In the present case it is easy to see that we automatically have the angle condition satis ed all along the asymptotes, so that ! = 3tan(45  ) = 3: The resulting gain to place poles at s =0 j3issimply the fourth power of the vector drawn from s = ;3tos =0+j3or K = h (3 2 +3 2 ) 1=2 i 4 = 18 2 = 324 These same basic ideas can be applied to more complicated examples with only a small increase in computational e ort. 3