Solution 5.8.1.3 G H + R CΣ Figure 1: Standard Closed Loop Con guration For the system of Figure 1 wehave GH = K (s +3) 3 : 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. The root locus on the real axis occurs only to the left of the triple pole at s = -3. The pole zero excess is p =3;0=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 The three asymptotes intersect the real axis at:  i = P real part of poles of GH ; P real part ofzeros of GH p = ;3;3;3;0 3 = ;3 Since the asymptotes meet at the triple pole, wecan 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 three asymptotes. 1 Re(s) Im(s) 3 Figure 2: Rootlocus We can verify this bycomputing dK ds = d ds h ;(s +3) 3 i = ;3(s +3) 2 Hence, the break-out occurs at s = ;3. We can nd the gain that places closed loop pole on the imaginary 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) 3 = s 3 +9s 2 +9s +27 (s +3) 3 Toobtain 1 + GH =0requires that s 3 +9s 2 +27s +27+K =0 Setting s = j! in this last equation yields ;j! 3 ;9! 2 + j27! +27+K = (27+ K;9! 2 )+j(;! 3 +27!) 2 If this expression is to equal 0+ j0wemust have 27 + K ;9! 2 =0 and ;! 3 +27! =0 The second equation is easily solved, yielding ! =  q (27) = 3 q (3) Substituting into the rst equation then yields K = 9! 2 ;27 = 216 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 erthefollowing alternativeanalysis which students usually nd unappealing because of its imprecise nature. We know that at every pointonthe root locus wehave GH =1 6 ;180  Thus, at those points where the root locus crosses the imaginary axis, the angle condition must be satis ed. In the presentcase it is easy to see that weautomatically havethe angle condition satis ed all along the asymptotes, so that ! = 3tan(60  ) = 3( p 3) = 5:196 The resulting gain to place poles at s =0j3 p (3) is simply the cube of the vector drawn from s = ;3tos =0+j3 p (3) or K = h (3 2 +(3 p 3) 2 ) 1=2 i 3 = 6 3 = 216 These same basic ideas can be applied to more complicated examples with only a small increase in computational e ort. Further, this approachis consistentwith understanding the Nyquist analysis. 3