5.8.2.4 For the system of Figure 1 let Σ G H + R C Figure 1: G(s)H(s)=K (s+1) s(s;1) The root locus is shown is shown in Figure 2. The pole zero excess(pze) is one, meaning one asymptote at  =180  .The gain to place poles on the imaginary axis can be found in a number of ways. One is to searchalong the imaginary axis until we nd where the angles contributions of the pole and zeros of GH sum to ;180  . Another is to solve for the for break-out point, either analytically or bytrialanderror,andthen recognize that the root locus o the real axis is part of a circle centered at the zero and whose radius is the distance along the real axis from the zero to the break-out point. In either case the basic formula is jGH(s)j=1;; (1) whichinthe presentcaseyields Kjs+1j jsjjs;2j =1;; or K = jsjjs;2j js+1j : The vector interpretation of equation 1 is shown in Figure 3. We simply pickpoints along the poxitive real axis to the left of the pole of GH at s =2. We know that all these points are closed loop poles for some value of K.Table 1 summarizes the searchusing the equation K = jsjjs;2j js+1j 1 X XO Re(s) Im(s) -1 2 Figure 2: Root locus X X O Re(s) Im(s) 2 -1 s + 1 s s - 2 Figure 3: Gain calculation along real axis s 0.72 0.73 0.74 0.731 0.732 K 0.53581 0.535896 0.53586 0.535898 0.535898 Table 1: Gain values along positiverealaxis 2 The break-out pointisnears =0:731. We can solveforthebreak-out pointanalytically by writing dK ds = dK ds ;s 2 +2s s+1 = s 2 ;2s (s+1) 2 + ;2s+2 s+1 = s 2 ;2s;2s 2 +2 (s+1) 2 = ;s 2 ;2s+2 (s+1) 2 : The critical points occur for s 2 +2s;2=0;; or s = ;1 p 3: The the break-out point will be at s = ;1+ p 3=0:7321;; and the break-in pointat s = ;1; q (3) = ;2:7321: Wecannow swing an arc of length p 3 using s = ;1asacenter and nd the crossing pointonthe imaginary axis to be ! = p 3;1= p 2: The gain to place poles on the imaginary axis is then K = js+1jjs+2j js;1j s=j p 2 = =2: Thus, the system is stable for 2 <K<1. 3