Solution 5.8.1.12 For the system of Figure 1 let G H +R C Figure 1: Standard Closed Loop Con guration GH = K(s +2) s 2 Speci c pole and zero locations are given, but as shown in Figure 2 the problem is essentially the same for anydouble pole with a zero to the left. The gure shows a circle drawn through the double pole with center at the θ Re(s) j Im(s) -2 γ φ φ Figure 2: Angles to PointonCircleCentered at s = ;2 zero. For anypointonthe circle, the triangle formed by drawing vectors from the poles and zeros is isosceles,with twosides being radii of the circle. drawn whose twosides are radii of the circle. The interior angles of the circle sum to 180 jcirc .Theinterior angles of the circle sum to 180  .Thus +2 =180  : 1 However,  =180  ;: Thus +2 = +2(180  ;) = +360  ;2 Thus ;2 +360  =180  ;; or ;2 = ;180  : Thus wehave 6 GH = ;2 = ;180  for anypointonthe circle. Things get a little more complicated if wehavetwo distinct poles to the rightofazero. Then analytical geometry works best. The key point here is that it is the angle condition that has to be satis ed. The gain never enters in to the proof. 2