Solution 4.6.3.3 The characteristic equation is 1+ K(s +1) (s;1)(s;2) =0;; which can be rewritten as s 2 ;3s +2+Ks+ K (s;1)(s;2) =0;; or s 2 +(K ;3)s +(K +2)=0: The initial Routh table is s 2 1 K +2 0 s 1 K ;3 0 0 s 0 b 1 b 2 0 . Then b 1 = ;Det " 1 K +2 K ;3 0 # K = ;(0;(K +2)(K ;3)) K ;3 = K +2;; b 2 = ;Det " 1 0 K ;3 0 # K ;3 = ;(0;0) K =0 The completed Routh table is s 2 1 K +2 0 s 1 K ;3 0 0 s 0 K +2 b 2 0 . For all the terms in the rst column to be positivewemust have K +2> 0 and (K ;3) > 0;; or, equivalently K>3: The constraint K>;2isrelated to the complementary root locus which wewill study in Chapter 5. 1