Solution 4.6.3.1 The characteristic equation is 1+ K(s +1) (s;2)(s +2) =0;; which can be rewritten as s 2 ;4+Ks+ K (s;2)(s+2) =0;; or s 2 + Ks+(K ;4) = 0: The initial Routh table is s 2 1 K-4 0 s 1 K 0 0 s 0 b 1 b 2 0 . Then b 1 = ;Det " 1 K ;4 K 0 # K = ;(0;K(K ;4)) K = K ;4;; b 2 = ;Det " 1 0 K 0 # K = ;(0;0) K =0 b 3 = ;Det " 1 0 K 0 # K = ;(0;0) K =0 The completed Routh table is s 2 1 K-4 0 s 1 K 0 0 s 0 K ;4 0 0 . For all the terms in the rst column to be positivewemust have K>0 and (K ;4) > 0;; or, equivalently K>4: 1