Solution 4.6.3.7 The characteristic equation is 1+ K s(s + 1)(s +50) =0;; which can be rewritten as s 3 +51s 2 +50s+ K s(s +1)(s +50) =0;; or s 3 +51s 2 +50s+ K =0: The initial Routh table is s 3 1 50 0 s 2 51 K 0 s 1 b 1 b 2 0 s 0 c 1 0 0 . Then b 1 = ;Det " 1 50 51 K # 51 = ;(K ;2550) 51 b 2 = ;Det " 1 0 51 0 # 51 = ;(0;0) 51 =0 The partially completed Routh table is s 3 1 50 0 s 2 51 K 0 s 1 ;(K;2550) 60 0 0 s 0 c 1 0 0 . Then c 1 = ;Det " 51 K ;(K;2550) 51 0 # ;(K;2550) 51 = K The completed Routh table is 1 s 3 1 50 0 s 2 51 K 0 s 1 ;(K;2550) 60 0 0 s 0 K 0 0 . For all the terms in the rst column to be positivewemust have ;(K ;2550) 60 > 0 and K>0;; or, equivalently 0 <K<2550: 2