Solution 4.6.3.8 The characteristic equation is 1+ K(s +1) s 2 (s +2)(s +50) =0;; which can be rewritten as s 4 +52s 3 +100s 2 + Ks+ K s 2 (s +2)(s +50) =0;; or s 4 +52s 3 +100s 2 + Ks+ K =0: The initial Routh table is s 4 1 100 K 0 s 3 52 K 0 0 s 2 b 1 b 2 0 0 s 1 c 1 c 2 0 0 s 0 d 1 0 0 0 . Then b 1 = ;Det " 1 100 52 K # 52 = ;(K ;5200) 52 b 2 = ;Det " 1 K 52 0 # 52 = ;(;52K) 52 = K The partially completed Routh table is s 4 1 100 K 0 s 3 52 K 0 0 s 2 ;(K;5200) 52 K 0 0 s 1 c 1 c 2 0 0 s 0 d 1 0 0 0 . 1 Then c 1 = ;Det " 52 K ;(K;5200) 52 K # ;(K;5200) 52 = ;(52 2 K + K (K;5200) 52 ) ;(K;5200) 52 = K 2 ;2496K) (K ;5200) The partially completed Routh table is s 4 1 100 K 0 s 3 52 K 0 0 s 2 ;(K;5200) 52 K 0 0 s 1 K 2 ;2496K (K;5200) 0 0 0 s 0 d 1 0 0 0 . Then d 1 = ;Det " 52 K K 2 ;2496K K;5200 0 # 80 = K The completed Routh table is s 4 1 100 K 0 s 3 52 K 0 0 s 2 ;(K;5200) 52 K 0 0 s 1 K 2 ;2496K (K;5200) 0 0 0 s 0 K 0 0 0 . For all the terms in the rst column to be positivewemust have ;(K ;5200) 52 > 0 and K 2 ;2496K (K ;5200) > 0and K>0;; From the rst inequality ;(K ;5200) 52 > 0;; 2 we see that K ;5200 < 0;; or K<5200: Applying the rst and the last inequalityto K 2 ;2496K (K ;5200) > 0and K>0;; we see that K ;5200 < 0;; and hence K 2 ;2496K<0: Since K>0wemust have K ;2496 < 0;; or K<2496: Thus, satisfying all three constraintrequires 0 <K<2496: 3