Solution 4.6.3.12 The characteristic equation is 1+ K(s +1) s 2 (s +10)(s+20) =0;; which can be rewritten as s 4 +30s 3 +200s 2 + Ks+ K s 2 (s +10)(s+20) =0;; or s 4 +30s 3 +200s 2 + Ks+ K =0: The initial Routh table is s 4 1 200 K 0 s 3 30 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 30 K # 30 = ;(K ;3000) 30 b 2 = ;Det " 1 K 30 0 # 30 = ;(;30K) 30 = K The partially completed Routh table is s 4 1 200 K 0 s 3 30 K 0 0 s 2 ;(K;3000) 30 K 0 0 s 1 c 1 c 2 0 0 s 0 d 1 0 0 0 . 1 Then c 1 = ;Det " 30 K ;(K;3000) 30 K # ;(K;3000) 30 = ;(30 2 K + K (K;3000) 30 ) ;(K;3000) 30 = K 2 ;2100K) (K ;3000) The partially completed Routh table is s 4 1 200 K 0 s 3 30 K 0 0 s 2 ;(K;3000) 30 K 0 0 s 1 K 2 ;2100K (K;3000) 0 0 0 s 0 d 1 0 0 0 . Then d 1 = ;Det " 30 K K 2 ;2100K K;3000 0 # K 2 ;2100K (K;3000) = K The completed Routh table is s 4 1 200 K 0 s 3 30 K 0 0 s 2 ;(K;3000) 30 K 0 0 s 1 K 2 ;2100K (K;3000) 0 0 0 s 0 K 0 0 0 . For all the terms in the rst column to be positivewemust have ;(K ;3000) 30 > 0 and K 2 ;2100K (K ;3000) > 0and K>0;; From the rst and third inequalities weseethatwemust have K>0 and K<3000: 2 Applying these two inequalties to the second inequalityweseethat K ;3000 < 0;; and consequently K 2 ;2100K<0: Since K>0, wemust have K<2100: Thus, for stabilitywemust have 0 <K<2100: 3