Solution 4.6.3.10 The characteristic equation is 1+ K(s +3) s(s + 2)(s+10)(s+50) =0;; which can be rewritten as s 4 +62s 3 +620s 2 + (1000+ K)s +3K s 2 (s +2)(s +50) =0;; or s 4 +62s 3 +620s 2 + (1000+ K)s +3K =0: The initial Routh table is s 4 1 620 3K 0 s 3 62 1000+ 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 " 1 620 62 1000+ K # 62 = ;(K +1000;62 2 (10)) 62 = ;(K ;37;;440) 62 b 2 = ;Det " 1 3K 62 0 # 52 = ;(;(3)(62)K) 62 = 3K The partially completed Routh table is 1 s 4 1 620 3K 0 s 3 62 1000+ K 0 0 s 2 ;(K;37;;440) 62 3K 0 0 s 1 c 1 c 2 0 0 s 0 d 1 0 0 0 . Then c 1 = ;Det " 62 100+ K ;(K;37;;440) 62 3K # ;(K;37;;440) 62 = 186K + (1000+ K)( (K;37;;440) 62 ) ;(K;37;;440) 62 = 11;;532K + K 2 +1000K;37;;440K;37;;440;;000 (K ;37;;440) = K 2 ;25;;908K;37;;440;;000 (K ;37;;440) The partially completed Routh table is The partially completed Routh table is s 4 1 620 3K 0 s 3 62 1000+ K 0 0 s 2 ;(K;37;;440) 62 3K 0 0 s 1 K 2 ;25;;908K;37;;440;;000 (K;37;;440) 0 0 0 s 0 d 1 0 0 0 . Then d 1 = ;Det " 52 K K 2 ;25;;908K;37;;440;;000 (K;37;;440) 0 # K 2 ;25;;908K;37;;440;;000 (K;37;;440) = K The completed Routh table is s 4 1 620 3K 0 s 3 62 1000+ K 0 0 s 2 ;(K;37;;440) 62 3K 0 0 s 1 c 1 c 2 0 0 s 0 d 1 0 0 0 . 2 For all the terms in the rst column to be positivewemust have ;(K ;37;;440) 62 > 0 and K 2 ;25;;908K;37;;440;;000 (K ;37;;440) > 0and K>0;; From the rst inequality ;(K ;37;;440) 62 > 0;; we see that K ;37;;440 < 0;; or K<37;;440: Applying the rst and the last inequalityto K 2 ;25;;908K;37;;440;;000 (K ;37;;440) > 0and K>0;; we see that K ;37;;440 < 0;; and hence K 2 ;25;;908K;37;;440;;000< 0: Since K>0wemust have K 2 ;25;;908K;37;;440;;000< 0;; or (K ;27;;280:4)(K+1372:4) < 0;; or K<27;;280:4 and K<;1372:4: Thus, satisfying all four constraints requires 0 <K<27;;280:4: 3