Solution 4.6.3.4 The characteristic equation is 1+ K s(s +10)(s+50) =0;; which can be rewritten as s 3 +60s 2 +500s + K s(s +10)(s+50) =0;; or s 3 +60s 2 +500s + K =0: The initial Routh table is s 3 1 500 0 s 2 60 K 0 s 1 b 1 b 2 0 s 0 c 1 0 0 . Then b 1 = ;Det " 1 500 60 K # 60 = ;(K ;30;;000) 60 b 2 = ;Det " 1 0 60 0 # 60 = ;(0;0) 60 =0 The partially completed Routh table is s 3 1 500 0 s 2 60 K 0 s 1 ;(K;30;;000) 60 0 0 s 0 c 1 0 0 . Then c 1 = ;Det " 60 K ;(K;30;;000) 60 0 # ;(K;30;;000) 60 = K The completed Routh table is 1 s 3 1 500 0 s 2 60 K 0 s 1 ;(K;30;;000) 60 0 0 s 0 K 0 0 . For all the terms in the rst column to be positivewemust have ;(K ;30;;000) 60 > 0 and K>0;; or, equivalently 0 <K<30;;000: 2