Solution 4.6.3.11 The characteristic equation is 1+ K s 2 (s +10) =0;; which can be rewritten as s 3 +10s 2 +0s + K s 3 +10s 2 =0;; or s 3 +10s 2 +0s + K =0: The initial Routh table is s 3 1 0 0 s 2 10 K 0 s 1 b 1 b 2 0 s 0 c 1 0 0 . Then b 1 = ;Det " 1 0 10 K # 10 = ;K 10 b 2 = ;Det " 1 0 10 0 # 52 = ;(0;0) 10 = 0 The partially completed Routh table is s 3 1 0 0 s 2 10 K 0 s 1 ;K 10 0 0 s 0 c 1 0 0 . 1 Then c 1 = ;Det " 10 K ;K 10 0 # ;K 10 = ;(0 + K 2 ) ;K 10 = 10K The completed Routh table is s 3 1 0 0 s 2 10 K 0 s 1 ;K 10 0 0 s 0 10K 0 0 . For all the terms in the rst column to be positivewemust have ;K 10 > 0 and 10K>0: Since K cannot be both greater than zero and less than zero, this system is unstable for all values of K.Wewill see whyinChapter 5. 2