Solution 4.6.3.5 The characteristic equation is 1+ K (s +3) 3 =0;; which can be rewritten as s 3 +9s 2 +27s +(K +27) (s +3) 3 =0;; or s 3 +9s 2 +27s +(K +27)=0: The initial Routh table is s 3 1 27 0 s 2 9 K +27 0 s 1 b 1 b 2 0 s 0 c 1 0 0 . Then b 1 = ;Det " 1 27 9 K +27 # 9 = ;(K ;216) 9 b 2 = ;Det " 1 0 9 0 # 9 = ;(0;0) 9 =0 The partially completed Routh table is s 3 1 27 0 s 2 9 K +27 0 s 1 ;(K;216) 9 0 0 s 0 c 1 0 0 . Then c 1 = ;Det " 9 K +27 ;(K;216) 9 0 # ;(K;216) 9 = K +27 The completed Routh table is 1 s 3 1 27 0 s 2 9 K +27 0 s 1 ;(K;216) 9 0 0 s 0 K +27 0 0 . For all the terms in the rst column to be positivewemust have ;(K ;216) 9 > 0 and K +27> 0;; or, equivalently ;27 <K<26: The lower bound comes from the complementary root locus whichwewill study in Chapter 5. 2