Solution 4.6.3.6 The characteristic equation is 1+ K (s +4) 4 =0;; which can be rewritten as s 4 +16s 3 +96s 2 +256s+(K +256) (s +4) 4 =0;; or s 4 +16s 3 +96s 2 +256s+(K +256)=0: The initial Routh table is s 4 1 96 K +256 0 s 3 16 256 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 96 16 256 # 16 = ;(256;(16)(96)) 16 = 80 b 2 = ;Det " 1 K+256 16 0 # 16 = ;(;(16)(K +256)) 16 = K +256 The partially completed Routh table is s 4 1 96 K +256 0 s 3 16 256 0 0 s 2 80 K +256 0 0 s 1 c 1 c 2 0 0 s 0 d 1 0 0 0 . 1 Then c 1 = ;Det " 16 256 80 K +256 # 80 = ;(16K +(16)(256);(80)(256) 80 = ;(16K16384) 80 The partially completed Routh table is s 4 1 96 K +256 0 s 3 16 256 0 0 s 2 80 K +256 0 0 s 1 ;(16K16384) 80 0 0 0 s 0 d 1 0 0 0 . Then d 1 = ;Det " 80 K +256 ;(16K;16384) 80 0 # 80 = K +256 The completed Routh table is s 4 1 96 K +256 0 s 3 16 256 0 0 s 2 80 K +256 0 0 s 1 ;(16K16384) 80 0 0 0 s 0 K +256 0 0 0 . For all the terms in the rst column to be positivewemust have ;(16K ;16384) 80 > 0 and K +256> 0;; or, equivalently ;256 <K<26 and K<1024 The lower bound comes from the complementary root locus whichwewill study in Chapter 5. 2