Solution 4.6.3.2 The characteristic equation is 1+ K (s;1)(s +5)(s +20) =0;; which can be rewritten as s 3 +24s 2 +75s +(K ;100) (s;1)(s +5)(s +20) =0;; or s 3 +24s 2 +75s +(K ;100) = 0: The initial Routh table is s 3 1 75 0 s 2 24 K ;100 0 s 1 b 1 b 2 0 s 0 c 1 0 0 . Then b 1 = ;Det " 1 75 24 K-100 # 24 = ;(K ;100;1800) 24 = ;(K ;1900) 24 b 2 = ;Det " 1 0 24 0 # 24 = ;(0;0) 24 =0 b 3 = ;Det " 1 0 K 0 # 24 = ;(0;0) 24 =0 The partially completed Routh table is s 3 1 75 0 s 2 24 K ;100 0 s 1 ; K;1900 24 0 0 s 0 c 1 0 0 . 1 Then c 1 = ;Det " 24 K ;100 ; K;1900 24 0 # ; K;1900 24 = K ;100 The completed Routh table is s 3 1 75 0 s 2 24 K ;100 0 s 1 ; K;1900 24 0 0 s 0 K ;100 0 0 . For all the terms in the rst column to be positivewemust have ; K ;1900 24 > 0 and (K ;100) > 0;; or, equivalently K>100 and K<1900: 2