Solution 4.6.2.9 The characteristic polynomial is p(s)=s 4 +24s 3 +196s 2 +624s + 640: The initial Routh table is s 4 1 196 640 0 s 3 24 624 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 " 1 196 24 624 # 24 = ;(624;4704) 24 =170 b 2 = ;Det " 1 640 24 0 # 24 = ;(0;(640)(24) 24 =640 b 3 = ;Det " 1 0 24 0 # 24 = ;(0;040) 24 =0 The partially completed Routh table is now s 4 1 196 640 0 s 3 24 624 0 0 s 2 170 640 0 0 s 1 c 1 c 2 0 0 s 0 d 1 0 0 0 . c 1 = ;Det " 24 624 170 640 # 170 = ;(15360;106080) 170 = 9072=17 1 c 2 = ;Det " 24 0 170 0 # 170 = ;(0;0) 170 = 0: The partially completed Routh table is now s 4 1 196 640 0 s 3 24 624 0 0 s 2 170 640 0 0 s 1 9072/17 0 0 0 s 0 d 1 0 0 0 . d 1 = ;Det " 170 640 9072=17 0 # 9072=17 = ;(0;(640)(9072=17)) 9072=17 = 640 d 2 = ;Det " 170 0 9072=17 0 # 9072=17 = ;(0;0) 9072=17 = 0: The completed Routh arrayis s 4 1 196 640 0 s 3 24 624 0 0 s 2 170 640 0 0 s 1 9072/17 0 0 0 s 0 640 0 0 0 . There are no sign changes in the rt column, and therefore no roots in the right half of the s plane. This is veri ed by the MATLAB dialogure: 2 EDU>p = [1 24 196 624 640] p= 1 24 196 624 640 EDU>roots(p) ans = -10.0000 -8.0000 -4.0000 -2.0000 EDU> 3