Solution 4.6.2.10 The characteristic polynomial is p(s)=s 4 +2s 3 ;2s 2 ;8s +;8: The initial Routh table is s 4 1 -2 -8 0 s 3 2 -8 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 ;2 2 ;8 # 2 = ;(;8+4) 2 =2 b 2 = ;Det " 1 ;8 2 0 # 2 = ;(0+ 16 2 = ;8 b 3 = ;Det " 1 0 2 0 # 2 = ;(0;0 2 =0 The partially completed Routh table is now s 4 1 -2 -8 0 s 3 2 -8 0 0 s 2 2 -8 0 0 s 1 c 1 c 2 0 0 s 0 d 1 0 0 0 . c 1 = ;Det " 2 ;8 2 ;8 # 2 = ;(;16 + 16) 2 = 0 1 c 2 = ;Det " 2 0 ;9 0 # 2 = ;(0;0) 2 = 0: Thus , wehaveazero rowThe auxilliary polynomial is 2s 3 ;8s;; Di erentiating yields d ds (2s 3 ;8s)=6s 2 ;8 Then the adjusted Routh table is now s 4 1 -2 -8 0 s 3 2 -8 0 0 s 2 6 -8 0 0 s 1 2 c 2 0 0 s 0 d 1 0 0 0 . Then c 1 = ;Det " 2 ;8 6 ;8 # 6 = ;(16 + 48) 6 = ;16=3 The partially completely Routh table is now s 4 1 -2 -8 0 s 3 2 -8 0 0 s 2 6 -8 0 0 s 1 -16/3 0 0 0 s 0 d 1 0 0 0 . 2 Then d 1 = ;Det " 6 ;8 ;16=3 0 # ;16=3 = ;(0;(;16=3)(;8)) ;16=3 = ;8 d 2 = ;Det " 6 0 ;16=3 0 # ;16=3 = ;(0;0) ;16=3 = 0: The completed Routh arrayis s 4 1 -2 -8 0 s 3 2 -8 0 0 s 2 6 -8 0 0 s 1 -16/3 0 0 0 s 0 -8 0 0 0 . There is one sign changes in the rt column, and therefore one roots in the right half of the s plane. This is veri ed by the MATLAB dialogue: EDU>p = [1 2-2-8-8] p= 1 2 -2 -8 -8 EDU>roots(p) ans = 2.0000 -2.0000 -1.0000+ 1.0000i 3 -1.0000- 1.0000i EDU> 4