Solution 4.6.2.8 The characteristic polynomial is p(s)=s 5 +0s 4 ;5s 3 +0s 2 +4s +0: Clearly one of the roots is s =0.Hencewecanconsider the polynomial p 0 (s)=s 4 +0s 3 ;5s 2 +0s +4: The initial Routh table is s 4 1 -5 4 0 s 3 0 0 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 . In this problem we start righto withazero row. The auxilliary equation is s 4 +0s 3 ;5s 2 +0s +4 Di erentiating weobtain d ds s 4 +0s 3 ;5s 2 +0s +4=4s 3 ;10s Then the modi ed Routh table is s 4 1 -5 4 0 s 3 4 -10 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 ;5 4 ;10 # 4 = ;(10 + 20) 4 = ;5=2 b 2 = ;Det " 1 4 4 0 # 4 = ;(0;16 4 =4 b 3 = ;Det " 1 0 4 0 # 4 = ;(0;040) 4 =0 1 s 4 1 -5 4 0 s 3 4 -10 0 0 s 2 ;5=2 4 0 0 s 1 c 1 c 2 0 0 s 0 d 1 0 0 0 . c 1 = ;Det " 4 ;10 ;5=2 4 # ;5=2 = ;(16;25) ;5=2 = 18=5 c 2 = ;Det " 4 0 ;5=2 0 # ;5=2 = ;(0;0) ;5=2 = 0: The partially completed Routh table is now s 4 1 -5 4 0 s 3 4 -10 0 0 s 2 ;5=2 4 0 0 s 1 18=5 0 0 0 s 0 d 1 0 0 0 . d 1 = ;Det " ;5=2 4 18=5 0 # 18=5 = ;(0;(18=5)(4)) 18=5 = 4 d 2 = ;Det " ;5=2 0 18=5 0 # 18=5 2 = ;(0;0) 18=5 = 0: The completed Routh arrayis s 4 1 -5 4 0 s 3 4 -10 0 0 s 2 ;5=2 4 0 0 s 1 18=5 0 0 0 s 0 4 0 0 0 . There are twosignchanges in the rt column, from negativetopositive and backtopositive. Thus, there are twopolesin the right half of the s plane. This is veri ed bytheMATLAB dialogure: EDU>p = [1 0-5040] p= 1 0 -5 0 4 0 EDU>roots(p) ans = 0 2.0000 -2.0000 1.0000 -1.0000 EDU> 3