Solution 4.6.2.4 The characteristic polynomial is p(s)=s 4 +5s 3 +0s 2 ;20s;16: The initial Routh table is s 4 1 0 -16 0 s 3 5 -20 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 0 5 ;20 # 5 = ;(;20;0) 5 =4 b 2 = ;Det " 1 ;16 5 0 # 5 = ;(0+ 80 5 = ;16 b 3 = ;Det " 1 0 5 0 # 5 = ;(0;040) 7 =0 The partially completed Routh table is now s 4 1 0 -16 0 s 3 5 -20 0 0 s 2 4 ;16 0 0 s 1 c 1 c 2 0 0 s 0 d 1 0 0 0 . c 1 = ;Det " 5 ;20 4 ;16 # 4 = ;(80;80) 4 = 0 1 c 2 = ;Det " 5 0 4 0 # 110=7 = ;(0;0) 110=7 = 0: Thus, wehaveazero row. The auxilliary equation is 4S 2 ;16 = 0;; yielding s = ;2 and s =2: Thus weknowtwoofthe four roots. Todetermine is there is more than one root in the right half plane, weletc 1 = ,toobtain s 4 1 0 -16 0 s 3 5 -20 0 0 s 2 4 ;16 0 0 s 1  0 0 0 s 0 d 1 0 0 0 . Then c 2 = ;Det " 4 ;16  0 #  = ;(0+16)  = ;1: The completed Routh arrayis s 4 1 0 -16 0 s 3 5 -20 0 0 s 2 4 ;16 0 0 s 1  0 0 0 s 0 ;1 0 0 0 . There is one sign change in the rt column, therefore one root in the left half of the s plane. This is veri ed bytheMATLAB dialogure: 2 EDU>p = [1 5 0 -20 -16] p= 1 5 0 -20 -16 EDU>roots(p) ans = -4.0000 2.0000 -2.0000 -1.0000 EDU> 3