Solution 4.6.2.3 The characteristic polynomial is p(s)=s 4 +7s 3 +24s 2 +58s +40: The initial Routh table is s 4 1 24 40 0 s 3 7 58 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 24 7 58 # 7 = ;(58;168) 7 = 110=7;; b 2 = ;Det " 1 40 7 0 # 7 = ;(0;740) 7 =40 b 3 = ;Det " 1 0 7 0 # 7 = ;(0;040) 7 =0 The partially completed Routh table is now s 4 1 24 40 0 s 3 7 58 0 0 s 2 110/7 1 0 0 s 1 c 1 c 2 0 0 s 0 d 1 0 0 0 . c 1 = ;Det " 7 58 110=7 1 # 110=7 = ;(7;6380=7) 110=7 = 6631 110 1 c 2 = ;Det " 7 0 110=7 0 # 110=7 = ;(0;0) 110=7 = 0: The partially nished Routh arrayatthis pointis s 4 1 24 40 0 s 3 7 58 0 0 s 2 110/7 1 0 0 s 1 6331/110 0 0 0 s 0 d 1 0 0 0 . d 1 = ;Det " 110=7 1 6331=110 0 # 6331=110 = ;[0;(6331=110)] 6331=110 = 1 d 2 = ;Det " 110=7 0 6331=110 0 # 6331=110 = ;(0;0) 6331=110 = 0: The nal Routh table is s 4 1 24 40 0 s 3 7 58 0 0 s 2 110/7 1 0 0 s 1 6331/110 0 0 0 s 0 1 0 0 0 . There are no sign changes in the rt column, therefore all the roots are in the left half of the s plane. This is veri ed by the MATLAB dialogure: 2 EDU>p = [1 7245840] p= 1 7 24 58 40 EDU>roots(p) ans = -4.0000 -1.0000+ 3.0000i -1.0000- 3.0000i -1.0000 EDU> 3