Solution 4.6.2.6 The characteristic polynomial is p(s)=s 4 +7s 3 +14s 2 +8s: Clearly wehaveonepole at s =0sowecanconsider the third order poly- nomial p 0 (s)=s 3 +7s 2 +14s +8: The initial Routh table is s 3 1 14 0 s 2 7 8 0 s 1 b 1 b 2 0 s 0 c 1 c;2 0 . Then b 1 = ;Det " 1 14 7 8 # 7 = ;(8;98) 7 =90=7 b 2 = ;Det " 1 0 7 0 # 7 = ;(00 7 =0 b 3 = ;Det " 1 0 7 0 # 7 = ;(0;040) 7 =0 The partially completed Routh table is now s 3 1 14 0 s 2 7 8 0 s 1 90/7 0 0 s 0 c 1 c 2 0 . c 1 = ;Det " 7 8 90=7 0 # 90=7 = ;(0;(90=7)8) (90=7) 1 = 8 c 2 = ;Det " 7 0 90=7 0 # 90=7 = ;(0;0) 90=7 = 0: The completed Routh table is now s 3 1 14 0 s 2 7 8 0 s 1 90/7 0 0 s 0 8 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: EDU>p = [1 7148] p= 1 7 14 8 EDU>roots(p) ans = -4.0000 -2.0000 -1.0000 EDU> 2