Solution 4.6.2.5 The characteristic polynomial is p(s)=s 4 +9s 3 +28s 2 +36s +16: The initial Routh table is s 4 1 28 16 0 s 3 9 36 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 28 9 36 # 9 = ;(36;252) 9 =24 b 2 = ;Det " 1 16 9 0 # 9 = ;(0;144 9 =16 b 3 = ;Det " 1 0 9 0 # 9 = ;(0;040) 7 =0 The partially completed Routh table is now s 4 1 28 16 0 s 3 9 36 0 0 s 2 24 16 0 0 s 1 c 1 c 2 0 0 s 0 d 1 0 0 0 . c 1 = ;Det " 9 36 24 16 # 24 = ;(144;864) 24 = 30 1 c 2 = ;Det " 9 0 24 0 # 24 = ;(0;0) 110=7 = 0: The partially completed Routh table is now s 4 1 28 16 0 s 3 9 36 0 0 s 2 24 16 0 0 s 1 30 0 0 0 s 0 d 1 0 0 0 . d 1 = ;Det " 24 16 30 0 # 30 = ;(0;1630) 30 = 16 c 2 = ;Det " 24 0 30 0 # 24 = ;(0;0) 30 = 0: The completed Routh arrayis s 4 1 28 16 0 s 3 9 36 0 0 s 2 24 16 0 0 s 1 30 0 0 0 s 0 16 0 0 0 . There are no 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: 2 EDU>p =[1 9 28 36 16] p= 1 9 28 36 16 EDU>roots(p) ans = -4.0000 -2.0000 -2.0000 -1.0000 EDU> 3