3)有理分式直接展开:|[r,P,]= residue(b,a) 式中:F(s)= b(s) bms"+bm-S+.+,s+bo (S)anS"+an1S"+…+a1S F()=(S)2r(1 r(2) (3) a(S)S-p(1)s-p(2)s-3)(0 +3s+5 例:F(S)= b=[1,3,5 s3+6s2+1ls+6 a=[1,6,11,6] 计算:[,P,k]= residue(b,a)3) 有理分式直接展开: 2 3 2 3 5 ( ) 6 11 6 s s F s s s s + + = + + + [ , , ] ) r p k res due = i (b,a 式中: 1 1 1 0 1 1 1 0 ( ) ( ) ( ) m m m m n n n n b S b S b s b s b F s a S a S a S a S a − − − − + + + + = = + + + + ( ) (1) (2) (3) ( ) ...... ( ) ( ) (1) (2) (3) b S r r r F s k s a S s p s p s p = = + + + − − − 例: b = [1,3,5] a = [1,6,11,6] 计算: [ , , ] ) r p k res due = i (b,a