(2)n=0 x(0=∑Rx(2x1x()k (z-1) 个二阶极点P12=1,又多了一个单极点P=0 Re Lz(-1)2」 (z-1)2 d Re (z-1)2 (z-1)2」,dz z(z-1)2 所以x()=1+(1)=0 x()=(n-1p(n-1)X 第 2 页 (2)n=0 ( ) (z ) z X z z 2 0 1 1 1 − = − ( ) 0 2 1 1 Res =       − z z z ( ) 1 2 1 1 Res =       − z z z 所以 x(0) = 1+ (−1) = 0 x(n) = (n−1)u(n−1) 1, 0 一个二阶极点 P1,2 = 又多了一个单极点 P1 =    = − = m z z m x X z z 0 1 (0) Res ( ) ( ) 0 2 1 1 =       − =  z z z z = 1 ( ) ( ) 1 2 2 1 1 1 d d =       − = −  z z z z z 1 2 1 = − = z z = −1