证:利用序列的移位,得 Z[x(n+1)-x(n)]=(-1)X(=) ∑[x(n+1)-x(m)=2[x(n+1)-x(m)n n=-00 =lim∑[xm+1)-x(m)m n→)00 m=-1 lim[(2-1)X(=)]=lim 2[x(m+1)-x(m)].1 n→)0 =lim{[x(0)-0]+[x(1)-x(0)]+[x(2)-x(1)]+ n→)0 +x(n+1-x(n=lim[x(n+D]=limx(n) n→0 x(∞)=lm(2-1)X(=)=Res[X(z)=1ZT[x(n 1)  x(n)]  (z 1)X (z) 证：利用序列的移位，得 1 1 [ ( 1) ( )] [ ( 1) ( )] lim [ ( 1) ( )] n n n n n m n m x n x n z x n x n z x m x m z                      1 1 lim[( 1) ( )] lim [ ( 1) ( )] 1 n m z n m z X z x m x m           lim{[ (0) 0] [ (1) (0)] [ (2) (1)] n x x x x x          [ ( 1) ( )]} lim[ ( 1)] lim ( ) n n x n x n x n x n         1 1 ( ) lim[( 1) ( )] Re [ ( )] z z x z X z s X z       