(4)对于因果序列x(n) k x()2 0 k=-m ZTLx(n-m)u(n=zX(z) ZTLx(n+m)u(n)== x(2)-2x(k)2 k=05 （4）对于因果序列x(n) ZT[x(n m)u(n)] z X(z) −m − = ( ) 0 1  = − =− − k m k  x k z       + == − − = − 1 0 [ ( ) ( )] ( ) ( ) m k m k ZT x n m u n z X z x k z