If the [[discrete-time signal]] $h(n)$ is the impulse response of a [[discrete-time LTI system]] with input $x(n)$, the output $y(n)$ can be computed using the [[convolution sum]]: $ y(n) = x(n) \ast h(n) = \sum_{k=-\infty}^{+\infty} x(k) h(n-k) $ The z-transform of the convolution is the product of the z-transforms of the signals: $ y(n)=x(n)\ast h(n) \xrightarrow[Z]{} Y(z)=H(z)X(z) $ This is a direct result of the application of the [[z-transform]] equation to the convolution sum: $ \begin{align} Y(z) &= \sum_{n = -\infty}^{+\infty} y(n) z^{-n} \\ &= \sum_{n = -\infty}^{+\infty} \sum_{k=-\infty}^{+\infty}x(n)h(n-k) z^{-n} \\ &= \sum_{n = -\infty}^{+\infty} x(n)\sum_{k=-\infty}^{+\infty}h(n-k) z^{-n} \\ &= \sum_{n = -\infty}^{+\infty} x(n)\sum_{m=-\infty}^{+\infty}h(m) z^{-m}z^{-n} \\ &= \sum_{n = -\infty}^{+\infty} x(n)z^{-n}\sum_{m=-\infty}^{+\infty}h(m) z^{-m} \\ &= X(z) H(z) \end{align} $