Z Transform of Convolution

From Eq. (5.5), we have that the output $y$ from a linear time-invariant filter with input $x$ and impulse response $h$ is given by the convolution of $h$ and $x$, i.e.,

$$y(n) = (h \ast x)(n) \protect$$ (7.3)

where ``$\ast$'' means convolution as before. Taking the z transform of both sides of Eq. (6.3) and applying the convolution theorem from the preceding section gives

$$Y(z) = H(z)X(z) \protect$$ (7.4)

where H(z) is the z transform of the filter impulse response. We may divide Eq. (6.4) by $X(z)$ to obtain

$$H(z) = \frac{Y(z)}{X(z)} \;\isdef \; \hbox{transfer function}.$$

This shows that, as a direct result of the convolution theorem, the z transform of an impulse response $h(n)$ is equal to the transfer function $H(z)=Y(z)/X(z)$ of the filter, provided the filter is linear and time invariant.