We introduced the
$z$ -transform before as
$$H\left(z\right)=\stackrel{\infty}{\sum _{k=-\infty}}h\left[k\right]{z}^{-k}$$
where
$z$ is a complex number. When
$H\left(z\right)$ exists (the sum converges), it can be
interpreted as the “response” of an LSI system with impulse response
$h\left[n\right]$ to the input of
${z}^{n}$ . The
$z$ -transform is useful mostly due to its ability
to simplify system analysis via the following result.
Theorem
If
$y=h*x$ , then
$Y\left(z\right)=H\left(z\right)X\left(z\right)$ .
Proof
First observe that
$$\begin{array}{cc}\hfill \stackrel{\infty}{\sum _{n=-\infty}}y\left[n\right]{z}^{-n}& =\stackrel{\infty}{\sum _{n=-\infty}}\left(\stackrel{\infty}{\sum _{k=-\infty}},x,\left[k\right],h,[n-k]\right){z}^{-n}\hfill \\ & =\stackrel{\infty}{\sum _{k=-\infty}}x\left[k\right]\left(\stackrel{\infty}{\sum _{n=-\infty}},h,[n-k],{z}^{-n}\right)\hfill \end{array}$$
Let
$m=n-k$ , and note that
${z}^{-n}={z}^{-m}\xb7{z}^{-k}$ . Thus we have
$$\begin{array}{cc}\hfill \stackrel{\infty}{\sum _{n=-\infty}}y\left[n\right]{z}^{-n}& =\stackrel{\infty}{\sum _{k=-\infty}}x\left[k\right]\left(\stackrel{\infty}{\sum _{n=-\infty}},h,\left[m\right],{z}^{-m}\right){z}^{-k}\hfill \\ & =\stackrel{\infty}{\sum _{k=-\infty}}x\left[k\right]H\left(z\right){z}^{-k}\hfill \\ & =H\left(z\right)\left(\stackrel{\infty}{\sum _{k=-\infty}},x,\left[k\right],{z}^{-k}\right)\hfill \\ & =H\left(z\right)X\left(z\right)\hfill \end{array}$$
This yields the “transfer function”
$$H\left(z\right)=\frac{Y\left(z\right)}{X\left(z\right)}.$$