# 2.3 The z-transform

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## The $z$ -transform

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

$\begin{array}{cc}\hfill \stackrel{\infty }{\sum _{n=-\infty }}y\left[n\right]{z}^{-n}& =\stackrel{\infty }{\sum _{n=-\infty }}\left(\stackrel{\infty }{\sum _{k=-\infty }},x,\left[k\right],h,\left[n-k\right]\right){z}^{-n}\hfill \\ & =\stackrel{\infty }{\sum _{k=-\infty }}x\left[k\right]\left(\stackrel{\infty }{\sum _{n=-\infty }},h,\left[n-k\right],{z}^{-n}\right)\hfill \end{array}$

Let $m=n-k$ , and note that ${z}^{-n}={z}^{-m}·{z}^{-k}$ . 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”

$H\left(z\right)=\frac{Y\left(z\right)}{X\left(z\right)}.$

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