The z-Transform

The z-transform is the Discrete-Time counterpart of the Laplace Transform.

$$\begin{eqnarray*} Laplace & : & F(s) = \int_{-\infty}^\infty f(t)e^{-st} dt \\ Z & : & F(z) = \sum_{n=-\infty}^{\infty} f[n] z^{-n} \end{eqnarray*}$$

It is

• Used in Digital Signal Processing
• Used to Define Frequency Response of Discrete-Time System.
• Used to Solve Difference Equations - use algebraic methods as we did for differential equations with Laplace Transforms; it is easier to solve the transformed equations since they are algebraic.

We will see that

1. Lines on the s-plane map to circles on the z-plane.

2. Role of $j \omega$-axis is replaced by unit circle, so
   1. The DT Fourier Transform exists for a signal if the ROC includes the unit circle.

   2. A stable system must have an ROC that contains the unit circle.
   3. A causal and stable system must have poles inside the unit circle.

Aside: You can relate the z- transform and Laplace transform directly when you are dealing with sampled signals:

Take a CT signal $f(t)$ and sample it:

$$f_s(t) = f(t) \sum_{n=-\infty}^{\infty} \delta(t-nT) = \sum_{n=- \infty}^{\infty} f(nT) \delta(t-nT)$$

The Laplace transform of the sampled signal is

$$\begin{eqnarray*} {\cal L}[{f_s(t)}] & = & \int_{-\infty}^{\infty} \left[ \sum... ...)e^{-st} dt\\ & = & \sum_{n=-\infty}^{\infty} f(nT) e^{-snT} \end{eqnarray*}$$

by the sifting property.

Let $f[n] = f(nT)$ and $z = e^{sT}$, then

$$\begin{eqnarray*} F(z) & = & \sum_{n=-\infty}^{\infty} f[n] z^{-n} \\ F(z)\ve... ...n=-\infty}^{\infty} f(nT) e^{-snT} \\ & = & {\cal L}[f_s(t)] \end{eqnarray*}$$

Thus, the z- transform with $z = e^{sT}$ is the same as the Laplace transform of a sampled signal! Of course, if the signal is already discrete, the notion of sampling is unnecessary for understanding and using the z- transform.