Lesson 12

The z-Transform

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

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

It is

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) \end{displaymath}

The Laplace transform of the sampled signal is

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

by the sifting property.

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

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

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.