Z-Transform

The discrete-time counterpart to the Laplace transform — turning a sequence of samples into a function of a complex variable, and the foundation of digital filter design.

A digital filter is stable exactly when every pole of its Z-transform sits inside the unit circle

Unit circle

Frequencies live on z = e^jω; the circle is the stability boundary.

Inside

Pole magnitude below 1 means the natural response decays.

Outside

Pole magnitude above 1 means repeated samples grow.

Definition

The Z-transform is the discrete-time counterpart to the Laplace transform: where Laplace turns a continuous signal $f(t)$ into a function $F(s)$ , the Z-transform turns a discrete sequence of samples $x[n]$ into a function of a complex variable $z$ :

$X(z) = \sum_{n=-\infty}^{\infty} x[n]\,z^{-n}$

It's the natural tool for analyzing digital filters and discrete-time systems — anything working on sampled data (audio processed by a computer, sensor readings, a stock price's daily closes) rather than a continuous function of time.

Transform of a simple sequence

For the sequence $x[n] = 1$ for $n \geq 0$ and $0$ otherwise (the discrete "step" signal):

$X(z) = \sum_{n=0}^\infty z^{-n} = \frac{1}{1-z^{-1}} = \frac{z}{z-1}$

(a geometric series, convergent for $|z| > 1$ ) — directly analogous to how the continuous unit step has Laplace transform $1/s$ .

Try it

Why does computing the Z-transform of an infinite sequence sometimes reduce to summing a geometric series?

Solution

Many useful sequences (like the decaying exponential $x[n] = a^n$ for $n\geq 0$ ) have a constant ratio between consecutive terms — exactly the defining property of a geometric series. Multiplying by $z^{-n}$ in the Z-transform sum preserves this geometric structure, so the transform often reduces to the same well-known closed-form sum used for ordinary geometric series, just with $az^{-1}$ playing the role of the common ratio.

Related concepts

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Series & Convergence

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