# The z-Transform

Let $$\delta(n)$$ be a digital impulse signal, that is, $$\delta(0) = 1$$ and is zero everywhere else. By scaling, shfting and summing $$\delta(n)$$ functions, we can construct any input signal. LTI filters preserve scaling, shifting and summing, thus the behaviour of such a filter on $$\delta(n)$$ determines the behaviour of the filter on any input signal.

We can easily compute the transfer function of a filter by looking at the output of the filter given $$\delta(n)$$.

Given a function $$x(n)$$ defined for all $$n \in \mathbb{Z}$$, define the $$z$$-transform of $$x$$ by

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

Let $$h(n)$$ be the output of an LTI filter applied to $$\delta(n)$$. Define the transfer function $$H(z)$$ to be the $$z$$-transform of $$h(n)$$.

Then by linearity and time-invariance, for any input signal $$x(n)$$ and corresponding output signal $$y(n)$$, we have $$Y(z) = H(z)X(z)$$ where $$X(z), Y(z)$$ are the $$z$$-transforms of $$x(n), y(n)$$ respectively. Note also $$y(n) = h(n) * x(n)$$, where $$*$$ denotes convolution.

Ben Lynn blynn@cs.stanford.edu 💡