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 💡