Transfer functions concisely describe the frequency response of a given filter.

Filters are linear time-invariant (LTI) systems.

Write $f(x)$ for the output of $x$. A system is linear if $f(x+x\prime )=f(x)+f(x\prime )$ and $f(ax)=af(x)$. Below we shall use the notation $x\to y$ to mean $y$ is the output of the system when the input is $x$, i.e. $y=f(x)$.

Time-invariance means that the behaviour of the system is independent of the current time; it only depends on the previous inputs.

LTI systems can be continuous (where the input is a function of time) or discrete (where the input is a sequence of values). Analog filters are continuous, whilst digital filters are discrete.

Write $x[n]$ for the $n$th input and $y[n]$ for the corresponding output.

Suppose the discrete input signal is $x[n]=z^n$ for some complex number $z$.

By linearity, $x[n]=z^n=zx[n-1]\to zy[n-1]$. Hence by induction, $y[n]=H{z}^n$ for some $H$ that depends on $z$.

Since $H$ depends on $z$, we write $H=H(z)$. It is called the transfer function. If the output is real when the input is real (which is always the case for audio filters), then it is easy to show that $H(z^{*})=H(z{)}^{*}$ (see the $z$ transform).

Points on the unit circle of $H(z)$ determine the filter’s effect on sinusoids: for all $a$, $\mid H(e^{\mathrm{ia}})\mid$ is the amount a sinusoid of the form $\cos (an+b)$ is amplified after passing through the filter (i.e. $\mid H(e^{\mathrm{ia}})\mid$ is the filter’s frequency response). As any periodic signal can be represented as a sum of sinusoids, $H(z)$ determines the effect of a filter on any periodic signal.

Proof: Suppose the input signal is

Then by linearity, and using the above observations,

Letting $H(e^{ia})={\mathrm{re}}^{i\theta }$, we see

For a low-pass filter we want $\mid H(e^{i\theta })\mid \approx 1$ when $\theta \approx 0$, so that low frequencies pass through unchanged, and $\mid H(e^{i\theta })\mid \approx 0$ when $\theta \approx \pi$, in order to cutoff higher frequencies.

Note that the phase response is determined by $\angle H(e^{ia})$.

Recursive filters, also known as Infinite Impulse Response (IIR) filters, are filters where the output sample is a linear combination of some number of previous inputs and outputs. In other words,

We have written the minus signs of the $b_i$ to follow most texts (it is more convenient when discussing the transfer function), but authors interchange $a_i$ and $b_i$. It can be shown that a digital filter is LTI and causal (outputs depend only on previous inputs and outputs) if and only if it can be described by the above equation.

First, consider the effect of this filter on an input of the form $x[n]=z^n$ for some $z$. Since it is an LTI system that we have $y[n]=z^{i}y[n-i]$. Hence

This means the transfer function must be

So to design a digital filter, we need only pick a transfer function with certain zeroes and poles such that $H(z)$ behaves in a certain way on the unit circle.

A common method is to design an analog filter first, and then transform it into a digital filter.

An digital impulse is the sequence $1,0,0,...$. A filter that only depends on previous inputs is called a Finite Impulse Response filter because if we feed an impulse to such a filter, after some time the output signal will be a sequence of zeroes.

If we allow the filter to involve previous outputs, then it is possible for the impulse signal to cause a response that never dies down, hence the term infinite impulse response.

Consider an analog filter, whose input at time $t$ is $x(t)$, and has corresponding output $y(t)$.

Suppose the input is the signal $x(t)=e^{st}$. Then by linearity,

Letting $d$ approach zero shows that $x\prime (t)\to y\prime (t)$. But since $x\prime (t)=s{e}^{st}$, we have

which means $y(t)=H{e}^{st}$ for some $H$ depending on $s$, so write $H=H(s)$. (Proof: $dy/dt=sy$ implies $dt/dy=1/sy$ which implies $t=(\ln y)/s+C$, for some constant $C$, in other words $y=H{e}^{st}$ for some constant $H$.)

The function $H(s)$ is the transfer function for the analog filter. As in the digital case, consider what happens to the input

If we let $H(ia)={\mathrm{re}}^{i\theta }$ then the output is $y(t)=r\cos (at+b+\theta )$, so we see that the effect of the filter on sinusoids is determined by the values of $H(s)$ on the imaginary axis (for example $H(2\pi i)$ determines the behaviour of 1Hz sinusoids).

For a low-pass analog filter, we want $\mid H(s)\mid \approx 1$ on the imaginary axis near the origin, and near zero as $s$ approaches infinity, so that $\cos (at)$ is cut off for high $a$ but let through unchanged for low $a$.

A filter is stable if the output tends to zero if the input becomes zero, otherwise it is unstable. It turns out a digital filter is stable if and only if the poles of its transfer function lie within the unit circle. Similarly an analog filter is stable if and only if the poles of its transfer function lie on the left side of the imaginary axis.