# Why is $\pi$ in the domain of filters?

I'm just beginning to learn (FIR) filter design and I was wondering, why is $\pi$ involved in the domain of such filters?

I assume it's related to some trigonometric functions, but why do I see e.g. cutoff frequency being related to $\pi$?

You can see for example here that the variable $\omega \in [-\pi, \pi]$ in function $z=e^{i \omega}$ of the transfer function of a FIR filter:

https://en.wikipedia.org/wiki/Finite_impulse_response#Transfer_function

• What do you mean by a 'filter's domain'? Also: look up angular frequency. – MBaz Jun 2 '16 at 19:34
• – mavavilj Jun 2 '16 at 19:44
• Because $e^{i \pi} = -1$ ? – hotpaw2 Jun 2 '16 at 20:11
• yeah, as @MBaz sez, look up angular frequency somewhere. the variable $\omega \in [-\pi, \pi]$ in function $z=e^{i \omega}$ is the number of radians per sample of a discrete-time sinusoid. why are angles in radians? because calculus likes it (radians are the natural unit of an angle). why are sinusoids used in signal processing? because, like exponentials, sinusoids are what we call "eigenfunctions" for Linear Time-Invariant (LTI) systems. – robert bristow-johnson Jun 3 '16 at 5:58
• i like the variant of Euler's formula: $e^{i\pi}+1=0$ it's a relationship that relates the five most important numbers using the operations of multiplication, addition, exponentiation, and equality. all packed into a very tight little expression. – robert bristow-johnson Jun 3 '16 at 6:00

In discrete time, distinct frequencies only have sense in a range of $2\pi$ (where the units are radians per sample).

A sinusoid (or complex exponential) of frequency $2\pi$ is indistinguishable from one with frequency 0. If in doubt, evaluate $x[n] = \cos(2\pi \times n)$ and $x[n] = \cos(0 \times n)$. Same thing with any two frequencies separated $2\pi$ or multiples of $2\pi$.

For example, you can easily show that $\cos( \theta_0 \times n ) = \cos((\theta_0 + 2k\pi) \times n)$ for any integer $k$.

So a range of size $2\pi$ is enough not to repeat frequencies.

In general, the range chosen is $-\pi \cdots \pi$, where $\pi$ is the maximum frequency ( $\cos(\pi \times n ) = (-1)^n$ ).

In the $z$ domain, the frequency response is the transfer function ($H(z)$) evaluated on the unit circle ($z = e^{j\theta}$), so here again you have the range $2\pi$ before repeating your path.

• Is this based on Fourier theory perhaps? I don't know any other body of theory that limits itself to basic trigonometric functions and concerns frequencies using them. – mavavilj Jun 2 '16 at 20:13

A FIR filter is a convolution. Fourier theory relates convolution in one domain to a linear multiplication (by a frequency response for example) in the other domain. Two typical domains, so related by the Fourier transform (or the FFT for vectors of samples), are the time domain and the sinusoidal frequency spectrum domain. Once you get to sinusoids, you get to pi.