# Why do linear phase filters have symmetric impulse responses?

It was given as a fact that linear phase filters have symmetric impulse responses, but I don't see why that has to be true. Can somebody please explain or prove this?

Actually, I think I see why.

$$X(j\Omega) = |X(j\Omega)|e^{-j\theta(\Omega)}$$

$|X(j\Omega)|$ is purely real, and therefore if we take the IFT it is even and symmetric.

$\theta(\Omega)= a\Omega$ since the phase is linear, so $e^{-ja\Omega}$ merely shifts the corresponding even and symmetric magnitude in the time domain, so the resulting impulse response is symmetric about $a$.

• yes, that's the explanation. – Maximilian Matthé Jan 25 '17 at 10:10
• Also take a look at this question and its answers. There are 4 types of linear phase FIR filters. You can also have an anti-symmetric impulse response (giving you an additional phase shift of $\pi/2$), and you have even length linear phase filters, where the symmetry is not about an integer sample number, but about a point in between two sample points. – Matt L. Jan 25 '17 at 12:47

Actually, it's not required, depending on how you define "filter". Clements and Pease derived an unimplementable filter that is linear phase, but does not have any symmetry.

The filters are useless as they are not able to be implemented, but it is an interesting thought problem.

• A very nice reference indeed. Thank you for putting it in light – Laurent Duval Jan 27 '17 at 11:05
• @LaurentDuval Yes, I liked it when I first came across it. It's not particularly useful for implementation purposes, but it does stretch one's brain a bit to see how they did it. :-) – Peter K. Jan 27 '17 at 13:35