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?


3 Answers 3


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 $a$.

  • 1
    $\begingroup$ yes, that's the explanation. $\endgroup$ Commented Jan 25, 2017 at 10:10
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    $\begingroup$ 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. $\endgroup$
    – Matt L.
    Commented Jan 25, 2017 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.

  • $\begingroup$ A very nice reference indeed. Thank you for putting it in light $\endgroup$ Commented Jan 27, 2017 at 11:05
  • $\begingroup$ @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. :-) $\endgroup$
    – Peter K.
    Commented Jan 27, 2017 at 13:35

As pointed out in Peter K.'s answer, it is not necessary that a discrete-time system with linear phase has any symmetry.

CAJ's argument goes in the right direction, but note that $e^{-jk\Omega}$ is only a simple discrete sample delay if $k \in \mathbb{Z}$. More importantly, $e^{-jk\Omega}$ doesn't always preserve the symmetry of the input signal.

One way to implement $e^{-jk\Omega}$ is to convert the discrete time input to continuous time with sinc interpolation, delay the equivalent time of $k$ samples, and resample the signal.

Here is an example of the signal $[...,\ 0,\ 1,\ 0,\ 1,\ 0,\ ...]$ "delayed" with different values for $k$. The first graph is the original signal, followed by the output of delays with $k=1$, $k=1.5$ and $k=1.75$ samples. The dashed lines are the sinc interpolations of each signal.

Delays examples

Notice that when $2k \notin \mathbb{Z}$, as in the last graph, the symmetry of the input signal is lost, so CAJ's argument is only correct for $2k \in \mathbb{Z}$.

Although the symmetry of the impulse response of a linear phase discrete time filter is not guaranteed, the symmetry of the signal's continuous sinc interpolation is guaranteed.

One example of a filter that has linear phase and no symmetry is precisely $e^{-jk\Omega}$ for any $2k\notin \mathbb{Z}$. It is interesting to note however that this frequency response is not a rational function on $z=e^{j\Omega}$, so it is not implementable with the typical IIR diference equation method.

It turns out that in the cases where the impulse response is not necessarily symmetric, so when $2k \notin \mathbb{Z}$, the impulse response is necessarily infinite in length. This means that for FIR filters, since it is necessary that $2k \in \mathbb{Z}$, symmetry is preserved and so CAJ's argument applies.

If you want to see some mathematical arguments for the claims in this answer regarding real FIR filters for your own study, you can see mine here. Keep in mind I am not an expert, so it may contain errors. If you find any, please let me know.

An important detail about the language is that when people say linear phase, they often don't mean strictly linear phase, as in $$ X(j\Omega) = |X(j\Omega)|e^{-jk\Omega}.$$

They usually mean generalised linear phase, as in $$ X(j\Omega) = A(j\Omega)e^{j(\beta - k\Omega)},$$ where $A(j\Omega)$ is real valued.

With generalised linear phase, if the filter response is real and finite, then it can be either symmetric or anti-symmetric.

As Matt L. points out in a comment, for further details on the different types of generalised linear phase FIR filters, you can read this question.


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