Signal Processing Stack Exchange is a question and answer site for practitioners of the art and science of signal, image and video processing. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Say I have an LTI system whose impulse response h[n] is:

$$ h[n] = \delta[n] + \delta[n-4] $$

I want to find the group delay of this system. I know the group delay ($grd$) is defined as: $$ grd = \frac{-d}{dw}(\angle H(e^{jw})) $$

So I get $\angle H(e^{jw}) = -4w$ and the $grd = 4$ but I believe this is incorrect.

For a more complicated example:

$$ h[n] = -\delta[n+1] + \delta[n] + 2*\delta[n-1] + 2*\delta[n-2] + \delta[n-3] - \delta[n-4] $$

What would my $\angle H(e^{jw})$ be? Thanks for helping me understand something fundamental that I am missing about group delay.

share|improve this question
Why do you believe 4 is incorrect? – Deve Feb 3 '13 at 9:07
up vote 5 down vote accepted

The z domain transfer function of the system is the z transform of the system impulse response, so start by taking the Z transform of h[n] ...

$$H[z] = -z^1 + 1 + 2z^{-1} + 2z^{-2} + z^{-3} -z^{-4}$$

You may be able to message this into a nicer form, but that isn't necessary.

Next, to get the the frequency response, replace z with $e^{jw}$

So this yields $-e^{jw} + 1 + 2e^{-jw} + 2e^{-2jw} + e^{-3jw} -e^{-4jw}$ which is a complex function with both a phase and a magnitude.

Find the phase of this expression and you are done. I don't have time to wrestle this into Magnitude*phase form, so I'll go to your first example which is a simple case.

$$H[z] = 1 + z^{-4} $$ or $$H[z] = z^{-2}(z^2 + z^{-2}) $$ then you have $$H[e^{jw}] = e^{-j2w}(e^{j2w} + e^{-j2w}) = 2cos(2w)e^{-j2w} $$

Your phase is the argument of the complex exponential $e^{-j2w}$, so...

$\angle H(e^{jw})$ is the expression $-2w$

share|improve this answer
+1 but the phase is just $-2\omega$ – Deve Feb 4 '13 at 7:55
Good point. The part you differentiat to get group delay is the argument of the complex exponential function exp[j(arg)] that represents the phase of the characteristic function, so arg = -2w. – user2718 Feb 4 '13 at 13:32
@Deve and others, thanks for the clarification! – user2718 Feb 4 '13 at 14:37

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.