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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.

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Why do you believe 4 is incorrect? –  Deve Feb 3 '13 at 9:07
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1 Answer 1

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 $-2w$

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+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
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