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How can I find this all pass filter's phase?

$$H(e^{jw}) =\frac{−a* + e^{-jw}}{1 - a(e^{-jw})}$$ where $a = ρe^{jϴ}$ is the pole.

Then I want to find group delay, I know that the group delay is the derivative of the phase with respect to w but i don't know how to find the phase

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phase of a filter is the angle (or "argument") of the complex number that is $H(e^{j\omega})$ at each value of $\omega$ . you may need to "unwrap" the phase by adding or subtracting integer multiples of $2 \pi$ .

BTW, to post math to these pages, surround the $\LaTeX$ expression with single dollar signs for inline math and double dollar signs for math expressions on their own line. if you need a quick guide for LaTeX, my suggestion is the one that lives at Wikipedia: https://en.wikipedia.org/wiki/Wikipedia:Math . i'm still learning how to do other markup here.

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  • $\begingroup$ Here is one of the best reference manuals for MathJax on SE: MathJax basic tutorial, although wiki is very detailed! $\endgroup$
    – jojeck
    Jun 19, 2014 at 7:07
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You can compute the phase of the function by taking inverse tangent of imaginary part divided by the real. Check out this link on how to compute the real and imaginary parts. The real part is computed by taking 1/2*(z + conj(z)) and the imaginary follows a similar process. Once you have those just divide them and take the inverse tangent.

Obtaining the group delay would require taking the derivative of that result. Feel free to use Wolfram Alpha to check your work.

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    $\begingroup$ there is a little more to computing phase than $\phi = \arctan \left( \frac{\Im\{H(e^{j\omega})\}}{\Re\{H(e^{j\omega})\}} \right)$. you have to get it into the correct quadrant. without that, your angle could be off by $\pi$ or 180°. $\endgroup$ Dec 29, 2013 at 4:35

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