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Trying to prove something about the integral of the phase of an FIR from, does this hold true at all? Or maybe for some?

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    $\begingroup$ yes, if the impluse response is real, then the phase is an odd-symmetry function. $\endgroup$ – robert bristow-johnson May 4 '17 at 16:56
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To expand on RBJ's comment from above, your statement is true iff the impulse response of the filter is real. Recall that the frequency response of a filter is the discrete time Fourier transform of its impulse response. That is:

$$ X(\omega) = \sum_{n=-\infty}^{\infty} x[n] e^{j\omega n} $$

Look at the relationship between $X(\omega)$ and $X(-\omega)$:

$$ X(-\omega) = \sum_{n=-\infty}^{\infty} x[n] e^{-j\omega n} $$ $$ X(-\omega) = \sum_{n=-\infty}^{\infty} x[n] \left(e^{j\omega n}\right)^* $$

If $x[n]$ is real, then we can absorb it into the conjugate:

$$ X(-\omega) = \sum_{n=-\infty}^{\infty} \left(x[n] e^{j\omega n}\right)^* $$

And since conjugation is a linear operation, we can pull it around the sum as well:

$$ X(-\omega) = \left(\sum_{n=-\infty}^{\infty} x[n] e^{j\omega n}\right)^* $$ $$ X(-\omega) = \left(X(\omega)\right)^* $$

So $X(\omega)$ exhibits Hermitian symmetry; its phase is oddly symmetric about $\omega = 0$, while its magnitude is evenly symmetric.

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  • $\begingroup$ the only thing i would do is change $X(\omega)$ to $X(e^{j \omega})$ and keep the DTFT pretty much compatible with the ZT. $\endgroup$ – robert bristow-johnson May 4 '17 at 18:01
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I would add this is not "so true", unless you are more precise about how you define the phase, not uniquely defined in general, up to $2\pi$. However, if you consider the principal value of the phase in $(-\pi,\pi]$ interval, or a periodization of $\mathbb{R}$ considered as a torus (identifying $-\pi$ to $\pi$), then you can derive the result in an intuitive way, using known properties for real signals:

  • the real part $\mathcal{R} $ of the Fourier transform $X$ of $x$ is even,
  • the imaginary part $\mathcal{I} $ of its Fourier transform is odd.

Hence $\mathcal{I}/\mathcal{R} $ is odd, and since the $\arctan$ function is odd, then :

$$ \operatorname{Arg}(X) = \operatorname{atan2}\Big(\mathcal{I}(X),\, \mathcal{R}(X)\Big) = \begin{cases} \arctan\left(\frac{\mathcal{I}(X)}{\mathcal{R}(X)}\right) &\text{if } \mathcal{R}(X) > 0, \\ \frac{\pi}{2} - \arctan\left(\frac{\mathcal{R}(X)}{\mathcal{I}(X)}\right) &\text{if } \mathcal{I}(X) > 0, \\ -\frac{\pi}{2} - \arctan\left(\frac{\mathcal{R}(X)}{\mathcal{I}(X)}\right) &\text{if } \mathcal{I}(X) < 0, \\ \arctan\left(\frac{\mathcal{I}(X)}{\mathcal{R}(X)}\right) + \pi &\text{if } \mathcal{R}(X) < 0, \ \mathcal{I}(X) \ge 0 \\ \arctan\left(\frac{\mathcal{I}(X)}{\mathcal{R}(X)}\right) - \pi &\text{if } \mathcal{R}(X) < 0, \ \mathcal{I}(X) < 0 \\ \text{undefined} &\text{if } \mathcal{R}(X) = 0, \ \mathcal{I}(X) = 0 \end{cases} $$

is odd.

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    $\begingroup$ if you're gonna be precise about phase, Laurent, you need to fix your $\operatorname{Arg}(X)$ equation. i might do it, but i hesitate to modify your answer without consent. $\endgroup$ – robert bristow-johnson May 6 '17 at 20:51
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    $\begingroup$ I trust your input on that one $\endgroup$ – Laurent Duval May 6 '17 at 21:05

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