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Does anybody know how to derive this phase function for that all-pass filter structure? Many papers re-print them, and corresponding chart as well.

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But nobody mentions that at normalized frequency 0.25, tan(pi/2) goes to +- infinity, and atan jumps from + to - pi/2 :

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Is phase shift periodic here by pi, so we can move the right curve down to glue them at 0.25. What does that mean in real filter operation?

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fixed picture for the comment

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First of all, that formula for the phase has a sign error, at least if $\alpha$ is defined as in the diagram of the filter structure. Second, your phase plot shows a jump from $-\pi$ to $\pi$, which is no jump at all because a phase of $\pi$ is the same as a phase of $-\pi$. The jump occurs only because the arctangent function returns the principal value of the phase in the interval $[-\pi,\pi]$. You can get rid of the jump by using the phase unwrapping function unwrap in Matlab/Octave. But note that this is just a cosmetic thing, there is no phase jump in the physical sense of the word.

Now for the derivation of the formula for the phase. From the given filter structure we get the following equation for the $\mathcal{Z}$-transforms of the input and output signals:

$$Y(z)=X(z)z^{-2}+\alpha\left(X(z)-Y(z)z^{-2}\right)\tag{1}$$

From this the transfer function can be computed as

$$H(z)=\frac{Y(z)}{X(z)}=\frac{\alpha+z^{-2}}{1+\alpha z^{-2}}=\frac{\alpha z+z^{-1}}{z+\alpha z^{-1}}=\frac{B(z)}{B\left(\frac{1}{z}\right)}\tag{2}$$

with $B(z)=\alpha z+z^{-1}$. $H(z)$ as given by $(2)$ is indeed an allpass transfer function with $|H(z)|=1$ for $|z|=1$. Note that the system described by $(2)$ is only stable for $-1<\alpha <1$.

On the unit circle $z=e^{j\omega}$, $H(z)$ can be written as

$$H(e^{j\omega})=\frac{B(e^{j\omega})}{B^*(e^{j\omega})}\tag{3}$$

where $^*$ denotes complex conjugation. Consequently, the phase of $H(e^{j\omega})$ is given by

$$\phi_H(\omega)=2\phi_B(\omega)\tag{4}$$

where $\phi_B(\omega)$ is the phase of the numerator $B(e^{j\omega})$:

$$\begin{align}\phi_B(\omega)&=\arg\big\{\alpha e^{j\omega}+e^{-j\omega}\big\}\\&=\arg\big\{(1+\alpha)\cos(\omega)-j(1-\alpha)\sin(\omega)\big\}\\&=-\arctan\left(\frac{1-\alpha}{1+\alpha}\tan(\omega)\right)\end{align}\tag{5}$$

The phase of $H(e^{j\omega})$ is thus given by

$$\phi_H(\omega)=-2\arctan\left(\frac{1-\alpha}{1+\alpha}\tan(\omega)\right)\tag{6}$$

which differs from the given formula in the sign of $\alpha$.

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  • $\begingroup$ unwrap helps! If I want to find frequencies of exact pi/2 phase shift, I have to compare with a single element FIR Z-1, which means in my case add 2*x in MatLab, as on the picture. Why its phase shift is linear, is any simple math which shows that? $\endgroup$ – K-man Nov 30 '18 at 18:12
  • $\begingroup$ @K-man: I'm not sure what you're trying to do, and what this has to do with your original question. $\endgroup$ – Matt L. Nov 30 '18 at 19:39
  • $\begingroup$ Two paths: one has the bi-quadratic filter which you described, another path contains only one delay element z-1. Phase difference between them will be close to pi/2, in some frequency range at proper α value. It will be exactly pi/2 on three frequencies, as matlab picture shows. I added ' + 2* x ' in formula, without real understanding why the phase of delay element is linear with frequency. Is any simple math, explaining this? $\endgroup$ – K-man Nov 30 '18 at 19:52
  • $\begingroup$ @K-man: A delay is just $z^{-1}$, so for $z=e^{j\omega}$ you get $e^{-j\omega}$, the phase of which is just $-\omega$. Subtracting this from the all-pass phase means adding $\omega$ (not $2\omega$!) to the all-pass phase. If they added $2\omega$, then they use a 2-sample delay $z^{-2}$. $\endgroup$ – Matt L. Nov 30 '18 at 21:06
  • $\begingroup$ Thank you, will get into math to figure out this ω or 2ω issue. If I add one ω, phase curve equals pi/2 only at one frequency: 0.25... $\endgroup$ – K-man Nov 30 '18 at 21:43

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