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I have the measurements of the group delay of a filter and the period of the signal is always smaller than the delay introduced and thus phase (frequency) wrapping is also introduced. Do you know any particular procedure followed for compensating for that phase wrapping?

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  • \$\begingroup\$ Do you know (at least roughly) the phase delay? Or is that what you're trying to measure? Is the delay really always more than the period (even at the lowest frequency you're measuring)? Can you extend your measurement frequency down to where it wouldn't be true? \$\endgroup\$ – The Photon Aug 25 '12 at 18:05
  • \$\begingroup\$ I thought and please correct me if i am wrong that the phase delay could be found by subtracting the group delay from the signal's period and converting the result in degrees.The group delay in the Frequency band that i am concerned is always more than the period of the signal like it is in the majority of the samples taken by the VNA.I am looking for a way to unwrapp the phase. \$\endgroup\$ – Giwrgos Rizeakos Aug 25 '12 at 18:34
  • \$\begingroup\$ What measurement instrument are you using? The way I know to get the group delay is to measure phase vs frequency, and take the derivative. In this case of course you'd already have the phase, so you wouldn't need to calculate it from group delay. \$\endgroup\$ – The Photon Aug 25 '12 at 18:40
  • \$\begingroup\$ If you subtract a delay from a period, you get something with units of time. How would you convert that to degrees? \$\endgroup\$ – The Photon Aug 25 '12 at 19:08
  • \$\begingroup\$ 320 degrees is one period,so you can estimate in which part of it you are. \$\endgroup\$ – Giwrgos Rizeakos Aug 26 '12 at 8:02
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The relationship between group delay and phase is

\$t_g(\omega) = -\dfrac{d\phi}{d\omega}\$.

So to get phase from group delay you don't subtract, you integrate. Since these are all real numbers (not complex), you can do this integration numerically by any standard method, such as trapezoidal or Simpson's rule. If you have access to a numerical math package like Mathematica, Matlab, or Octave, it will have a built in routine to do this integration.

Once you do this integration numerically, there's nothing that says the phase you extract will be banded between 0 and 2\$\pi\$. So there's no reason you should have to unwrap the phase.

If you do the measurement the other way around (that is, measure phase and then calculate phase delay and group delay), then you typically measure phase between 0 and 2\$\pi\$ and you have to unwrap it to obtain a group delay without glitches where the raw phase measurement is discontinuous.

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  • \$\begingroup\$ I have the measurements of the group delay so i assumed that whenever the group delay is larger than the signals period that would introduce phase wrapping.Because the signal would be delayed for more than one period and so the phase would 'go back' in a initial position.Moreover, i dont understand,i have my data (group delay measurements in a numerical form eg 1.6*10^-8 secs) will i calculate the phase by trapezoidal or Simpson's rule?Thank you for your reply \$\endgroup\$ – Giwrgos Rizeakos Aug 26 '12 at 8:09
  • \$\begingroup\$ "will I calculate the phase by trapezoidal or Simpson's rule?" --- I already answered this: "you can do this integration numerically by any standard method." Writing an entire book chapter on numerical integration methods is beyond what I'm willing (or qualified) to do for this answer. Have a look at a book like Numerical Recipes if you want to write the code yourself. Or just use the built-in methods from your numerical math package like I also already suggested. \$\endgroup\$ – The Photon Aug 26 '12 at 16:53
  • \$\begingroup\$ "the signal would be delayed for more than one period and so the phase would 'go back' in a initial position." --- not in the case you're discussing. If you integrate the group delay, the resulting phase will simply keep increasing from 0 to 2*pi to 4*pi and so on. If you want to wrap (not unwrap) these values back into the range [0, 2*pi], just divide by 2*pi and keep the remainder. \$\endgroup\$ – The Photon Aug 26 '12 at 16:58

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