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Yet another group delay vs. phase delay question! Though this is a question that's been asked several times I don't feel like it's been fully discussed so I'll post an example that I can't quite seem to wrap my head around.

Assume some system whose phase response is shown below;

enter image description here

where the phase is piecewise affine.

For frequencies 0 < f < 0.3 the phase is purely linear and, as is expected, the phase and group delay are identical.

For frequencies 0.35 < f < 0.55 the phase is linear (excluding the discontinuous jump) and, again, in this range the phase and group delay are identical.

However, for f > 0.6 things get confusing for me. Here, the phase is affine (i.e. a straight line that doesn't necessarily pass through the origin). Here, the group delay is constant (as is obvious by its definition) but the phase delay varies with f. As the group delay is constant I would expect a band-limited signal, say in the range 0.7 < f < 0.9, to exhibit no dispersion, i.e. the output from the system is a (possibly frequency varying scaled) time-shifted version of the input. However, if we look at the phase delay it seems like the output should in fact show dispersion (as for some sinusoidal input sin(0.7*t) the delay is on the order of 2.1, while for some sinusoidal input sin(0.9*t) the delay is on the order of 2.5).

If anyone has a way of explaining this apparent paradox I'd really appreciate it!

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Note that a constant group delay is not sufficient for a band-limited signal to exhibit no dispersion. It is the phase delay that needs to be constant. If the phase is affine, i.e., if we have

$$\phi(\omega)=a+b\omega,\qquad \omega>0\tag{1}$$

the group delay is constant

$$\tau_g(\omega)=-\frac{d\phi(\omega)}{d\omega}=-b\tag{2}$$

but the phase delay is not constant if $a\neq 0$:

$$\tau_p(\omega)=-\frac{\phi(\omega)}{\omega}=-\frac{a}{\omega}-b\tag{3}$$

And it is the phase delay that determines the delay of a sinusoid at the output of an LTI system. For $x(t)=\sin(\omega_0t)$, the output of an LTI system with magnitude response $M(\omega)$ and phase delay $\tau_p(\omega)$ is given by

$$y(t)=M(\omega_0)\sin(\omega_0(t-\tau_p(\omega_0)))\tag{4}$$

Only for a (strictly) linear phase (i.e., for $a=0$ in Eq. $(1)$) does the phase delay equal the group delay. In that case, the delay experienced by a sinusoid at the input of an LTI system becomes independent of frequency.

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  • $\begingroup$ and remember, with phase delay, you gotta properly unwrap phase. with group delay, if you do things correctly, phase wrapping is not an issue. if you do things wrong, you will get some dirac impulses in the group delay. $\endgroup$
    – robert bristow-johnson
    Dec 1, 2016 at 18:29
  • $\begingroup$ Thanks for the reply! All of what you're saying makes sense, but seems to clash with what I've previously learned. I was under the assumption that a constant group delay in some range was a sufficient condition for ensuring that there is no dispersion, but obviously (as you just derived) that's not the case. Is there some intuitive explanation as to how the system will act with a constant group delay and varying phase delay? Is group delay ever an interesting metric, as constant group delay doesn't seem to guarantee anything at all? $\endgroup$
    – lerneaen_hydra
    Dec 2, 2016 at 7:36
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    $\begingroup$ @lerneaen_hydra: Take an ideal differentiator or Hilbert transformer as an example. Add some delay if you like. In both cases the phase is linear (or, without delay, constant) but it is discontinuous at $\omega=0$. This means that apart from $\omega=0$, the group delay is constant. However, the delay of a sinusoid at the input is frequency dependent. You get a constant phase shift (plus delay), but not a constant delay. The term group delay is only meaningful for narrow-band signals, and it corresponds to the delay of the envelope of the narrow-band signal. $\endgroup$
    – Matt L.
    Dec 2, 2016 at 7:57
  • $\begingroup$ Ah, I think this is starting to make sense! So for an affine phase response (say a differentiator with delay) the obvious output is a delayed pi/2 phase-shifted (and frequency-dependent scaled) copy of the input. In the case of a differentiator the group delay should be constant, so regardless of the bandwidth of the signal the envelope will be delayed by some amount, that said, each frequency component is phase-shifted by some constant angle, leading to a varying phase-delay. Could one loosely say that group delay -> magnitude delay, while phase delay -> magnitude and phase delay? $\endgroup$
    – lerneaen_hydra
    Dec 2, 2016 at 9:44
  • $\begingroup$ @lerneaen_hydra: For narrow-band signals, group delay is the delay of the envelope, and phase delay is the delay of the carrier. For wideband signals, both terms have no meaning. Also take a look at this answer to a related question. $\endgroup$
    – Matt L.
    Dec 3, 2016 at 12:43

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