Quelle est la différence entre le retard de phase et le retard de groupe?

J'étudie des DSP et j'ai du mal à comprendre la différence entre retard de phase et retard de groupe .

Il me semble qu'ils mesurent tous les deux le temps de retard des sinusoïdes passées à travers un filtre.

• Ai-je raison de penser cela?
• Si oui, en quoi les deux mesures diffèrent?
• Quelqu'un pourrait-il donner un exemple d'une situation dans laquelle une mesure serait plus utile que l'autre?

MISE À JOUR

À lire dans Introduction aux filtres numériques de Julius Smith , j'ai trouvé une situation dans laquelle les deux mesures donnent au moins des résultats différents: les filtres à phase affine . C'est une réponse partielle à ma question, je suppose.

Tout d'abord les définitions sont différentes:

• Retard de phase: (le négatif de) Phase divisée par la fréquence
• Retard de groupe: (le négatif de) Première dérivée de la phase par rapport à la fréquence

En mots cela signifie:

• Retard de phase: Angle de phase à ce point de fréquence
• Group delay: Taux de changement de phase autour de ce point en fréquence.

When to use one or the other really depends on your application. The classical application for group delay is modulated sine waves, for example AM radio. The time that it takes for the modulation signal to get through the system is given by the group delay not by the phase delay. Another audio example could be a kick drum: This is mostly a modulated sine wave so if you want to determine how much the kick drum will be delayed (and potentially smeared out in time) the group delay is the way to look at it.

They don't both measure how much a sinusoid is delayed. Phase delay measures exactly that. Group delay is a little more complicated. Picture a short sine wave with an amplitude envelope applied to it so that it fades in and fades out, say, a gaussian multiplied by a sinusoid. This envelope has a shape to it, and in particular, it has a peak that represents the center of that "packet." Group delay tells you how much that amplitude envelope will be delayed, in particular, how much the peak of that packet will move by.

I like to think about this by going back to the definition of group delay: it's the derivative of phase. The derivative gives you a linearization of the phase response at that point. In other words, at some frequency, the group delay is telling you approximately how the phase response of the neighboring frequencies relate to the phase response at that point. Now, remember how we're using an amplitude-modulated sinusoid. The amplitude modulation will take the sinusoid's peak, and introduce sidebands at neighboring frequencies. So, in a way, the group delay is giving you information about how the sidebands will be delayed relative to that carrier frequency, and applying that delay will change the shape of the amplitude envelope in some way.

The crazy thing? Causal filters can have negative group delay! Take your gaussian multiplied by a sinusoid: you can build an analog circuit such that when you send that signal through, the envelope's peak will appear in the output before the input. It seems like a paradox, since it would appear that the filter has to "see" into the future. It's definitely weird, but a way to think about it is that since the envelope has a very predictable shape, the filter already has enough information to anticipate what is going to happen. If a spike were inserted in the middle of the signal, the filter would not anticipate that. Here's a really interesting article about this: http://www.dsprelated.com/showarticle/54.php

For those who still cannot chalk the difference here is an simple example

Take long transmission line with simple sine signal with an amplitude envelope, $v(t)$, at its input

$$v(t) \cdot \sin(\omega t)$$

If you measure this signal at the transmission line end, it might come somewhere like this:

$$v(t-\tau_g) \cdot \sin(\omega t + \phi)\\ = v(t-\tau_g) \cdot \sin(\omega (t - \tau_\phi))$$

where $\phi$ is phase difference from input to output.

If you want how much time in it takes the phase of the sinusoid, $\sin(\omega t)$ transmission from input to end then $\tau_\phi = -\tfrac{\phi}{\omega}$ is your answer in seconds.

If you want how much time in it takes the envelope, $v(t)$, of the sinusoid transmission from input to end then $\tau_g = -\tfrac{d\,\phi}{d\,\omega}$ is your answer in seconds.

Phase delay is just traveling time for a single frequency while group delay is measure of amplitude distortion if array of multiple frequencies are applied.

The phase delay of any filter is the amount of time delay each frequency component suffers in going through the filters (If a signal consists of several frequencies.)

The group delay is the average time delay of the composite signal suffered at each component of frequency.

I know this is a pretty old question, but I've been looking for a derivation of the expressions for group delay and phase delay on the internet. Not many such derivations exist on the net so I thought I'd share what I found. Also, note that this answer is more of a mathematical description than an intuitive one. For intuitive descriptions, please refer to the above answers. So, here goes:

Let's consider a signal

$$a(t) = x(t)cos(\omega_0 t)$$ and pass this through an L.T.I. system with frequency response $$H(j\omega) = e^{j\phi(\omega)}$$ We have considered the gain of the system to be unity because we are interested in analyzing how the system alters the phase of the input signal, rather than the gain. Now, given that multiplication in time domain corresponds to convolution in frequency domain, the Fourier Transform of the input signal is given by $$A(j\omega) = {1 \over 2\pi}X(j\omega) * (\pi\delta(\omega - \omega_0) + \pi\delta(\omega + \omega_0))$$ which amounts to $$A(j\omega) = {X(j(\omega-\omega_0))+X(j(\omega+\omega_0)) \over 2}$$ Therefore, the output of the system has a frequency spectrum given by $$B(j\omega) = {e^{j\phi(\omega)} \over 2} (X(j(\omega-\omega_0))+X(j(\omega+\omega_0)))$$ Now, to find the inverse Fourier Transform of the above expression, we need to know the exact analytical form for $\phi(\omega)$. So, to simplify matters, we assume that the frequency content of $x(t)$ includes only those frequencies which are significantly lower than the carrier frequency $\omega_0$. In this scenario, the signal $a(t)$ can be viewed as an amplitude modulated signal, where $x(t)$ represents the envelope of the high frequency cosine signal. In the frequency domain, $B(j\omega)$ now contains two narrow bands of frequencies centered at $\omega_0$ and $-\omega_0$ (refer to the above equation). This means that we can use a first order Taylor series expansion for $\phi(\omega)$. $$\phi(\omega) = \phi(\omega_0) + \frac{d\phi}{d\omega}(\omega_0)(\omega - \omega_0) = \alpha + \beta\omega$$ where $$\alpha = \phi(\omega_0) - \omega_0\frac{d\phi}{d\omega}(\omega_0)$$ $$\beta = \frac{d\phi}{d\omega}(\omega_0)$$ Plugging this in, we can calculate the Fourier transform of the first half of $B(j\omega)$ as $$\frac{1}{2\pi}\int_{-\infty}^{\infty} \frac{1}{2} X(j(\omega - \omega_0)) e^{j(\omega t + \alpha + \beta\omega)} d\omega$$ Substituting $\omega - \omega_0$ for $\omega'$, this becomes $$\frac{1}{2\pi}\int_{-\infty}^{\infty} \frac{1}{2} X(j(\omega')) e^{j((\omega' + \omega_0) (t + \beta) + \alpha)} d\omega'$$ which simplifies to $$x(t + \beta)\frac{e^{j(\omega_0 t + \omega_0\beta + \alpha)}}{2}$$ Plugging in the expressions for $\alpha$ and $\beta$, this becomes $$x(t + \beta)\frac{e^{j(\omega_0 t + \phi(\omega_0))}}{2}$$ Similarly the other half of the inverse Fourier Transform of $B(j\omega)$ can be obtained by replacing $\omega_0$ by $-\omega_0$. Noting that for real signals, $\phi(\omega)$ is an odd function, this becomes $$x(t + \beta)\frac{e^{-j(\omega_0 t + \phi(\omega_0))}}{2}$$ Thus, adding the two together, we get $$b(t) = x(t + \frac{d\phi}{d\omega}(\omega_0))cos(\omega_0(t + \frac{\phi(\omega_0)}{\omega_0}))$$ Notice the delays in the envelope $x(t)$ and the carrier cosine signal. Group delay $(\tau_g)$ corresponds to the delay in the envelope while phase delay $(\tau_p)$ corresponds to the delay in the carrier. Thus, $$\tau_g = -\frac{d\phi}{d\omega}(\omega_0)$$ $$\tau_p = -\frac{\phi(\omega_0)}{\omega_0}$$