# What is the difference between phase delay and group delay?

I'm studying some DSP and I'm having trouble understanding the difference between phase delay and group delay.

It seems to me that they both measure the delay time of sinusoids passed through a filter.

• Am I correct in thinking this?
• If so, how do the two measurements differ?
• Could someone give an example of a situation in which one measurement would be more useful than the other?

UPDATE

Reading ahead in Julius Smith's Introduction to Digital Filters, I've found a situation where the two measurements at least give different results: affine-phase filters. That's a partial answer to my question, I guess.

• You might find this page useful. It explains group delay, and its effects, without any math. – user5108_Dan Jan 30 '16 at 16:01
• the wikipedia page spells out the definitions and difference mathematically. if you have a linear-phase filter, group delay and phase delay are the same value and are simply the throughput delay of the filter. for any general filter that has some gain at DC (i.e. not an HPF nor BPF with $-\infty$ dB at DC) and does not have a polarity reversal at DC, the group delay and phase delay are the same value at and close to DC. – robert bristow-johnson Jan 10 '17 at 23:23

## 5 Answers

First of all the definitions are different:

• Phase delay: (the negative of) Phase divided by frequency
• Group delay: (the negative of) First derivative of phase vs frequency

In words that means:

• Phase delay: Phase angle at this point in frequency
• Group delay: Rate of change of the phase around this point in frequency.

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.

• "Absolute phase at this point in frequency" Wouldn't that just be called "phase"? – endolith Feb 23 '13 at 16:50
• I meant "absolute" as compared to "relative", but I see that this can be confused with "absolute value". I'll edit it – Hilmar Jun 20 '13 at 13:44
• one last important difference: the phase delay at some frequency $f$ is the time delay of the phase of the quasi-sinusoidal signal of frequency $f$ passed through the filter. the group delay is the time delay of the envelope or "group" of the quasi-sinusoid. – robert bristow-johnson Jan 10 '17 at 23:29

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

• When you say "picture a...", an actual image would be really helpful here. – Gabriel Staples Dec 7 '18 at 17:20

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, $$\tau_g = -\frac{d\phi}{d\omega}(\omega_0)$$ $$\tau_p = -\frac{\phi(\omega_0)}{\omega_0}$$