# What is meant by "Group delay"?in simple words?

Although likes of this question have been asked many times on DSP SE

But i am unable to understand those and i wish to have a crystal clear explanation in simple words with example

The only thing that i am able to understand from those questions is that group delay is negative derivative of phase with respect to frequency. But what does negative derivative means here and why not positive derivative here?

• The other similar questions explain quite well that it's a bit of a fuzzy world, but they do it with examples and clear words. Mar 17 '20 at 16:51
• the wikipedia article doesn't do a half bad job. some people might be able to see fingerprints of someone's writing in it. Mar 17 '20 at 16:57
• if people want, i can copy the introductory section from wikipedia to here and make it a legit answer. Mar 17 '20 at 17:04
– Peter K.
Mar 17 '20 at 17:10
• as if i didn't have something better to do... as a matter of fact... Mar 17 '20 at 17:24

Here is my simplest explanation:

The group delay, as the negative derivative of phase, predicts the time delay of the amplitude envelope of a pulse, as shown in the hand-drawn graphic below. The upper part of the sketch shows a sinusoidal waveform varied in amplitude by its envelope. The lower one is showing this same envelope before and after a system that has group delay.

This applies when the phase of the frequency response can be approximated as linear for the "group" of frequencies within the pulse envelope. Thus for non-linear phase systems, this applies to generally narrower band signals such as the pulse I show where the amplitude transition is gradual.

Consider a single sine-wave with the amplitude envelope such as I show. The time delay of the sine wave itself would be predicted directly from the phase of the frequency response (by dividing by the frequency of the sine-wave: with $$\phi = angle(H(j\omega))$$, the time delay is $$-\phi(\omega)/\omega$$), while the time delay of the pulse envelope is predicted from the negative derivative of the phase with respect to frequency ($$-d\phi(\omega)/d\omega$$)).

Calculate the time delay introduced by group delay for IIR-Filters

https://electronics.stackexchange.com/questions/135475/physical-significance-of-group-delay

And most helpful to what otherwise seems like a paradox of causality for positive group delay is this paper https://www.researchgate.net/publication/253463703_Causality_and_Negative_Group_Delays_in_a_Simple_Bandpass_Amplifier referenced by Max in this post Physical Meaning of Negative Group Delay for causal LTI systems which I bottom line as causality is not violated but due to the bandwidth restrictions above we create a condition that causes the pulse envelope at the output to precede the input: The output pulse does not appear until the input amplitude varies (and if we have gain in the system the output envelope grows faster) and due to the bandwidth constraint and the result of destructive summation of the input pulse the output will start to decrease before the input does. Very cool DSP magic trick. See this post that illustrates this with a specific example.

• I agree with this provided the bandwidth of interest is sufficiently small relative to the frequency of interest Mar 18 '20 at 2:06
• Mr Boschen, I wasn’t referring to linear phase, mostly considering that given linear phase the group delay would be constant, and could just as well be called delay. I worked out a derivation for group delay a while back, and my recollection is something like this: you define your input as a product of an envelope sinusoid and a carrier sinusoid. The maths will give the delay of the envelope as the group delay equation so long as the amplitude response is uniform. In general, this is always true as the bandwidth approaches zero. It has been a while for me though. Mar 18 '20 at 2:38
• So it may have been more appropriate for me to say: so long as the amplitude response is sufficiently uniform over the given bandwidth... Mar 18 '20 at 2:40
• Even my last statement is a bit misleading, because it would seem incorrect for linear phase... apologies, I don’t think I’m putting my thoughts to words effectively. I was attempting to include non-linear phase more generally. Mar 18 '20 at 2:45
• @DanSzabo See my update, I believe this may be closer to what you were getting at in our earlier conversation with regards to bandwidth. Apr 4 '20 at 16:25

To understand group delay, it is important to first understand phase delay.

Phase delay is the amount of phase lag for certain frequency. The units are in degrees. But there is a weird relationship between phase delay (units = degrees) and time delay (units = seconds). Let me explain:

Say I have a 1Hz signal that goes through a filter and it experiences 90 degree phase delay. 90 degrees is 1/4 of a full 360 degree cycle. Thus for a 1Hz signal (which has 1 second period) the time delay experienced is 1 second / 4 = 0.25 seconds. Essentially the output lags the input by 0.25 seconds.

Now lets say I have a 2Hz signal. The 2Hz signal has a cycle period of 0.5 seconds. Let say I feed this signal through a filter and it also experiences 90 degree phase delay. Again, 90 degrees is 1/4 of a full 360 cycle. Thus for a 2hz signal, the time delay experienced is 0.5 second / 4 = 0.125 seconds. Now the output lags the input signal by 0.125 seconds.

What this says is that constant phase delay does not equal constant time delay! Deriving the time delay from phase delay is dependent on the frequency itself. The only way for all frequencies to get delayed by the same time delay is if the phase response is linear.

When the phase response is linear, we know all the frequencies get time delayed by the same amount. Thus if all frequencies are delayed the same amount, we have this notion of a "group" delay. Group refers to all frequencies.

Let's look at this visually.

If we feed an input signal into a filter with a constant group delay, all frequencies will be time delayed the same amount. Referring to the picture below, the outputted signal matches the input signal except it is slightly delayed.

If instead we fed that input signal into a filter with a non-constant group delay. The frequencies will time delay different amounts resulting in an output signal that looks nothing like the inputted signal.

So even though each filter is low-pass, one filter distorts the signal such that it doesn't resemble the inputted signal. This is why linear-phase (constant group delay) filters are desirable in some applications.

• "Essentially the output lags the input by 0.25 seconds"What do you mean? Do you mean here that if in input we have peak at 0 second,then in output peak will occur at 0.25 second??
– engr
Apr 4 '20 at 16:24
• Correct. Visualize it like this: sengpielaudio.com/Sinusoidal%20Wave.gif
– Izzo
Apr 4 '20 at 21:11
• @engr I added some visuals which demonstrate the characteristics of constant group delay. And I fixed a couple typos.
– Izzo
Apr 4 '20 at 21:51
• what concept, you are trying to convey through the link in your comment sengpielaudio.com/Sinusoidal%20Wave.gif
– engr
Apr 9 '20 at 10:48
– engr
Apr 9 '20 at 10:51

Group delay is a useful measure of time distortion, and is calculated by differentiating, with respect to frequency, the phase response of the device under test (DUT): the group delay is a measure of the slope of the phase response at any given frequency. Variations in group delay cause signal distortion, just as deviations from linear phase cause distortion.

In linear time-invariant (LTI) system theory, control theory, and in digital or analog signal processing, the relationship between the input signal, $$x(t)$$, to output signal, $$y(t)$$, of an LTI system is governed by a convolution operation:

$$y(t) = (h*x)(t) \ \triangleq \ \int_{-\infty}^{\infty} x(u) h(t-u) \, \mathrm{d}u$$

Or, in the frequency domain,

$$Y(s) = H(s) X(s) \,$$

where

$$X(s) = \mathscr{L} \Big\{ x(t) \Big\} \ \triangleq \ \int_{-\infty}^{\infty} x(t) e^{-st}\, \mathrm{d}t$$

$$Y(s) = \mathscr{L} \Big\{ x(t) \Big\} \ \triangleq \ \int_{-\infty}^{\infty} y(t) e^{-st}\, \mathrm{d}t$$

and

$$H(s) = \mathscr{L} \Big\{ x(t) \Big\} \ \triangleq \ \int_{-\infty}^{\infty} h(t) e^{-st}\, \mathrm{d}t$$

Here $$h(t)$$ is the time-domain impulse response of the LTI system and $$X(s)$$, $$Y(s)$$, $$H(s)$$, are the Laplace transforms of the input $$x(t)$$, output $$y(t)$$, and impulse response $$h(t)$$, respectively. $$H(s)$$ is called the transfer function of the LTI system and, like the impulse response $$h(t)$$, fully defines the input-output characteristics of the LTI system.

Suppose that such a system is driven by a quasi-sinusoidal signal, that is a sinusoid having an amplitude envelope $$a(t)>0$$ that is slowly changing relative to the frequency $$\omega$$ of the sinusoid. Mathematically, this means that the quasi-sinusoidal driving signal has the form

$$x(t) = a(t) \cos(\omega t + \theta)$$

and the slowly changing amplitude envelope $$a(t)$$ means that

$$\left| \frac{d}{dt} \log \big( a(t) \big) \right| \ll \omega \ .$$

Then the output of such an LTI system is very well approximated as

$$y(t) = \big| H(i \omega) \big| \ a(t - \tau_g) \cos \big( \omega (t - \tau_\phi) + \theta \big) \; .$$

Here $$\tau_g$$ and $$\tau_\phi$$, the group delay and phase delay respectively, are given by the expressions below (and potentially are functions of the angular frequency $$\omega$$). The sinusoid, as indicated by the zero crossings, is delayed in time by phase delay, $$\tau_\phi$$. The envelope of the sinusoid is delayed in time by the group delay, $$\tau_g$$.

In a linear phas system (with non-inverting gain), both $$\tau_g$$ and $$\tau_\phi$$ are constant (i.e. independent of $$\omega$$) and equal, and their common value equals the overall delay of the system; and the unwrapped phase shift of the system (namely $$-\omega \tau_\phi$$) is negative, with magnitude increasing linearly with frequency $$\omega$$.

More generally, it can be shown that for an LTI system with transfer function $$H(s)$$ driven by a complex sinusoid of unit amplitude,

$$x(t) = e^{i \omega t}$$

the output is

\begin{align} y(t) & = H(i \omega) \ e^{i \omega t} \ \\ & = \left( \big| H(i \omega) \big| e^{i \phi(\omega)} \right) \ e^{i \omega t} \ \\ & = \big| H(i \omega) \big| \ e^{i \left(\omega t + \phi(\omega) \right)} \ \\ \end{align} \

where the phase shift $$\phi$$ is

$$\phi(\omega) \ \triangleq \arg \left\{ H(i \omega) \right\} \;$$

Additionally, it can be shown that the group delay, $$\tau_g$$, and phase delay, $$\tau_\phi$$, are frequency-dependent, and they can be computed from the properly unwrapped phase shift $$\phi$$ by

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

$$\tau_\phi(\omega) = - \frac{\phi(\omega)}{\omega}$$

It stems from the definition of the Laplace/fourier transforms using $$e^{-st}$$ or $$e^{-j\omega t}$$. This can be checked intuitively by looking at the transform pair of a delayed impulse, compare the time domain delay to the frequency domain phase. If you modified the transform to use $$e^{+st}$$instead, it would be the other way round.