Hi people,
I previously posted a question regarding group delay. Thanks to a hint
from one of you, I realized that if I filtered a pure (single
frequency) sinewave with a filter, and then compared the displacement
of zero crossings, that this was the phasedelay introduced by the
filter.
Now, I know the theoretical difference between phasedelay and
groupdelay. I know that the phasedelay is the phase divided by the
frequency (the angle of a straight line towards the origin (or towards
the phase at f=0 Hz ?)), whereas the groupdelay is minus the
derivative of the phasefunction.
I also know that the groupdelay may be interpreted as 'the time delay
of the amplitude envelope of a sinusoid at frequency w' (where the
bandwidth of the amplitude envelope must be restricted to a frequency
interval over which the phase response is approximately linear).
The problem is, I still have trouble understanding this explaination.
So, let's turn it around:
If I plot the groupdelay of a filter, and the graph says that at an
quarter of the sampling frequency (1/2 * Nyquist/Normalized frequency)
the group delay is 5 samples, how can I interprete that ?
Obviously, it doesn't mean that a sinewave of that frequency is
delayed for 5 samples. Does it perhaps mean that the composite of
frequencies, which amplitude envelope has a frequency of 1/2 *
Nyquist, will be delayed for 5 samples ? But in that case, what are
the composite frequencies ... ?
I hope somebody is able to shed some light on the issue.
Thanks in advance !
Nico
it means that the low frequency (or slowly changing) _envelope_ that
may be attached to that sinusoid is delayed by 5 samples. that's all
that it means.
r bj
If you trigger an input with a pair of tones, the propagation time
through the filter will be different for the two tones if they
differ in frequency. The difference is group delay.
>Obviously, it doesn't mean that a sinewave of that frequency is
>delayed for 5 samples.
Why not?
>Does it perhaps mean that the composite of
>frequencies, which amplitude envelope has a frequency of 1/2 *
>Nyquist, will be delayed for 5 samples ? But in that case, what are
>the composite frequencies ... ?
>
>I hope somebody is able to shed some light on the issue.
>
>Thanks in advance !
>
>Nico

Floyd L. Davidson <http://www.ptialaska.net/~floyd>
Ukpeagvik (Barrow, Alaska) fl...@barrow.com
>dum...@gmx.net (Nico) wrote:
>>
>>I also know that the groupdelay may be interpreted as 'the time delay
>>of the amplitude envelope of a sinusoid at frequency w' (where the
>>bandwidth of the amplitude envelope must be restricted to a frequency
>>interval over which the phase response is approximately linear).
>>
>>The problem is, I still have trouble understanding this explaination.
>>So, let's turn it around:
>>
>>If I plot the groupdelay of a filter, and the graph says that at an
>>quarter of the sampling frequency (1/2 * Nyquist/Normalized frequency)
>>the group delay is 5 samples, how can I interprete that ?
>
>If you trigger an input with a pair of tones, the propagation time
>through the filter will be different for the two tones if they
>differ in frequency. The difference is group delay.
>
No, the group delay is the actual time it takes for a tine to
propagate through a filter. If two different tones take different
amounts of time, then the group delay is not flat.
>>Obviously, it doesn't mean that a sinewave of that frequency is
>>delayed for 5 samples.
>
>Why not?
Absolutely. The units for group delay are seconds  and it is exactly
that. In the digital domain it is more convenient to think in terms of
number of clock cycles, but it means the same thing for a given
implementation.
>
>>Does it perhaps mean that the composite of
>>frequencies, which amplitude envelope has a frequency of 1/2 *
>>Nyquist, will be delayed for 5 samples ? But in that case, what are
>>the composite frequencies ... ?
>>
>>I hope somebody is able to shed some light on the issue.
>>
>>Thanks in advance !
>>
>>Nico
Just think in these terms. The signal goes in, and you have to wait a
little while before it comes out. That is group delay, and it is not
necessarily the same at every frequency, although for audio filters
efforts are normally made to ensure that it is.
d
_____________________________
Telecommunications consultant
http://www.pearce.uk.com
What is the derivative of the phase phi of its output with respect to
radian frequency? d phi/d freq is that delay, you will find.
Now imagine sending a pulse centered around some frequency. The
properties of the filter not near that frequency don't matter;
all that matters is d phi/d freq at that frequency.
Nothing now prevents you from adding other filters for other
frequencies with other delays, and producing a frequency dependent
group delay for the total summed filter.
Finally, imagine every filter to have been constructed that way,
with differing weights for each frequency's filter.
The group delay at each frequency is d phi/d freq and physically
it's how long a pulse at that frequency is delayed.

Ron Hardin
rhha...@mindspring.com
On the internet, nobody knows you're a jerk.
> dum...@gmx.net (Nico) wrote:
>>
>>I also know that the groupdelay may be interpreted as 'the time delay
>>of the amplitude envelope of a sinusoid at frequency w' (where the
>>bandwidth of the amplitude envelope must be restricted to a frequency
>>interval over which the phase response is approximately linear).
>>
>>The problem is, I still have trouble understanding this explaination.
>>So, let's turn it around:
>>
>>If I plot the groupdelay of a filter, and the graph says that at an
>>quarter of the sampling frequency (1/2 * Nyquist/Normalized frequency)
>>the group delay is 5 samples, how can I interprete that ?
>
> If you trigger an input with a pair of tones, the propagation time
> through the filter will be different for the two tones if they
> differ in frequency. The difference is group delay.
>
>>Obviously, it doesn't mean that a sinewave of that frequency is
>>delayed for 5 samples.
>
> Why not?
cause it's _group_ delay, not _phase_ delay. if the system or filter is
phase linear, then the two numbers are the same, but not so otherwize.
>>Does it perhaps mean that the composite of
>>frequencies, which amplitude envelope has a frequency of 1/2 *
>>Nyquist, will be delayed for 5 samples ? But in that case, what are
>>the composite frequencies ... ?
>>
>>I hope somebody is able to shed some light on the issue.
>>
>>Thanks in advance !
>>
>>Nico
>
>
>Floyd L. Davidson <http://www.ptialaska.net/~floyd>
>Ukpeagvik (Barrow, Alaska) fl...@barrow.com
^^^^^^ ^^^^^^ (GACK!)
wow! what's the sunbathing like up there?

r bj
Wave Mechanics, Inc.
45 Kilburn St.
Burlington VT 054014750
tel: 802/9519700 ext. 207 http://www.wavemechanics.com/
fax: 802/9519799 rob...@wavemechanics.com

It is meaningless to discuss the group delay of
a single pure sine wave. There is no such concept
for a single pure sine wave. You must have a group
(hint  group delay) of different frequencies travelling
together, which would (normally) be obtained by
modulating one sine wave with another, in the case
of the training that you're undergoing.
Nico <dum...@gmx.net> wrote in message
news:a1a88ba8.02011...@posting.google.com...
>(First of all, I realize that this might seem to be a FAQ, but even
>after some time spend searching on Deja/Google I couldn't find a
>satisfying answer)
>I previously posted a question regarding group delay. Thanks to a hint
>from one of you, I realized that if I filtered a pure (single
>frequency) sinewave with a filter, and then compared the displacement
>of zero crossings, that this was the phasedelay introduced by the
>filter.
>Now, I know the theoretical difference between phasedelay and
>groupdelay. I know that the phasedelay is the phase divided by the
>frequency (the angle of a straight line towards the origin (or towards
>the phase at f=0 Hz ?)), whereas the groupdelay is minus the
>derivative of the phasefunction.
>I also know that the groupdelay may be interpreted as 'the time delay
>of the amplitude envelope of a sinusoid at frequency w' (where the
>bandwidth of the amplitude envelope must be restricted to a frequency
>interval over which the phase response is approximately linear).
Physics doesn't have group delay but it has group velocity, which
is related to group delay. In the case of a continuous function,
say the value of the electric field in a medium as a function
of time and position, as a wave propagates. The phase velocity
is w/k and group velocity is dw/dk.
(That should be an omega, but I don't have one here.)
In the case of a pure sine wave, sin(kxwt), both the phase and
group velocity are the same. If you take an amplitude modulated
sine wave travelling through a material, you are usually interested
in the velocity of the information (modulation), which is the group
velocity, where the propagation of wave crests and valleys is the
phase velocity. Consider a piece of glass where the index of refraction
varies slowly with frequency. One can then determine the phase and
group velocities for pulses that relatively narrow in frequency
centered around a certain frequency, but also reasonably well defined
in time.
If you have a material where the velocity (index of refraction) changes
sharply with frequency, such as near a resonance, phase and group
velocity are not very meaningful. It is likely that what comes out
doesn't look much like what went it, or it might not come out at all.
(Group velocity is pretty much the second term in a Taylor series,
the assumption is that the subsequent terms can be ignored. If they
can't be then group velocity isn't meaningful.)
In the above cases, signal travelling through a continuous medium,
velocity is delay per unit thickness of the material. For the
discrete case, it is just delay, but the problems are similar.
If the filter is relatively smooth, you will likely find that the
signal coming out looks similar but is delayed by some amount of
time. If the filter is too sharp, the signal coming out may not
look at all like what went in, and group delay isn't well defined.
 glen
(Note also that there are matrials where the phase velocity is
greater than the speed of light. There are also materials where
the index of refraction is less than one, or even negative, at
specific frequencies. Those are the same frequencies where
group velocity isn't well defined.)
glen herrmannsfeldt wrote:
>
>
> (Note also that there are matrials where the phase velocity is
> greater than the speed of light. There are also materials where
> the index of refraction is less than one, or even negative, at
> specific frequencies. Those are the same frequencies where
> group velocity isn't well defined.)
I've done work with metallic reflections where the iondex of refraction
is complex.
For example with aluminum and HeNe red light, n=1+4.45i
Clay
Pete Gianakopoulos KE9OA
Formerly of RockwellCollins
Now back home, Chicago, Il.
Clay S. Turner <phy...@bellsouth.net> wrote in message
news:3C3F5DF1...@bellsouth.net...
Except that for a pure sine wave, that is, a given frequency,
group delay is only defined module the period.
So, yes, group delay is a function of frequency, but it should be
a slowly varying function of frequency. For a pulse, centered
around some frequency it should be reasonably constant over
the width of the pulse.
I don't believe that it must be an integer multiple of the
sampling frequency, either, though I couldn't prove that.
 glen
>Ron Hardin <rhha...@mindspring.com> writes:
>
>>Imagine a filter that just delays everything by T.
>
>>What is the derivative of the phase phi of its output with respect to
>>radian frequency? d phi/d freq is that delay, you will find.
>
>>Now imagine sending a pulse centered around some frequency. The
>>properties of the filter not near that frequency don't matter;
>>all that matters is d phi/d freq at that frequency.
>
>>Nothing now prevents you from adding other filters for other
>>frequencies with other delays, and producing a frequency dependent
>>group delay for the total summed filter.
>
>>Finally, imagine every filter to have been constructed that way,
>>with differing weights for each frequency's filter.
>
>>The group delay at each frequency is d phi/d freq and physically
>>it's how long a pulse at that frequency is delayed.
>
>Except that for a pure sine wave, that is, a given frequency,
>group delay is only defined module the period.
>
If you are being VERY strict in your pure sine wave definition, then
group delay has no meaning. All you can see is a phase shift, which as
you say is modulo the period  it goes back to zero when you reach 360
degrees.
>So, yes, group delay is a function of frequency, but it should be
>a slowly varying function of frequency. For a pulse, centered
>around some frequency it should be reasonably constant over
>the width of the pulse.
>
N, this isn't right. Group delay can be whatever you want. In a
chirped RADAR, a very fastchanging group delay is used to compress
the received pulse in time in order to maximise positional resolution.
And of course a pulse is not centred on a frequency  it has widely
dispersed frequencies. If you want to maintain the shape of the pulse,
then the group delay must be constant over all those frequencies.
>I don't believe that it must be an integer multiple of the
>sampling frequency, either, though I couldn't prove that.
>
> glen
No need at all for group delay to be an integer multiple. In fact
because of the analogue reconstruction filter you can guarantee it
won't be.
>
> >>The group delay at each frequency is d phi/d freq and physically
> >>it's how long a pulse at that frequency is delayed.
> >
> >Except that for a pure sine wave, that is, a given frequency,
> >group delay is only defined module the period.
> >
> Pearce
> If you are being VERY strict in your pure sine wave definition, then
> group delay has no meaning. All you can see is a phase shift, which as
> you say is modulo the period  it goes back to zero when you reach 360
> degrees.
Bob writes
Even for a pure sine wave, group delay has meaning. The formula:
d phi/d freq, refers to only one frequency. Group delay physicaly is:
the time it takes a single frequency (pure sine wave), to pass from
point A to point B.
For example, the group delay for 100 feet of ideal cable is: 0.10167
uSec. A pure sine wave will take 0.10167 uSec to pass through that
cable.
(To measure group delay always requires using two frequencys, but the
end result is: the delay at one frequency.)
Bob Stanton
I know this, Bob  but I did stipulate "very" strict. To qualify under
those terms the sine wave has no beginning or end, so you don't have
any access to information beyond the modulo 360 degree phase shift.
You cannot tell how long a true sine wave has taken to pass through.
As soon as you have any kind of transient event  like turning the
signal on  you no longer have a true sine wave.
You are right about needing two frequencies so you can get the first
differential of phase with frequency  which is the group delay.
>So, yes, group delay is a function of frequency, but it should be
>a slowly varying function of frequency. For a pulse, centered
>around some frequency it should be reasonably constant over
>the width of the pulse.
Why? If you want the pulse undistorted, indeed, but try
putting something through a Cauer Elliptic IIR filter and see
what happens when you work past one of the inband ripples close
to the transition, ...

Copyright j...@research.att.com 2001, all rights reserved, except transmission
by USENET and like facilities granted. This notice must be included. Any
use by a provider charging in any way for the IP represented in and by this
article and any inclusion in print or other media are specifically prohibited.
>In article <a1rek8$g...@gap.cco.caltech.edu>,
>glen herrmannsfeldt <g...@ugcs.caltech.edu> wrote:
>>Ron Hardin <rhha...@mindspring.com> writes:
>>>The group delay at each frequency is d phi/d freq and physically
>>>it's how long a pulse at that frequency is delayed.
>>So, yes, group delay is a function of frequency, but it should be
>>a slowly varying function of frequency. For a pulse, centered
>>around some frequency it should be reasonably constant over
>>the width of the pulse.
>Why? If you want the pulse undistorted, indeed, but try
>putting something through a Cauer Elliptic IIR filter and see
>what happens when you work past one of the inband ripples close
>to the transition, ...
It is somewhat subjective, but if the pulse comes out completely
different shape than it goes in, how do you find the position.
The case I was describing, for a linear continuous system, comes
when you are right on the edge of a resonance peak. I would
expect sharp digital filters to have a similar effect.
Note, for example, that for an optical filter there are regions
where the group and/or phase velocity can be negative,
according to the w/k and dw/dk definition. A similar condition
should allow a negative group delay for a digital filter.
Those are the conditions I was trying to describe. You could even
have one where the group delay was positive for part of the
pulse and negative for another part.
 glen
>
> I know this, Bob  but I did stipulate "very" strict. To qualify under
> those terms the sine wave has no beginning or end, so you don't have
> any access to information beyond the modulo 360 degree phase shift.
> You cannot tell how long a true sine wave has taken to pass through.
> As soon as you have any kind of transient event  like turning the
> signal on  you no longer have a true sine wave.
>
> You are right about needing two frequencies so you can get the first
> differential of phase with frequency  which is the group delay.
>
> d
Bob writes:
Group delay is slope of the phase curve. If the phase shift is
nonlinear, the slope line can be tangent to the phase curve only at
single points.
X
X
X .
X .
X .
X.
X .
X .
X .
My old high school geometry teacher once said, If a line is tangent to
a circle, the line only touchs the circle at *one* point.
Your definition of group delay seems to be: the slope of two points on
the phase curve. (Two points that are very close to each other in
frequency.) No matter how close the points are in frequency, there
will always be some error in the slope number. The closer the two
points are, the smaller will be the error. The (two points) definition
of group delay will not give the *exact* slope (group delay). Close,
but no cigar.
My definition of group delay is: the slope of a line tangent to the
phase shift curve. That definition is consistant with the formula Tgd
= d B/ d w . The (tangent) slope line, can only touch the phase shift
curve at *one* point. The point of intersection will be at *one*
frequency (not two). Group delay is therefore, a number refering to
one frequency.
One frequency = pure sine wave.
Bob Stanton
Hoping that is as clear to you, as it is to me!
Sure, but that doesn't answer the point that it can't be measured
using a single, pure sine wave. There is not enough information at a
single point to tell you what the slope is  you have assumed what it
is.
Also, consider that even with a twofrequency measurement you are
actually sampling in the fervency domain and must guard against
aliasing. If you made measurements at, say 10kHz and 11kHz and saw a
degree of difference you would make an assumption about the length of
delay that represented. Well, that could just as easily have been 361
or 721 degrees of difference  you have no way (purely from the
measurement) of knowing which it is. That would of course mean a much
longer delay.
So in an actual measurement, it is hard to reach unambiguous results 
although reasonable answers are easy. But you certainly can't measure
slope by looking at a single point (fervency).
Don Pearce wrote:
Wouldn't it be easily measured with 2 sine waves starting on the same frequency and moving one up in
frequency and noting the first frequency that there is a 90 degree, or whatever you want to look for, shift
?
Regards
Gary
Now you're talking. That would do nicely.
I didn't follow your explanation. Maybe you won't follow mine, but here
goes:
Look at the wake of a boat. You will see that it is made up of wavelets.
Look closely, and you will see that the wavelets move faster than the
wake as a whole. Wavelets rise spontaneously at the training edge of the
wake, move to the front, and then die out. The wavelets move at the
phase velocity. The group of wavelets (the wake itself) moves at the
group velocity.
Second try. Consider a transmission medium in which the phase and group
velocities differ. Any medium with dispersion, such as a line in which
l/c (that's an ell, not a one) isn't equal to r/g, or a waveguide. Pulse
on a carrier, with suitably limited rise and fall times so that there
are no frequencies very far from the carrier. Observe that the pulse
travels down the medium at one speed, while the carrier itself travels
at another.* How is this possible? Observing at different points (or
delays), you can see that the carrier moves through the pulse, just as
in the wake above. The pulse moves at the group velocity, the velocity
that information and energy moves at. (In a waveguide, the phase
velocity always exceeds the speed of light, but the group velocity is
less than C. In detail, V_p * V_g = C^2.)
Jerry
_______________________
* You can compute the carrier speed easily from the wavelength and
frequency. It says little about group velocity.

Engineering is the art of making what you want from things you can get.

When you speak of a pulse on a carrier  do you mean something like
RADAR? If so, then the pulse is a modulation which by definition
spreads the frequency of the carrier by producing sidebands. A
nonlinear group delay  or if you like a different propagation
velocity for different frequencies  will act on different parts of
that signal cluster  carrier plus sidebands  by changing the
relative phase. The result is a smeared pulse. This must be corrected
in either of two ways. You can equalise the group delay, or apply a
complementary error to the original pulse.
Your equation relating phase velocity, group velocity and the speed of
light is quite correct, but since all the information is carried in
the group, not the phase, there are no problems with the phase term
coming out superluminal.
All true. The purpose of limiting the rise and fall times of the pulse
is, as I wrote, to limit the pulse to frequencies near the carrier, thus
assuring a group velocity nearly equal for all components; equal enough
so that the illustration can work. It was emphasized earlier in the
thread that although it isn't possible to measure group velocity with a
singlefrequency probe, there is a group velocity at every frequency.
Useful communication channels have reasonably constant group velocities
over the bands of intended use. Solitons are a special case where
uniform group velvety is enforced by nonlinearity.
Jerry
>
> Sure, but that doesn't answer the point that it can't be measured
> using a single, pure sine wave. There is not enough information at a
> single point to tell you what the slope is  you have assumed what it
> is.
> ...
> So in an actual measurement, it is hard to reach unambiguous results 
> although reasonable answers are easy. But you certainly can't measure
> slope by looking at a single point (fervency).
>
> d
>
Bob writes:
I agree (and have agreed) there is *no way* to measure group delay
using a continuous sine wave. That doesn't leave much room for
disagreement, but I'm going to try. :)
The dictionary defines group delay as: "In a modulated signal, a delay
of the transmission of data." By that defination, you are correct in
saying group delay must always involve two signals.
I was thinking of "group delay" in terms of its' common usage. For
example, in filter books we commonly see plots of delay
characteristics, with the ordinate labled "group delay". Any *single
point* on those curves is the delay at a single frequency. Each point,
on these curves, corresponds to a single point on a phase curve.
Perhaps by the strict dictionary defination, the ordinate should not
be labled "group delay", but that is common usage.
Here is a circuit of a resistor and capacitor:
_________ R = 1000 _____________


C = 1uF

________________________________
This circuit has a delay of 500 usec at the frequency of 159.1549 Hz.
*I didn't get this number by measuring the phase shift at two
frequencies.* This number came from the slope of the phase curve, (at
one frequency). By the strict dictionary defination, I did not
calculate group delay, but "group delay" is what it is commonly
called.
Can we say a continueous sine wave takes a certain time to pass
through a circuit? If I go down to KMart, buy a flash light, and
shoot a continueous beam of light at the moon, I know the light will
take 1.28 seconds to reach the moon, even though the light beam is a
continueous sine wave. The photons that make up the sine wave take
1.28 sec to get to the moon.
Even thought we can't say a continueous (159.1549 Hz) sine wave takes
a 500 usec to pass through this filter we know, from group delay
calculations, the energy that the sine wave carries takes 500 usec to
pass through the filter.
Bob Stanton
>All true. The purpose of limiting the rise and fall times of the pulse
>is, as I wrote, to limit the pulse to frequencies near the carrier, thus
>assuring a group velocity nearly equal for all components; equal enough
>so that the illustration can work. It was emphasized earlier in the
>thread that although it isn't possible to measure group velocity with a
>singlefrequency probe, there is a group velocity at every frequency.
Well, there may be a dw/dk at every frequency, but it may or may not
be a useful group velocity. You can use two different frequencies
in the limit as they get closer and closer together, and hope it is
a continuous function.
>Useful communication channels have reasonably constant group velocities
>over the bands of intended use. Solitons are a special case where
>uniform group velvety is enforced by nonlinearity.
>Engineering is the art of making what you want from things you can get.
I agree with this one!
 glen
Group velocity is interesting primarily when it is substantially
constant over a band that has interesting width. Generally, that
requires a relatively flat frequency response. The group velocity at the
halfpower point of an RC rolloff doesn't usually qualify.
Jerry

Engineering is the art of making what you want from things you can get.

Bob_Stanton wrote:
With the flash light you know the distance and you know the propagation
speed. You know more than one element. With the resistor and capacitor you
know that the phase shift will be 90 degrees or less. Throw in an unknown
number of resistor  capacitor elements and you then do not know how many
cycles it will take before the single sine wave comes out the other side.
Regards
Gary
My copy of O+S (1975 edition) says it's the additive inverse of that:
d phi/d freq
Any ideas about the difference?
Allan.