Physical meaning of Group-Delay ?

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Nico

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Jan 10, 2002, 9:23:47 AM1/10/02
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(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)

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) sine-wave with a filter, and then compared the displacement
of zero crossings, that this was the phase-delay introduced by the
filter.

Now, I know the theoretical difference between phase-delay and
group-delay. I know that the phase-delay 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 group-delay is minus the
derivative of the phase-function.

I also know that the group-delay 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 group-delay 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

robert bristow-johnson

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Jan 11, 2002, 2:11:04 AM1/11/02
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dum...@gmx.net (Nico) wrote in message news:<a1a88ba8.02011...@posting.google.com>...

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 b-j

Floyd Davidson

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Jan 11, 2002, 7:05:07 AM1/11/02
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dum...@gmx.net (Nico) wrote:
>
>I also know that the group-delay 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 group-delay 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?

>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

Don Pearce

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Jan 11, 2002, 8:05:43 AM1/11/02
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On 11 Jan 2002 03:05:07 -0900, Floyd Davidson <fl...@ptialaska.net>
wrote:

>dum...@gmx.net (Nico) wrote:
>>
>>I also know that the group-delay 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 group-delay 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

Ron Hardin

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Jan 11, 2002, 9:21:29 AM1/11/02
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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.
--
Ron Hardin
rhha...@mindspring.com

On the internet, nobody knows you're a jerk.

robert bristow-johnson

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Jan 11, 2002, 11:49:37 AM1/11/02
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In article <87zo3lg...@barrow.com> , Floyd Davidson
<fl...@ptialaska.net> wrote:

> dum...@gmx.net (Nico) wrote:
>>
>>I also know that the group-delay 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 group-delay 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 b-j

Wave Mechanics, Inc.
45 Kilburn St.
Burlington VT 05401-4750

tel: 802/951-9700 ext. 207 http://www.wavemechanics.com/
fax: 802/951-9799 rob...@wavemechanics.com

--

The Electricitist

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Jan 11, 2002, 12:32:58 PM1/11/02
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You're confused.

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...

glen herrmannsfeldt

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Jan 11, 2002, 4:08:14 PM1/11/02
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dum...@gmx.net (Nico) writes:

>(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) sine-wave with a filter, and then compared the displacement
>of zero crossings, that this was the phase-delay introduced by the
>filter.

>Now, I know the theoretical difference between phase-delay and
>group-delay. I know that the phase-delay 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 group-delay is minus the
>derivative of the phase-function.

>I also know that the group-delay 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(kx-wt), 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.)

Clay S. Turner

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Jan 11, 2002, 4:49:37 PM1/11/02
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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

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Jan 12, 2002, 7:00:56 PM1/12/02
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In the radio design field, when characterizing a filter, group delay is
specified over a band of frequencies, instead of just one frequency. I
believe the same approach is used for audio filters, amplifiers, etc,
because when designing the loop filters for frequency synthesizers, these
filters are operating in the audio range.
Group delay is specified as a delta function, meaning that from F_low to
F_high, the phase shift will be no more than plus or minus X degrees over
that frequency range.
In the audio field, an amplifier may be specified as having no more than
plus or minus 3 degrees of phase shift, from 10Hz to 40KHz. This group
delay characteristic is especially important in the IF subsystem in an FM
stereo receiver; it can adversely affect stereo separation. In digital
radio systems, the Bit Error Rate, or BER will suffer.

Pete Gianakopoulos KE9OA
Formerly of Rockwell-Collins
Now back home, Chicago, Il.

Clay S. Turner <phy...@bellsouth.net> wrote in message
news:3C3F5DF1...@bellsouth.net...

glen herrmannsfeldt

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Jan 13, 2002, 2:56:56 AM1/13/02
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Ron Hardin <rhha...@mindspring.com> writes:

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

Don Pearce

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Jan 13, 2002, 5:03:17 AM1/13/02
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On 13 Jan 2002 07:56:56 GMT, g...@ugcs.caltech.edu (glen
herrmannsfeldt) wrote:

>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 fast-changing 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.

Bob_Stanton

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Jan 14, 2002, 7:26:23 AM1/14/02
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Don Pearce <don...@pearce.uk.com> wrote in message


>
> >>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.

> > http://www.pearce.uk.com

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

Don Pearce

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Jan 14, 2002, 8:17:33 AM1/14/02
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On 14 Jan 2002 04:26:23 -0800, rsta...@stny.rr.com (Bob_Stanton)
wrote:

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.

jj, DBT thug and skeptical philalethist

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Jan 14, 2002, 12:14:36 PM1/14/02
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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 in-band 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.

glen herrmannsfeldt

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Jan 14, 2002, 3:06:37 PM1/14/02
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j...@research.att.com (jj, DBT thug and skeptical philalethist) writes:

>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 in-band 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

Bob_Stanton

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Jan 14, 2002, 6:53:10 PM1/14/02
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Don Pearce <don...@pearce.uk.com> wrote in message news:<gcm54uokrfpj0vi1b...@4ax.com>...


>
> 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
non-linear, 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

Mark Disher

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Jan 14, 2002, 9:38:52 PM1/14/02
to
If by "physical meaning" I can substitute a physical example, lets try
this. (You gurus are informally invited to theoretically zap me, as you
see fit).
A speaker driver has a Zero Delay Plane (ZDP) whose location is where
there is no delay due to phase shift at a given point in the frequency
domain.
---A physical exzmple of group delay would be a pictorial of the
sideview of a driver (oriented as seen in an enclosure)
---A horizontal axis for time and a vertical axis for frequency are
superimposed on the sideview. The ZDP will be a vertical line running
through, or very close to through, the voice coil. Frequency increases
from low to high from bottom to top.
---You get a plot where the low frequencies radiate well behind the
driver and progressively higher frequencies radiate from points closer and
closer to the ZDP. That plot is the goup delay!

Hoping that is as clear to you, as it is to me!

Don Pearce

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Jan 15, 2002, 3:49:54 AM1/15/02
to
On 14 Jan 2002 15:53:10 -0800, rsta...@stny.rr.com (Bob_Stanton)
wrote:

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 two-frequency 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).

Gary Schafer

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Jan 15, 2002, 1:08:47 PM1/15/02
to

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

Don Pearce

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Jan 15, 2002, 1:37:58 PM1/15/02
to
On Tue, 15 Jan 2002 18:08:47 GMT, Gary Schafer <gsch...@mediaone.net>
wrote:

Now you're talking. That would do nicely.

Jerry Avins

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Jan 15, 2002, 2:19:11 PM1/15/02
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Mark,

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.
-----------------------------------------------------------------------

Don Pearce

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Jan 15, 2002, 2:47:04 PM1/15/02
to

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
non-linear 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.

Jerry Avins

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Jan 15, 2002, 2:58:22 PM1/15/02
to

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
single-frequency 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

Mark Disher

unread,
Jan 15, 2002, 4:20:39 PM1/15/02
to
Jerry,
Your explanation seems much better than mine, not least because you're talking
about a VISIBLE physical phenomena. Well said!

Bob_Stanton

unread,
Jan 15, 2002, 6:15:56 PM1/15/02
to
Don Pearce <don...@pearce.uk.com> wrote in message

>

> 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 K-Mart, 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

glen herrmannsfeldt

unread,
Jan 15, 2002, 8:27:26 PM1/15/02
to
Jerry Avins <j...@ieee.org> writes:

>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
>single-frequency 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

Jerry Avins

unread,
Jan 15, 2002, 8:57:22 PM1/15/02
to
glen herrmannsfeldt wrote:
>
> Jerry Avins <j...@ieee.org> writes:
>
> >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
> >single-frequency 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 velocity 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
half-power point of an RC rolloff doesn't usually qualify.

Jerry
--

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

-----------------------------------------------------------------------

Gary Schafer

unread,
Jan 15, 2002, 9:21:36 PM1/15/02
to

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

Allan Herriman

unread,
Jan 15, 2002, 11:34:19 PM1/15/02
to
A number of posters in this thread have said that the group delay is
d phi/d freq

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.

Randy Yates

unread,
Jan 15, 2002, 11:44:39 PM1/15/02
to
Allan Herriman wrote:
>
> A number of posters in this thread have said that the group delay is
> d phi/d freq
>
> My copy of O+S (1975 edition) says it's the additive inverse of that:
> -d phi/d freq

That's the way I've always seen it defined (-d phi/ d freq).

> Any ideas about the difference?

Take a sine wave at f Hz, sin(2*pi*f*t) and delay it by tau seconds:

sin(2*pi*f*(t - tau)) = sin(2*pi*f*t - 2*pi*f*tau)

As f increases, the phase decreases, and therefore the slope is negative for a
positive delay. Thus the negative sign on d phi/ d freq.
--
% Randy Yates % "...the answer lies within your soul
%% Fuquay-Varina, NC % 'cause no one knows which side
%%% 919-577-9882 % the coin will fall."
%%%% <ya...@ieee.org> % 'Big Wheels', *Out of the Blue*, ELO
http://personal.rdu.bellsouth.net/~yatesc

Don Pearce

unread,
Jan 16, 2002, 2:22:12 AM1/16/02
to
On 15 Jan 2002 15:15:56 -0800, rsta...@stny.rr.com (Bob_Stanton)
wrote:

I agree with what you say, Bob. The problem is one of non-commutation.
Although a knowledge of the group delay of a circuit allows you to
determine the resulting phase shift on a sine wave, knowledge of the
phase shift doesn't tell you the group delay - there is no unique
solution. You do need that bandwidth - from an event, modulation or
whatever to get at the delay.

jj, DBT thug and skeptical philalethist

unread,
Jan 16, 2002, 12:11:41 PM1/16/02
to
In article <67ef4d54.02011...@posting.google.com>,

Bob_Stanton <rsta...@stny.rr.com> wrote:
>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. :-)

Heh, but you CAN measure it with a slow sweep :)

Just unwrap the phase :)

Jerry Avins

unread,
Jan 16, 2002, 12:56:25 PM1/16/02
to
Allan Herriman wrote:
>
> A number of posters in this thread have said that the group delay is
> d phi/d freq
>
> 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.
>
...

I can't say. I'm accustomed to thinking in terms of the velocities of
traveling waves. If we describe the wave as exp[j(wt - ßz)] (ß a
function of frequency; w standing in for omega), then running along side
the wave at a speed dz/dt = w/ß, the phase will remain constant. When ß
varies with frequency, then a signal containing a number of frequencies
will vary in shape as it propagates; it will disperse. When there is
only little dispersion, a modulated carrier can exhibit an undispersed
envelope. Consider just the sidebands of an AM signal (we can include
the carrier, but there's nothing gained from the extra term):
sin(w_0 + dw)t + sin(w_0 - dw)t at the origin. Along the line, this
becomes sin[(w_0 + dw)t - (ß_0 + dß)z] + sin[(w_0 - dw)t - (ß_0 - dß)z].
Simplifying, we get 2cos(dwt - dßz) * sin(wt - ßz). In other words, to
keep in step with the carrier phase, we need to run at dz/dt = w/ß, as
before; To keep in step with the modulation, we need to run at dz/dt =
dw/dß.

Delay is distance divided by velocity, not simply the reciprocal of
velocity. The changed sign might be due to the assumed implicit
direction of positive distance. Does that muddy the waters even more?

Chuck

unread,
Jan 16, 2002, 3:28:25 PM1/16/02
to
Hi Don,

Language is a very poor form of communication. ;-)

We tend to be sloppy with our terminology, and lax about qualifiers, which
makes a big problem (clear communication) a lot bigger. For example, I've
worked in half dozen labs where a "coder" was a person who translated a
detailed software design into a high-level computer language. An
analog-to-digital converter was an "ADC", or "A2D", or sometimes "A-to-D
converter," and D2A converters had like designations. Though the labs,
their work, and their terminology's varied, there were some constants, one
being the meaning of the word "coder." It's a low level technical position,
less than a programmer, but more than a gopher. Part of the "culture" was
that Ada Lovelace was the first coder. As Charles Baggage's lover, she
supposedly coded his "calculating engine" based on equations that he
supplied. If this is true, then she was the first coder, given the
"accepted" definition of the term (but of course we know it all depends on
who's doing the accepting). Now some of the labs I'm talking about were
doing bleeding-edge research, and all were on the cutting edge in one way or
another, so there was a tendency to respect the people working there, and
accept their statements as being well informed. In general, they were. I
made a real arse out of myself once, when someone called an audio DAC a
"coder." It started with what I thought an obvious suggestion. Audiophiles
usually call them DACs, and it's best to use the terminology of the culture
being addressed. I also pointed out that "coder" was a job title, not an
electronic device, a comment that offended the other party, and lead to a
heated and totally unnecessary discussion that we could have probably both
done without.

This kind of thing happens even in the simplest conversations, when both
parties agree on the meanings of the terms they use. A few months ago a
friend got new speakers, and was having trouble with the setup. His system
was now bi-amped, something he hadn't done before, and from his description
of the problem I guessed that one of his amps might have inverted output
(180-degrees out of phase with its input) so I suggested that he reverse the
connections at the woofers, to see if changing their phase relationship to
the tweeters would solve his problem. It would have fixed his problem if I
hadn't used the word "phase" but he thought I was telling him to connect the
two woofers out of phase with each other, and of course that didn't help at
all.

In audio we face an additional hurdle in that we have all had or seen
statements, that were correct and accurate, twisted into pretzels by
golden-eared pseudo-scientists. We sometimes mistake an honest
misunderstanding for an attempt to discredit fact. Sometimes we get
argumentative too quickly because of past bad experiences.

Anyway, I've got to throw in my two cents worth, and if I only get a penny
for my thoughts, I want to know where the other penny goes. ;-) I'm
familiar with the term "group delay" as it's used in electronics, but have
seen it used far more often (a culture thing) to refer to groups of
mathematical routines, performed either in hardware, software, or a mix of
the two, when used in real-time systems. An implementation of an equation
has a group delay, because each component of the group (each low level
equation) has some delay, and the total delay (usually given as min, max,
and avg) of the group is often referred to as the "group delay." If the
group delay is too long for the application, you can either work on the top
level equation, or you can work on its component parts, changing hardware
for software, or whatever you need to do to get the speed up and the delay
down.

In many ways, this is the same thing as the group delay of an electronic
circuit, because it's the time delay caused by the DUT. There is even an
analog to phase (a good example of the way we use the same term to mean
different things), in that the output data may be fed back to the input
(closed-loop operation), and too much lag in arrival time (group delay) can
cause instability in the system. There are so many parallels that it might
be easy to confuse the two at times, but they do difference of course.
Feedback is still feedback, and group delay is still group delay, even if
we're talking about the human nervous system, but subtle differences seem to
trip us up sometimes. Qualifiers help, but it's often hard to see the need
for them, because *we* know what we mean, and miss the ambiguities.

The closest we can come to avoiding confusion is to address our audience
appropriately, but I've found that this is impossible on the Web (for me at
least) because the audience is infinitely varied, and sometimes
misrepresents its background and level of understanding. Not generally a
problem here with the RATs, but a problem with most of the other audio
related Web resources. I'm still new to the group, and only get to read the
posts once every few days, but I like what you guys have done with this
newsgroup. Best audio discussions on the Web.
Thanks RATs.

Now I'm puzzled a bit though. Someone, I think it was you Don, said that
group delay (in an analog system) was so named because you tested with a
group of signals. It's true that we do test group delay with a group of
data or signals that are representative of the data the DUT will be expected
to handle. However, I've never seen the term (and all that means is that
I've never seen it, not that it's never used this way) used to describe the
delay of a single element of a system. It's always used to describe the
total delay of a group of elements, and the group delay (often a curve if
the delay varies depending on the characteristics of the input) was often
calculated by summing the time delay of the individual "devices" in the
data/signal path. I could obviously pull a book off the shelf and look up a
"formal" definition, but that would only give me one technocrats
understanding of the usage, and that could be as wrong as the next fellows.
;-)

I'd like to see something more than "customary usage" here, because I've
been in a culture where the custom was to call the delay of a group of
devices the group delay, even if the group only carries a serial bit-stream
at a fixed rate. Here there is only one "signal" to test with, but you
still have a group delay. Would it be fair to say (I'm asking, not arguing)
that if it is customary for the audio community to use the term "group
delay" because it's tested with a group of signals, that this interpretation
is in reality a slight misuse of the term?

I've enjoyed the thread.

Thanks Don and Bob.

Party on,

Chuck

Don Pearce <don...@pearce.uk.com> wrote in message

news:2daa4usjudfgm2pmq...@4ax.com...

Jerry Avins

unread,
Jan 16, 2002, 4:29:18 PM1/16/02
to
Chuck wrote:
>
>
> ... I'm

> familiar with the term "group delay" as it's used in electronics, but have
> seen it used far more often (a culture thing) to refer to groups of
> mathematical routines, performed either in hardware, software, or a mix of
> the two, when used in real-time systems. An implementation of an equation
> has a group delay, because each component of the group (each low level
> equation) has some delay, and the total delay (usually given as min, max,
> and avg) of the group is often referred to as the "group delay." If the
> group delay is too long for the application, you can either work on the top
> level equation, or you can work on its component parts, changing hardware
> for software, or whatever you need to do to get the speed up and the delay
> down.

I suppose there are some circles where "group delay" describes the extra
time needed to get a bunch of people onto the bus, compared to jumping
in the car and going. The delay you described above is known as
"processing time" or "latency". I know what group delay means on a
transmission line (it's the line length divided by the group velocity),
and there are useful analogs of that in lumped-constant and certain
digital filters. Beyond that, it's beyond me.


>
> In many ways, this is the same thing as the group delay of an electronic
> circuit, because it's the time delay caused by the DUT. There is even an
> analog to phase (a good example of the way we use the same term to mean
> different things), in that the output data may be fed back to the input
> (closed-loop operation), and too much lag in arrival time (group delay) can
> cause instability in the system. There are so many parallels that it might
> be easy to confuse the two at times, but they do difference of course.
> Feedback is still feedback, and group delay is still group delay, even if
> we're talking about the human nervous system, but subtle differences seem to
> trip us up sometimes. Qualifiers help, but it's often hard to see the need
> for them, because *we* know what we mean, and miss the ambiguities.
>

Amen to that. My standard for my own technical writing (which I often
fall far short of) is to write so that even the most malicious reader
can't construe a meaning I don't intend. The "I know what that means so
you should too" trap is more than a pitfall. It often seems to be a
whirlpool actively trying to suck me in.

One has no business talking about group delay except when it can be
distinguished from phase delay. Otherwise, it's just technojargon for
"delay", and confuses a discussion with false erudition.

...

Don Pearce

unread,
Jan 16, 2002, 5:27:46 PM1/16/02
to
On Wed, 16 Jan 2002 15:28:25 -0500, "Chuck" <dv...@mindspring.com>
wrote:

OK - I think.

I would like to sum up group delay this way.
A) You put a signal in
B) you get the signal out.
Group delay is how long after A that B happens.

Incidentally - it was Charles Babbage, not Baggage. And I think Ada
Lovelace was Lord Byron's niece (or something similar).

And it wasn't me that said the thing about the delay of groups.

Gary Schafer

unread,
Jan 16, 2002, 5:24:06 PM1/16/02
to

Jerry Avins wrote:

> One has no business talking about group delay except when it can be
> distinguished from phase delay. Otherwise, it's just technojargon for
> "delay", and confuses a discussion with false erudition.
>
> ...
>
> Jerry
> --
> Engineering is the art of making what you want from things you can get.
> -----------------------------------------------------------------------

This is probably a simple definition of it.
"For linear phase responses, the group delay and the phase delay are identical,
and each may be
interpreted as time delay."

Regards
Gary


Bob_Stanton

unread,
Jan 16, 2002, 6:29:54 PM1/16/02
to
Gary Schafer <gsch...@mediaone.net> wrote in message

> 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

Bob writes

That is true.

If we have an unknown circuit inside a black box, the only way to know
the group delay, is to measure dB/dW.

Bob Stanton

Allan Herriman

unread,
Jan 16, 2002, 8:52:30 PM1/16/02
to

Err, thanks Jerry.

I was actually questioning why the minus sign had been dropped in this
thread, rather than asking why it should be there. (I'm never very
good at expressing simple things, so that the reader really knows what
I mean.)

I think the answer lies in the message header: this thread appears in
some groups that aren't as anally retentive as comp.dsp, and the odd
minus sign isn't seen as an issue.

Regards,
Allan.

Bob_Stanton

unread,
Jan 16, 2002, 10:05:03 PM1/16/02
to
Randy Yates <ya...@ieee.org> wrote in message
>
> That's the way I've always seen it defined (-d phi/ d freq).
>
> > Any ideas about the difference?
>
> Take a sine wave at f Hz, sin(2*pi*f*t) and delay it by tau seconds:
>
> sin(2*pi*f*(t - tau)) = sin(2*pi*f*t - 2*pi*f*tau)
>
> As f increases, the phase decreases, and therefore the slope is negative for a
> positive delay. Thus the negative sign on d phi/ d freq.

Bob writes

In most circuits the phase decreases as frequency increases. In some
circuits the phase seems to go positive, for part of the band.

I checking this out on several circuits, on a circuit analysis
program. One circuit was an elipitical lowpass filter. Near the
notches of the stopband, the phase increased.

Another circuit was an audio tone control. The phase increased from 20
to 280 Hz, decrease from 280 to 3000 Hz, then it again increased from
3500 to 20K. (I'm not 100% sure the circuit analysis program was
correct, but it did show phase increasing and decreasing with
frequency.)

A negative sign for "d phi", in the case where the phase is
increasing, would indicate *negative time*. Since a signal can not
leave a circuit before it enters, there is no such thing as negative
time, or it there?

If I'm wrong about there being no negative time, please answer this
question before I write it. :-) Thank you.

Bob Stanton

Chuck

unread,
Jan 16, 2002, 10:24:58 PM1/16/02
to
LOL, if the usage "Baggage" was consistent we'll have to blame it on a bad
click while running the spelling checker.

Yea, I think the Lord Byron bit is correct as well.

I love talking about the ladies. ;-)

Take care,

Chuck

Don Pearce <don...@pearce.uk.com> wrote in message

news:ravb4ugk9s2oik76m...@4ax.com...

robert bristow-johnson

unread,
Jan 17, 2002, 1:14:44 AM1/17/02
to
Jerry Avins <j...@ieee.org> wrote in message news:<3C45BEC9...@ieee.org>...

> Allan Herriman wrote:
> >
> > A number of posters in this thread have said that the group delay is
> > d phi/d freq
> >
> > My copy of O+S (1975 edition) says it's the additive inverse of that:
> > -d phi/d freq
> >
> > Any ideas about the difference?

the one with the negative sign is right (providing that "freq" is
measured as radians per unit time).

Jerry Avins wrote:

> One has no business talking about group delay except when it can be
> distinguished from phase delay. Otherwise, it's just technojargon for
> "delay", and confuses a discussion with false erudition.

and the two are distinguished from each other for any non-linear phase
filter.

this thread has appeared multiple times in the past and i'll repeat
what i've said before:


let x(t) = a(t)*cos(w0*t)

be input to a linear filter with complex transfer function H(s). a(t)
is a slowly moving envelope that is bandlimited to much less than w0.
that is


|A(j*w)| ~= 0 for all |w|>B where B << w0

if that is the case, then the output of the filter is

y(t) = |H(j*w0)| * a(t-Tg) * cos(w0*(t-Tp))

where Tp = -arg{H(j*w0)}/w0

and Tg = - d( arg{H(j*w)} )/dw evaluated at w = w0

that is the phase of the sinusoid is delayed by Tp, the "phase delay"
and the envelope of the sinusoid is delayed by Tg, the "group delay".

that is the only salient physical meaning of group delay vs. phase
delay that i can think of.

r b-j

Bob_Stanton

unread,
Jan 17, 2002, 10:44:01 AM1/17/02
to
Randy Yates <ya...@ieee.org> wrote in message

> > Any ideas about the difference?
>
> Take a sine wave at f Hz, sin(2*pi*f*t) and delay it by tau seconds:
>
> sin(2*pi*f*(t - tau)) = sin(2*pi*f*t - 2*pi*f*tau)
>
> As f increases, the phase decreases, and therefore the slope is negative for a
> positive delay. Thus the negative sign on d phi/ d freq.

Bob writes

(The last message I wrote didn't come up, if it does than this message
will be redundant.)


I have seen some active filters, (looking on a circuit analysis
program) where the phase shift is positive in a portion of the
stopband. (It may be that my circuit analysis program is wrong.) If
the analysis program is correct however, there should be no negative
sign on "d phi/d freq".

A negative sign on "d phi/d freq", would imply that time is running
backwards, at those frequenies where the phase shift is positive.

Bob Stanton

Randy Yates

unread,
Jan 17, 2002, 5:46:38 PM1/17/02
to
Bob_Stanton wrote:
>
> Randy Yates <ya...@ieee.org> wrote in message
> >
> > That's the way I've always seen it defined (-d phi/ d freq).
> >
> > > Any ideas about the difference?
> >
> > Take a sine wave at f Hz, sin(2*pi*f*t) and delay it by tau seconds:
> >
> > sin(2*pi*f*(t - tau)) = sin(2*pi*f*t - 2*pi*f*tau)
> >
> > As f increases, the phase decreases, and therefore the slope is negative for a
> > positive delay. Thus the negative sign on d phi/ d freq.
>
> Bob writes
>
> In most circuits the phase decreases as frequency increases. In some
> circuits the phase seems to go positive, for part of the band.
>
> I checking this out on several circuits, on a circuit analysis
> program. One circuit was an elipitical lowpass filter. Near the
> notches of the stopband, the phase increased.
>
> Another circuit was an audio tone control. The phase increased from 20
> to 280 Hz, decrease from 280 to 3000 Hz, then it again increased from
> 3500 to 20K. (I'm not 100% sure the circuit analysis program was
> correct, but it did show phase increasing and decreasing with
> frequency.)
>
> A negative sign for "d phi", in the case where the phase is
> increasing, would indicate *negative time*.

That is indeed what the group delay would be in that context.

> Since a signal can not
> leave a circuit before it enters, there is no such thing as negative
> time, or it there?

You're interpreting group delay wrong. It does NOT indicate absolute
time delay unless the phase response is linear. In the cases you
specify, the phase response is not linear.
--
Randy Yates
DSP Engineer, Sony Ericsson Mobile Communications
Research Triangle Park, NC, USA
randy...@ericsson.com, 919-472-1124


.

Jerry Avins

unread,
Jan 18, 2002, 11:01:02 AM1/18/02
to
robert bristow-johnson wrote:
>
...

>
> let x(t) = a(t)*cos(w0*t)
>
> be input to a linear filter with complex transfer function H(s). a(t)
> is a slowly moving envelope that is bandlimited to much less than w0.
> that is
>
> |A(j*w)| ~= 0 for all |w|>B where B << w0
>
> if that is the case, then the output of the filter is
>
> y(t) = |H(j*w0)| * a(t-Tg) * cos(w0*(t-Tp))
>
> where Tp = -arg{H(j*w0)}/w0
>
> and Tg = - d( arg{H(j*w)} )/dw evaluated at w = w0
>
> that is the phase of the sinusoid is delayed by Tp, the "phase delay"
> and the envelope of the sinusoid is delayed by Tg, the "group delay".
>
> that is the only salient physical meaning of group delay vs. phase
> delay that i can think of.
>
> r b-j

Hallelujah!

Jerry Avins

unread,
Jan 18, 2002, 11:16:49 AM1/18/02
to
Don Pearce wrote:
>
...

>
> I would like to sum up group delay this way.
> A) You put a signal in
> B) you get the signal out.
> Group delay is how long after A that B happens.
>
> Incidentally - it was Charles Babbage, not Baggage. And I think Ada
> Lovelace was Lord Byron's niece (or something similar).
>
> And it wasn't me that said the thing about the delay of groups.
>
> d
>
> _____________________________
> Telecommunications consultant
> http://www.pearce.uk.com

Augusta Ada Byron King, Countess of Lovelace, was the poet Lord Byron's
legitimate daughter. Her parents separated soon after her birth and she
never knew either of them. "Niece or something similar" is a pretty (but
I imagine inadvertently) good description.

Bob_Stanton

unread,
Jan 18, 2002, 12:47:29 PM1/18/02
to
Randy Yates <eus...@rtp.ericsson.com> wrote in message

> You're interpreting group delay wrong. It does NOT indicate absolute
> time delay unless the phase response is linear. In the cases you
> specify, the phase response is not linear.
> --
> Randy Yates

Bob writes:

I think you may have it backwards:

Tpd = - B/W
Tpd is the absolute time delay, *only* if the phase response is linear.

T = -(dB/dW)
T is the absolute time delay, whether the phase response is linear or non-linear.

Bob Stanton

robert bristow-johnson

unread,
Jan 18, 2002, 12:49:53 PM1/18/02
to
In article <3C4846BE...@ieee.org> , Jerry Avins <j...@ieee.org> wrote:

>
> Hallelujah!
>

and Puhraze the Laward and pass that ammunition!

> Jerry

:-\

r b-j


Randy Yates

unread,
Jan 18, 2002, 1:01:34 PM1/18/02
to
Bob_Stanton wrote:
>
> Randy Yates <eus...@rtp.ericsson.com> wrote in message
>
> > You're interpreting group delay wrong. It does NOT indicate absolute
> > time delay unless the phase response is linear. In the cases you
> > specify, the phase response is not linear.
> > --
> > Randy Yates
>
> Bob writes:
>
> I think you may have it backwards:

I am absolutely positive that I do not have it backwards.

Dave Martindale

unread,
Jan 18, 2002, 1:45:23 PM1/18/02
to
rsta...@stny.rr.com (Bob_Stanton) writes:

>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.

There's a difference between saying you can't *measure* group delay
using a single frequency, and saying that you can't *know* group delay
at a single frequency. The former doesn't imply the latter.

>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.

If you know the components in the circuit, you know the *complete* shape
of the phase curve, at infinitely many frequencies. Using calculus, you
can calculate the slope at any single frequency. But that's because you
already know the phase at *all* frequencies. It's when you don't know
the shape of the phase curve, and you're trying to measure it, that you
need more than one measurement.

>Can we say a continueous sine wave takes a certain time to pass
>through a circuit? If I go down to K-Mart, 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.

Yeah, but that's not a continuous waveform. You turned on the beam
abruptly, so you can tell when the same photons return. The equivalent
in an electronic circuit is abruptly connecting a signal source. But
that isn't the "continuous" unmodulated sine wave that the discussion
has been about.

>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.

I don't see why you can't say that the sine wave takes 500 us to pass
through - the delay at any given frequency is well-defined. You just
can't *measure* it using a single-frequency unmodulated waveform.

Dave

Bob_Stanton

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Jan 18, 2002, 8:57:57 PM1/18/02
to
Randy Yates <eus...@rtp.ericsson.com> wrote in message news:<3C4862FE...@rtp.ericsson.com>...
> Bob_Stanton wrote:
> >

> > Bob wrote:
> >
> > I think you may have it backwards:
>

> Randy: I am absolutely positive that I do not have it backwards.

Bob writes

OK, let's look at a circuit. The one I put up before, a low pass
filter consisting of: a 1000 Ohm resistor and a 1uF capacitor.

The group delay at 159.15 Hz was 500 usec. This group delay number
tells us how long the signal (on the modulation envelope) is delayed.
If there was a pure sine wave going through the filter and I increased
the level, it would take 500 usec before the level at the output of
the filter starts to increase. The information (data) the I imparted
to the sine wave, took 500 usec to travel through the filter.


Now let's look at phase delay:

Tpd = - (ang/360)*(1/F)
The phase shift at 159.15 Hz. was -45 degs.
Tpd = -(-45/360) * 1/159.15
Tpd = 0.125 * 0.0062834
Tpd = 785.4 usec.

I believe you think of absolute delay as: the time it takes the
carrier to pass through the filter.

If it took the modulation information only 500 usec to pass through
the filter, why did it take the carrier 785.4 usec? Information can't
go faster than the carrier that is carrying it.

If *information* can pass through this filter in 500 usec, than the
carrier of that information must also be passing through the filter in
500 usec, not in 785.4 usec.


Bob Stanton

Randy Yates

unread,
Jan 18, 2002, 10:09:38 PM1/18/02
to
Bob_Stanton wrote:
>
> If *information* can pass through this filter in 500 usec, than the
> carrier of that information must also be passing through the filter in
> 500 usec, not in 785.4 usec.

Using your logic, information at certain carrier frequencies will come through
some filters (the ones which have a temporarily positive phase function slope)
*ahead* of the time it came in.

I think you're thinking about this too hard. Simply apply the definitions.
A single sine wave at 159.15 Hz will indeed have an input-to-output delay
of 785 usec.

Bob_Stanton

unread,
Jan 19, 2002, 2:19:04 PM1/19/02
to
Randy Yates <ya...@ieee.org> wrote in message news:<3C48E372...@ieee.org>...

> Bob_Stanton wrote:
> >
> > If *information* can pass through this filter in 500 usec, than the
> > carrier of that information must also be passing through the filter in
> > 500 usec, not in 785.4 usec.
>
> Using your logic, information at certain carrier frequencies will come through
> some filters (the ones which have a temporarily positive phase function slope)
> *ahead* of the time it came in.
>

Bob writes

I did not say information could come out before it goes in. (But, that
would be nice.) My point was, dB/dW tells how long it takes for
information to *pass through* a device. If you put information in
point A, (dB/dW)sec latter it comes out point B.

For example, suppose you put a 53.05 Hz square wave into the filter.
(A square wave is composed of a fundamental and an infinite number of
odd haromics.) The fudamental (53.05 Hz), will take 900 usec to pass
through. The third harmonic (159.15 Hz), will arive at the output
after 500 usec. The 5th harmonic (265.25 Hz), will reach the output
after 265 usec, etc.

Note that the 159.15 Hz (3rd harmonic), takes 500 usec to pass through
the filter (not 785 usec).

The 785 usec number has only *one* physical meaning. It means if you
are look a 159.15 Hz sine wave, on a dual trace scope, the output
trace will be 785 usec after the input trace. This number (785 usec)
is *not* the time it takes a signal to pass through the filter.

As for positive going phase shifts: I was saying that information *CAN
NOT* come out before it goes in. Therefore, a negative sign in front
of (dB/dW) may give *wrong* results.

> Randy wrote


> I think you're thinking about this too hard.

Bob writes
That reminds me of what Yogi Berra once said: "If you don't think too
good, don't think too much." :-) We all think too much.

Bob Stanton

Randy Yates

unread,
Jan 19, 2002, 3:36:13 PM1/19/02
to
Bob_Stanton wrote:
>
> Randy Yates <ya...@ieee.org> wrote in message news:<3C48E372...@ieee.org>...
> > Bob_Stanton wrote:
> > >
> > > If *information* can pass through this filter in 500 usec, than the
> > > carrier of that information must also be passing through the filter in
> > > 500 usec, not in 785.4 usec.
> >
> > Using your logic, information at certain carrier frequencies will come through
> > some filters (the ones which have a temporarily positive phase function slope)
> > *ahead* of the time it came in.
> >
>
> Bob writes
>
> I did not say information could come out before it goes in. (But, that
> would be nice.) My point was, dB/dW tells how long it takes for
> information to *pass through* a device. If you put information in
> point A, (dB/dW)sec latter it comes out point B.

What is B? What is W? What is point A? What is point B? Is point
B the same as the B in your differential? What is "information?"

Bob, your semantic integrity is falling completely apart. This
makes it impossible to assign any meaning to your statements.

> For example, suppose you put a 53.05 Hz square wave into the filter.
> (A square wave is composed of a fundamental and an infinite number of
> odd haromics.) The fudamental (53.05 Hz), will take 900 usec to pass
> through. The third harmonic (159.15 Hz), will arive at the output
> after 500 usec. The 5th harmonic (265.25 Hz), will reach the output
> after 265 usec, etc.
>
> Note that the 159.15 Hz (3rd harmonic), takes 500 usec to pass through
> the filter (not 785 usec).

Would you care to put a wager on that?

I'm not sure where your confusion lies, but you are indeed confused.
Do you not know that the 3rd harmonic of the aforementioned square
wave is indeed a sine? As long as the circuit is linear, then
that sine will behave exactly the same as part of a square wave
as it does alone.

My assertion stands, and I can prove it if I need to.

Randy Yates

unread,
Jan 19, 2002, 3:52:25 PM1/19/02
to
Bob_Stanton wrote:

> The 785 usec number has only *one* physical meaning. It means if you
> are look a 159.15 Hz sine wave, on a dual trace scope, the output
> trace will be 785 usec after the input trace. This number (785 usec)
> is *not* the time it takes a signal to pass through the filter.

It's not? You mean it's actually taking 500 usec to pass through the
filter but somehow, magically, the scope is tracing it as though it
were taking 785 usec? How does the scope know this is really the input
and output of this filter and not the input and output of a 500 usec
delay line?

Bob_Stanton

unread,
Jan 19, 2002, 10:47:03 PM1/19/02
to
Randy Yates <ya...@ieee.org> wrote in message news:<3C49D8BD...@ieee.org>...

> Bob_Stanton wrote:
> >
> > Randy Yates <ya...@ieee.org> wrote in message news:<3C48E372...@ieee.org>...
> > > Bob_Stanton wrote:
> > > >
> > > > If *information* can pass through this filter in 500 usec, than the
> > > > carrier of that information must also be passing through the filter in
> > > > 500 usec, not in 785.4 usec.
> > >
> > > Using your logic, information at certain carrier frequencies will come through
> > > some filters (the ones which have a temporarily positive phase function slope)
> > > *ahead* of the time it came in.
> > >
> >
> > Bob writes
> >
> > I did not say information could come out before it goes in. (But, that
> > would be nice.) My point was, dB/dW tells how long it takes for
> > information to *pass through* a device. If you put information in
> > point A, (dB/dW)sec latter it comes out point B.
>
> What is B? What is W? What is point A? What is point B? Is point
> B the same as the B in your differential? What is "information?"
>
> Bob, your semantic integrity is falling completely apart. This
> makes it impossible to assign any meaning to your statements.
>


Bob writes

The "B" is supposed to be the greek leter Beta. If you ever saw the
greek leter Beta you would notice it looks very similar to the english
letter "B"

The "W" was supposed to be the greek leter "lower case omega". If you
ever saw lower case omega, you would notice it looks very much like
the english letter "w". (I use english letters, because my keyboard
doesn't have greek letters.)

Some filter books use the formula: "Tgd = -dB/dw". It is an alternate
form for group delay.
w = 2 pi f
B is the phase shift in radians/sec

As far as using the words "point A to point B": In the future I will
use: "the input to the output". I'm sorry, I thought it was obvious
from the context, what "from point A to point B" meant.

"Information", in the context of group delay, is carried in the
envelope waveform.

The delay of the envelope waveform, is called group delay. Conversely,
this group delay is how long it takes envelope "information" to pass
through the filter.

> > For example, suppose you put a 53.05 Hz square wave into the filter.


> > (A square wave is composed of a fundamental and an infinite number of
> > odd haromics.) The fudamental (53.05 Hz), will take 900 usec to pass
> > through. The third harmonic (159.15 Hz), will arive at the output
> > after 500 usec. The 5th harmonic (265.25 Hz), will reach the output
> > after 265 usec, etc.
> >
> > Note that the 159.15 Hz (3rd harmonic), takes 500 usec to pass through
> > the filter (not 785 usec).
>
> Would you care to put a wager on that?
>
> I'm not sure where your confusion lies, but you are indeed confused.
> Do you not know that the 3rd harmonic of the aforementioned square
> wave is indeed a sine?

I know that all the odd harmonics, which make up a square wave, are
sine waves. B.T.W. The 3rd harmonic (sine wave) takes 500 usec to go
from the input to the output of the filter, not 785 usec.

> As long as the circuit is linear, then
> that sine will behave exactly the same as part of a square wave
> as it does alone.

The statement above does not apply. The filter is *not phase linear*.
Because the filter is *non-phase linear*, the formula Tpd = -B/w will
not give the correct results for the delay time of the harmonics of a
square wave.

>
> My assertion stands, and I can prove it if I need to.

Which assertion?

Bob Stanton

Randy Yates

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Jan 20, 2002, 8:22:01 AM1/20/02
to
Bob_Stanton wrote:

> B is the phase shift in radians/sec

Phase shift, that is, a shift in phase, should have
purely angular units, not angle per time. Your
definition makes no sense.

> B.T.W. The 3rd harmonic (sine wave) takes 500 usec to go
> from the input to the output of the filter, not 785 usec.

Ahh, I see - proof by assertion! Bob, do you believe that
if you repeat the statement enough times it will somehow
magically come true?

> > As long as the circuit is linear, then
> > that sine will behave exactly the same as part of a square wave
> > as it does alone.
>
> The statement above does not apply. The filter is *not phase linear*.

Did I say "phase linear?" No. I said "linear." These are two entirely
different concepts. If you don't understand the difference, I would
suggest you do some reading on linear system theory. The introductory
chapters in Oppenheim et al.'s "Signals and Systems" is an excellent
place to start.

> Because the filter is *non-phase linear*, the formula Tpd = -B/w will
> not give the correct results for the delay time of the harmonics of a
> square wave.

If B(w) is the phase response of the system at frequency w (w = 2*pi*f),
then -B(w)/w will be the delay, in seconds, of a sine wave of radian
frequency w.

Bob, why don't you try it? That is, why don't you build the circuit
and measure it?

> > My assertion stands, and I can prove it if I need to.
>
> Which assertion?

If a system is linear with phase response B(w), then -B(w)/w is
the amount of time it takes for a sine wave of frequency w to
propagate through the system, whether that system is phase
linear or not.

Bob_Stanton

unread,
Jan 20, 2002, 7:05:24 PM1/20/02