Thread Subject: Physical meaning of Group-Delay ?

(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

dummy21@gmx.net (Nico) wrote in message news:<a1a88ba8.0201100623.3957a061@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)
>
> 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.

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

dummy21@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) floyd@barrow.com

On 11 Jan 2002 03:05:07 -0900, Floyd Davidson <floyd@ptialaska.net>
wrote:

>dummy21@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

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
rhhardin@mindspring.com

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

In article <87zo3lgkjw.fld@barrow.com> , Floyd Davidson
<floyd@ptialaska.net> wrote:

> dummy21@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) floyd@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 robert@wavemechanics.com

--

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 <dummy21@gmx.net> wrote in message
news:a1a88ba8.0201100623.3957a061@posting.google.com...
> 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.

dummy21@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.)

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

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 <physics@bellsouth.net> wrote in message
news:3C3F5DF1.9F727662@bellsouth.net...
>
>
> 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
>

Ron Hardin <rhhardin@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.

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

On 13 Jan 2002 07:56:56 GMT, gah@ugcs.caltech.edu (glen
herrmannsfeldt) wrote:

>Ron Hardin <rhhardin@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.

d

_____________________________
Telecommunications consultant
http://www.pearce.uk.com

Don Pearce <donald@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

On 14 Jan 2002 04:26:23 -0800, rstanton2@stny.rr.com (Bob_Stanton)
wrote:

>Don Pearce <donald@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

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

_____________________________
Telecommunications consultant
http://www.pearce.uk.com

In article <a1rek8$gsg@gap.cco.caltech.edu>,
glen herrmannsfeldt <gah@ugcs.caltech.edu> wrote:
>Ron Hardin <rhhardin@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 jj@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.

jj@research.att.com (jj, DBT thug and skeptical philalethist) writes:

>In article <a1rek8$gsg@gap.cco.caltech.edu>,
>glen herrmannsfeldt <gah@ugcs.caltech.edu> wrote:
>>Ron Hardin <rhhardin@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

Don Pearce <donald@pearce.uk.com> wrote in message news:<gcm54uokrfpj0vi1b80toe1coai2a79pvm@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

  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!
Nico wrote:

> (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

On 14 Jan 2002 15:53:10 -0800, rstanton2@stny.rr.com (Bob_Stanton)
wrote:

>Don Pearce <donald@pearce.uk.com> wrote in message news:<gcm54uokrfpj0vi1b80toe1coai2a79pvm@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

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).

d

_____________________________
Telecommunications consultant
http://www.pearce.uk.com

Don Pearce wrote:

> On 14 Jan 2002 15:53:10 -0800, rstanton2@stny.rr.com (Bob_Stanton)
> wrote:
>
> >Don Pearce <donald@pearce.uk.com> wrote in message news:<gcm54uokrfpj0vi1b80toe1coai2a79pvm@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
>
> 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).
>
> d
>

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

On Tue, 15 Jan 2002 18:08:47 GMT, Gary Schafer <gschafer@mediaone.net>
wrote:

>
>
>Don Pearce wrote:
>
>> On 14 Jan 2002 15:53:10 -0800, rstanton2@stny.rr.com (Bob_Stanton)
>> wrote:
>>
>> >Don Pearce <donald@pearce.uk.com> wrote in message news:<gcm54uokrfpj0vi1b80toe1coai2a79pvm@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
>>
>> 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).
>>
>> d
>>
>
>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.

d

_____________________________
Telecommunications consultant
http://www.pearce.uk.com

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.
-----------------------------------------------------------------------
Mark Disher wrote:
>
> 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!
> Nico wrote:
>
> > (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

On Tue, 15 Jan 2002 14:19:11 -0500, Jerry Avins <jya@ieee.org> wrote:

>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.

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.

d

_____________________________
Telecommunications consultant
http://www.pearce.uk.com

Don Pearce wrote:
>
> On Tue, 15 Jan 2002 14:19:11 -0500, Jerry Avins <jya@ieee.org> wrote:
>
> >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.
>
> 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.
>
> d
>
> _____________________________
> Telecommunications consultant
> http://www.pearce.uk.com

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

Jerry,
  Your explanation seems much better than mine, not least because you're talking
about a VISIBLE physical phenomena. Well said!

Jerry Avins wrote:

> 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.
> -----------------------------------------------------------------------
> Mark Disher wrote:
> >
> > 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!
> > Nico wrote:
> >
> > > (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

Don Pearce <donald@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

Jerry Avins <jya@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

glen herrmannsfeldt wrote:
>
> Jerry Avins <jya@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.
-----------------------------------------------------------------------

Bob_Stanton wrote:

> Don Pearce <donald@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

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

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.

On 10 Jan 2002 06:23:47 -0800, dummy21@gmx.net (Nico) wrote:

>(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

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."
%%%% <yates@ieee.org> % 'Big Wheels', *Out of the Blue*, ELO
http://personal.rdu.bellsouth.net/~yatesc

On 15 Jan 2002 15:15:56 -0800, rstanton2@stny.rr.com (Bob_Stanton)
wrote:

>Don Pearce <donald@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

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.

d

_____________________________
Telecommunications consultant
http://www.pearce.uk.com

In article <67ef4d54.0201151515.151660ff@posting.google.com>,
Bob_Stanton <rstanton2@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 :)

--
Copyright jj@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.

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

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 <donald@pearce.uk.com> wrote in message
news:2daa4usjudfgm2pmqrsjg3el3rf0ubo8l9@4ax.com...
> On 15 Jan 2002 15:15:56 -0800, rstanton2@stny.rr.com (Bob_Stanton)
> wrote:
>
> >Don Pearce <donald@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
>
> 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.
>
> d
>
> _____________________________
> Telecommunications consultant
> http://www.pearce.uk.com

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.

  ...

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

On Wed, 16 Jan 2002 15:28:25 -0500, "Chuck" <dvman@mindspring.com>
wrote:

>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 <donald@pearce.uk.com> wrote in message
>news:2daa4usjudfgm2pmqrsjg3el3rf0ubo8l9@4ax.com...
>> On 15 Jan 2002 15:15:56 -0800, rstanton2@stny.rr.com (Bob_Stanton)
>> wrote:
>>
>> >Don Pearce <donald@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
>>
>> 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.
>>
>> d
>>
>> _____________________________
>> Telecommunications consultant
>> http://www.pearce.uk.com
>
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.

d

_____________________________
Telecommunications consultant
http://www.pearce.uk.com

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

Gary Schafer <gschafer@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

On Wed, 16 Jan 2002 12:56:25 -0500, Jerry Avins <jya@ieee.org> wrote:

>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?

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.

Randy Yates <yates@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

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 <donald@pearce.uk.com> wrote in message
news:ravb4ugk9s2oik76mk27b78v8p299b6rh3@4ax.com...
> On Wed, 16 Jan 2002 15:28:25 -0500, "Chuck" <dvman@mindspring.com>
> wrote:
>
> >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 cultu
re
> >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 <donald@pearce.uk.com> wrote in message
> >news:2daa4usjudfgm2pmqrsjg3el3rf0ubo8l9@4ax.com...
> >> On 15 Jan 2002 15:15:56 -0800, rstanton2@stny.rr.com (Bob_Stanton)
> >> wrote:
> >>
> >> >Don Pearce <donald@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
> >>
> >> 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.
> >>
> >> d
> >>
> >> _____________________________
> >> Telecommunications consultant
> >> http://www.pearce.uk.com
> >
> 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.
>
> d
>
> _____________________________
> Telecommunications consultant
> http://www.pearce.uk.com

Jerry Avins <jya@ieee.org> wrote in message news:<3C45BEC9.7DC6840E@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

Randy Yates <yates@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

Bob_Stanton wrote:
>
> Randy Yates <yates@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.yates@ericsson.com, 919-472-1124
















































.

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

Randy Yates <euschya@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

In article <3C4846BE.947A63C8@ieee.org> , Jerry Avins <jya@ieee.org> wrote:

>
> Hallelujah!
>

and Puhraze the Laward and pass that ammunition!

> Jerry

:-\

r b-j

Bob_Stanton wrote:
>
> Randy Yates <euschya@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.
--
Randy Yates
DSP Engineer, Sony Ericsson Mobile Communications
Research Triangle Park, NC, USA
randy.yates@ericsson.com, 919-472-1124

rstanton2@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

Randy Yates <euschya@rtp.ericsson.com> wrote in message news:<3C4862FE.5B54FB54@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

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.
--
% Randy Yates % "...the answer lies within your soul
%% Fuquay-Varina, NC % 'cause no one knows which side
%%% 919-577-9882 % the coin will fall."
%%%% <yates@ieee.org> % 'Big Wheels', *Out of the Blue*, ELO
http://personal.rdu.bellsouth.net/~yatesc

Randy Yates <yates@ieee.org> wrote in message news:<3C48E372.40176D4C@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

Bob_Stanton wrote:
>
> Randy Yates <yates@ieee.org> wrote in message news:<3C48E372.40176D4C@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 % "...the answer lies within your soul
%% Fuquay-Varina, NC % 'cause no one knows which side
%%% 919-577-9882 % the coin will fall."
%%%% <yates@ieee.org> % 'Big Wheels', *Out of the Blue*, ELO
http://personal.rdu.bellsouth.net/~yatesc

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?
--
% Randy Yates % "...the answer lies within your soul
%% Fuquay-Varina, NC % 'cause no one knows which side
%%% 919-577-9882 % the coin will fall."
%%%% <yates@ieee.org> % 'Big Wheels', *Out of the Blue*, ELO
http://personal.rdu.bellsouth.net/~yatesc

Randy Yates <yates@ieee.org> wrote in message news:<3C49D8BD.E5A2F791@ieee.org>...
> Bob_Stanton wrote:
> >
> > Randy Yates <yates@ieee.org> wrote in message news:<3C48E372.40176D4C@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

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.
--
% Randy Yates % "...the answer lies within your soul
%% Fuquay-Varina, NC % 'cause no one knows which side
%%% 919-577-9882 % the coin will fall."
%%%% <yates@ieee.org> % 'Big Wheels', *Out of the Blue*, ELO
http://personal.rdu.bellsouth.net/~yatesc

Randy Yates <yates@ieee.org> wrote in message news:<3C4AC479.2DB87E12@ieee.org>...
> 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.
>

It's not 'my' definition. "Tpd = -B/w" comes from "Electronic Filter
Design Handbook", by Arthur B. Williams. McGraw-Hill Book Company.
Copyright 1981. Pg.2-20

The fact you are unfamiliar with this common formula, and that it
"mades no sence" to you, indicates a certain lack of understanding.

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

Oh yes! It's 500 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.
> >
> Did I say "phase linear?" No. I said "linear."
 
Right. I somehow misread your statement as: phase linear.
And I agree with your statement, the 3rd harmonic will behave exactly
as a sine 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.
>

What if a black box contained an ideal autotransformer, wired for 180
phase reversal. You look at the input and output waves, and see that
the output wave 'lags' the input wave by 180 degrees. Could you use
your formula to calculate the delay at 159.15 Hz? What would it be?
 
> Bob, why don't you try it? That is, why don't you build the circuit
> and measure it?
>

If I built the circuit, I would see, on a dual trace oscilloscope, the
output wave laging the input by 785 usec. So I don't need to build it.

The 785 usec is not how long envelope information (on a 159.15
carrier) takes to travel through the filter.


> > > 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.

You said if a system is "linear with phase response". Do you mean by
that "phase linear"?

Bob Stanton

Bob_Stanton wrote:
>
> Randy Yates <yates@ieee.org> wrote in message news:<3C4AC479.2DB87E12@ieee.org>...
> > 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.
> >
>
> It's not 'my' definition. "Tpd = -B/w" comes from "Electronic Filter
> Design Handbook", by Arthur B. Williams. McGraw-Hill Book Company.
> Copyright 1981. Pg.2-20
>

Given the limitations of plain ASCII text (but the standard ISO-8859-1
allows ß, ±, ², and a few others), I assume that "Tpd" stands for time
of phase delay. If so, it is correct. It is also _not_ group delay.

> The fact you are unfamiliar with this common formula, and that it
> "mades no sence" to you, indicates a certain lack of understanding.
>
  ...
>
> If I built the circuit, I would see, on a dual trace oscilloscope, the
> output wave laging the input by 785 usec. So I don't need to build it.

The pope refused to look through Galileo's telescope, insisting it would
show nothing. Try it: you may be surprised.
>
> The 785 usec is not how long envelope information (on a 159.15
> carrier) takes to travel through the filter.
>
> > > > 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.
>
> You said if a system is "linear with phase response". Do you mean by
> that "phase linear"?

"Linear with phase response" is not the kind of nonsense description I
expect from Randy. Are those really his words? Are there words around it
that could give it valid meaning? "Phase linear" is another thing
altogether, not related to the kind of linearity related to harmonic
generation and such. A linear system is one in which superposition
holds. A property of such systems is that it responds to a pure sinusoid
with a pure sinusoid of the same frequency. Linearity does not determine
the amplitude or phase of the response.
>
> Bob Stanton

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

Bob_Stanton wrote:
>
> Randy Yates <yates@ieee.org> wrote in message news:<3C4AC479.2DB87E12@ieee.org>...
> > 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.
> >
>
> It's not 'my' definition. "Tpd = -B/w" comes from "Electronic Filter
> Design Handbook", by Arthur B. Williams. McGraw-Hill Book Company.
> Copyright 1981. Pg.2-20

That much I agree with. The part that doesn't make sense is
the units you stated, radians/sec. Tpd should have units of
time, right? So if B is in units of radians/sec, and w is
in radians/sec, then the units of -B/w are radians/sec over
radians/sec, or no units.

You made an error if B is supposed to represent phase. The
units should be simply radians. Then the units of the expression
-B/w would be radians/radians/sec = sec, which makes sense.

I stand by my comment: you definition makes no sense.

> The fact you are unfamiliar with this common formula, and that it
> "mades no sence" to you, indicates a certain lack of understanding.

This is absolutely ludricrous. You're accusing me of possessing
a lack of understanding for pointing out an error in your
understanding. Amazing.

> > Ahh, I see - proof by assertion! Bob, do you believe that
> > if you repeat the statement enough times it will somehow
> > magically come true?
>
> Oh yes! It's 500 usec! :-)

Which proves NOTHING.

> > > > 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.
> > >
> > Did I say "phase linear?" No. I said "linear."
>
> Right. I somehow misread your statement as: phase linear.

Somehow? Hmmm.

> And I agree with your statement, the 3rd harmonic will behave exactly
> as a sine wave.

Oh really? So now you're doing an about face? How about some accountability
then for the last 4 or 5 levels of this thread? Something like, "I'm sorry,
I was wrong and was calling you ignorant when I really was the one that was
ignorant."?

> > 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.
> >
>
> What if a black box contained an ideal autotransformer, wired for 180
> phase reversal. You look at the input and output waves, and see that
> the output wave 'lags' the input wave by 180 degrees. Could you use
> your formula to calculate the delay at 159.15 Hz? What would it be?
>
> > Bob, why don't you try it? That is, why don't you build the circuit
> > and measure it?
> >
>
> If I built the circuit, I would see, on a dual trace oscilloscope, the
> output wave laging the input by 785 usec. So I don't need to build it.
>
> The 785 usec is not how long envelope information (on a 159.15
> carrier) takes to travel through the filter.

Did I ever say it was? No. Part of my point in arduously discussing
this with you since last week is that the phase response at frequency
w ALWAYS (indirectly) indicates the amount of time for a sine wave of
frequency w to propagate through the filter, WHETHER OR NOT THAT FILTER
IS PHASE LINEAR. This is to refute the following erroneous statement which
you made a few posts back:

  Tpd is the absolute time delay, *only* if the phase response is linear.

> > > > 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.
>
> You said if a system is "linear with phase response". Do you mean by
> that "phase linear"?

No.

What I said was: "If a system is linear with phase response B(w), then
[...]". What I meant was this: Let a system be linear in the sense that
it possesses the properties of homogeneity and additivity. Then this
system has a phase response and I will denote that phase response as
B(w). I am not saying whether the phase response is linear or nonlinear.

It is exceedingly obvious that you have a hole in your understanding
of signal processing systems when it comes to linear system theory.
The following web site gives a brief introduction to the salient points
of the theory:

http://white.stanford.edu/~heeger/linear-systems/linear-systems.html
--
% Randy Yates % "...the answer lies within your soul
%% Fuquay-Varina, NC % 'cause no one knows which side
%%% 919-577-9882 % the coin will fall."
%%%% <yates@ieee.org> % 'Big Wheels', *Out of the Blue*, ELO
http://personal.rdu.bellsouth.net/~yatesc

rstanton2@stny.rr.com (Bob_Stanton) writes:
(snip)
>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.

The definition of derivative requires two points, though infinitely
close together. If you have a function that is not continuous
anywhere then you can't calculate the derivative.

>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.

A light beam isn't a continuous sine wave, in the sense implied here,
unless it exists from t=-infinity to t=+infinity. In optics, the
coherence length or coherence time are used to measure this.
The coherence length, usually of a laser, tells over what path
length difference a laser beam can interfere with itself.

>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.

Oh well.

-- glen

Randy Yates <yates@ieee.org> wrote in message news:<3C4B8E09.5040A0BB@ieee.org>...
> Bob_Stanton wrote:
> >
> > Randy Yates <yates@ieee.org> wrote in message news:<3C4AC479.2DB87E12@ieee.org>...
> > > 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.
> > >
> >
> > It's not 'my' definition. "Tpd = -B/w" comes from "Electronic Filter
> > Design Handbook", by Arthur B. Williams. McGraw-Hill Book Company.
> > Copyright 1981. Pg.2-20
>
> That much I agree with. The part that doesn't make sense is
> the units you stated, radians/sec. Tpd should have units of
> time, right? So if B is in units of radians/sec, and w is
> in radians/sec, then the units of -B/w are radians/sec over
> radians/sec, or no units.
>
> You made an error if B is supposed to represent phase. The
> units should be simply radians. Then the units of the expression
> -B/w would be radians/radians/sec = sec, which makes sense.
>
> I stand by my comment: you definition makes no sense.
>

Bob writes


All I did was quote a page from a book. A book which makes perfect
sence to me, and "makes no sence" to you. I'm sorry you can't
understand. If you have access to a technical library, read it for
yourself. I'm not going to try to explain it to you, step by simple
step.

> This is absolutely ludricrous. You're accusing me of possessing
> a lack of understanding for pointing out an error in your
> understanding. Amazing.
>

Bob writes

I think you have a high opinion of yourself. :-)


> > > Ahh, I see - proof by assertion! Bob, do you believe that
> > > if you repeat the statement enough times it will somehow
> > > magically come true?
> >
> > Oh yes! It's 500 usec! :-)
>
> Which proves NOTHING.

It is indeed 500 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.
> > > >

> > > Did I say "phase linear?" No. I said "linear."
> >
> > Right. I somehow misread your statement as: phase linear.
>
> Somehow? Hmmm.

I thought you were trying to say something less obvious, so I read it
wrong.

>
> > And I agree with your statement, the 3rd harmonic will behave exactly
> > as a sine wave.
>
> Oh really? So now you're doing an about face? How about some accountability
> then for the last 4 or 5 levels of this thread? Something like, "I'm sorry,
> I was wrong and was calling you ignorant when I really was the one that was
> ignorant."?
>

I didn't call you "ignorant". (Even if you may be, just a little bit.)


> >
> > What if a black box contained an ideal autotransformer, wired for 180
> > phase reversal. You look at the input and output waves, and see that
> > the output wave 'lags' the input wave by 180 degrees. Could you use
> > your formula to calculate the delay at 159.15 Hz? What would it be?
> >

No answer for this question Randy?

> ..... my point in arduously discussing
> this with you since last week is that the phase response at frequency
> w ALWAYS (indirectly) indicates the amount of time for a sine wave of
> frequency w to propagate through the filter, WHETHER OR NOT THAT FILTER
> IS PHASE LINEAR. This is to refute the following erroneous statement which
> you made a few posts back:

The slope of the phase curve, *directly* indicates the amount of time
for a sine wave, of frequency w, to propagate.


>
> Tpd is the absolute time delay, *only* if the phase response is linear.
>

My statement was and is correct.


Bob Stanton

gah@ugcs.caltech.edu (glen herrmannsfeldt) wrote in message news:<a2g582$4u5@gap.cco.caltech.edu>...
> rstanton2@stny.rr.com (Bob_Stanton) writes:
> (snip)
> >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.
>
> The definition of derivative requires two points, though infinitely
> close together. If you have a function that is not continuous
> anywhere then you can't calculate the derivative.
>

If we use two points that are infinitely close together, than the
error in our slope calculation is zero. If the error is zero, than we
know the slope at one point.

But, I'll tell you the truth. I have to brush the dust of my old
calculus book, before I can make further comments on finding the
derivative of the slope of the phase curve.
 
> >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.
>
> A light beam isn't a continuous sine wave, in the sense implied here,
> unless it exists from t=-infinity to t=+infinity. In optics, the
> coherence length or coherence time are used to measure this.
> The coherence length, usually of a laser, tells over what path
> length difference a laser beam can interfere with itself.
>

If I shine a light on the wall, isn't the light bean continuous from
the time after it hits the wall, until I turn the light off? During
that time, energy is moving across the room, at the speed of light.

Bob Stanton

Jerry Avins <jya@ieee.org> wrote in message news:<3C4B6DBF.B194619B@ieee.org>...
> Bob_Stanton wrote:
> >

> > It's not 'my' definition. "Tpd = -B/w" comes from "Electronic Filter
> > Design Handbook", by Arthur B. Williams. McGraw-Hill Book Company.
> > Copyright 1981. Pg.2-20
> >
>
> Given the limitations of plain ASCII text (but the standard ISO-8859-1
> allows ß, ±, ², and a few others), I assume that "Tpd" stands for time
> of phase delay. If so, it is correct. It is also _not_ group delay.
>

Right

> >
> > If I built the circuit, I would see, on a dual trace oscilloscope, the
> > output wave laging the input by 785 usec. So I don't need to build it.
>
> The pope refused to look through Galileo's telescope, insisting it would
> show nothing. Try it: you may be surprised.
> >

Actually, I have a telescope, but I don't have a dual trace
oscilloscope. My circuit analysis program says the phase shift of the
filter, at 159.15 Hz is 45 degs. Is that good enough?


> > > 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.
> >
> > You said if a system is "linear with phase response". Do you mean by
> > that "phase linear"?
>
> "Linear with phase response" is not the kind of nonsense description I
> expect from Randy. Are those really his words?

You figured it out, they are not really Randys' words! Somebody else
was typing on his computer and pretending to the him.

> Are there words around it
> that could give it valid meaning? "Phase linear" is another thing
> altogether, not related to the kind of linearity related to harmonic
> generation and such. A linear system is one in which superposition
> holds. A property of such systems is that it responds to a pure sinusoid
> with a pure sinusoid of the same frequency. Linearity does not determine
> the amplitude or phase of the response.

We were writing about a simple R-C low pass filter. It was obvious
that the filter was amplititude linear, and phase non-linear.

Bob Stanton

Bob_Stanton wrote:

> The slope of the phase curve, *directly* indicates the amount of time
> for a sine wave, of frequency w, to propagate.

> My statement was and is correct.
>
> Bob Stanton

Bob,
Your above statement is half true. The negative of the slope of the
phase curve gives the group delay. But is not the time that a single
sine takes to go through the filter.

For your RC circuit

E_out 1
----- = ------- where s=jw and w is freq in rads/sec
E_in 1 + sRC

Phase shift = arg(E_out/E_in) = -atan(wRC)

group delay = -d phase/d w = (RC) / (1+(wRC)^2)

With R=1000 Ohms and C=1E-6 Farads

The phase shift at 1000 rads/sec = -pi/4 radians

The Group delay at 1000 rads/sec = 500 uSec

If you put a dual trace scope on the circuit with one trace looking at
the input and the other looking at the output, you will see the 1000
radian per second signal has a -pi/4 phase shift between the input and
output - this is 785 uSec of delay not 500uSec!

On the other hand if you put two very closely spaced frequencies say at
1000 +- e (e is an epsilon which will become small later), you will see
these two combine to produce a modulated signal. The "carrier" is at
1000 rads/sec and the modulation is at e rads/sec.

Use the following classical trig identity to convince yourself of this

cos(a)+cos(b) = 2 cos((a+b)/2)*cos((a-b)/2)

Now what is the delay of the modulation?

Since each original freq is phase shifted, the output is

cos((1000-e)t+p1)+cos(1000+e)t+p2) where p1 and p2 are the phase shifts

The input is of course cos((1000-e)t)+cos((1000+e)t)

By the aforementioned trig identity the following carrier-modulation
form is found for the filter's output:

2*cos(1000t+(p1+p2)/2)*cos(et+(p2-p1)/2)

This can be interpreted as a carrier at the average of the two original
frequencies with a phase shift that is also the average of the original
phase shifts modulated by a sinusoid whose frequency is one half of the
frequency difference and has a phase shift that is one half of the
differences in the phase shifts of the original two frequencies.

Now look at the modulation portion cos(et+(p2-p1)/2)

It has a frequency of e radians per sec. and a phase shift of (p2-p1)/2
rads.

Thus the modulation has a delay of -(p2-p1)/(2e) seconds.

This is simply -delta(phase)/delta(frequency)

If you let e->0 (delta frequency) in a limiting fashion, you can see the
group delay is just the negative derivative of phase with resect to
frequency!

The group delay is the delay of the modulation resulting from a mix of
frequencies. A single sinusoid's delay can and will be usually
different from the group delay except for the case of linear phase.


I hope this clears up some of the confusion.

Clay

Bob_Stanton wrote:
>
  ...
>
> If I shine a light on the wall, isn't the light bean continuous from
> the time after it hits the wall, until I turn the light off? During
> that time, energy is moving across the room, at the speed of light.
>
> Bob Stanton

Again, it's subject to interpretation. The light is the combined
radiation of many atoms, each lasting a short time, and starting and
stopping independently. The phase at any location or time can't be
predicted from prior knowledge. The frequency range is broad; that's why
it looks white. What we call "light" continues, but no wave that
comprises it is continuous. There's a difference between "continuous"
and "enduring", just as there is between "simultaneous" and
"concurrent". Thinking gets fuzzy without distinctions like that.

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

"Clay S. Turner" <physics@bellsouth.net> wrote in message news:<3C4D88DD.5E7090B3@bellsouth.net>...
> Bob_Stanton wrote:
>
> > The slope of the phase curve, *directly* indicates the amount of time
> > for a sine wave, of frequency w, to propagate.
>
> > My statement was and is correct.
> >
> > Bob Stanton
>
> Bob,
> Your above statement is half true. The negative of the slope of the
> phase curve gives the group delay. But is not the time that a single
> sine takes to go through the filter.


.....


> This is simply -delta(phase)/delta(frequency)
>
> If you let e->0 (delta frequency) in a limiting fashion, you can see the
> group delay is just the negative derivative of phase with resect to
> frequency!
>
> The group delay is the delay of the modulation resulting from a mix of
> frequencies. A single sinusoid's delay can and will be usually
> different from the group delay except for the case of linear phase.
>
>
> I hope this clears up some of the confusion.
>
> Clay

Bob writes

I don't disagree with your numbers. Earlier in this thread I put out
the same numbers. Phase delay = 785 usec. Group delay = 500 usec.

With a modulated carrier, the information contained in the envelope
propagates through the system (a filter in this case) in 500 usec
(only if the carrier frequency is 159.15 Hz.).

Physically, a sine wave modulated carrier consists of three waves. The
carrier sinewave, the upper side band sinewave, and the lower side
band sinewave.

In order for the group delay number (500 usec) to be accurate, the
upper and lower side bands must be very close in frequency to the
carrier. Let's use (in this example) the side bands that are only
0.001 Hz above and below the carrier.

We know the *envelope wave* travels through the filter in 500 usec.
The two side bands that create the envelope, must also be traveling at
the same rate (through the filter in 500 usec).

If the two side bands, which are at 159.151 Hz and 159.149 Hz, pass
through the filter in about 500 usec, it is unreasonable to think that
the carrier, at virtually the same frequency, would take 785 usec.


Bob Stanton

Bob_Stanton wrote:
>
> Jerry Avins <jya@ieee.org> wrote in message news:<3C4B6DBF.B194619B@ieee.org>...
> > Bob_Stanton wrote:
> > >
>
> > > It's not 'my' definition. "Tpd = -B/w" comes from "Electronic Filter
> > > Design Handbook", by Arthur B. Williams. McGraw-Hill Book Company.
> > > Copyright 1981. Pg.2-20
> > >
> >
> > Given the limitations of plain ASCII text (but the standard ISO-8859-1
> > allows ß, ±, ², and a few others), I assume that "Tpd" stands for time
> > of phase delay. If so, it is correct. It is also _not_ group delay.
> >
>
> Right
>
> > >
> > > If I built the circuit, I would see, on a dual trace oscilloscope, the
> > > output wave laging the input by 785 usec. So I don't need to build it.
> >
> > The pope refused to look through Galileo's telescope, insisting it would
> > show nothing. Try it: you may be surprised.
> > >
>
> Actually, I have a telescope, but I don't have a dual trace
> oscilloscope. My circuit analysis program says the phase shift of the
> filter, at 159.15 Hz is 45 degs. Is that good enough?

It is indeed 45 degrees. I don't need a program; that it is the
half-power point of a single-break circuit is enough. 45 degrees is not
the time equivalent of group delay, however. It is the time equivalent
of phase delay, which can only be measured in the steady state. Phase
delay doesn't apply to transients.
>
> > > > 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.
> > >
> > > You said if a system is "linear with phase response". Do you mean by
> > > that "phase linear"?
> >
> > "Linear with phase response" is not the kind of nonsense description I
> > expect from Randy. Are those really his words?
>
> You figured it out, they are not really Randys' words! Somebody else
> was typing on his computer and pretending to the him.

I looked through past posts and couldn't find it. Can you point me to
the message?
>
> > Are there words around it
> > that could give it valid meaning? "Phase linear" is another thing
> > altogether, not related to the kind of linearity related to harmonic
> > generation and such. A linear system is one in which superposition
> > holds. A property of such systems is that it responds to a pure sinusoid
> > with a pure sinusoid of the same frequency. Linearity does not determine
> > the amplitude or phase of the response.
>
> We were writing about a simple R-C low pass filter. It was obvious
> that the filter was amplititude linear, and phase non-linear.
>
> Bob Stanton

The R-C network you drew has uniform output neither of phase nor
amplitude. It is linear in the sense of not generating harmonics. What
do you mean by "amplitude linear"? When I learn that, maybe we can
straighten this whole mess out.

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

Bob_Stanton wrote:
>
  ...
>
> The slope of the phase curve, *directly* indicates the amount of time
> for a sine wave, of frequency w, to propagate.

You better find the book again and read the very restrictive conditions
under which that statement applies. Earlier, you wrote

Tpd (time related to phase delay) = -ß/w, which is true i the steady
state. Now, you claim (correctly) that propagation time is -dß/dw, the
slope of the curve, rather than the cord from the origin. This is true
for transients. You keep flopping back and forth between phase and group
delay, and seem to use them indiscriminately.. Read a little more and
straighten it out.
>
> >
> > Tpd is the absolute time delay, *only* if the phase response is linear.
> >
>
> My statement was and is correct.

Yes. When the delay is the same for all frequencies, the phase delay is
also the group delay. Then and only then is your statement correct.
>
> Bob Stanton

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

Bob_Stanton wrote:
>
>
>
> Physically, a sine wave modulated carrier consists of three waves. The
> carrier sinewave, the upper side band sinewave, and the lower side
> band sinewave.

Actually you are thinking of the old style AM radio where the carrier is
only modulated by multiplying by positive numbers. Double balanced
modulators do not suffer this problem. Look at the cosine theorem
again:


cos(a) + cos(b) = 2 cos(c)cos(d) where c=(a+b)/2 d=(a-b)/2

Notice on the right hand side - there are only two terms not three.


>
> In order for the group delay number (500 usec) to be accurate, the
> upper and lower side bands must be very close in frequency to the
> carrier. Let's use (in this example) the side bands that are only
> 0.001 Hz above and below the carrier.
>
> We know the *envelope wave* travels through the filter in 500 usec.
> The two side bands that create the envelope, must also be traveling at
> the same rate (through the filter in 500 usec).

NO! Set up the circuit and use a scope. The group delay time is due to
the difference in phase shifts of the constituent frequencies.



>
> If the two side bands, which are at 159.151 Hz and 159.149 Hz, pass
> through the filter in about 500 usec, it is unreasonable to think that
> the carrier, at virtually the same frequency, would take 785 usec.

The two frequencies have different phase shifts, so where they combine
to form a peak is different than where either had a peak alone.

Clay


>
> Bob Stanton

Jerry Avins wrote:
> Bob Stanton wrote:
> > Tpd is the absolute time delay, *only* if the phase response is linear.
> >
> > My statement was and is correct.
>
> Yes.

No.

Let's break it down into a logic problem. Let "A" denote "Tpd is
the absolute time delay" and let "B" denote "the phase response is
linear."

Then Bob wrote "A only if B." This is logically equivalent to
"A iff (if and only if) B" which is in turn logically equivalent
to the two statements:

  1. "If A then B"
  2. "If B then A"

Statement 1 is not true: "If Tpd is the absolute time delay, then
the phase response is linear." The fact is, Tpd is the absolute
time delay WHETHER OR NOT the phase response is linear.

Are you with me, Jerry, or do you actually dispute this assertion?
--
Randy Yates
DSP Engineer, Sony Ericsson Mobile Communications
Research Triangle Park, NC, USA
randy.yates@ericsson.com, 919-472-1124

Randy Yates wrote:
>
> Jerry Avins wrote:
> > Bob Stanton wrote:
> > > Tpd is the absolute time delay, *only* if the phase response is linear.
> > >
> > > My statement was and is correct.
> >
> > Yes.
>
> No.
>
> Let's break it down into a logic problem. Let "A" denote "Tpd is
> the absolute time delay" and let "B" denote "the phase response is
> linear."
>
> Then Bob wrote "A only if B." This is logically equivalent to
> "A iff (if and only if) B" which is in turn logically equivalent
> to the two statements:
>
> 1. "If A then B"
> 2. "If B then A"
>
> Statement 1 is not true: "If Tpd is the absolute time delay, then
> the phase response is linear." The fact is, Tpd is the absolute
> time delay WHETHER OR NOT the phase response is linear.
>
> Are you with me, Jerry, or do you actually dispute this assertion?
> --
> Randy Yates
> DSP Engineer, Sony Ericsson Mobile Communications
> Research Triangle Park, NC, USA
> randy.yates@ericsson.com, 919-472-1124

Of course you're right. I shouldn't be elliptical in a discussion like
this. Phase delay equals group delay when the phase is linear. That is
to say, those delays are the same when all delays are the same. (I think
we agree on that.) I meant to say that that's all he was saying. The
other thing that's gobbling up his mind is the notion that there is a
phase delay that one can see. There isn't. There is a phase shift one
can measure once the transients have died out, and dividing phase in
radians by frequency in radians per second, there is a time we calculate
that we call phase delay. But, unlike information or energy travel,
there is no phenomenon that lasts for -- or requires -- that time. It's
just a number. Under some circumstances, the phase delay from here to
there is less than the time it takes light to go from here to there. So
what?

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

Jerry Avins wrote:
>
> Randy Yates wrote:
> >
> > Jerry Avins wrote:
> > > Bob Stanton wrote:
> > > > Tpd is the absolute time delay, *only* if the phase response is linear.
> > > >
> > > > My statement was and is correct.
> > >
> > > Yes.
> >
> > No.
> >
> > Let's break it down into a logic problem. Let "A" denote "Tpd is
> > the absolute time delay" and let "B" denote "the phase response is
> > linear."
> >
> > Then Bob wrote "A only if B." This is logically equivalent to
> > "A iff (if and only if) B" which is in turn logically equivalent
> > to the two statements:
> >
> > 1. "If A then B"
> > 2. "If B then A"
> >
> > Statement 1 is not true: "If Tpd is the absolute time delay, then
> > the phase response is linear." The fact is, Tpd is the absolute
> > time delay WHETHER OR NOT the phase response is linear.
> >
> > Are you with me, Jerry, or do you actually dispute this assertion?
> > --
> > Randy Yates
> > DSP Engineer, Sony Ericsson Mobile Communications
> > Research Triangle Park, NC, USA
> > randy.yates@ericsson.com, 919-472-1124
>
> Of course you're right.

I was beginning to wonder!

> I shouldn't be elliptical in a discussion like
> this.

Your "ellipticalness" is my paranoia.

> Phase delay equals group delay when the phase is linear. That is
> to say, those delays are the same when all delays are the same. (I think
> we agree on that.)

Yes, of course. That has been the consensus since several days ago.

> I meant to say that that's all he was saying.

You mean you're trying to overlook what he wrote and infer what he
meant?

Normally (especially on first interpretation and response) I would say that
is a good thing. However, when the person becomes adamant about the truthfulness
of the statement after several attempts to point out the error, I'd say it's
time to put it under a spotlight and interpret it exactly as it is stated.

Besides, there seems to be little to interpret here. I mean, it's almost like
saying "I think you mean yes even though you wrote no." If he meant yes, then
he should have written it, at least after a post or two attempting to clarify.
--
Randy Yates
DSP Engineer, Sony Ericsson Mobile Communications
Research Triangle Park, NC, USA
randy.yates@ericsson.com, 919-472-1124









































.

Randy Yates wrote:
>
  ...
>
> You mean you're trying to overlook what he wrote and infer what he
> meant?
>
> Normally (especially on first interpretation and response) I would say that
> is a good thing. However, when the person becomes adamant about the truthfulness
> of the statement after several attempts to point out the error, I'd say it's
> time to put it under a spotlight and interpret it exactly as it is stated.
>
> Besides, there seems to be little to interpret here. I mean, it's almost like
> saying "I think you mean yes even though you wrote no." If he meant yes, then
> he should have written it, at least after a post or two attempting to clarify.
> --

The difference is, you're peeved (as well you might be, taking the brunt
of it for a while) while some of us are still trying to get his head on
straight. I may give up soon, but that will induce pity, not anger.

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

"Clay S. Turner" <physics@bellsouth.net> writes:

>> Physically, a sine wave modulated carrier consists of three waves. The
>> carrier sinewave, the upper side band sinewave, and the lower side
>> band sinewave.

>Actually you are thinking of the old style AM radio where the carrier is
>only modulated by multiplying by positive numbers. Double balanced
>modulators do not suffer this problem. Look at the cosine theorem
>again:

>cos(a) + cos(b) = 2 cos(c)cos(d) where c=(a+b)/2 d=(a-b)/2

>Notice on the right hand side - there are only two terms not three.

How about:

cos(c) (1 + cos(d)) = cos(c) + cos(c)cos(d)
                    = cos(c) + (1/2) cos(a) + (1/2) cos(b)

using the same identity at 100% modulation.

-- glen

Hello Glen,
I was showing how the sum of two equal amplitude cosines yield a
product of cosines. This was to be interpreted in light of the group
delay confusion and not double sideband AM.

Clay

glen herrmannsfeldt wrote:
>
> "Clay S. Turner" <physics@bellsouth.net> writes:
>
> >> Physically, a sine wave modulated carrier consists of three waves. The
> >> carrier sinewave, the upper side band sinewave, and the lower side
> >> band sinewave.
>
> >Actually you are thinking of the old style AM radio where the carrier is
> >only modulated by multiplying by positive numbers. Double balanced
> >modulators do not suffer this problem. Look at the cosine theorem
> >again:
>
> >cos(a) + cos(b) = 2 cos(c)cos(d) where c=(a+b)/2 d=(a-b)/2
>
> >Notice on the right hand side - there are only two terms not three.
>
> How about:
>
> cos(c) (1 + cos(d)) = cos(c) + cos(c)cos(d)
> = cos(c) + (1/2) cos(a) + (1/2) cos(b)
>
> using the same identity at 100% modulation.
>
> -- glen

even so, Clay, since delaying the "1" term along with the cos(d) term
by the "group delay" does not change that "1" term, glen's setup for
showing what group delay is (in justaposition to phase delay), is also
legit.

AGAIN, I'M NOT JUST TRYING TO DRAW ATTENTION FOR THE SAKE OF DRAWING
ATTENTION, but (clears throat) i wanna end the argument:

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

guzzinta a linear (not necessarily "phase-linear") filter with Laplace
transfer function H(s). a(t) is slow moving and bandlimited to much
less than w0.

what come out is:

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

where Tp = - arg{H(j*w0)}/w0 and called "phase delay"

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

and that's called "group delay".

the envelope (or group), a(t), is delayed by Tg whether a(t) = 1 -
cos(d) or not.

this can be worked out. it is mathematically true (approximately true
and becomes less approximate the more bandlimited a(t) is) and should
end the blankity-blank debate.

fine~!

r b-j


"Clay S. Turner" <physics@bellsouth.net> wrote in message news:<3C4E05E8.36D08038@bellsouth.net>...
> Hello Glen,
> I was showing how the sum of two equal amplitude cosines yield a
> product of cosines. This was to be interpreted in light of the group
> delay confusion and not double sideband AM.
>
> glen herrmannsfeldt wrote:
> >
> > "Clay S. Turner" <physics@bellsouth.net> writes:
> >
> > >> Physically, a sine wave modulated carrier consists of three waves. The
> > >> carrier sinewave, the upper side band sinewave, and the lower side
> > >> band sinewave.
>
> > >Actually you are thinking of the old style AM radio where the carrier is
> > >only modulated by multiplying by positive numbers. Double balanced
> > >modulators do not suffer this problem. Look at the cosine theorem
> > >again:
>
> > >cos(a) + cos(b) = 2 cos(c)cos(d) where c=(a+b)/2 d=(a-b)/2
>
> > >Notice on the right hand side - there are only two terms not three.
> >
> > How about:
> >
> > cos(c) (1 + cos(d)) = cos(c) + cos(c)cos(d)
> > = cos(c) + (1/2) cos(a) + (1/2) cos(b)
> >
> > using the same identity at 100% modulation.

Jerry Avins <jya@ieee.org> wrote in message news:<3C4DB875.B1C67F57@ieee.org>...
> Bob_Stanton wrote:
> >
> ...
> >
> > The slope of the phase curve, *directly* indicates the amount of time
> > for a sine wave, of frequency w, to propagate.
>
> Jerry
> You better find the book again and read the very restrictive conditions
> under which that statement applies.

Bob writes

I have done as you asked, and looked at the books I have available. I
saw *no* limiting conditions. Please tell me under what condition, the
slope of the phase curve is not the group delay.

Perhaps you are thinking of the limits when making group delay
*measurements*?

...Jerry
Earlier, you wrote
>
> Tpd (time related to phase delay) = -ß/w, which is true i the steady
> state. Now, you claim (correctly) that propagation time is -dß/dw, the
> slope of the curve, rather than the cord from the origin. This is true
> for transients. You keep flopping back and forth between phase and group
> delay, and seem to use them indiscriminately.. Read a little more and
> straighten it out.

Bob writes

I have looked back at all my messages in this thread, and found *no*
place where I said "phase delay" is equivalent to "group delay". I
always used the terms "phase delay" and "group delay" correctly, and
not indiscriminately.

We agree that propagation time = -dB/dw. But, this is true for both
transients and *steady state* signals. For example: A carrier
modulated by a sinewave is a steady state signal. The propagation time
of the envelope sinewave is the "group delay".

I'm supposed to go back and straighten what out? What is the "it" I'm
supposed to straighten out?

> > >
> > > Bob
> > > Tpd is the absolute time delay, *only* if the phase response is linear.
> > >
> >
>
> Jerry
> Yes. When the delay is the same for all frequencies, the phase delay is
> also the group delay. Then and only then is your statement correct.

Bob writes
When the phase response is linear, the delay *is the same* for all
frequencies. So, you just said the same thing I did. I'm glad you
agree.

Bob Stanton

Jerry Avins wrote:
>
> The difference is, you're peeved (as well you might be, taking the brunt
> of it for a while) while some of us are still trying to get his head on
> straight. I may give up soon, but that will induce pity, not anger.

Good luck. I also want to help people overcome their misconceptions, but
here I think it's a case of "throwing pearls before swine."
--
% Randy Yates % "...the answer lies within your soul
%% Fuquay-Varina, NC % 'cause no one knows which side
%%% 919-577-9882 % the coin will fall."
%%%% <yates@ieee.org> % 'Big Wheels', *Out of the Blue*, ELO
http://personal.rdu.bellsouth.net/~yatesc

robert bristow-johnson wrote:
>
> even so, Clay, since delaying the "1" term along with the cos(d) term
> by the "group delay" does not change that "1" term, glen's setup for
> showing what group delay is (in justaposition to phase delay), is also
> legit.
>
> AGAIN, I'M NOT JUST TRYING TO DRAW ATTENTION FOR THE SAKE OF DRAWING
> ATTENTION, but (clears throat) i wanna end the argument:
>
> x(t) = a(t) * cos(w0*t)
>
> guzzinta a linear (not necessarily "phase-linear") filter with Laplace
> transfer function H(s). a(t) is slow moving and bandlimited to much
> less than w0.
>
> what come out is:
>
> y(t) = |H(j*w0)| * a(t-Tg) * cos(w0*(t-Tp)
>
> where Tp = - arg{H(j*w0)}/w0 and called "phase delay"
>
> and Tg = - d (arg{H(j*w)})/dw evaluated at w = w0
>
> and that's called "group delay".
>
> the envelope (or group), a(t), is delayed by Tg whether a(t) = 1 -
> cos(d) or not.
>
> this can be worked out. it is mathematically true (approximately true
> and becomes less approximate the more bandlimited a(t) is) and should
> end the blankity-blank debate.
>
> fine~!
>
> r b-j
>
We should be so lucky! The debate won't end until the participants
realize that although phase delay carries the unit "time", it doesn't
represent the duration of any observable or imaginable phenomenon.

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

Bob_Stanton wrote:
>
> Jerry Avins <jya@ieee.org> wrote in message news:<3C4DB875.B1C67F57@ieee.org>...
> > Bob_Stanton wrote:
> > >
> > ...
> > >
> > > The slope of the phase curve, *directly* indicates the amount of time
> > > for a sine wave, of frequency w, to propagate.
> >
> > Jerry
> > You better find the book again and read the very restrictive conditions
> > under which that statement applies.
>
> Bob writes
>
> I have done as you asked, and looked at the books I have available. I
> saw *no* limiting conditions. Please tell me under what condition, the
> slope of the phase curve is not the group delay.

The speed at which a sine wave propagates can't be determined directly.
The speed at which a group propagates is given by the slope.
>
> Perhaps you are thinking of the limits when making group delay
> *measurements*?

No.
>
> ...Jerry
> Earlier, you wrote
> >
> > Tpd (time related to phase delay) = -ß/w, which is true i the steady
> > state. Now, you claim (correctly) that propagation time is -dß/dw, the
> > slope of the curve, rather than the cord from the origin. This is true
> > for transients. You keep flopping back and forth between phase and group
> > delay, and seem to use them indiscriminately.. Read a little more and
> > straighten it out.
>
> Bob writes
>
> I have looked back at all my messages in this thread, and found *no*
> place where I said "phase delay" is equivalent to "group delay". I
> always used the terms "phase delay" and "group delay" correctly, and
> not indiscriminately.
>
> We agree that propagation time = -dB/dw. But, this is true for both
> transients and *steady state* signals. For example: A carrier
> modulated by a sinewave is a steady state signal. The propagation time
> of the envelope sinewave is the "group delay".
>
> I'm supposed to go back and straighten what out? What is the "it" I'm
> supposed to straighten out?

Earlier, unless I remember wrong, you cited a formula Tpd = -ß/w as
providing group delay, that is, the time for a signal to actually
propagate. It has nothing to do with propagation; it is simply phase
shift divided by frequency, and useful primarily with transmission lines
and waveguides for determining where the nodes will be. Some filters are
analogs of transmission lines, and phase delay can be useful in that
analogy, too.
>
  ...

There's been far too much "he said, I said". I think the best way to
continue this discussion id to start over.

Phase delay is phase angle divided by frequency. It is an interesting
number with little (if any) physical significance. The best significance
I can find is that it represents the integral of delays I can measure.

Group delay is the slope of phase delay curve. It is the time it takes a
signal to traverse a network. The flatter the phase delay curve, the
broader a the band of frequencies that can be considered to share a
group delay.

Can we start from there?

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

Jerry Avins wrote:
> [...]
> Phase delay is phase angle divided by frequency. It is an interesting
> number with little (if any) physical significance.

The fact that it indicates the propagation time through the filter at
that frequency is not of any physical significance? I think it's extremely
significant. As an example, this (as well as group delay) can be a good
indicator when a filter will or will not preserve the shape of a waveform.

> The best significance
> I can find is that it represents the integral of delays I can measure.

Not sure I follow you here, Jerry.
--
Randy Yates
DSP Engineer, Sony Ericsson Mobile Communications
Research Triangle Park, NC, USA
randy.yates@ericsson.com, 919-472-1124

Randy Yates wrote:
>
> Jerry Avins wrote:
> > [...]
> > Phase delay is phase angle divided by frequency. It is an interesting
> > number with little (if any) physical significance.
>
> The fact that it indicates the propagation time through the filter at
> that frequency is not of any physical significance? I think it's extremely
> significant. As an example, this (as well as group delay) can be a good
> indicator when a filter will or will not preserve the shape of a waveform.

It would be very significant if it were true, but it's not. (There are
structures for which the phase delay is less than the time for a light
beam to traverse it.) Phase delay applies to the steady state. Transit
time has no meaning then.
>
> > The best significance
> > I can find is that it represents the integral of delays I can measure.
>
> Not sure I follow you here, Jerry.

Ignore the red herring. I was wrong. There's no point to explaining what
I meant that isn't so.
> --
> Randy Yates
> DSP Engineer, Sony Ericsson Mobile Communications
> Research Triangle Park, NC, USA
> randy.yates@ericsson.com, 919-472-1124

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

Jerry Avins wrote:
>
> Randy Yates wrote:
> >
> > Jerry Avins wrote:
> > > [...]
> > > Phase delay is phase angle divided by frequency. It is an interesting
> > > number with little (if any) physical significance.
> >
> > The fact that it indicates the propagation time through the filter at
> > that frequency is not of any physical significance? I think it's extremely
> > significant. As an example, this (as well as group delay) can be a good
> > indicator when a filter will or will not preserve the shape of a waveform.
>
> It would be very significant if it were true, but it's not. (There are
> structures for which the phase delay is less than the time for a light
> beam to traverse it.)

Jerry, I can't understand these examples you give using light. Can we
please just stick with signals definable as a function x(t)?

> Phase delay applies to the steady state.

Agreed.

> Transit
> time has no meaning then.

Transit time? I'm not sure what you mean by that.

For a wide class of signals, representation as a sum of sinusoids
exists. Note such a representation does not limit us to "steady-state"
signals (i.e., periodic). Thus, for the signals that can be so
represented, phase delay is relevent as it determines how
each of the component sines travel through the filter.

If you have a signal x(t) composed as follows:

  x(t) = cos(2*pi*100*t) + cos(2*pi*103*t)

and you pass it through a filter, then that filter's
phase response (or phase delay) at 100 Hz and 103 Hz
will tell you whether the output waveform will look
like the input one or not. The same is true for a "transient"
signal, but the expression of an example of such a signal
using text in a usenet news article is too intractable.
--
Randy Yates
DSP Engineer, Sony Ericsson Mobile Communications
Research Triangle Park, NC, USA
randy.yates@ericsson.com, 919-472-1124

Randy Yates wrote:
>
> Jerry Avins wrote:
> >
  ...
> > [Phase delay being in-to-out time would be very significant if it
> > were true, but it's not. (There are
> > structures for which the phase delay is less than the time for a light
> > beam to traverse it.)
>
> Jerry, I can't understand these examples you give using light. Can we
> please just stick with signals definable as a function x(t)?

I was thinking of waveguides.
>
> > Phase delay applies to the steady state.
>
> Agreed.
>
> > Transit
> > time has no meaning then.
>
> Transit time? I'm not sure what you mean by that.

The time from when a signal goes in till when it comes out. On a
transmission line or a wave guide, or in a speaking tube, the concept is
clear to me. With filters, it's a bit muddier. In a waveguide, the phase
velocity is always faster than light if energy is being carried.
Faster-than-sound phase velocity is possible in a speaking tube, but not
with the usual longitudinal mode. The phase velocity of waves beaching
obliquely is faster than the speed of waves through the water. These
easy examples all involve distance, so we can think in terms of
velocity. Of course, these examples can all be expressed in terms of
time, too, a.k.a. phase delay. It is clear that for them, the times
involved have no physical significance. The equations for their phase
delays are the same as those for the signals we've been discussing, and
have no more physical significance.
>
> For a wide class of signals, representation as a sum of sinusoids
> exists. Note such a representation does not limit us to "steady-state"
> signals (i.e., periodic). Thus, for the signals that can be so
> represented, phase delay is relevent as it determines how
> each of the component sines travel through the filter.

Once the steady state has been reached, who can tell when the wave
started? All we know is the phase shift between input and output, and
the frequency. Dividing gives a number with dimension "time". To
construe that number as an interval, we need to propose events that
define its start and end. The assumption of steady state makes that
impossible. I don't know what it means.
>
> If you have a signal x(t) composed as follows:
>
> x(t) = cos(2*pi*100*t) + cos(2*pi*103*t)
>
> and you pass it through a filter, then that filter's
> phase response (or phase delay) at 100 Hz and 103 Hz
> will tell you whether the output waveform will look
> like the input one or not. The same is true for a "transient"
> signal, but the expression of an example of such a signal
> using text in a usenet news article is too intractable.
> --
> Randy Yates
> DSP Engineer, Sony Ericsson Mobile Communications
> Research Triangle Park, NC, USA
> randy.yates@ericsson.com, 919-472-1124

When the phase delay and the group delay differ but the slope of the
phase delay curve is straight over the range of pertinent frequencies,
then a modulated pulse on a carrier will emerge from our black box with
the shape of the envelope intact, but the phase of the carrier with
respect to the edge of the pulse will be altered. If the delay mechanism
is a transmission line that can be observed at intervals along its
length, or if it is an iterative L-C-R-G structure that can be observed
at various taps, it will be seen that the phase of the carrier relative
to the envelope varies from place to place. If our test signal is quiet
wherever there is no envelope (as Clay's suppressed-carrier example
would have it), how can you convince me -- or yourself! -- that the
carrier and envelope travel at different speeds? The group delay is the
delay of all frequencies in the band, including the carrier. The phase
delay is just an interesting number.

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

Jerry Avins <jya@ieee.org> wrote in message news:<3C4EFF05.40BB9B5E@ieee.org>...
> Bob_Stanton wrote:
> >
> > Jerry Avins <jya@ieee.org> wrote in message news:<3C4DB875.B1C67F57@ieee.org>...
> > > Bob_Stanton wrote:
> > > >
> ...
> > > >
> > > > The slope of the phase curve, *directly* indicates the amount of time
> > > > for a sine wave, of frequency w, to propagate.
> > >
> > > Jerry
> > > You better find the book again and read the very restrictive conditions
> > > under which that statement applies.
> >
> > Bob writes
> >
> > I have done as you asked, and looked at the books I have available. I
> > saw *no* limiting conditions. Please tell me under what condition, the
> > slope of the phase curve is not the group delay.
>
> The speed at which a sine wave propagates can't be determined directly.
> The speed at which a group propagates is given by the slope.
> >
Bob writes

Yes. The speed at which a group propagates is given by the slope.

Now let's think about the physical meaning. I believe you think of the
"group" as the envelope information, riding down the carrier. Right?
So do I.

But, there is another physical way to think of an amplititude
modulated wave: As three sinewaves. The sinewave in the center is the
carrier. The ones above and below the carrier are the sidebands. They
go down a cable and through devices together, like three Caballeros.
They all propagate at the same speed. They are even phase locked.
These three sinewaves propagate at Time(group delay).

> >
> > ...Jerry
> > Earlier, you wrote
> > >
> > > ..Read a little more and straighten it out.
> >
> > Bob writes
> >
> > I have looked back at all my messages in this thread, and found *no*
> > place where I said "phase delay" is equivalent to "group delay.
> >
> > I'm supposed to go back and straighten what out? What is the "it" I'm
> > supposed to straighten out?
> Jerry
> Earlier, unless I remember wrong, you cited a formula Tpd = -ß/w as
> providing group delay, that is, the time for a signal to actually
> propagate.

Bob
I don't actually think I said that. Tpd is definately *not* the time
it takes for a signal to propagate.

> Jerry
> It has nothing to do with propagation; it is simply phase
> shift divided by frequency, and useful primarily with transmission lines
> and waveguides for determining where the nodes will be. Some filters are
> analogs of transmission lines, and phase delay can be useful in that
> analogy, too.
> >
>

Bob
I agree with what you said above.


>
> There's been far too much "he said, I said". I think the best way to
> continue this discussion id to start over.
>
> Phase delay is phase angle divided by frequency. It is an interesting
> number with little (if any) physical significance. The best significance
> I can find is that it represents the integral of delays I can measure.
>
> Group delay is the slope of phase delay curve. It is the time it takes a
> signal to traverse a network. The flatter the phase delay curve, the
> broader a the band of frequencies that can be considered to share a
> group delay.
>
> Can we start from there?
>

Bob
You just said everything I have been trying to say in this thread, and
you said it better and clearer then I did. I agree with your statement
above 100%.

Bob Stanton

"Clay S. Turner" <physics@bellsouth.net> wrote in message news:<3C4DBE5B.E2BE847E@bellsouth.net>...
> Bob_Stanton wrote:
> >
> >
> >
> > Physically, a sine wave modulated carrier consists of three waves. The
> > carrier sinewave, the upper side band sinewave, and the lower side
> > band sinewave.
>
> Actually you are thinking of the old style AM radio where the carrier is
> only modulated by multiplying by positive numbers. Double balanced
> modulators do not suffer this problem.

Bob writes

Most text books use the old style amplititude modulated sinewave, when
showing the effects of group delay. Group delay is not a function of
the kind of waveform you use to measure it. Group delay is the slope
of the phase curve and should not be defined as being the effect on a
certain type of waveform.

I could for example, say that group delay is the amount of ringing on
a square wave. But, that is the effect of group delay not the
definition of it. Likewise the effect on the envelope of an
amplititude modulated sinewave is not the definition of group delay.
It just so happens that the propagation time of a sinewave is equal to
the propagation time of the envelope. So, the propagation time of the
envelope waveform becomes a very useful number.

>
> >
> > In order for the group delay number (500 usec) to be accurate, the
> > upper and lower side bands must be very close in frequency to the
> > carrier. Let's use (in this example) the side bands that are only
> > 0.001 Hz above and below the carrier.
> >
> > We know the *envelope wave* travels through the filter in 500 usec.
> > The two side bands that create the envelope, must also be traveling at
> > the same rate (through the filter in 500 usec).
>
> NO! Set up the circuit and use a scope. The group delay time is due to
> the difference in phase shifts of the constituent frequencies.
>

No! You got it backwards. The phase shift of the consitituent
frequencies is caused by the delay of the device.

>
>
> >
> > If the two side bands, which are at 159.151 Hz and 159.149 Hz, pass
> > through the filter in about 500 usec, it is unreasonable to think that
> > the carrier, at virtually the same frequency, would take 785 usec.
>
> The two frequencies have different phase shifts, so where they combine
> to form a peak is different than where either had a peak alone.
>

Do you think the side bands travel through a device slower than the
envelope waveform. If they do, that's going to be bad news. Eventually
the envelope will be traveling all by itself, with the carrier and the
sidebands left behind. :-)


Bob Stanton

Bob_Stanton wrote:

> "Clay S. Turner" <physics@bellsouth.net> wrote in message news:<3C4DBE5B.E2BE847E@bellsouth.net>...
> > Bob_Stanton wrote:
> > >
> > >
> > >
> > > Physically, a sine wave modulated carrier consists of three waves. The
> > > carrier sinewave, the upper side band sinewave, and the lower side
> > > band sinewave.
> >
> > Actually you are thinking of the old style AM radio where the carrier is
> > only modulated by multiplying by positive numbers. Double balanced
> > modulators do not suffer this problem.
>
> Bob writes
>
> Most text books use the old style amplititude modulated sinewave, when
> showing the effects of group delay. Group delay is not a function of
> the kind of waveform you use to measure it. Group delay is the slope
> of the phase curve and should not be defined as being the effect on a
> certain type of waveform.
>
> I could for example, say that group delay is the amount of ringing on
> a square wave. But, that is the effect of group delay not the
> definition of it. Likewise the effect on the envelope of an
> amplititude modulated sinewave is not the definition of group delay.
> It just so happens that the propagation time of a sinewave is equal to
> the propagation time of the envelope. So, the propagation time of the
> envelope waveform becomes a very useful number.
>
> >
> > >
> > > In order for the group delay number (500 usec) to be accurate, the
> > > upper and lower side bands must be very close in frequency to the
> > > carrier. Let's use (in this example) the side bands that are only
> > > 0.001 Hz above and below the carrier.
> > >
> > > We know the *envelope wave* travels through the filter in 500 usec.
> > > The two side bands that create the envelope, must also be traveling at
> > > the same rate (through the filter in 500 usec).
> >
> > NO! Set up the circuit and use a scope. The group delay time is due to
> > the difference in phase shifts of the constituent frequencies.
> >
>
> No! You got it backwards. The phase shift of the consitituent
> frequencies is caused by the delay of the device.
>
> >
> >
> > >
> > > If the two side bands, which are at 159.151 Hz and 159.149 Hz, pass
> > > through the filter in about 500 usec, it is unreasonable to think that
> > > the carrier, at virtually the same frequency, would take 785 usec.
> >
> > The two frequencies have different phase shifts, so where they combine
> > to form a peak is different than where either had a peak alone.
> >
>
> Do you think the side bands travel through a device slower than the
> envelope waveform. If they do, that's going to be bad news. Eventually
> the envelope will be traveling all by itself, with the carrier and the
> sidebands left behind. :-)
>
> Bob Stanton

I think that you guys are talking about two different types of filters and trying to draw a common
conclusion. The resistor capacitor filter has a constantly changing phase shift with frequency. driving
it with an AM signal is actually a signal at 3 different frequencies so the phase shift is going to be
different for each.

If you put that same AM signal into a band pass filter with a flat response at the frequencies that the
AM signal occupies the phase shift should be the same for all 3 frequencies. The group delay will also
be the same for all 3 frequencies.

Regards
Gary

Gary Schafer <gschafer@mediaone.net> wrote in message

>
> I think that you guys are talking about two different types of filters and

> trying to draw a common
> conclusion. The resistor capacitor filter has a constantly changing phase

> shift with frequency. driving
> it with an AM signal is actually a signal at 3 different frequencies so the

> phase shift is going to be
> different for each.
>

Bob writes

When measuring group delay, the phase shift must be measured over a
very small change in frequency. The change in frequency must be so
small, that the phase shift between frequencies is virtually linear.
Of course it will never be completely linear for an non-phase linear
device, such as a filter. There will always be some error in the
measurement, but the error can be made very small.

> If you put that same AM signal into a band pass filter with a flat response
> at the frequencies that the AM signal occupies the phase shift should be the

 > same for all 3 frequencies. The group delay will also be the same
for all 3 > frequencies.
>

That is exactly correct.

Bob Stanton

In article <67ef4d54.0201231856.1e7d9a94@posting.google.com>,
Bob_Stanton <rstanton2@stny.rr.com> wrote:
>> The two frequencies have different phase shifts, so where they combine
>> to form a peak is different than where either had a peak alone.

Very true.
 
>Do you think the side bands travel through a device slower than the
>envelope waveform. If they do, that's going to be bad news. Eventually
>the envelope will be traveling all by itself, with the carrier and the
>sidebands left behind. :-)

You don't see the point, Bob. ONE sideband may go at a different rate
than the other, and the combined peak will shift (not so predictably)
as a result.

And the question of having the envelope w/o the carrier and sidebands
isn't meaningful here. I mean, if it was, you could have information
transferred w/o energy transfer. THAT would be neat, eh? :)
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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 <67ef4d54.0201240400.a67d962@posting.google.com>,
Bob_Stanton <rstanton2@stny.rr.com> wrote:
>When measuring group delay, the phase shift must be measured over a
>very small change in frequency.

You can measure or calculate (depending on your system) the phase
shift AT ANY GIVEN FREQUENCY. That's NOT at "a variety of
frequencies" but A GIVEN FREQUENCY...
Now, phase shift measured that way with quasi-continuous signals
will only show up as -pi to pi.

BUT if you do something called "phase wrapping" you can calculate the
real phase shift, by doing many frequencies, not just one.

Real phase shift can be much larger tha pi.
--
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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.

jj@research.att.com (jj, DBT thug and skeptical philalethist) wrote in message news:<GqGHF0.FuI@research.att.com>...
> In article <67ef4d54.0201231856.1e7d9a94@posting.google.com>,
> Bob_Stanton <rstanton2@stny.rr.com> wrote:
> >> The two frequencies have different phase shifts, so where they combine
> >> to form a peak is different than where either had a peak alone.
>
> Very true.
>

If the device that the wave is passing through is phase linear, the
sidebands and carrier move together.

If the device is not phase linear, (such as a typical passive bandpass
filter) the sidebands and carrier will travel through (propagate) at
different rates.

> >Do you think the side bands travel through a device slower than the
> >envelope waveform. If they do, that's going to be bad news. Eventually
> >the envelope will be traveling all by itself, with the carrier and the
> >sidebands left behind. :-)
>
> You don't see the point, Bob. ONE sideband may go at a different rate
> than the other, and the combined peak will shift (not so predictably)
> as a result.
>

One sideband can indeed go at a different rate and that will result in
waveform distortion.

> And the question of having the envelope w/o the carrier and sidebands
> isn't meaningful here. I mean, if it was, you could have information
> transferred w/o energy transfer. THAT would be neat, eh? :)

Yes, neat.

When I said the envelope would outrun the sidebands, I was only trying
(unsucessfully perhaps) to be funny. Of course, if one (wrongly)
assumes that the envelope propagates faster than the carriers or
sidebands, well ......
Can you imagin an envelope in the air, with no carriers? I can't or
can I?

I quess once the envelope got ahead of the carriers, it could leave
the cable entirely and fly through the air by itself. Maybe it could
go faster than the speed of light and backwards through time. Maybe
it could bring back pictures of what life was like in the days of the
Romans.

I think this group delay thread is making me crazy. Their coming to
take me away, Ha Ha, Ho Ho, Hee Hee. Their coming to take me away, Hee
Hee, Ho Ho, Ha Ha!

Bye Bye.

In article <67ef4d54.0201241857.1020c5e4@posting.google.com>,
Bob_Stanton <rstanton2@stny.rr.com> wrote:
>If the device that the wave is passing through is phase linear, the
>sidebands and carrier move together.

It is conventional to back out the constant delay term
in most filters, however, so if you have varying phase shift,
one usually assumes that the delay term has been separated.
It's a very useful thing to do most of the time.

>If the device is not phase linear, (such as a typical passive bandpass
>filter) the sidebands and carrier will travel through (propagate) at
>different rates.

You mean "pure delay" or "linear phase", I think. Phase Linear
was an amplifier from the 1970's. :)

>When I said the envelope would outrun the sidebands, I was only trying
>(unsucessfully perhaps) to be funny. Of course, if one (wrongly)
>assumes that the envelope propagates faster than the carriers or
>sidebands, well ......

Heh. I thought it was a joke.

>Can you imagin an envelope in the air, with no carriers? I can't or
>can I?

I can also IMAGINE time travel :)

>I think this group delay thread is making me crazy. Their coming to
>take me away, Ha Ha, Ho Ho, Hee Hee. Their coming to take me away, Hee
>Hee, Ho Ho, Ha Ha!

Heh. It seems to me to be much ado about nothing.
--
Copyright jj@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.

On Fri, 25 Jan 2002 07:29:09 GMT, jj@research.att.com (jj, DBT thug
and skeptical philalethist) wrote:

>
>You mean "pure delay" or "linear phase", I think. Phase Linear
>was an amplifier from the 1970's. :)
>

"phase linear Chiefly a European phrase meaning "linear phase." Any
system which accurately preserves phase relationships
between frequencies, i.e., that exhibits pure delay. See: group delay"

http://www.rane.com/par-p.html

Put a cork in it, JJ.

jj@research.att.com (jj, DBT thug and skeptical philalethist) wrote in
>
> Heh. It seems to me to be much ado about nothing.


Bob writes

Perhaps. There seems to be general agreement about the group delay
equations. It's always interesting to (try to) relate these equations
to what is physically happening.

I will admit however, that no matter how we picture group delay in our
minds, the darned equations still give the same answers.

Bob Stanton

In article <3c5153aa.8744276@news.dircon.co.uk>,
Goofball_star_dot_etal <who@needs.email.anyhow> wrote:
>"phase linear Chiefly a European phrase meaning "linear phase." Any
>system which accurately preserves phase relationships
>between frequencies, i.e., that exhibits pure delay. See: group delay"

>http://www.rane.com/par-p.html

>Put a cork in it, JJ.

I see one of my stalkers has wandered over here. Go back home to
r.a.o, you became wearing yearrs ago. You saw the :-), I'm sure.

--
Copyright jj@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.

Bob_Stanton wrote:
>
> jj@research.att.com (jj, DBT thug and skeptical philalethist) wrote in
> >
> > Heh. It seems to me to be much ado about nothing.
>
> Bob writes
>
> Perhaps. There seems to be general agreement about the group delay
> equations. It's always interesting to (try to) relate these equations
> to what is physically happening.
>
> I will admit however, that no matter how we picture group delay in our
> minds, the darned equations still give the same answers.
>
> Bob Stanton

It seems to me that group delay has a fairly obvious and intuitive
meaning. The sticky concept is phase delay. It has the dimension of
time, and people keep trying to relate it to some interval. There's no
way to do that.

To forestall those who would accuse me of heresy, consider radio Morse
code sent as CW. (That the carrier is turned on and off is allowed by
"continuous wave". The alternative to CW is spark gap, still allowed at
sea when I was too young to care. Does anyone remember "logarithmic
decrement"?) The rise and fall times of CW pulses are controlled to keep
the sidebands within a few hundred Hz (or less) of the carrier. I don't
care to calculate the group and phase delays of am IF strip consisting
of several double-tuned transformers, each with different Q, coupling
coefficient, and center frequency (the whole making up a Chebychev flat
passband), but it can be done (and more easily measured).

If the receiver is any good, a dash coming out looks like it did going
in, provided we look only at the envelope. Looking closely, you can see
that the phase of the carrier relative to the edge of the pulse is
messed up. It's easy, with a dual-trace scope, to measure the time delay
through the strip if we match the pulse and ignore the carrier. That's
easy to do if the sweep is slow enough not to resolve the carrier.

Except near the edges of the dash, there aren't any sidebands; it's all
carrier. So when we measured the transit time of our signal through the
IF strip, it was mostly carrier that we measured. It's a real delay.
Now, without changing the signal (dah dah dah ...), change the sweep
speed and look at the phase shift between input and output when there's
signal. (Between dahs, there's nothing to measure.) Forget about the
two-pi ambiguity. Assume that there are ways to unwrap the phase.)
Calculate the time represented by the measured phase angle. Divide by
the frequency: it's a time! (Cigars all around.) In general, that
bouncing baby time won't be the same as the IF delay we measured before.
If anyone knows what it means, please tell me.

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