Jim Carr wrote:

"Bob Cain" wrote in message
...

When you push on air, it moves and pushes on the air in
front of it but with some delay in the transfer. That's
what causes the speed of sound. The push propegates outward
from this bit of air to the bit in front of it and that's a
wave. Same when you pull on it.

Does that help?


Sorta. From what you're saying. the *origin* of each individual wave can
take place at any point within the throw of the diaphragm. Is that correct?

Not sure even how to define the origin of the wave in those
terms. Thanks for that. I just realized that the
assumptions which are being made about that are the flaw in
the intuitive description of "Doppler distortion."

Something that is ocuring dynamically is being described in
terms of a static piston in one sense and dynamically in
another. That doesn't work. The distance from the piston
to the the sensor isn't relevant to the argument if it is
riding the wave. In a way, it's effect is being included
twice if you do that. That's a no-no that will lead to
false prediction.

The flaw in the common argument for "Doppler distortion" has
proven very elusive but I think that this nails it. It
really is subtle which explains why it's been around so long.

I added rec.audio.pro to this because it's highly relevant
to the thread on this subject that is happening there.


Bob
--

"Things should be described as simply as possible, but no
simpler."

A. Einstein

"Bob Cain" wrote in message
...

Sorta. From what you're saying. the *origin* of each individual wave can
take place at any point within the throw of the diaphragm. Is that
correct?

Not sure even how to define the origin of the wave in those
terms. Thanks for that. I just realized that the
assumptions which are being made about that are the flaw in
the intuitive description of "Doppler distortion."

I'm having a hard time envisioning just one wave being started by one thrust
of the piston. Maybe if I fully understood that rather than the quite
satisfactory "it just does and that's how a speaker works" mentality I'v
always had, I could argue intelligently one way or the other.

I added rec.audio.pro to this because it's highly relevant
to the thread on this subject that is happening there.

Great. Feed me to the wolves. Hey, RAP folks: I hold no degrees in
electronics, physics, acoustics, etc. I do not work with audio as a
profession. I just find the topic interesting and do not purport myself to
be an expert. As I noted earlier in the thread, which was not cross-posted,
if I seem condescending, it is because I am trying to explain things in
simple terms to myself. Since my logic is usually sound, my guess is that a
basic premise somewhere is wrong or incomplete, hence the detailed and
simplist explanations.

With that said, help me out here. I can't get myself away from the
assumption that since a speaker diaphragm has a throw of a certain distance,
then the waves started by the diaphragm may be started from any point in
that throw. As such two waves which are created a certain time apart may end
up traveling different distances to reach my stationary ear, thus a Doppler
shift. Measurable? I dunno. Discernible to my ear? Probably not.

"Jim Carr"

With that said, help me out here. I can't get myself away from the
assumption that since a speaker diaphragm has a throw of a certain
distance,
then the waves started by the diaphragm may be started from any point in
that throw. As such two waves which are created a certain time apart may
end
up traveling different distances to reach my stationary ear, thus a
Doppler
shift.


** A time delay or advance is just that - it is not Doppler. Any such
delay or advance depends solely on the position of the cone - not its
*velocity*. If a cone is displaced by 10mm, that will introduce a time
error of 29 uS or a phase shift of 50 degrees at 5 kHz.

Any attempt to measure Doppler frequency shifts must allow for this -
most have not.



.............. Phil

"Phil Allison" wrote in message

"Jim Carr"

With that said, help me out here. I can't get myself away from the
assumption that since a speaker diaphragm has a throw of a certain
distance, then the waves started by the diaphragm may be started
from any point in that throw. As such two waves which are created a
certain time apart may end up traveling different distances to reach
my stationary ear, thus a Doppler shift.


** A time delay or advance is just that - it is not Doppler. Any
such delay or advance depends solely on the position of the cone -
not its *velocity*. If a cone is displaced by 10mm, that will
introduce a time error of 29 uS or a phase shift of 50 degrees at 5
kHz.

Any attempt to measure Doppler frequency shifts must allow for this
- most have not.

That's because this time shift, more specifically the time rate of change of
this time shift, is the cause of Doppler.

Arny,

This information is probably in this thread somewhere, but it has
gotten so long and convoluted that it's much easier just to ask: Are
you asking whether FM (Doppler) modulation at the high frequency is
the ONLY effect that results when that high frequency in addition to a
low frequency (purposely left undefined since the actual values depend
on a number of factors in the physical setup) are reproduced in the
same transducer, or is there some amount of AM modulation as well?

"The Ghost" gave me an idea for determining this without requiring any
measurement of the instantaneous cone displacement. Perform an FM
discrimination of the received (microphone) signal at the high
frequency "carrier." Call the discriminated signal m(t). Regenerate a
perfect FM signal using the modulating signal m(t) and subtract that
from the original signal. The result is the residual modulation on the
signal, which could then be AM-detected to determine if AM is present.

Three practical issues which must be dealt with come to mind:

1) How to synchronize the regenerated FM carrier amplitude to the
original FM amplitude? Easy answer: emit a signal consisting of the
high frequency tone alone for a length of time adequate to measure the
amplitude.

2) What modulation index, or depth of modulation, should be used in
the regenerated FM signal? Said another way, what gain (if any) should
be applied to m(t) when regenerating the FM signal?

3) How do you synchronize the regenerated signal in time with the
original signal? There are actually two synchronization tasks to be
done: phase synchronization of the carriers, and delay in the modulating
signal, i.e., tau in A*m(t-tau). (A is the parameter in question 2).

Does this make any sense?
--
% Randy Yates % "My Shangri-la has gone away, fading like
%% Fuquay-Varina, NC % the Beatles on 'Hey Jude'"
%%% 919-577-9882 %
%%%% % 'Shangri-La', *A New World Record*, ELO
http://home.earthlink.net/~yatescr

"Randy Yates" wrote in message


Arny,

This information is probably in this thread somewhere, but it has
gotten so long and convoluted that it's much easier just to ask: Are
you asking whether FM (Doppler) modulation at the high frequency is
the ONLY effect that results when that high frequency in addition to a
low frequency (purposely left undefined since the actual values depend
on a number of factors in the physical setup) are reproduced in the
same transducer, or is there some amount of AM modulation as well?

I'm not asking that question, because I know the answer, and I knew it
walking in the door last week.

The results of playing multiple tones through something as dirty as a
speaker produces copious amonts of both AM and FM. As a rule, the AM
dominates.

"The Ghost" gave me an idea for determining this without requiring any
measurement of the instantaneous cone displacement. Perform an FM
discrimination of the received (microphone) signal at the high
frequency "carrier." Call the discriminated signal m(t). Regenerate a
perfect FM signal using the modulating signal m(t) and subtract that
from the original signal. The result is the residual modulation on the
signal, which could then be AM-detected to determine if AM is present.

I've tried that, and a lot of other things. It has the usual problems with
nulling in the real world. You can get roughly a 2:1 to 10:1 reduction of
the unwanted distortion by that means.

Three practical issues which must be dealt with come to mind:

1) How to synchronize the regenerated FM carrier amplitude to the
original FM amplitude?

Pretty easy to do an fair job of in the digital domain.

Easy answer: emit a signal consisting of the
high frequency tone alone for a length of time adequate to measure the
amplitude.

If you've looked at the raw data page posted at
http://www.pcavtech.com/techtalk/doppler/ you'd know that finding that out
with pretty fair precision is a matter of reading numbers off a screen.

2) What modulation index, or depth of modulation, should be used in
the regenerated FM signal? Said another way, what gain (if any) should
be applied to m(t) when regenerating the FM signal?

At this point I should point out that since the AM dominates, it might make
sense to apply an AM signal to null the AM part out, leaving the FM.

3) How do you synchronize the regenerated signal in time with the
original signal? There are actually two synchronization tasks to be
done: phase synchronization of the carriers, and delay in the
modulating signal, i.e., tau in A*m(t-tau). (A is the parameter in
question 2).

Well, we know quite a bit about the signal that we are trying to clean up.

Does this make any sense?

Been there, done that. Seriously, I come back to this problem of separating
AM and FM from a real world signal every once and while, and learn a bit
more about solving it.

This time I realized that ideally, AM distortion related sidebands are
indepenendent of of the carrier frequency, but increase in amplitude with
carrier frequency for FM. Trouble is, this practical example is so heavily
dominated by the AM distortion. I hope to go back to studying jitter, and
play this card there.

I suspect that lots of people have been misidentifying AM distortion
products as jitter.

"Arny Krueger"
"Phil Allison"
"Jim Carr"

With that said, help me out here. I can't get myself away from the
assumption that since a speaker diaphragm has a throw of a certain
distance, then the waves started by the diaphragm may be started
from any point in that throw. As such two waves which are created a
certain time apart may end up travelling different distances to reach
my stationary ear, thus a Doppler shift.


** A time delay or advance is just that - it is not Doppler. Any
such delay or advance depends solely on the position of the cone -
not its *velocity*. If a cone is displaced by 10mm, that will
introduce a time error of 29 uS or a phase shift of 50 degrees at 5
kHz.

Any attempt to measure Doppler frequency shifts must allow for this
- most have not.

That's because this time shift, more specifically the time rate of change
of
this time shift, is the cause of Doppler.



** So this is what all the Doppler Distortion fuss is about ????

A tiny bit of phase jitter, which at 5 kHz rarely amounts to more than a
few degrees ??


I was looking at it on my scope yesterday:


1. A 5 inch woofer, in box, driven by an amp fed from with two sine wave
generators with outputs summed.

2. A condenser mic feeding a pre-amp and followed by a 12 dB/oct HPF at
2 kHz thence to the scope.

3. The high frequency generator output is also linked to the scope which
operates in X-Y mode.

4. Park mic in front of woofer fed with a circa 5000 Hz sine wave at
about 10 watts. ( I used ear muffs)

5. Adjust scope and exact mic position to get a straight, diagonal line
traced on the scope screen - note that adjusting the 5000 Hz amplitude
affects the angle of the diagonal line only (ie makes it easy to visually
distinguish amplitude modulation ).

6. Turn up low frequency generator, set to say 40 Hz, and watch the line
open out to form a narrow ellipse indicating that the phase is changing as
the cone moves closer and further away from the mic.

7. Sweep low frequency generator up and down and note that cone excursion
alone controls the size of the ellipse - it never opens out more than
about 15 degrees for a linear cone excursion of 3 mm.

8. Try hard to imagine that this is the notorious, evil, Doppler
distortion before your eyes.


Wow.



........... Phil

"Phil Allison" wrote in message

"Arny Krueger"
"Phil Allison"
"Jim Carr"

With that said, help me out here. I can't get myself away from the
assumption that since a speaker diaphragm has a throw of a certain
distance, then the waves started by the diaphragm may be started
from any point in that throw. As such two waves which are created a
certain time apart may end up travelling different distances to
reach my stationary ear, thus a Doppler shift.


** A time delay or advance is just that - it is not Doppler. Any
such delay or advance depends solely on the position of the cone -
not its *velocity*. If a cone is displaced by 10mm, that will
introduce a time error of 29 uS or a phase shift of 50 degrees at 5
kHz.

Any attempt to measure Doppler frequency shifts must allow for
this
- most have not.

That's because this time shift, more specifically the time rate of
change of this time shift, is the cause of Doppler.



** So this is what all the Doppler Distortion fuss is about ????

A tiny bit of phase jitter, which at 5 kHz rarely amounts to more
than a few degrees ??

It's not a lot. The most important thing is that its swamped by all teh AM
distortion.

I was looking at it on my scope yesterday:

1. A 5 inch woofer, in box, driven by an amp fed from with two sine
wave generators with outputs summed.

2. A condenser mic feeding a pre-amp and followed by a 12 dB/oct
HPF at 2 kHz thence to the scope.

3. The high frequency generator output is also linked to the scope
which operates in X-Y mode.

4. Park mic in front of woofer fed with a circa 5000 Hz sine wave
at about 10 watts. ( I used ear muffs)

5. Adjust scope and exact mic position to get a straight, diagonal
line traced on the scope screen - note that adjusting the 5000 Hz
amplitude affects the angle of the diagonal line only (ie makes it
easy to visually distinguish amplitude modulation ).

6. Turn up low frequency generator, set to say 40 Hz, and watch
the line open out to form a narrow ellipse indicating that the phase
is changing as the cone moves closer and further away from the mic.

7. Sweep low frequency generator up and down and note that cone
excursion alone controls the size of the ellipse - it never opens
out more than about 15 degrees for a linear cone excursion of 3 mm.

8. Try hard to imagine that this is the notorious, evil, Doppler
distortion before your eyes.

I never said it was notorious or evil. But net it out -we're saying pretty
much the same thing, Phil. The Doppler distortion is there but its small.

I think the guy who brought up Doppler as some kind of a serious problem did
so a few weeks ago. He used Doppler distortion as a justification for not
liking long-excursion woofers. In the end he admitted that he used 2-way
monitors with either 6.5 or 8" woofers, and no subwoofer. Ironic enough?

In article ,
"Phil Allison" wrote:

"Arny Krueger"
"Phil Allison"
"Jim Carr"

With that said, help me out here. I can't get myself away from the
assumption that since a speaker diaphragm has a throw of a certain
distance, then the waves started by the diaphragm may be started
from any point in that throw. As such two waves which are created a
certain time apart may end up travelling different distances to reach
my stationary ear, thus a Doppler shift.


** A time delay or advance is just that - it is not Doppler. Any
such delay or advance depends solely on the position of the cone -
not its *velocity*. If a cone is displaced by 10mm, that will
introduce a time error of 29 uS or a phase shift of 50 degrees at 5
kHz.

Any attempt to measure Doppler frequency shifts must allow for this
- most have not.

That's because this time shift, more specifically the time rate of change
of
this time shift, is the cause of Doppler.



** So this is what all the Doppler Distortion fuss is about ????

A tiny bit of phase jitter, which at 5 kHz rarely amounts to more than a
few degrees ??


I was looking at it on my scope yesterday:


1. A 5 inch woofer, in box, driven by an amp fed from with two sine wave
generators with outputs summed.

2. A condenser mic feeding a pre-amp and followed by a 12 dB/oct HPF at
2 kHz thence to the scope.

3. The high frequency generator output is also linked to the scope which
operates in X-Y mode.

4. Park mic in front of woofer fed with a circa 5000 Hz sine wave at
about 10 watts. ( I used ear muffs)

5. Adjust scope and exact mic position to get a straight, diagonal line
traced on the scope screen - note that adjusting the 5000 Hz amplitude
affects the angle of the diagonal line only (ie makes it easy to visually
distinguish amplitude modulation ).

6. Turn up low frequency generator, set to say 40 Hz, and watch the line
open out to form a narrow ellipse indicating that the phase is changing as
the cone moves closer and further away from the mic.

7. Sweep low frequency generator up and down and note that cone excursion
alone controls the size of the ellipse - it never opens out more than
about 15 degrees for a linear cone excursion of 3 mm.

8. Try hard to imagine that this is the notorious, evil, Doppler
distortion before your eyes.

A dynamic loudspeaker is a mechanical system operated above resonance.
That means that the instantaneous position of the cone is *not*
represented by the voltage across the voice coil at that instant -- in
other words, there is a phase shift between the driving voltage and the
driven cone.

Compared to the displacement of the cone when driven by a DC voltage of
a certain amplitude, at cone resonance, the phase shift is 90 degrees;
well above that frequency it approaches 180 degrees.

To visualize how the driving force and the cone excursion are not in
phase, experiment with a weight on the end of a rubber band, the weight
heavy enough to cause the band to be significantly stretched. Put
several bands in series to make it easier -- say 18" or so long when
stretched.

Hold the upper end of the string of bands in your hand, and move your
hand up and down very slowly. The weight follows along, in phase. This
is below resonance.

Now move your hand up and down fast. The weight goes up when your hand
goes down. This is well above resonance.

If you are careful, you can find resonance, and note that the motion of
the weight moves in quadrature to the position of your hand.

Notice also that when above resonance, the peak-to-peak displacement of
the weight goes down as the driving frequency goes up, if you hold the
excursion of your hand constant. This is exactly the way that a
loudspeaker maintains constant SPL over frequency when operated in its
"passband". It's automatic, an inevitable result of a mechanical system
being operated above resonance.

Could this explain the phase shift you are seeing? What happens if,
instead of changing cone excursion by changing the frequency, you keep
the frequency constant and adjust the amplitude?

Isaac

Arny Krueger wrote:

That's because this time shift, more specifically the time rate of change of
this time shift, is the cause of Doppler.

Doesn't exist, Arny. Look he

http://www.silcom.com/~aludwig/Physi..._of_sound.html

Tellingly, as deep as the discussion goes, no mention is
made of "Doppler distortion" and if you read it you will see
why such nonsense wouldn't even have been considered.

It also directly supports what I have said recently that
distance from an oscilating piston, for the purposes of the
physics of piston interaction with air is the distance to
the rest position. I must say, that I found this link just
minutes ago, oddly enough looking for links to IR's for
Acoutic Modeler. I fingered it out earlier all by m'self.

Now about that data you posted...



Bob
--

"Things should be described as simply as possible, but no
simpler."

A. Einstein

"Bob Cain" wrote in message

Arny Krueger wrote:

That's because this time shift, more specifically the time rate of
change of this time shift, is the cause of Doppler.

Doesn't exist, Arny. Look he

http://www.silcom.com/~aludwig/Physi..._of_sound.html

Tellingly, as deep as the discussion goes, no mention is
made of "Doppler distortion" and if you read it you will see
why such nonsense wouldn't even have been considered.

It also directly supports what I have said recently that
distance from an oscilating piston, for the purposes of the
physics of piston interaction with air is the distance to
the rest position. I must say, that I found this link just
minutes ago, oddly enough looking for links to IR's for
Acoutic Modeler. I fingered it out earlier all by m'self.

Now about that data you posted...

Sorry Bob, but I'm not buying.

"Phil Allison" wrote in message
...

** A time delay or advance is just that - it is not Doppler. Any such
delay or advance depends solely on the position of the cone - not its
*velocity*.

I disagree. The time delay or advance *is* Doppler. The speed of sound is
constant in a given medium. In the classic example of the train whistle the
velocity of the train changes the distance/time between waves, but those
waves travel just as fast as if the train were still. It seems like you are
saying Doppler has to do with adding velocities together, which is untrue.

"Jim Carr"
"Phil Allison"

** A time delay or advance is just that - it is not Doppler. Any
such
delay or advance depends solely on the position of the cone - not its
*velocity*.

I disagree. The time delay or advance *is* Doppler.


** Doppler frequency shift is proportional to source velocity - so they
are not the same.


The speed of sound is constant in a given medium.


** If the medium is moving at some speed then that adds to, or subtracts
from, the speed of sound in still air (ie 343 m/S)


In the classic example of the train whistle the
velocity of the train changes the distance/time between waves,


** Yes, because the train is moving through the air.


but those waves travel just as fast as if the train were still.

It seems like you are
saying Doppler has to do with adding velocities together, which is untrue.



** Not at all - but the magnitude of the Doppler shift is proportional to
the velocity of the source compared to the surrounding air. A woofer cone
takes a small volume of with it for the ride.




............. Phil

"Phil Allison" wrote in message
...
I disagree. The time delay or advance *is* Doppler.


** Doppler frequency shift is proportional to source velocity - so they
are not the same.

Maybe we just have a failure to communicate. I say the actual frequency does
not shift. Assume a constant frequency at the source. If the distance
between the source and the receiver changes, then the receiver cannot
reliably determine the frequency. This is because the distance between the
sound waves (wavelength) emitted by the source changes. The receiver
determines the frequency by measuring the wavelength, which is the distance
between a given point on a wave and the corresponding point on the next
cycle of the wave.

Now, our ears don't measure distance, they are measuring time to put it
loosely. They don't care how fast the wave is moving. They sense the time
interval (a function of distance) between waves. Since movement of either
the source, observer or both can change that distance, there is an
*apparent* shift in frequency, not a "real" shift. We know this because we
already agreed the source emitted a constant frequency. It can be expressed
like this:

fo = fs . (v - vo) / (v - vs)
fo is the apparent frequency of the observer.
fs is the frequency of the source
v is the speed of sound
vo and vs are the velocities of the source and observer.

But then again, you seemed to disagree about the speed of sound being
constant when you wrote...

** If the medium is moving at some speed then that adds to, or subtracts
from, the speed of sound in still air (ie 343 m/S)

To which I say, care to cite a reference? The apparent shift is certainly a
function of the velocity of the source and/or receiver, but the speed of
sound is constant. If it's not, there is no Doppler.

Let's just reason it out by using the whistle on the train and the changing
speed of sound which you cite. Suppose further that the listener is 3,430
meters away. The sound wave has to travel 10 seconds to arrive at the
observer, right?

Train #1 is stationary and blows the whistle. A second train is moving at
100m/S. At the exact moment is parallel with train #1 it blows its whistle.
This happens to be the exact same time train #1 blew its whistle. Based on
your statement above, the first sound of the whistle is moving at 343m/S.
The second is traveling at 443m/S.

Are you saying that the second whistle will arrive at the receiver 2.25
seconds earlier and that we would hear two separate whistles?

If they did, they would sound exactly the same anyway, which would mean no
Doppler shift.

Refer to the formula above. For Doppler to work, v (speed of sound) must
remain constant in the medium. If the velocities add together, then the
distance between each wave would therefore remain constant. Hence, no shift.

On Sat, 14 Aug 2004 21:04:25 -0700, "Jim Carr"
wrote:

** A time delay or advance is just that - it is not Doppler. Any such
delay or advance depends solely on the position of the cone - not its
*velocity*.

I disagree. The time delay or advance *is* Doppler. The speed of sound is
constant in a given medium. In the classic example of the train whistle the
velocity of the train changes the distance/time between waves, but those
waves travel just as fast as if the train were still. It seems like you are
saying Doppler has to do with adding velocities together, which is untrue.

Doppler has to do with adding velocities together.

Chris Hornbeck

How come every time I look at a message in this thread (and I really
try to avoid it, but I'm a junkie and I can't be cured) it seems to be
cross-posted to more and more newsgroups?

Don't we have enough experts in rec.audio.pro?

--
I'm really Mike Rivers )
However, until the spam goes away or Hell freezes over,
lots of IP addresses are blocked from this system. If
you e-mail me and it bounces, use your secret decoder ring
and reach me he double-m-eleven-double-zero at yahoo

On 15 Aug 2004 09:26:57 -0400, (Mike Rivers)
wrote:


How come every time I look at a message in this thread (and I really
try to avoid it, but I'm a junkie and I can't be cured)

If you think it's bad for you, look at us participants. Of course,
we could go back to politics... those threads appear to have died
down, and I actually avoided reading most of them.

it seems to be
cross-posted to more and more newsgroups?

Don't we have enough experts in rec.audio.pro?

I think it was Bob doing that, or some of it. I think he's looking
for input (or something?) from the other groups. The "original"
original thread(s) weren't even on RAP...
-----
http://mindspring.com/~benbradley

Jim Carr wrote:

With that said, help me out here. I can't get myself away from the
assumption that since a speaker diaphragm has a throw of a certain distance,
then the waves started by the diaphragm may be started from any point in
that throw. As such two waves which are created a certain time apart may end
up traveling different distances to reach my stationary ear, thus a Doppler
shift. Measurable? I dunno. Discernible to my ear? Probably not.

To recapitulate, the problem with that intuitive view, which
is the whole basis of believing that there is "Doppler
distortion" is that it assumes that the distance from the
driver is the distance from the instantaneous position of
the piston. That's wrong. The distance from the driver,
since it is riding the wave it is creating, is the distance
from its zero or rest position, the position about which it
oscillates. That doesn't change with the nature of the
signal unless there is a DC component.

If the distance from the driver is not changing, there is no
Doppler shift. None of the proposed scenarios which have
the face of the driver oscillating about a rest position
will produce Doppler shift despite intuition.

Whew!


Bob
--

"Things should be described as simply as possible, but no
simpler."

A. Einstein

In alt.music.home-studio,rec.audio.tech,rec.audio.pro,
Bob Cain wrote:



Jim Carr wrote:

With that said, help me out here. I can't get myself away from the
assumption that since a speaker diaphragm has a throw of a certain distance,
then the waves started by the diaphragm may be started from any point in
that throw. As such two waves which are created a certain time apart may end
up traveling different distances to reach my stationary ear, thus a Doppler
shift. Measurable? I dunno. Discernible to my ear? Probably not.

To recapitulate, the problem with that intuitive view, which
is the whole basis of believing that there is "Doppler
distortion" is that it assumes that the distance from the
driver is the distance from the instantaneous position of
the piston. That's wrong. The distance from the driver,
since it is riding the wave it is creating, is the distance
from its zero or rest position, the position about which it
oscillates. That doesn't change with the nature of the
signal unless there is a DC component.

Oh, goody. You want a DC component, I'll GIVE you a DC component!
insert emoticon here

Let's go to extremes. Say we got one of these big honking
high-power woofers (that I recall reading Arny's writings about a
while back) with X-max of one or two inches or so. Superimpose a 1kHz
tone (probably the highest frequency it will reasonably reproduce)
onto a "DC component", say a 1/2 Hz sine wave that slooowly moves the
cone in and out a total distance of two inches, all the while it's
also putting this 1kHz tone into the air. Don't think of it as a 0.5Hz
sine wave, think of it as a varying DC component (that's obviously
what it is, you can see the cone move back and forth with your eyes).
What will you say is the acoustic source of the 1kHz, the driver
frame, which does not move, or the cone, which DOES move?

And presuming you see this as a possible cause of doppler
distortion, how is this "DC component" any different from a higher
frequency (say 20Hz or 50Hz) that also causes substantial cone
displacement?

If the distance from the driver is not changing, there is no
Doppler shift.

The distance from which part of the driver? The frame? The cone?
Something else?

None of the proposed scenarios which have
the face of the driver oscillating about a rest position
will produce Doppler shift despite intuition.
Whew!

If you think it's frustrating for you, imagine how I feel with Phil
agreeing with me!

Bob

-----
http://mindspring.com/~benbradley

Ben Bradley wrote:


Let's go to extremes. Say we got one of these big honking
high-power woofers (that I recall reading Arny's writings about a
while back) with X-max of one or two inches or so. Superimpose a 1kHz
tone (probably the highest frequency it will reasonably reproduce)
onto a "DC component", say a 1/2 Hz sine wave that slooowly moves the
cone in and out a total distance of two inches, all the while it's
also putting this 1kHz tone into the air. Don't think of it as a 0.5Hz
sine wave, think of it as a varying DC component (that's obviously
what it is, you can see the cone move back and forth with your eyes).
What will you say is the acoustic source of the 1kHz, the driver
frame, which does not move, or the cone, which DOES move?

Doesn't matter how slow the oscilation is, it won't produce
Doppler shift.

If the distance from the driver is not changing, there is no
Doppler shift.


The distance from which part of the driver? The frame? The cone?
Something else?

The rest position, the one it will settle to when the
driving signal is removed. If the driving signal contains a
DC component, and the piston is not restrained by a
compliance, then and only then will Doppler shift occurs.
Hard to swallow, I know but it is the truth.




None of the proposed scenarios which have
the face of the driver oscillating about a rest position
will produce Doppler shift despite intuition.
Whew!


If you think it's frustrating for you, imagine how I feel with Phil
agreeing with me!

That really must be rough. I sympathize. :-)

Look here if you aquire the math to understand it.


http://www.silcom.com/~aludwig/Physi..._of_sound.html

I just found it and everything I've said is in it if not in
the same context. It doesn't consider "Doppler distortion"
because there is no reason to.


Bob
--

"Things should be described as simply as possible, but no
simpler."

A. Einstein

On Sun, 15 Aug 2004 00:37:44 -0700, Bob Cain
wrote:

If the distance from the driver is not changing, there is no
Doppler shift.


I wrote:

The distance from which part of the driver? The frame? The cone?
Something else?

The rest position, the one it will settle to when the
driving signal is removed. If the driving signal contains a
DC component, and the piston is not restrained by a
compliance, then and only then will Doppler shift occurs.
Hard to swallow, I know but it is the truth.

Okay, Bob, I'm going to add a DC component, but I'm not going to
tell you that this DC component is really a millihertz-frequency sine
wave. How will you know the difference? Just what IS the difference
over a time period of one second?

Bob

-----
http://mindspring.com/~benbradley

To recapitulate, the problem with that intuitive view, which
is the whole basis of believing that there is "Doppler
distortion" is that it assumes that the distance from the
driver is the distance from the instantaneous position of
the piston. That's wrong. The distance from the driver,
since it is riding the wave it is creating, is the distance
from its zero or rest position, the position about which it
oscillates. That doesn't change with the nature of the
signal unless there is a DC component.

If the distance from the driver is not changing, there is no
Doppler shift. None of the proposed scenarios which have
the face of the driver oscillating about a rest position
will produce Doppler shift despite intuition.

Sorry, Bob, but I disagree. There is my thought experiment (which I consider
proof). And there is also the claim by an anonymous poster outlining his test
procedure and claiming he measured it.

William Sommerwerck wrote:

To recapitulate, the problem with that intuitive view, which
is the whole basis of believing that there is "Doppler
distortion" is that it assumes that the distance from the
driver is the distance from the instantaneous position of
the piston. That's wrong. The distance from the driver,
since it is riding the wave it is creating, is the distance
from its zero or rest position, the position about which it
oscillates. That doesn't change with the nature of the
signal unless there is a DC component.


If the distance from the driver is not changing, there is no
Doppler shift. None of the proposed scenarios which have
the face of the driver oscillating about a rest position
will produce Doppler shift despite intuition.


Sorry, Bob, but I disagree.

Can you be more specific. I don't mean by offering another
scenario which I must find specific disagreement with but
doing it yourself with what I've offered. What part of it
is wrong?


Bob
--

"Things should be described as simply as possible, but no
simpler."

A. Einstein

If the distance from the driver is not changing, there is no
Doppler shift. None of the proposed scenarios which have
the face of the driver oscillating about a rest position
will produce Doppler shift despite intuition.


Sorry, Bob, but I disagree.

Can you be more specific. I don't mean by offering another
scenario which I must find specific disagreement with but
doing it yourself with what I've offered. What part of it
is wrong?

Consider the Principle of Superposition. QED.

I think I'm closing in on the truth here and, as usual, we
are (mostly) all right in some regards and are all wrong in
some.

First, my analysis of what happens in an infinite tube
containing a piston is wrong in the case of a constant
velocity component.

If the piston contains no constant velocity component then
the propegating wave will be a reproduction of the velocity
of the piston because its position is always in proper
correspondence to the velocity it is imparting so as to
impart no error. The usual intuitive rationalization for
Doppler distortion is just wrong. The process is linear and
real in this case.

However, my belief that in the tube a piston of constant
velocity imparts no constant velocity to the wave was wrong
as shown he

http://www.silcom.com/~aludwig/Physi...collisions.htm

The upshot is that there is no Doppler distortion in an
infinite tube with a driving piston in it for any signal
(until the piston smacks into the receiver if moving toward
it. :-)

In all other configurations, no Doppler distortion will
occur among components within the portion of its passband
that is fairly flat. It is the difference in the coupling
between the piston and the air at different frequencies that
produces Doppler distortion in the far field. Yes, a
speaker swinging back and forth on a rope will evidence
Doppler distortion because the low frequence swing does not
couple to the air signifigantly. This means that the
piston's position is never appropriate to the superimposed
velocities and a Doppler shift that varies with the lower
frequency velocity will occur. (I still believe the
spectrum will have a flat top if it is emitting a single
tone but that's another story.)

A Doppler distortion will occur for any two frequencies that
couple differently to the air but in the region of fairly
flat transduction, there will be none.

This is a bit more elaborate and may well contain error but
I believe it is the case and accomodates all the cases of
whistles and trains or little speakers moving in big ways.
It does say, however, that the vernacular belief in Doppler
distortion from a speaker due to components within its flat
passband is simply wrong.

Whatcha think?


Bob
--

"Things should be described as simply as possible, but no
simpler."

A. Einstein

"Bob Cain" wrote in message
...

I think I'm closing in on the truth here and, as usual, we
are (mostly) all right in some regards and are all wrong in
some.

First, my analysis of what happens in an infinite tube
containing a piston is wrong in the case of a constant
velocity component.

If the piston contains no constant velocity component then
the propegating wave will be a reproduction of the velocity
of the piston because its position is always in proper
correspondence to the velocity it is imparting so as to
impart no error. The usual intuitive rationalization for
Doppler distortion is just wrong. The process is linear and
real in this case.

However, my belief that in the tube a piston of constant
velocity imparts no constant velocity to the wave was wrong
as shown he

http://www.silcom.com/~aludwig/Physi...collisions.htm

The upshot is that there is no Doppler distortion in an
infinite tube with a driving piston in it for any signal
(until the piston smacks into the receiver if moving toward
it. :-)

In all other configurations, no Doppler distortion will
occur among components within the portion of its passband
that is fairly flat. It is the difference in the coupling
between the piston and the air at different frequencies that
produces Doppler distortion in the far field. Yes, a
speaker swinging back and forth on a rope will evidence
Doppler distortion because the low frequence swing does not
couple to the air signifigantly. This means that the
piston's position is never appropriate to the superimposed
velocities and a Doppler shift that varies with the lower
frequency velocity will occur. (I still believe the
spectrum will have a flat top if it is emitting a single
tone but that's another story.)

A Doppler distortion will occur for any two frequencies that
couple differently to the air but in the region of fairly
flat transduction, there will be none.

This is a bit more elaborate and may well contain error but
I believe it is the case and accomodates all the cases of
whistles and trains or little speakers moving in big ways.
It does say, however, that the vernacular belief in Doppler
distortion from a speaker due to components within its flat
passband is simply wrong.

Whatcha think?


If you mean that a whistle riding on a moving train, or a speaker
swinging back and forth (or spinning round and round like in a Leslie
speaker system) in a repeating oscillation cycle will produce Doppler shift,
but a stationary speaker reproducing a complex waveform containing a mixed
LF and HF tone (or any multiple combination of tones, as would be the case
in a complex musical waveform) won't produce Doppler shift, then, by golly,
I think you're right!

Porky wrote:

If you mean that a whistle riding on a moving train, or a speaker
swinging back and forth (or spinning round and round like in a Leslie
speaker system) in a repeating oscillation cycle will produce Doppler shift,
but a stationary speaker reproducing a complex waveform containing a mixed
LF and HF tone (or any multiple combination of tones, as would be the case
in a complex musical waveform) won't produce Doppler shift, then, by golly,
I think you're right!

I even finally agree that in many cases it has the
properties that can be called Doppler distortion, so it is a
real phenomenon in qualified conditions but not for the
reasons usually given and not from the systems for which it
has been claimed.


Bob
--

"Things should be described as simply as possible, but no
simpler."

A. Einstein

"Bob Cain" wrote in message
...


Porky wrote:

If you mean that a whistle riding on a moving train, or a speaker
swinging back and forth (or spinning round and round like in a Leslie
speaker system) in a repeating oscillation cycle will produce Doppler
shift,
but a stationary speaker reproducing a complex waveform containing a
mixed
LF and HF tone (or any multiple combination of tones, as would be the
case
in a complex musical waveform) won't produce Doppler shift, then, by
golly,
I think you're right!

I even finally agree that in many cases it has the
properties that can be called Doppler distortion, so it is a
real phenomenon in qualified conditions but not for the
reasons usually given and not from the systems for which it
has been claimed.


I would go so far as to say that anything generating sound and in motion
relative to the listener will generate Doppler shift, but in the special
case of a speaker reproducing a complex waveform consisting of more than one
pure tone, the velocity of the source relative to the listener is
effectively zero, and therefore no Doppler shift will be generated, because:
1) there is one complex waveform driving the speaker and producing the
energy necessary to generate the sound, not some higher frequency or
frequencies "riding" on some lower frequency.
2) the effective sound source is not the speaker cone, but some point or
plane which does not move relative to the listener. The moving speaker cone
just provides the mechanical energy which is transformed into acoustical
energy by it's interaction with the surrounding air. Each instantaneous
point in the complex motion of the cone is in direct co-relation to the
corresponding instantaneous point of compression or rarefaction in the
complex waveform and the notion that the source has motion relative to the
listener is an illusion. The sound wave can be pictured as standing still
with each bit coming into existance as the cone passes through that space.
The sound wave might be represented as:
-- - - -- - - --
with the space between the dots representing compression and rarefaction of
the wave, and the cone's motion producing the sound can be represented as:
]-- - - -- - - --
]- - - -- - - --
] - - -- - - --
]- - - -- - - --
]-- - - -- - - --
]- - - -- - - --
] - - -- - - --
]- - - -- - - --
Note that while the cone is moving, the apparent source is not moving, and
this applies no matter how complex the waveform is. This can actually be
measured with a sensitive pressure meter and graphed, showing that this is
really what is happening.
3) the effective sum of all velocities is zero relative to the listener
4) due to the nature of how a speaker produces sound, the listener is
effectively "riding on the train"
5) In order to produce Doppler shift, there must be a sound source in motion
relative to the listener and in the case of a stationary speaker there is a
complex sound source, but no motion relative to the listener
6) If a speaker did generate Doppler distortion, simply turning it so that
the listener was looking at the edge of the speaker would eliminate the
motion toward and away from the listener (the cone would move side to side,
but would stay at a constant distance from the listener), and thus would
eliminate Doppler distortion. I don't believe that this is the case in the
real world.
7) the empirical measurements I made indicated that there was no audible or
measurable Doppler shift when the Doppler equations predicted that there
would be audible and measurable shift, therefore the Doppler equations do
not apply to this special case.
8) I just have a gut feeling about it
9) none of the above
or
10) any of the above
Personally, I prefer number two.

Porky wrote:


I would go so far as to say that anything generating sound and in motion
relative to the listener will generate Doppler shift, but in the special
case of a speaker reproducing a complex waveform consisting of more than one
pure tone, the velocity of the source relative to the listener is
effectively zero, and therefore no Doppler shift will be generated, because:

Tell me if this is equivalent or not: There is Doppler type
mixing between two frequencies if and only if the pressure
in the far field due to them is a different function of the
velocity of the piston. Where the transfer function is flat
in the Fourier sense, nothing mixes. That is what it all
boils down to in the end. That is not at all the same as
the standard argument because it won't happen in a tube and
the standard argument says it will.

Also, a single tone cannot produce Doppler distortion so I
am definitely wrong about that. Yikes! I see why. The
definition I've been bandying about for a linear system is
actually the definiton of a linear, time invariant system.
The system we are considering must be time variant in terms
of the impedences involved. This is getting wierd.

Can this yield what I've been asking for, a general
expression for far field pressure as a function of piston
velocity that includes the Doppler distortion? I'm not sure
yet, but I'll be thinking about it. It is straightforward
for any two frequencies, however, and is left as an exercise
for the student. :-)


Bob
--

"Things should be described as simply as possible, but no
simpler."

A. Einstein

With all this sturm and drang no one has yet produced a
mathematical expression for the sound pressure at some
chosen distance from a velocity controled piston in a tube
(to keep the situation as simple as possible) given a signal
containing the sum of some chosen pair of frequencies at a
chosen relative magnitude. Odd, that. Sounds simple enough.

Several have said they could if they wanted to but don't.
Odd, that.

I have, but almost no one likes it. :-)


Bob
--

"Things should be described as simply as possible, but no
simpler."

A. Einstein

Bob Cain writes:

With all this sturm and drang no one has yet produced a mathematical
expression for the sound pressure at some chosen distance from a
velocity controled piston in a tube (to keep the situation as simple
as possible) given a signal containing the sum of some chosen pair of
frequencies at a chosen relative magnitude. Odd, that. Sounds simple
enough.

You know Bob, you are an irritating ****. This is in every physics book
and dozens if not hundreds of web sites. To terminate this drivel, I'll
type it in for you here.

The perceived frequency, or Doppler frequency, fd, of a frequency f due
to a relative velocity v between source and observer is given as

fd = f*c/(c + v),

where c is the speed of the medium (about 1100 ft/second for sound at
reasonable temperatures at sea level).

The instantaneous velocity v(t) of a speaker cone that is reproducing
a sine wave A*sin(2*pi*fl*t) at frequency fl (f low) and at an excursion
of A (meters, inches or whatever) is

v(t) = A*2*pi*fl*cos(2*pi*fl*t).

Put these two facts together and you get the dynamic doppler shift
in a speaker:

fd = f*c/(c + A*2*pi*fl*cos(2*pi*fl*t)).

Of course units have to match, but that's up to the person applying
this equation.

Now what????
--
% Randy Yates % "Remember the good old 1980's, when
%% Fuquay-Varina, NC % things were so uncomplicated?"
%%% 919-577-9882 % 'Ticket To The Moon'
%%%% % *Time*, Electric Light Orchestra
http://home.earthlink.net/~yatescr

Randy Yates wrote:

Bob Cain writes:


With all this sturm and drang no one has yet produced a mathematical
expression for the sound pressure at some chosen distance from a
velocity controled piston in a tube (to keep the situation as simple
as possible) given a signal containing the sum of some chosen pair of
frequencies at a chosen relative magnitude. Odd, that. Sounds simple
enough.


You know Bob, you are an irritating ****.

As always, irritation is a personal choice. I prefer the
more objective word, tenacious. :-)

This is in every physics book
and dozens if not hundreds of web sites. To terminate this drivel, I'll
type it in for you here.

The perceived frequency, or Doppler frequency, fd, of a frequency f due
to a relative velocity v between source and observer is given as

fd = f*c/(c + v),

Randy, that equation is only defined for a static v.


where c is the speed of the medium (about 1100 ft/second for sound at
reasonable temperatures at sea level).

The instantaneous velocity v(t) of a speaker cone that is reproducing
a sine wave A*sin(2*pi*fl*t) at frequency fl (f low) and at an excursion
of A (meters, inches or whatever) is

v(t) = A*2*pi*fl*cos(2*pi*fl*t).

Put these two facts together and you get the dynamic doppler shift
in a speaker:

fd = f*c/(c + A*2*pi*fl*cos(2*pi*fl*t)).

Of course units have to match, but that's up to the person applying
this equation.

Now what????

Try again, 'cause it's not correct.

You cannot simply subsitiute a v(t) into an equation for a
static v, especially without considering the conditions
under which it holds, and expect correct results. Did you
see my post to William where I explained that? I'm sure it
would be no news to you.

To get a dynamic equation you must do a fully dynamic
analysis of a specific system, including some set of boundry
conditions. That's why I like the tube which is what is
assumed in what follows. The boundry conditions lead to a
much easier analysis. It's just a section of the infinite
plane wave and a piston creating it.

Instead of starting from the acoustic wave equation and
solving it with the given boundry conditions (something I
could have done once but which would be a real struggle now
if I could do it at all), I chose to assume a solution and
see if it leads to any contradictions. This approach is not
uncommon in physics or math. It is further assumed that the
air remains in its linear regime, i.e. it's instantaneous
velocity is always signifigantly less than the speed of
sound. If that is not true the relationship between the
pressure and velocity of air will yield all kinds of
disortions without appeal to Doppler.

My assumed solution is:

p(x(t)+d,t+d/C) = v(x(t),t)*Ra

where v() and x() are the instantaneous controlled velocity
and resulting position of the piston at the same time t, d
is any displacement from the rest position of the piston, Ra
is the characteristic impedence of air, C is the speed of
sound and p() is instantaneous pressure at d. This is a
formulaic way of stating that the position of the piston is
always appropriate to the velocity being imparted so as to
propegate that condition down the tube as a wave at a speed
C. It says that the piston follows the pressure/position
profile of the traveling wave at its position in space.

If instead of driving the wave at that point, a test piston
were placed there in the path of the same wave traveling by,
it would have the same pressure/position profile in time.

If the assumed solution is absurd, here's the place to say
why.

From form alone, does the above not require that

p'(X+d,t+d/C) = v'(X,t)*Ra

for any X in the tube including the changing position of the
piston?

If so, we can let X=0 and get:

p'(d,t+d/c) = v'(0,t)*Ra

This says that the pressure at a point d from the piston is
the delayed volume velocity at the rest position of the
piston times the characteristic impedence of the air. It
remains to connect the volume velocity at the rest postion
to the velocity of the piston as it moves and here I appeal
to the reciprocity argument I stated above.

The final expression is independant of the nature of the
controled velocity and is purely linear.

If its consequences as I've stated them are logically
correct, does my assumption lead to any contradictions in
physics that can be stated formulaically? I can find none.
I invite anyone else to find one.

This is all to simply show that the usual argument for why
Doppler distortion exists is fundamentally wrong. It's
wrong because it predicts it in this configuration and the
above analysis says it won't happen.


Bob
--

"Things should be described as simply as possible, but no
simpler."

A. Einstein

Randy, that equation is only defined for a static v.

It is? So if the train happens to speed up or down as it passes me, the Doppler
shift ISN'T consistent with the train's instantaneous speed?

Hello? knock knock knock Anybody home?


Sudden profound insight! For trains, v isn't constant -- even when the train is
traveling at a constant velocity! Its relative velocity varies, dropping as the
train approaches the listener (reaching zero as the whistle passes opposite the
listener), then increases as the train moves away.

So the formula DOESN'T apply to train whistles? Isn't it amazing what we can
learn when we give the math precedence and ignore what happens in the real
world?


This is my absolutely final posting on the topic.

"Think, people, think!" -- Lex Luthor, "Superman -- the Motion Picture"

William Sommerwerck wrote:

Randy, that equation is only defined for a static v.


It is? So if the train happens to speed up or down as it passes me, the Doppler
shift ISN'T consistent with the train's instantaneous speed?

As I've often explained, that is because the low frequency
component is almost negligably coupled to the air in
comparison to the whistle and thus Randy's equation is a
close approximation. In the case where they are equally
coupled, the tube, no Doppler distortion occurs so long as
we remain in the linear regime of the air and that equation
does not at all describe what happens.

Once again, at this point I do not deny the phenomenon. I
don't know why you would think that if you've been following
along. I just feel the actual reason for it should be
understood because it gives considerably more insight into
when and where it might be a problem and offers the strong
possibility of a more quantitative formulation that can
indicate the real degree of the phenomenon for real systems.

The usual simple argument of a little, fast sinusoid riding
a big slow one on a cone, while intutitively attractive, is
wrong. This implies that the phenomenon occurs at the
piston/air interface when in fact it evolves over space as
the transfer function of frequency changes. In the audible
range, there is probably little or no Doppler distortion in
the very near field of real speakers because the low
frequencies have not yet begun to decouple.

I'm sure that to many, why the lightbulb goes on when you
flip the switch is of no interest or value but I'm not one
of them, and I presume that at least some others involved in
audio aren't either. That's why I became an
engineer/scientist in the first place. I doubt that anybody
with the other perspective is still with this discussion anyway.


Bob
--

"Things should be described as simply as possible, but no
simpler."

A. Einstein

In alt.music.home-studio,rec.audio.tech,rec.audio.pro, Bob Cain
wrote:



Randy Yates wrote:

This is in every physics book
and dozens if not hundreds of web sites. To terminate this drivel, I'll
type it in for you here.

The perceived frequency, or Doppler frequency, fd, of a frequency f due
to a relative velocity v between source and observer is given as

fd = f*c/(c + v),

Randy, that equation is only defined for a static v.

So if you start changing v, the doppler effect stops until you
leave v alone for a while? How does the doppler effect know to stop
and start up again?

Seriously (or you can answer the above question seriously if you
like), do you have any reference for the equation being defined only
for v being static?

Bob

-----
http://mindspring.com/~benbradley

Ben Bradley wrote:
In alt.music.home-studio,rec.audio.tech,rec.audio.pro, Bob Cain
wrote:



Randy Yates wrote:


This is in every physics book
and dozens if not hundreds of web sites. To terminate this drivel, I'll
type it in for you here.

The perceived frequency, or Doppler frequency, fd, of a frequency f due
to a relative velocity v between source and observer is given as

fd = f*c/(c + v),

Randy, that equation is only defined for a static v.


So if you start changing v, the doppler effect stops until you
leave v alone for a while? How does the doppler effect know to stop
and start up again?

I didn't say it stops, it just isn't described by that
formula any more. The Doppler effect doesn't know anything.
It is just produced in differing amounts in different
situations.


Seriously (or you can answer the above question seriously if you
like), do you have any reference for the equation being defined only
for v being static?

Not really. That's just the assumption wherever it is
presented. As an example of why an expression that contains
static values won't generally remain valid when simply
substituting a dynamic one, consider a black box containing
some arbitrary network of R, L anc C elements, distributed
and lumped, linear and non-linear, which connect two inputs
to one output.

If we hold one of those inputs X constant and characterize
the output by applying dynamic signals at input Y, then
change the value of the X input to another constant value
and again characterize the output, and repeat this with
enough different values on the constant input we will, in
general, be able to write down an expression that relates
output to input under those conditions. If now instead of
holding the input X constant we apply a time varying signal,
will the characterization we developed describe what the
output will do? Consider what would happen if there were a
parallel RC in series with X before going elsewhere in the
network.

The expression given for Doppler shift is the consequence of
keeping v constant and cannot be generalized to the dynamic
case by substitution. In the general dynamic case it will
be a whole lot more complicated and geometry dependant.
Depending on the geometry it will only be approximated by
the given expression even in the static case. It is really
only generally valid for a point source emitter and receiver
in an unbounded free space filled with nothing but air.


Bob
--

"Things should be described as simply as possible, but no
simpler."

A. Einstein

Ben Bradley wrote:

fd = f*c/(c + v),

Randy, that equation is only defined for a static v.


So if you start changing v, the doppler effect stops until you
leave v alone for a while? How does the doppler effect know to stop
and start up again?

C'mon, Ben. Where did I imply that it stops and starts?
That equation just can't tell the whole story. Consider
that for v constant none of its motion is being imparted to
the wave that reaches the Rx but if it is oscilating, some
of it is. That has to make some difference in the net
effect beyond the predicted warble. That difference is
missing from the equation because it is a term which drops
out for dv/dt=0. Can I derive that yet, no. Am I sure
there are additional terms dependant on rate of change, or
multiplied by w if two tones, yes.


Seriously (or you can answer the above question seriously if you
like), do you have any reference for the equation being defined only
for v being static?

I'm still awiating Pierce's book wherin it is claimed that
it is derived for the fully dynamic case giving the same
result. All the derivations I somewhat remember from long
ago university freshman physics definitely assumed constant
v as a premise.

The main reason I'm working out the proof of why Doppler
mixing doesn't happen with a piston in a tube is that the
equation above will thus be violated. After it has ramped
up from a stationary position to where it is oscilationg
with a constant motion superimposed on it, and after that
ramping up has passed an observer at some distance from the
piston, he will see no change in frequency but instead the
same oscilation superimposed on a constant air velocity
(until the piston smacks him up 'long side the head if the
constant motion is toward him.)

I'm pretty sure I now have that proof but am sitting with it
a while instead of possibly jumping the gun again and I've
asked a few folks to sanity check it. Would you care to?


Bob
--

"Things should be described as simply as possible, but no
simpler."

A. Einstein

Ben Bradley wrote:

fd = f*c/(c + v),

Randy, that equation is only defined for a static v.


So if you start changing v, the doppler effect stops until you
leave v alone for a while? How does the doppler effect know to stop
and start up again?

C'mon, Ben. Where did I imply that it stops and starts?
That equation just can't tell the whole story. Consider
that for v constant none of its motion is being imparted to
the wave that reaches the Rx but if it is oscilating, some
of it is. That has to make some difference in the net
effect beyond the predicted warble. That difference is
missing from the equation because it is a term which drops
out for dv/dt=0. Can I derive that yet, no. Am I sure
there are additional terms dependant on rate of change, or
multiplied by w if two tones, yes.


Seriously (or you can answer the above question seriously if you
like), do you have any reference for the equation being defined only
for v being static?

I'm still awiating Pierce's book wherin it is claimed that
it is derived for the fully dynamic case giving the same
result. All the derivations I somewhat remember from long
ago university freshman physics definitely assumed constant
v as a premise.

The main reason I'm working out the proof of why Doppler
mixing doesn't happen with a piston in a tube is that the
equation above will thus be violated. After it has ramped
up from a stationary position to where it is oscilationg
with a constant motion superimposed on it, and after that
ramping up has passed an observer at some distance from the
piston, he will see no change in frequency but instead the
same oscilation superimposed on a constant air velocity
(until the piston smacks him up 'long side the head if the
constant motion is toward him.)

I'm pretty sure I now have that proof but am sitting with it
a while instead of possibly jumping the gun again and I've
asked a few folks to sanity check it. Would you care to?


Bob
--

"Things should be described as simply as possible, but no
simpler."

A. Einstein

Bob Cain wrote in message ...
fd = f*c/(c + v),

Randy, that equation is only defined for a static v.

You say that with such authority, but you most certainly don't have
the authority required to make such a bold assertion. Can you
provide a reference to the technical literature to support such a
claim? The answer is no, because no such reference exists. The fact
of the matter is that your assertion is nothing more than a personal
belief, which you have accepted without questioning its validity.
Had you looked into it, as I have, you would have discovered that the
equation applies under both constant velocity and dynamic velocity
conditions. You will find the derivation in Allan Pierce's book
entitled "Acoustics: An Introduction to Its Physical Principles and
Applications." In connection with the derivation, Allan Pierce states
"The result holds regardless of the detailed time history of the
trajectory. The Doppler-shifted frequency at a given time and
position is affected only by the source's velocity and frequency at
the instant of generation of the wavelet currently being received.
The source does not have to be traveling with constant velocity or in
a straight line for the equation to apply."

Bob, I've said this many times before and I am going to say it again.
Your level of ignorance never ceases to amaze me.

The Ghost wrote:

You will find the derivation in Allan Pierce's book
entitled "Acoustics: An Introduction to Its Physical Principles and
Applications."

Thanks for the reference. Do you happen to know of one that
derives this and which costs somewhat less that $200 used?


Bob
--

"Things should be described as simply as possible, but no
simpler."

A. Einstein