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  #521   Report Post  
Bob Cain
 
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Chris Hornbeck wrote:

On Thu, 19 Aug 2004 01:23:56 -0700, Bob Cain
wrote:


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.



No, but the distance between them may vary....


Darn, you did it again. Why not? I'm looking at the case
where there is Doppler mixing but no relative motion. Think
of the compound system of a speaker emitting a tone and
swinging on a long rope. The rest position of that system
is when the speaker is off and it is sitting at bottom dead
center. When there is no motion between the rest position
of that compound Tx and a Rx , Doppler mixing will still
occur and I maintain that is just because of the low
coupling of the LF swing to the Rx.



Where the transfer function is flat
in the Fourier sense, nothing mixes.



No, that's AM.


Not talking about that here. Flat is the condition of a
piston in a tube which is the only condition (other than an
oscilating infinite plane) when Doppler mixing won't occur.
In any other configuration there will be a coupling
function that is not flat between Tx ane Rx and that is the
source of the mixing in the far field.


Bob
--

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

A. Einstein
  #522   Report Post  
Ben Bradley
 
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On Tue, 31 Aug 2004 23:28:15 -0700, Bob Cain
wrote:



Chris Hornbeck wrote:

On Thu, 19 Aug 2004 10:08:22 -0700, Bob Cain
wrote:


I'm trying to
understand why a system that appears linear in the case of
any pure sinusoid would produce mixing when presented with
superpositions. That defies my understanding at the moment
but I plan on fixing that. :-)



What you want is in Terman. Really.


Is there more than one by him?


Yes. Type terman in bookfinder.com and scroll to the
electronics/radio books to get (most/all?) his titles.

I've got Pierce on order but
wouldn't mind at all beefing up my shelf.


I have at least two Terman books, I need to decide between reading
and selling them. I'm in the middle of moving, so it may take a
while...

Bob


-----
http://mindspring.com/~benbradley
  #523   Report Post  
Chris Hornbeck
 
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On Tue, 31 Aug 2004 22:46:07 -0700, Bob Cain
wrote:

In reading these threads from out of town for two weeks, I've seen
no refutation of the real argument.


Which one might that be? :-)


That relative motion modulates relative distance which
modulates relative time, because the speed of sound is
constant.

Chris Hornbeck
  #524   Report Post  
Chris Hornbeck
 
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On Tue, 31 Aug 2004 23:21:08 -0700, Bob Cain
wrote:

That error is Doppler shift.



In a word, no.


How about a few more of them, then? :-)


The Doppler effect occurs downstream of transmitting
wavefront. In fact, you've shown very clearly that no
Doppler effect occurs close enough to the transmitter
for the wave to still be a plane.

Or, to say it in your newer and better way, close
enough to the transmitter that there are no frequency
dependent errors in the path to the receiver.

Chris Hornbeck
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Chris Hornbeck
 
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On Tue, 31 Aug 2004 23:23:07 -0700, Bob Cain
wrote:

Chris, Mike's thinking has moved considerably forward from
when he made that post.


Sorry; I'm way behind.

Chris Hornbeck


  #526   Report Post  
Chris Hornbeck
 
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On Tue, 31 Aug 2004 23:26:56 -0700, Bob Cain
wrote:

Perzactly. The Doppler effect is a reciprocity failure between
driving and terminating pistons.


That is a _very_ nice perspective that I haven't seen yet.
Good on ya.


I only post to rec.audio.pro and don't crosspost. This is
Scott Dorsey's original idea, which I've only adopted and
regurgitated several times. Crossposting has its bad side.

Chris Hornbeck
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Bob Cain
 
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Chris Hornbeck wrote:

On Tue, 31 Aug 2004 22:46:07 -0700, Bob Cain
wrote:


In reading these threads from out of town for two weeks, I've seen
no refutation of the real argument.


Which one might that be? :-)



That relative motion modulates relative distance which
modulates relative time, because the speed of sound is
constant.


Did you see my proof of why a piston in a tube will evidence
no such behavior with any signal including constant
translation of the piston? I believe it is a refutation of
that (unless, of course either I still don't understand what
you are saying or someone can refute my proof.)


Bob
--

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

A. Einstein
  #528   Report Post  
Bob Cain
 
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Chris Hornbeck wrote:

On Tue, 31 Aug 2004 23:21:08 -0700, Bob Cain
wrote:


That error is Doppler shift.


In a word, no.


How about a few more of them, then? :-)



The Doppler effect occurs downstream of transmitting
wavefront. In fact, you've shown very clearly that no
Doppler effect occurs close enough to the transmitter
for the wave to still be a plane.

Or, to say it in your newer and better way, close
enough to the transmitter that there are no frequency
dependent errors in the path to the receiver.


That makes three people that seem to understand. :-)


Bob
--

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

A. Einstein
  #529   Report Post  
Chris Hornbeck
 
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On Wed, 01 Sep 2004 13:56:57 -0700, Bob Cain
wrote:

relative motion modulates relative distance which
modulates relative time, because the speed of sound is
constant.


Did you see my proof of why a piston in a tube will evidence
no such behavior with any signal including constant
translation of the piston? I believe it is a refutation of
that (unless, of course either I still don't understand what
you are saying or someone can refute my proof.)


Isn't this (and the case of coupled infinitely extended
planes) just a special case with no relative motion?
Chris Hornbeck
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Bob Cain
 
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Chris Hornbeck wrote:

On Wed, 01 Sep 2004 13:56:57 -0700, Bob Cain
Did you see my proof of why a piston in a tube will evidence
no such behavior with any signal including constant
translation of the piston? I believe it is a refutation of
that (unless, of course either I still don't understand what
you are saying or someone can refute my proof.)



Isn't this (and the case of coupled infinitely extended
planes) just a special case with no relative motion?
Chris Hornbeck


Relative motion as we usually think of it is just the DC
component of the soundfield created by the Tx. That it is
fully coupled in the tube (or infinite plane) means it is
received as is by the Rx without giving rise to any Doppler
effect. Art's step function derivation shows that. That it
is not coupled at all to an Rx in the far, free field is
what allows simplification to the standard Doppler equation
(which is only valid for a constant v and a single frequency
in the Tx spectrum.)


Bob
--

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

A. Einstein


  #531   Report Post  
Chris Hornbeck
 
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On Wed, 01 Sep 2004 16:40:43 -0700, Bob Cain
wrote:

Isn't this (and the case of coupled infinitely extended
planes) just a special case with no relative motion?


Relative motion as we usually think of it is just the DC
component of the soundfield created by the Tx.


But for our purposes wouldn't it be a better model to
think of the relative motion as a separate thing, and
outside of the Tx and Rx soundfields?

Within those soundfields, especially if they're rigidly
coupled, nothing interesting happens.


That it is
fully coupled in the tube (or infinite plane) means it is
received as is by the Rx without giving rise to any Doppler
effect. Art's step function derivation shows that. That it
is not coupled at all to an Rx in the far, free field is
what allows simplification to the standard Doppler equation
(which is only valid for a constant v and a single frequency
in the Tx spectrum.)


Yeah, this is the part that Scott calls reciprocity failure.
It's a photographer's pun, but it works.

Chris Hornbeck
  #532   Report Post  
Bob Cain
 
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Chris Hornbeck wrote:


I only post to rec.audio.pro and don't crosspost.


Just as well. From the technical competence of its regulars
I've come to realize that there is a _very_ good reason why
alt.sci.physics.acoustics is in the alt. subtree.


Bob
--

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

A. Einstein
  #533   Report Post  
Bob Cain
 
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Chris Hornbeck wrote:

Relative motion as we usually think of it is just the DC
component of the soundfield created by the Tx.



But for our purposes wouldn't it be a better model to
think of the relative motion as a separate thing, and
outside of the Tx and Rx soundfields?


I don't thing so because it is a general phenomenon and
where people have gotten into trouble with it is precisely
for thinking there is something intrinsically definitive
about constant motion and then erroneously applying that
simplified special case solution to the more general case.


Within those soundfields, especially if they're rigidly
coupled, nothing interesting happens.


Yes.


Yeah, this is the part that Scott calls reciprocity failure.
It's a photographer's pun, but it works.


Yeah, I like the intuitive feel it gives while at the same
time being accurate, which as we've just found out is not
true in general with intuition.


Bob
--

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

A. Einstein
  #534   Report Post  
Bob Cain
 
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I withdraw the proof. It is flawed. In considering a
challenge by Art Ludwig, I realized that my proof begins
with an unrealistic assumption which allows no conclusions
to be drawn, much less a proof of anything. The flaw is
that I began with:
http://www.silcom.com/~aludwig/Physi...collisions.htm

having a value of zero for delta-t. For any medium with
non-zero mass density that requires infinite force to
achieve a step change in velocity. Obviously, nothing that
follows from the assumption of a step change in velocity can
be valid for a system that disallows that.

Back to the drawing board. There are heuristic proofs
involving reciprocity, conservation of energy and time
reversal (which are actually more satisfying to a physicist)
but I want to find a more direct one.


Sorry,

Bob
--

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

A. Einstein
  #535   Report Post  
Bob Cain
 
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Bob Cain wrote:


Did you see my proof of why a piston in a tube will evidence no such
behavior with any signal including constant translation of the piston?
I believe it is a refutation of that (unless, of course either I still
don't understand what you are saying or someone can refute my proof.)


I just refuted it myself. Only the particular method of
proof, not what I lamely attempted to prove. Tryin' again.


Bob
--

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

A. Einstein


  #536   Report Post  
Chris Hornbeck
 
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On Wed, 01 Sep 2004 18:35:20 -0700, Bob Cain
wrote:

But for our purposes wouldn't it be a better model to
think of the relative motion as a separate thing, and
outside of the Tx and Rx soundfields?


I don't thing so because it is a general phenomenon and
where people have gotten into trouble with it is precisely
for thinking there is something intrinsically definitive
about constant motion and then erroneously applying that
simplified special case solution to the more general case.


I've just finally realized that you're after the general case.
Sorry, never was too bright. But I'll make it up to you by
sending you my spare copy of Beranek, 1954. If your address
of a year ago isn't good, let me know quick.

You'll appreciate it's complete non-mention of Doppler effect!


Yeah, this is the part that Scott calls reciprocity failure.
It's a photographer's pun, but it works.


Yeah, I like the intuitive feel it gives while at the same
time being accurate, which as we've just found out is not
true in general with intuition.


Makes life worthwhile.

Chris Hornbeck
  #537   Report Post  
Bob Cain
 
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Chris Hornbeck wrote:


I've just finally realized that you're after the general case.
Sorry, never was too bright. But I'll make it up to you by
sending you my spare copy of Beranek, 1954. If your address
of a year ago isn't good, let me know quick.


Actually it has so hold on a bit.


You'll appreciate it's complete non-mention of Doppler effect!


Ah, virgin territory!


Bob
--

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

A. Einstein
  #538   Report Post  
Bob Cain
 
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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
  #539   Report Post  
Bob Cain
 
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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
  #540   Report Post  
Bob Cain
 
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Bob Cain wrote:

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.)


Ouch! That's dead wrong. Compass drift. With this problem
it is really difficult staying in the correct frame of
reference and when I wrote that I'd partially stepped off a
stationary one onto one that was moving, one foot still
firmly in each. :-)


Bob
--

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

A. Einstein


  #541   Report Post  
Bob Cain
 
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Bob Cain wrote:

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.)


Ouch! That's dead wrong. Compass drift. With this problem
it is really difficult staying in the correct frame of
reference and when I wrote that I'd partially stepped off a
stationary one onto one that was moving, one foot still
firmly in each. :-)


Bob
--

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

A. Einstein
  #542   Report Post  
Randy Yates
 
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Bob Cain writes:

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?


When you stated

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

I had the same impression as Ben.
--
% Randy Yates % "...the answer lies within your soul
%% Fuquay-Varina, NC % 'cause no one knows which side
%%% 919-577-9882 % the coin will fall."
%%%% % 'Big Wheels', *Out of the Blue*, ELO
http://home.earthlink.net/~yatescr
  #543   Report Post  
Randy Yates
 
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Bob Cain writes:

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?


When you stated

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

I had the same impression as Ben.
--
% Randy Yates % "...the answer lies within your soul
%% Fuquay-Varina, NC % 'cause no one knows which side
%%% 919-577-9882 % the coin will fall."
%%%% % 'Big Wheels', *Out of the Blue*, ELO
http://home.earthlink.net/~yatescr
  #544   Report Post  
Bob Cain
 
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Randy Yates wrote:

Bob Cain writes:


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?



When you stated

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

I had the same impression as Ben.


Ah, I see. The word "valid" would have been much clearer
than "defined" then. A point to remember, thanks.

I hope it's clear now that I don't mean that the effect
stops, but rather that the common equation describing it
stops being the correct one.


Bob
--

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

A. Einstein
  #545   Report Post  
Bob Cain
 
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Randy Yates wrote:

Bob Cain writes:


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?



When you stated

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

I had the same impression as Ben.


Ah, I see. The word "valid" would have been much clearer
than "defined" then. A point to remember, thanks.

I hope it's clear now that I don't mean that the effect
stops, but rather that the common equation describing it
stops being the correct one.


Bob
--

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

A. Einstein


  #546   Report Post  
Bob Cain
 
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The Ghost wrote:

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."


To be more precise it says on page 453, "The source does not
have to be traveling with constant velocity or in a straight
line for Eq. (5) to apply; however, determination of the
point on trajectory from which the wavelet originates
requires additional labor to match the kinematics, possibly
a graphical solution if the motion is not rectilinear."

I would add "and not constant."

He disqualifies the whole section for direct application to
a local analysis of a superimposed HF and LF signal in the
third sentence of the section. I'll leave it as an exercise
for the student to figure out what is wrong with the way fd
= f*c/(c + v) has been applied to that analysis. For now
that is. :-)


Bob
--

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

A. Einstein
  #547   Report Post  
Bob Cain
 
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The Ghost wrote:

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."


To be more precise it says on page 453, "The source does not
have to be traveling with constant velocity or in a straight
line for Eq. (5) to apply; however, determination of the
point on trajectory from which the wavelet originates
requires additional labor to match the kinematics, possibly
a graphical solution if the motion is not rectilinear."

I would add "and not constant."

He disqualifies the whole section for direct application to
a local analysis of a superimposed HF and LF signal in the
third sentence of the section. I'll leave it as an exercise
for the student to figure out what is wrong with the way fd
= f*c/(c + v) has been applied to that analysis. For now
that is. :-)


Bob
--

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

A. Einstein
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