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#521
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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
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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
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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
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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 |
#525
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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
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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 |
#527
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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
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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
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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 |
#530
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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
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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
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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
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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
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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
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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
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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
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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
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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
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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
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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
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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
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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
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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
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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
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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
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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
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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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