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Basic Acoustic Derivation/Proof Needed
Bob Cain wrote in message ...
Bob Cain wrote: snip.....snip The answer turns out to be simple. If the velocity and position of the particles moving about their rest position at the origin are Vp(t) = Sp'(t) snip......snip The Doppler distortion in a plane wave is a consequence of the non-linear relationship between the signal motion imparted to the air by the driver and the flow velocity (pressure) at a point about which that motion is manifest and measured. For a two tone Vd(t), one at 40 Hz and another at 2 kHz and allowing a motion of 2 cm (a reasonable Xmax for a two way system) the RMS IM distortion sidebands about the 2 kHz fundamental near the speaker face are on the order of .2% of that fundamental. While of some signifigance compared to other portions of an audio chain, it probably isn't for a loudspeaker given its other distortion mechanisms. Bob Does your prediction of 0.2% distortion apply to the situation of a piston in a tube or a piston in an infinite baffle, or doesn't it matter? |
#2
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The Ghost wrote: Does your prediction of 0.2% distortion apply to the situation of a piston in a tube or a piston in an infinite baffle, or doesn't it matter? First, that prediction is incorrect (I realized this morning.) I accidentally plugged the 2 cm in as velocity rather than displacement when doing the calculation but .2% is correct for a magnitude .01 m/sec LF velocity. The correct figure for 40 Hz at 2.53 m/s (2 cm throw) and 2 kHz at .05 m/sec (approximately equal acceleration, as is closer to the situation with a driver in a cabinet) is about 25.4% RMS IM distortion. I find this rather startling. The sidebands around the 2 kHz signal are -15 dB, -36 dB, -60 dB, -87 dB. These figures would obtain in a tube were it possible to generate LF displacements that large when so loaded, and they would not be distance dependant. This simple plane wave analysis says nothing at all about a piston in a baffle or a cabinet. The assumption of constant characteristic impedence that it makes is no longer the case even at the piston/air interface. A correct analysis would have to take into account the radiation impedence seen by the piston. I'm not sure whether the effect would be more or less than in a tube when that difference is accounted for. That's what I should look at next but I've fallen far enough behind in what I really should be doing that I must defer that. As far as dependancy on distance from a speaker in a baffle, I now don't think there will be much, if any, but thats only a SWAG based on thinking that the frequencies of the IM terms are close enough to the fundamental that they will propegate from the plane of the speaker pretty much the same as the fundamental so that the distortion as a percent of the fundamental should not change much with distance. Yes, this is a complete reversal of my thinking. The trap I fell into was the false belief that the flow velocity at a plane and the particle velocity about that plane were the same. Having finally thought my way out of that misconception and seeing that in fact they aren't even related linearly (in free space as well as near a driver), I see that I was wrong in what I believed followed from it. This brings me back to the reason I began trying to understand this to begin with, the quantitative determination of how signifigant the effect really is. It now seems to be more signifigant than has been assumed, not less. Whether or not that remains true for more realistic speaker couplings is an open question. Bob -- "Things should be described as simply as possible, but no simpler." A. Einstein |
#3
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The Ghost wrote: Bob Cain wrote in message ... Bob Cain wrote: snip.....snip Just realized that you copied only part of what I said in sci.physics to everywere else this discussion had become active (except your home group alt.sci.physics.acoustics oddly enough), so to rectify that the full sci.physics post follows. Bob Cain wrote: For the case of a propegating sinousiodal plane wave (or more generally oscillating plane waves if at all possible) where v(d,t) = Vp cos(w(t-d/c)) or a general v(t-d/c) describes the planar particle velocity at a position d relative to some origin, I need a rock solid proof/derivation of the state equations, position and velocity, of the particles which are at rest in a plane d when the field is quiescent. I cannot believe how elusive this has remained for me and need help. The answer turns out to be simple. If the velocity and position of the particles moving about their rest position at the origin are Vp(t) = Sp'(t) then the flow velocity of the particles in the plane of the origin is Vf(t) = Vp(t + Sp(t)/c) and the pressure there is P(t) = Ra*Vf(t) This implies that there is a non-linear relationship between particle velocity about a point and the flow velocity at that point. In acoustics texts this relationship is swept away in the approximation that Vf(t) = Vp(t) when motion is small relative to the wavelength of the sound. If there is a driver at the origin which imparts its motion Vd(t) and Sd(t) to the air then Vp(d,t) = Vd(t-d/c) Sp(d,t) = Sd(t-d/c) where d is the distance from the driver's rest position to the point about which the motion is of interest. The flow velocity (and proportional pressure) at that point is Vf(t,d) = Vd(t-(d-Sd(t-d/c))/c) The Doppler distortion in a plane wave is a consequence of the non-linear relationship between the signal motion imparted to the air by the driver and the flow velocity (pressure) at a point about which that motion is manifest and measured. For a two tone Vd(t), one at 40 Hz and another at 2 kHz and allowing a motion of 2 cm (a reasonable Xmax for a two way system) the RMS IM distortion sidebands about the 2 kHz fundamental near the speaker face are on the order of .2% of that fundamental. While of some signifigance compared to other portions of an audio chain, it probably isn't for a loudspeaker given its other distortion mechanisms. Bob -- "Things should be described as simply as possible, but no simpler." A. Einstein |
#4
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Bob Cain wrote in
: The Ghost wrote: Does your prediction of 0.2% distortion apply to the situation of a piston in a tube or a piston in an infinite baffle, or doesn't it matter? First, that prediction is incorrect (I realized this morning.) I accidentally plugged the 2 cm in as velocity rather than displacement when doing the calculation but .2% is correct for a magnitude .01 m/sec LF velocity. Yes, I know. I asked the question because I knew the prediction was incorrect for a tube and because I wanted you to confirm that it was for a tube before I called attention to what was a very glaring and clearly incorrect prediction. The correct figure for 40 Hz at 2.53 m/s (2 cm throw) and 2 kHz at .05 m/sec (approximately equal acceleration, as is closer to the situation with a driver in a cabinet) is about 25.4% RMS IM distortion. I find this rather startling. The sidebands around the 2 kHz signal are -15 dB, -36 dB, -60 dB, -87 dB. Since you failed to mention prior work on this problem, or acknowledge its validity, I will simply state that the -15dB sideband level is in good agreement with the prediction of Art Ludwig's prior analysis of Doppler distortion in a tube. |
#5
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Bob Cain wrote in
: Just realized that you copied only part of what I said in sci.physics to everywere else this discussion had become active (except your home group alt.sci.physics.acoustics oddly enough), so to rectify that the full sci.physics post follows. I didn't cross-post it to alt.sci.physics.acoustics because I didn't want to be responsible for initiating another firestorm of cross-posting of ad hominem attacks from you and your scum-bag buddies who admire and respect me so much. Also, I didn't cross-post it to alt.sci.physics.acoustics because of the generally sloppy presentation involving undefined terms, the use of non-standard terminology and a a complete lack of coherence. We all familiar what can happen (statistically speaking) if enough typewriters are put in the hands of enough monkeys. Until such time as you clean up your so-called analysis, I will remain unconvinced that the analysis is valid and that your predictions are anything other than accidental and/or the haphazard result of an even number of favorable errors. |
#6
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Bob Cain wrote in
: What I find most strange is that with all the specialists that had comments on all of this (other than in sci.physics), the simple analysis I finally saw relating flow velocity at a plane to particle velocity about it which leads to the general, quantitative answer was never offered by anyone else. Had it been, the issue could have been put to bed much sooner. There is no such thing as a problem that has only one method of solution. Art Ludwig provided the analysis and the solution weeks ago, but you rejected his analysis because you clearly couldn't understand it. You now claim to have obtained the same result by different means. Even if your sloppy, alternative analysis is correct, it adds nothing in the way of understanding. It does however provide you with the personal gratification that you did it yourself.....which is clearly what this has been about all along. The general, quantitative solution is as I've written it. It is exact and can be applied to any physically realizable signal. I will not drag this on any further unless there are questions to answer (and until/unless I can come back with a quantitative solution for a piston in a baffle.) Bob What do you mean by "flow velocity?" It's not discussed or defined in any of my reference books on acoustics, and I have most if not all of them. Define your terms........Sp'(t), Ra, Vd, etc. If you have a closed form solution, present it. |
#7
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The Ghost wrote: What do you mean by "flow velocity?" It's not discussed or defined in any of my reference books on acoustics, and I have most if not all of them. Define your terms........Sp'(t), Ra, Vd, etc. If you have a closed form solution, present it. I'm not an acoustics specialist, as is obvious, and am not terribly concerned about satisfying them with terms. What I mean is the velocity of the acoustic air flow passing a plane in units of distance/time. If there is a more correct term I'd be happy to know it. Definition of terms: Vp(t) is the velocity of a particle normally at rest at the origin as an acoustic wave goes through it in MKS units of meter/second. Sp(t) is the position of that particle in meters. Vf(t) = Vp(t+Sp(t)/c) is the flow velocity at the origin as a function of the particle velocity Vp(t) and position Sp(t) defined above. This relationship holds everywhere, not just near a driver. Vd(t) is the velocity of a driver face at rest at the origin. Sd(t) is it the position of the driver face. Vp(d,t) = Vd(t-d/c) is the velocity of a particle normally at rest at the position d as an acoustic wave created by a driver at the origin goes past it. It is also in MKS units of meter/second. Sp(d,t) = Sd(t-d/c) is the positon of that same particle about position d in units of the meter. Ra is the acoustic impedence of air in Pascal*sec/meter. In a tube it is homogenious and isoptropic with an approximate value of 300 Pascal*sec/meter. Vf(d,t) = Vd(t-(d-Sp(d,t))/c) is the velocity of particles passing a plane at d in meter/sec. P(d,t) = Ra*Vf(d,t) is the pressure at that plane. On closed forms: If the driver velocity function of time has a closed form indefinite integral, the application of my expression for Vf(d,t) to obtain a closed form for the flow velocity and pressure should be obvious. For example if the velocity function of the driver is Vd(t) = Al*sin(Wl*t) + Ah*sin(Wh*t) then Sd(t) = -((Al/Wl)cos(Wl*t) + (Ah/Wh)cos(Wh*t)) and Sp(d,t) = -((Al/Wl)cos(Wl*(t-d/c)) + (Ah/Wh)cos(Wh*(t-d/c)) Vp(d,t) = Al*sin(Wl*(t-(d-Sp(d,t))/c)) + Ah*sin(Wh*(t-(d-Sp(d,t))/c)) I haven't bothered to do the algebraic substitution of the penultimate equation into the final one. With that, it is an exact closed form solution for two sinusoids. Bob -- "Things should be described as simply as possible, but no simpler." A. Einstein |
#8
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Bob Cain wrote in message ...
Hi, Bob, Good to see that you're still sticking with it. Since you failed to mention prior work on this problem, or acknowledge its validity, I will simply state that the -15dB sideband level is in good agreement with the prediction of Art Ludwig's prior analysis of Doppler distortion in a tube. Presuming it built on them, I had not gone past the unmotivated equations 4) and 5) in his note to see that he had proposed something similar for two sinusoids as an approximation further on without regard to those equations. What I wrote was determined without refering to anything in that note following equations 4) and 5). The discussion in sci.physics with Zigoteau was invaluable in leading me to the answer and that discussion is on the record if not crossposted to every group in which the discussion found itself. extracted from previous post For a two tone Vd(t), one at 40 Hz and another at 2 kHz and allowing a motion of 2 cm (a reasonable Xmax for a two way system) the RMS IM distortion sidebands about the 2 kHz fundamental near the speaker face are on the order of .2% of that fundamental. While of some signifigance compared to other portions of an audio chain, it probably isn't for a loudspeaker given its other distortion mechanisms. I presume that these are the values given by Art Ludwig that I asked for. Thanks. I've improved my value for the acoustic impedance of air, from values given on http://hypertextbook.com/facts/2000/RachelChu.shtml and http://www.rfcafe.com/references/gen..._still_air.htm The values c=343 m.s^-1 and rho=1.25 kg.m^-3 give Z=429 Pa.m^-1.s Art must like it loud. At 40Hz, let's say the speaker diaphragm has an rms amplitude of 1e-2 m. Its rms velocity excursion v is 2.4 m.s^-1, and hence the power flux is Z*v^2 = 2471 W.m^-2. Sound intensities are normally expressed in dB wrt 1 pW.m^-2. This works out at 154 dB. This may be compared to the following table from http://www.nidcd.nih.gov/health/educ...cibel_text.asp Decibel level What we hear 10 dB Normal breathing 20 dB Rustling leaves, mosquito 30 dB Whisper 40 dB Stream, refrigerator humming 50-60 dB Quiet office 50-65 dB Normal conversation 60-65 dB Laughter 70 dB Vacuum cleaner, hair dryer 75 dB Dishwasher 78 dB Washing machine 80 dB Garbage disposal, city traffic noise Prolonged exposure to any noise above 90 dB can cause gradual hearing loss. 84 dB Diesel truck 70-90 dB Recreational vehicle 88 dB Subway, motorcycle 85-90 dB Lawnmower 100 dB Train, garbage truck 97 dB Newspaper press 98 dB Farm tractor Regular exposure of more than 1 minute risks permanent hearing loss. 103 dB Jet flyover at 100 feet 105 dB Snowmobile 110 dB Jackhammer, power saw, symphony orchestra 120 dB Thunderclap, discotheque/boom box 110-125 dB Stereo 110-140 dB Rock concerts 130 dB Jet takeoff, shotgun firing 145 dB Boom cars Cheers, Zigoteau. |
#9
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Your abuse of the English language is disgusts me. When's the last time you
heard music performed inside a tube? Never? Exactly. We all familiar what can happen (statistically speaking) if enough typewriters are put in the hands of enough monkeys. |
#10
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zigoteau wrote: Bob Cain wrote in message ... Hi, Bob, Good to see that you're still sticking with it. I think I got it, finally, and I thank you again for the fruitful discussion. For a two tone Vd(t), one at 40 Hz and another at 2 kHz and allowing a motion of 2 cm (a reasonable Xmax for a two way system) the RMS IM distortion sidebands about the 2 kHz fundamental near the speaker face are on the order of .2% of that fundamental. While of some signifigance compared to other portions of an audio chain, it probably isn't for a loudspeaker given its other distortion mechanisms. I presume that these are the values given by Art Ludwig that I asked for. Thanks. Those aren't his. I used the parameters I did because they more closely relate to real world speakers in enclosures (even though the analysis is for a plane wave.) The .2% distortion figure I gave was an error caused by my accidentally plugging .01 as the velocity rather than the position magnitude. With that corrected, the distortion becomes a whopping 25%. I thought I had posted his for you. Sorry if I failed to do that (ah, I see that post still in my drafts folder.) He has recently used 100 Hz, 8000 Hz, .04 m/s, .0004 m/s. In a plane wave that would be 119 dB SPL and 77 dB SPL in a tube. That only gives a LF excurison of 6.4 E-5 meters. The ratio of RMS IM distortion to the HF signal for those parameters is 0.66%. Even this small amount is signifigant compared to other factors in the recording/reproduction chain were it the only source of speaker distortion. I've improved my value for the acoustic impedance of air, from values given on http://hypertextbook.com/facts/2000/RachelChu.shtml and http://www.rfcafe.com/references/gen..._still_air.htm The values c=343 m.s^-1 and rho=1.25 kg.m^-3 give Z=429 Pa.m^-1.s Thank you. I took my approximate values from Pierce's text. Not sure why he'd have used Z=300 Pa.m^-1.s which is quite different. Art must like it loud. At 40Hz, let's say the speaker diaphragm has an rms amplitude of 1e-2 m. Its rms velocity excursion v is 2.4 m.s^-1, and hence the power flux is Z*v^2 = 2471 W.m^-2. Sound intensities are normally expressed in dB wrt 1 pW.m^-2. This works out at 154 dB. Again, it was I who plugged in that number because it is a reasonable excursion in real world situations due to poor LF coupling to the air. I am hardly sure, however, that mixing apples with oranges by using a plane wave analysis with nearly point source conditions leads to anything meaningful so that 25% figure needs be taken with a grain of salt. I.E. I'm not sure how well LF piston velocity in an enclosure is coupled to motion of the air immediately in front of it. Thanks, Bob -- "Things should be described as simply as possible, but no simpler." A. Einstein |
#11
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"zigoteau" wrote in message om... Bob Cain wrote in message ... [snips] Art must like it loud. At 40Hz, let's say the speaker diaphragm has an rms amplitude of 1e-2 m. Its rms velocity excursion v is 2.4 m.s^-1, and hence the power flux is Z*v^2 = 2471 W.m^-2. Sound intensities are normally expressed in dB wrt 1 pW.m^-2. This works out at 154 dB. This may be compared to the following table from First off, this is not the example I include on my site. I use a velocity of 1 m/s at 50 Hz, and this is just to generate some convenient numbers which can easily be scaled. Second, the relationship you use is valid for a piston in a tube, but for the real world case of a piston in a baffle the sound output is much less because the real part of the impedance seen by the diaphragm is much less than your Z. At 50 Hz a 1 m/s velocity corresponds to a peak displacement of 3 millimeters which is quite realistic for a woofer. I invite any educated physicist to look at my site and decide for him or her self if my derivation is valid. It can be seen at http://www.silcom.com/~aludwig/Physi...on/dopdist.htm As you pointed out, I assume the linear wave equation in this derivation, and for very high velocities, or for propagation in a tube, the non-linearity of air can be important. I have now analyzed the solution to the nonlinear acoustic equations as well. That analysis is at http://www.silcom.com/~aludwig/Physi..._acoustics.htm Art Ludwig |
#12
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"Daydream Electric" wrote in
: Your abuse of the English language is disgusts me. When's the last time you heard music performed inside a tube? Never? Exactly. We all familiar what can happen (statistically speaking) if enough typewriters are put in the hands of enough monkeys. "Your abuse of the English language IS disgusts me?" Right on, moron! The topic of discussion involves a theoretically ideal situation of a piston in a tube. Since you clearly have **** for brains, I don't expect you to understand why that theoretically ideal situation is being discussed. Nonetheless, why don't you do both of us a favor and simply not read my posts, you stupid idiot. |
#13
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"Art Ludwig" wrote in message news:O__7d.3313$gk.2647@okepread01...
Hi, Art, "zigoteau" wrote in message om... Bob Cain wrote in message ... [snips] Art must like it loud. At 40Hz, let's say the speaker diaphragm has an rms amplitude of 1e-2 m. Its rms velocity excursion v is 2.4 m.s^-1, and hence the power flux is Z*v^2 = 2471 W.m^-2. Sound intensities are normally expressed in dB wrt 1 pW.m^-2. This works out at 154 dB. This may be compared to the following table from First off, this is not the example I include on my site. I use a velocity of 1 m/s at 50 Hz, and this is just to generate some convenient numbers which can easily be scaled. Second, the relationship you use is valid for a piston in a tube, but for the real world case of a piston in a baffle the sound output is much less because the real part of the impedance seen by the diaphragm is much less than your Z. At 50 Hz a 1 m/s velocity corresponds to a peak displacement of 3 millimeters which is quite realistic for a woofer. I invite any educated physicist to look at my site and decide for him or her self if my derivation is valid. It can be seen at http://www.silcom.com/~aludwig/Physi...on/dopdist.htm As you pointed out, I assume the linear wave equation in this derivation, and for very high velocities, or for propagation in a tube, the non-linearity of air can be important. I have now analyzed the solution to the nonlinear acoustic equations as well. That analysis is at http://www.silcom.com/~aludwig/Physi..._acoustics.htm I hope my response didn't sound too disrespectful, but Bob was complaining that my formula gave distortion figures which were too high. My point was that the values plugged in to the formula correspond to extremely high acoustic intensities, so large distortion is not unreasonable. I take your point about the difference between a loudpeaker in a conventional enclosure and a highly idealized one in a tube. I am interested if you have managed to solve the nonlinear wave equations. I haven't got time to digest it just now, but will look at your website ASAP. Cheers, Zigoteau, |
#14
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After so many words spent on the effect in regard to speaker, when are
you to start examining it in regard to (dynamic) microphone. And don't forget the air. Molecules are moving. There must be some dopler there, too. |
#16
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The Ghost wrote: Thank you for pointing out the fact that Bob Cain used your formula to predict the Doppler IM distortion levels that he provided. Knowing that fact certainly clears up my confusion over Bob Cains ability to provide seemingly correct distortion numbers and his total inability to present either a sensible and coherent verbal or mathematical explanation of how the distiotion numbers were calculated. Please notice the difference between what I wrote and what Zigoteau wrote. His approach, a first order approximation using a M-T power series, yields (using common symbols and frames of reference): Vp(d,t) = Vd(t - (d - Sd(t-d/c))/(c - Vd(t-d/c))) Mine, which involves no approximation, yields: Vp(d,t) = Vd(t - (d - Sd(t-d/c))/c) I answered you as to precisely what my symbols mean and how to get from that relationship to a closed form solution for two sinusoids. Rather than attempting an expansion of that to yield an approximation for the distortion products, I simply generate a band limited signal from that two sinusoid solution and use a combination of Matlab and tools designed for audio signal manipulation to measure what's in that signal in both the frequency and time domain. I am interested in quantitative answers to the question rather than in demonstrating mathematical prowess. Bob -- "Things should be described as simply as possible, but no simpler." A. Einstein |
#17
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The Ghost wrote: Bob Cain can't define or explain what he means because he doesn't know what he means. You have obviously not grasped the simplicty of what I wrote and the clarification of it that you asked for. I doubt you've even read it. It merely answers the question "what is the pressure at a point in a plane wave as a function of the motion of the particles about that point" in a simple and direct way. One that even you could understand if you wished to. It's not at all complicated, Gary, and is the answer that should have been given long ago when I first began asking about it. I challenged you any number of times to produce it if you knew it. Enough of this. I've got what I came for. Bob -- "Things should be described as simply as possible, but no simpler." A. Einstein |
#18
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Bob Cain wrote in : The Ghost wrote: Bob Cain can't define or explain what he means because he doesn't know what he means. You have obviously not grasped the simplicty of what I wrote and the clarification of it that you asked for. I doubt you've even read it. It merely answers the question "what is the pressure at a point in a plane wave as a function of the motion of the particles about that point" in a simple and direct way. One that even you could understand if you wished to. You are correct, I have not grasped the simplicity of what you wrote, and the reason is because what you wrote is little more than nonsensical gibberish. The fact of the matter is that the problem is not with my inability to grasp what you have done, the problem is with your inability to present what you have done in a coherent and sensible manner. It's not at all complicated, Gary, and is the answer that should have been given long ago when I first began asking about it. I challenged you any number of times to produce it if you knew it. You have been insulting me in alt.sci.physics.acoustics for over four years, and you find it odd that I would not give you the time of day if your life depended on it? Get a reality check and stop confusing capability with motivation. Art Ludwig gave you the analysis that you requested and you spit in his face, walked away and came here begging for someone else to provide you with an analysis that you might be able to understand. If I were you, I wouldn't expect any help on other problems from Art anytime soon. Enough of this. I've got what I came for. Bob That you have, and apparently to hell with anyone else reading this thread who might want to benefit from the help that you have received and from what you have done (whatever that might be). |
#19
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Bob Cain wrote in
: I answered you as to precisely what my symbols mean and how to get from that relationship to a closed form solution for two sinusoids. You answered NOTHING. I asked you to define what you meant by flow felocity and all you did was to restate your equation for so-called flow velocity, Vf(t), which you seem to have pulled out of thin air. Where did you get that equation? Did you derive it or did you make it up? If you derived it, show the derivation. In other places you equate your so-called flow velocity to pressure. Which is it. Pressure is force per unit area. Velocity is distance per unit time. They are not the same. Your terminology is not only unconventional, it is nonsensical. Rather than attempting an expansion of that to yield an approximation for the distortion products, I simply generate a band limited signal from that two sinusoid solution and use a combination of Matlab and tools designed for audio signal manipulation to measure what's in that signal in both the frequency and time domain. I am interested in quantitative answers to the question rather than in demonstrating mathematical prowess. Bob Great! We now know that the distortion numbers that you gave resulted from both an undefined analysis as well as an undefined simulation. No doubt you have a reason for keeping it all a secret? Perhaps that's because you are just pretending to have an analysis and simulation, and that the distortion numbers that you are providing are really from Art Ludwig's analysis and the MatLab program that he sent to you. If not, I suggest that you post your analysis and that you offer to provide your simulation to anyone interested, as Art has done. You have recently called me a fraud. Perhaps it is you who is the real fraud? If not, prove it. |
#20
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"The Ghost" wrote in message
. 29... You have recently called me a fraud. Perhaps it is you who is the real fraud? If not, prove it. Hmmm...Celebrating in Art's shadow and insulting people does not help you shed the overcoat of fraud. You've still offered nothing beyond putting together an experiment that demonstrated what everyone in the discussion agreed upon already. |
#21
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"Jim Carr" wrote in
newson8d.12144$mS1.11114@fed1read05: "The Ghost" wrote in message . 29... You have recently called me a fraud. Perhaps it is you who is the real fraud? If not, prove it. Hmmm...Celebrating in Art's shadow and insulting people does not help you shed the overcoat of fraud. Y Screw you, asswipe. All you've ever offered is mindless, ignorant arm- chair criticism. So, you don't even qualify for the fraud test. You've still offered nothing beyond putting together an experiment that demonstrated what everyone in the discussion agreed upon already. Quite the contrary, At the time everyone in your mindless camp was denying the mere existence of dynamic Doppler shift. Shortly after I presented my measurement results, the tune immediately changed. The new tune became that Doppler distortion did exist, but not in a tube. Now that two people have independently presented theoretical predicitions of the levels of Doppler distiortion in a tube, the tune has once again changed. Kiss my ass you technically-inept piece of ****. You are so full of it that it is flowing out your ears. |
#22
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The Ghost wrote: Bob Cain wrote: I answered you as to precisely what my symbols mean and how to get from that relationship to a closed form solution for two sinusoids. You answered NOTHING. I asked you to define what you meant by flow felocity and all you did was to restate your equation for so-called flow velocity, Vf(t), I fully explained my terms in a response to you and I gave a detailed answer to what I mean by flow velocity in response to Edward Green. which you seem to have pulled out of thin air. Where did you get that equation? Did you derive it or did you make it up? If you derived it, show the derivation If you will bother to understand my definitions of Vf, Vp and Xp, then if Vf(t) = Vp(t + Sp(t)/c) defined at the origin of a cordinate system is not obvious to you by inspection there is no hope of your understanding anything else. I suppose I could draw some pretty pictures for you if that is the level at which you require explanation. Having understood that, everything else is just straithtforward translation to move the origin of the frame of reference to a driver an arbitrary distance d from the point where conditions are to be described. In that translation, there is an assumption I suppose you could argue with but I don't think you will because it is correct. That assumption is Vp(d,t) = Vd(t-d/c). I.E. the motion of particls about position d is the same as that of the driver face about its rest position at a time earlier by d/c. In other places you equate your so-called flow velocity to pressure. Which is it. I've explained this. It is the velocity at a point that is related to the pressure at that point by the acoustic impedence. Pressure is force per unit area. Velocity is distance per unit time. They are not the same. Of course not. The are related by a constant of proprionality called characteristic impedence or more generally acoustic impedence which is in units of Pascal.s.m^-1 and your pretending that you didn't understand my explanation of that in the definition of terms I gave is nothing but baiting. Great! We now know that the distortion numbers that you gave resulted from both an undefined analysis as well as an undefined simulation. You are simply ignoring the explicit formula I gave you that I used to generate a discrete time signal and pretending ignorance of commonly understood methods of discrete time signal analysis. That is your problem, not mine. So that you cannot continue to pretend you haven't seen it, here again, copied from a prior response to you with a bit more elaboration of what I call flow velocity, is the definition of the terms I use and the results I obtain. Definition of terms: Vp(t) is the velocity of a particle normally at rest at the origin as an acoustic wave goes through it in MKS units of meters/second. Sp(t) is the position of that particle in meters. Vf(t) = Vp(t+Sp(t)/c) is the flow velocity at the origin as a function of the particle velocity Vp(t) and position Sp(t) defined above. This relationship holds everywhere, not just near a driver. What I call flow velocity is the velocity of whatever particle happens to be at the origin at any point in time. Vd(t) is the velocity of a driver face at rest at the origin. Sd(t) is it the position of the driver face. Vp(d,t) = Vd(t-d/c) is the velocity of a particle normally at rest at the position d as an acoustic wave created by a driver at the origin goes past it. It is also in MKS units of meter/second. Sp(d,t) = Sd(t-d/c) is the positon of that same particle about position d in units of the meter. Vf(d,t) = Vd(t-(d-Sp(d,t))/c) is the velocity of particles passing a plane at d in meter/sec. P(d,t) = Ra*Vf(d,t) is the pressure at that plane. Ra is the acoustic impedence of air in Pascal*sec/meter. In a tube it is homogenious and isoptropic with an approximate value of 429 Pascal*sec/meter. On closed forms: If the driver velocity function of time has a closed form indefinite integral, the application of my expression for Vf(d,t) to obtain a closed form for the flow velocity and pressure should be obvious. For example if the velocity function of the driver is Vd(t) = Al*sin(Wl*t) + Ah*sin(Wh*t) then Sd(t) = -((Al/Wl)cos(Wl*t) + (Ah/Wh)cos(Wh*t)) and Sp(d,t) = -((Al/Wl)cos(Wl*(t-d/c)) + (Ah/Wh)cos(Wh*(t-d/c)) Vp(d,t) = Al*sin(Wl*(t-(d-Sp(d,t))/c)) + Ah*sin(Wh*(t-(d-Sp(d,t))/c)) I haven't bothered to do the algebraic substitution of the penultimate equation into the final one. With that, it is an exact closed form solution for two sinusoids. Final note; the last two formulas are what I used to generate a discrete time signal (by, of all things, calculating it for discrete values of time) for analysis using standard tools with which I am familiar. There are no secrets in doing this that even a near novice to discrete time signal processing would fail to know. Bob -- "Things should be described as simply as possible, but no simpler." A. Einstein |
#23
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The Ghost wrote: If I were you, I wouldn't expect any help on other problems from Art anytime soon. Do you speak for Art Ludwig now? Have you, at any point, spoken _as_ Art Ludwig? Bob -- "Things should be described as simply as possible, but no simpler." A. Einstein |
#24
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"The Ghost" wrote in message . 29... "Jim Carr" wrote in newson8d.12144$mS1.11114@fed1read05: "The Ghost" wrote in message . 29... You have recently called me a fraud. Perhaps it is you who is the real fraud? If not, prove it. Hmmm...Celebrating in Art's shadow and insulting people does not help you shed the overcoat of fraud. Y Screw you, asswipe. All you've ever offered is mindless, ignorant arm- chair criticism. So, you don't even qualify for the fraud test. You've still offered nothing beyond putting together an experiment that demonstrated what everyone in the discussion agreed upon already. Quite the contrary, At the time everyone in your mindless camp was denying the mere existence of dynamic Doppler shift. Shortly after I presented my measurement results, the tune immediately changed. The new tune became that Doppler distortion did exist, but not in a tube. Now that two people have independently presented theoretical predicitions of the levels of Doppler distiortion in a tube, the tune has once again changed. Kiss my ass you technically-inept piece of ****. You are so full of it that it is flowing out your ears. I would suggest that you take a bottle or two of Immodium AD, your oral diarrhea is really getting out of control again. You're losing the race, the human race, that is... |
#25
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"Vladan" wrote in message ... After so many words spent on the effect in regard to speaker, when are you to start examining it in regard to (dynamic) microphone. And don't forget the air. Molecules are moving. There must be some dopler there, too. In a microphone, the diaphragm excursion is so small that any possible Doppler shift would be negligable, even assuming that there is Doppler shift in a mic. BTW, if there is Doppler shift in a microphone because of diaphragm excursion, then there must also be Doppler shift in the human ear because of the excursion of the eardrum, and if that is true, our hearing mechanism must have compensation built in, so, since the amount of excursion in the mic and the eardrum are similar, if there is any Doppler shift in a mic, it can be disregarded. |
#26
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"The Ghost" wrote in message
. 29... Screw you, asswipe. All you've ever offered is mindless, ignorant arm- chair criticism. So, you don't even qualify for the fraud test. I have stated from day one that I understand the concept of Doppler distortion in a speaker as repeatedly described. I also commented that if it happens as described, then microphones suffer from it. Instruments like guitars and pianos must also suffer from it as well. However, because of the short distances in the vibrations, it must not be much. What bothered me was that I could not (and still cannot) see how a speaker really works. Yeh, I can describe the mechanics involved, but I still don't fully understand the exact physics where the diaphragm creates the sound wave. Is it at the start of the throw? The end? The middle? If it's in the middle of a long throw for a loud low frequency, how does it make the higher frequencies at the same time? The speaker pushes air and makes a breeze, but it also transfers energy to individual molecules which start bouncing into each other in a wave, which is what we ultimately perceive as sound. I accept that it works, but the physics escape me. I can live with that. So, when Bob suggested that there might be something more involved with Doppler distortion due the physics described above, I thought that was a fair question to ask. He conducted himself quite well and took the humble step of expanding the discussion out of a.m.h-s and into groups where people with expertise greater than his might answer the question. It was going quite well until some home-schooled, insecure mamma's boy (that would be you) started venting his years of frustration and feelings of inadequacy. It's been a fun little game watching you yap like a poodle while several of us kicked the fence. FYI, Porky and I are far from being pals. Check the Google archives if you don't believe me. I have stated repeatedly that I have neither the training nor the experience to even begin to answer the questions myself. Quite frankly I couldn't care less if it's there or not. First, I cannot hear it. Second, even if I could, there's nothing I could do about it anyway. Of the issues in the processes required for me to produce good music on my home computer, Doppler distortion is at the bottom of the list. So, there is no fraud test for me to pass or fail. Well, maybe there is. Maybe I am lying and really do have a degree in physics, and I'm just playing dumb. Sorta like you, I guess. As for your blustering about threats, give it a rest. There's no way you have the balls to say in person any of the things you've written. You might have when you were younger, but now you're too big to hide in your mother's apron when things get tough. You know deep down I'm right, so don't even bother playing the "if we met in person" game. Chances are it would never happen. If it did, we all know how you'd behave. Kiss my ass you technically-inept piece of ****. You are so full of it that it is flowing out your ears. Try something new, will ya? That one is getting old. |
#27
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"The Ghost" wrote in message . 6... "Daydream Electric" wrote in : Your abuse of the English language is disgusts me. When's the last time you heard music performed inside a tube? Never? Exactly. We all familiar what can happen (statistically speaking) if enough typewriters are put in the hands of enough monkeys. "Your abuse of the English language IS disgusts me?" Right on, moron! The topic of discussion involves a theoretically ideal situation of a piston in a tube. Since you clearly have **** for brains, I don't expect you to understand why that theoretically ideal situation is being discussed. Nonetheless, why don't you do both of us a favor and simply not read my posts, you stupid idiot. The original discussion was about possible Doppler shift in a loudspeaker. Theoretical ideals do not necessarily equate to real world situations, and even if they do, the relationship is often tenuous at best. The whole reason for using a piston in a tube is to reduce a fiendishly complex issue to simplest terms so that a mathematical solution might be considered (and from the length of this discussion, it's still extremely complex). In a real listening room with a real loudspeaker, one must not only consider the possible phase shifts in the soundwaves that travel directly from the speaker to the ear, but also the phase relationships of all the soundwaves that are reflected off the walls, floor, ceiling, and any item in the room that is capable of reflecting sound toward the ear, and then refraction also must be taken into account. Since even the largest anechoic chambers aren't anechoic at very low frequencies, the situation is still hoplessly complex even there. I would suggest that until someone builds a computer model of a real loudspeaker reproducing real music in a real room, comes up with the necessary algorythms and runs the simulation, no accurate, or even approximate solution for the real world issue is going to be found. From a practical standpoint, the question isn't whether Doppler shift exists in a loudspeaker, the real issue is whether or not it's audible, and if it is audible, whether or not our hearing has a mechanism to compensate for it. |
#28
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Porky wrote: "Vladan" wrote in message ... After so many words spent on the effect in regard to speaker, when are you to start examining it in regard to (dynamic) microphone. And don't forget the air. Molecules are moving. There must be some dopler there, too. In a microphone, the diaphragm excursion is so small that any possible Doppler shift would be negligable, even assuming that there is Doppler shift in a mic. Nonetheless, even if the mic can measure pressure or velocity at a point with zero excursion there is still a non-linear relationship between the motion of a tiny zero-mass test particle normally at rest at that point and the pressure/velocity measured there. Remaining tubular for the nonce, if we had a mixture of a 40 Hz and a 2 kHz wave each at 94 dB SPL and if the motion of the test particle is that of the signal then the pressure at its rest position will show about 0.024% IM distortion relative to the HF fundamental and sidebands about -75 dB down from it. The thing that concerns me, now that I can put numbers to at least the tube conditions and see that they can get big, is that if conditions are such that in the very near field of a speaker a signifigant percentage of its LF velocity is coupled to the air then the Doppler effect can get _very_ signifigant because of the large excursions needed to offset the poor far field coupling. My hope is that a retarded wave cancels most of that up close so that it won't grow to too great a signifigance. The degree of near field LF coupling for a given far field measure from a speaker in an enclosure is the next thing I want to know. Bob -- "Things should be described as simply as possible, but no simpler." A. Einstein |
#29
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"Porky" wrote in message
t... From a practical standpoint, the question isn't whether Doppler shift exists in a loudspeaker, the real issue is whether or not it's audible, and if it is audible, whether or not our hearing has a mechanism to compensate for it. How could the ear/brain possibly compensate for it? There's no way to analyze a sound to know if it's distorted unless you know how the sound is supposed to be in the first place. Take the classic whistle on a train. Unless you know the pitch is supposed to stay constant, how can you know the sound is affected by Doppler? You can't. As for the "real" issue the discussion always been esoteric. I don't think anyone really thinks there is anything practical that can be done about it. |
#30
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"Porky" wrote in message ... In a microphone, the diaphragm excursion is so small that any possible Doppler shift would be negligable, even assuming that there is Doppler shift in a mic. BTW, if there is Doppler shift in a microphone because of diaphragm excursion, then there must also be Doppler shift in the human ear because of the excursion of the eardrum, and if that is true, our hearing mechanism must have compensation built in, so, since the amount of excursion in the mic and the eardrum are similar, if there is any Doppler shift in a mic, it can be disregarded. What a load of crap. Whilst I agree that any doppler in a microphone would be minute, it would not be compensated for by the ear/brain just because it may be similar magnitude to the eardrum. TonyP. |
#31
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"Bob Cain" wrote in message ... Porky wrote: "Vladan" wrote in message ... After so many words spent on the effect in regard to speaker, when are you to start examining it in regard to (dynamic) microphone. And don't forget the air. Molecules are moving. There must be some dopler there, too. In a microphone, the diaphragm excursion is so small that any possible Doppler shift would be negligable, even assuming that there is Doppler shift in a mic. Nonetheless, even if the mic can measure pressure or velocity at a point with zero excursion there is still a non-linear relationship between the motion of a tiny zero-mass test particle normally at rest at that point and the pressure/velocity measured there. Remaining tubular for the nonce, if we had a mixture of a 40 Hz and a 2 kHz wave each at 94 dB SPL and if the motion of the test particle is that of the signal then the pressure at its rest position will show about 0.024% IM distortion relative to the HF fundamental and sidebands about -75 dB down from it. If we get into Doppler shift due to motion in air molecules, I suspect we're getting down to "the bumble bee doesn't really fly because the math says it can't" point. The thing that concerns me, now that I can put numbers to at least the tube conditions and see that they can get big, is that if conditions are such that in the very near field of a speaker a signifigant percentage of its LF velocity is coupled to the air then the Doppler effect can get _very_ signifigant because of the large excursions needed to offset the poor far field coupling. My hope is that a retarded wave cancels most of that up close so that it won't grow to too great a signifigance. The degree of near field LF coupling for a given far field measure from a speaker in an enclosure is the next thing I want to know. If the equations show that all that much Doppler distortion in a speaker, why can't we hear it? By Occam's Rasor, either our hearing mechanism has built in compensation, so Doppler distortion doesn't matter, or the math is wrong and it needs to be revised. That isn't to say that it doesn't happen in the piston-in-an-infinite-tube model, it just means that the speaker/room model is a totally different animal. |
#32
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"TonyP" wrote in message u... "Porky" wrote in message ... In a microphone, the diaphragm excursion is so small that any possible Doppler shift would be negligable, even assuming that there is Doppler shift in a mic. BTW, if there is Doppler shift in a microphone because of diaphragm excursion, then there must also be Doppler shift in the human ear because of the excursion of the eardrum, and if that is true, our hearing mechanism must have compensation built in, so, since the amount of excursion in the mic and the eardrum are similar, if there is any Doppler shift in a mic, it can be disregarded. What a load of crap. Whilst I agree that any doppler in a microphone would be minute, it would not be compensated for by the ear/brain just because it may be similar magnitude to the eardrum. It seems that you missed the point, which was that if there is Doppler shift in a mic because of diaphragm motion, then there must also be Doppler shift in the human ear because of eardrum motion. If that is true then the hearing mechanism must have a method of compensation for it. |
#33
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"Jim Carr" wrote in message news:tVq8d.12196$mS1.2969@fed1read05... "Porky" wrote in message t... From a practical standpoint, the question isn't whether Doppler shift exists in a loudspeaker, the real issue is whether or not it's audible, and if it is audible, whether or not our hearing has a mechanism to compensate for it. How could the ear/brain possibly compensate for it? There's no way to analyze a sound to know if it's distorted unless you know how the sound is supposed to be in the first place. Take the classic whistle on a train. Unless you know the pitch is supposed to stay constant, how can you know the sound is affected by Doppler? You can't. The point is that everyone hears the Doppler shift in a train whistle, but when comparing a "live" sound to it's replica being reproduced by a very accurate loudspeaker system, under the closest to ideal conditions possible, there is a great deal of disagreement as to whether there is any audible distortion produced by the speaker system, and even more as to whether any of that is due to Doppler shift. J B Lansing, Altech and others have done considerable experimentation with "live vs reproduced" sound, and the matter of realism seems to get very close under the best conditions, so close that even highly trained expert listeners have trouble telling the difference. If it were otherwise, we wouldn't be having this discussion! :-) That argues that either our hearing mechanism compensates for any Doppler distortion that might be present, or that it isn't audible even if it does exist. As for the "real" issue the discussion always been esoteric. I don't think anyone really thinks there is anything practical that can be done about it. As to it being an esoteric discussion, it has certainly become one, but the original question was "Do speakers create Doppler distortion when producing both a HF tone and an LF tone at the same time?", and there is nothing esoteric about that question. If audible Doppler distortion does occur in a speaker, then there must a number of ways to minimize it, so this discussion does have some possible practical application, assuming that anyone can prove that we actually hear Doppler distrotion in a speaker when listening to music. |
#34
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Bob Cain wrote in message ...
Hi, Bob, Just a minor correction. Please notice the difference between what I wrote and what Zigoteau wrote. His approach, a first order approximation using a M-T power series, yields (using common symbols and frames of reference): Vp(d,t) = Vd(t - (d - Sd(t-d/c))/(c - Vd(t-d/c))) Mine, which involves no approximation, yields: Vp(d,t) = Vd(t - (d - Sd(t-d/c))/c) I am sorry, Bob, but the second equation does involve an approximation. Cheers, Zigoteau. |
#35
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"Porky" wrote in message ... It seems that you missed the point, which was that if there is Doppler shift in a mic because of diaphragm motion, then there must also be Doppler shift in the human ear because of eardrum motion. If that is true then the hearing mechanism must have a method of compensation for it. It seems you've missed my point. The brain "compensates" for the auditory system itself, because you have NO other point of reference. TonyP. |
#36
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"TonyP" wrote in message u... "Porky" wrote in message ... It seems that you missed the point, which was that if there is Doppler shift in a mic because of diaphragm motion, then there must also be Doppler shift in the human ear because of eardrum motion. If that is true then the hearing mechanism must have a method of compensation for it. It seems you've missed my point. The brain "compensates" for the auditory system itself, because you have NO other point of reference. Sure you do, ever hear of bone conduction? That bypasses the eardrum totally. There is also considerable evidence that we can "hear" the audio in audio modulated RF at certain frequencies, which would seem to bypass the physical hearing mechanism entirely. However, my point was that if one can tell the difference between a "live" sound and the same sound reproduced on a very high quality sound system, then either there is no audible distortion present, or our ears have a mechanism that compensated for whatever distortion is present, including any Doppler distortion. |
#37
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Porky wrote: If we get into Doppler shift due to motion in air molecules, I suspect we're getting down to "the bumble bee doesn't really fly because the math says it can't" point. Huh? That's exactly what Doppler distortion is due to. There is a mild non-linear relationship everywhere in a soundfield between particle velocity about a point and the fluid velocity at that point. That was my epiphany. Not much as epiphanies go but, hey, they get fewer and fewer every year. :-) If the equations show that all that much Doppler distortion in a speaker, why can't we hear it? Who says we can't? It's sorta hard to get rid of to do an ABX test on. By Occam's Rasor, either our hearing mechanism has built in compensation, so Doppler distortion doesn't matter, or the math is wrong and it needs to be revised. That isn't to say that it doesn't happen in the piston-in-an-infinite-tube model, it just means that the speaker/room model is a totally different animal. Sure it is but it should get signifigantly worse with a driver in an enclosure rather than in a tube because of the large excursions demanded at the low frequencies to couple anything from an enclosed speaker to a room. You can get real high SPL low frequencies in a tube without much excursion, which is what causes it, but not so with an enclosed speaker in a room. You have to push a whole lot of LF air up close to get what's in the signal to reach you at any distance. Dunno what you mean by built in compensation nor why there would be anything like that. It's not the kind of thing evolution would have devoted much energy to. There weren't many broadband sound sources to work with even if it had been deemed important for some reason. Bob -- "Things should be described as simply as possible, but no simpler." A. Einstein |
#38
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Porky wrote: It seems that you missed the point, which was that if there is Doppler shift in a mic because of diaphragm motion, then there must also be Doppler shift in the human ear because of eardrum motion. If that is true then the hearing mechanism must have a method of compensation for it. There is none of _any_ signifigance with either. Excursions in either case are more than a few orders of magnitude away from signifigance. Bob -- "Things should be described as simply as possible, but no simpler." A. Einstein |
#39
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"Porky" wrote in message ... "TonyP" wrote in message It seems you've missed my point. The brain "compensates" for the auditory system itself, because you have NO other point of reference. Sure you do, ever hear of bone conduction? That bypasses the eardrum totally. Not at all. It is concurrent with the eardrum. Audioligists must use large masking signals just to get some figure for bone conduction, but it is less than via the eardrum. I know of no way to seperate the two for comparison purposes, do you? (The worlds audiologists await your reply :-) There is also considerable evidence that we can "hear" the audio in audio modulated RF at certain frequencies, which would seem to bypass the physical hearing mechanism entirely. Some level of diode detection has been demonstated in some cases, usually connected with metal fillings. This couples audio signals via bone conduction. However, my point was that if one can tell the difference between a "live" sound and the same sound reproduced on a very high quality sound system, then either there is no audible distortion present, or our ears have a mechanism that compensated for whatever distortion is present, including any Doppler distortion. ??? TonyP. |
#40
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[snips] "zigoteau" wrote in message om... Bob Cain wrote in message ... Mine, which involves no approximation, yields: Vp(d,t) = Vd(t - (d - Sd(t-d/c))/c) I am sorry, Bob, but the second equation does involve an approximation. Zigoteau. I agree completely, and I would like to point out that if you evaluate the above equation at d=0 you get an result identical to my equation (8) in my analysis at http://www.silcom.com/~aludwig/Physi...on/dopdist.htm, which has been posted for quite a while now. Rod Elliot presented the same equation before I did. I present this equation as an approximate analysis, but show that it is quite accurate. I also show that this equation is completely equivalent to the good old fashioned "Doppler distortion" that folks have been using for years. Just follow the link under "interesting issues" for a proof of this statement. As far as I am aware you are not claiming your analysis is novel, Zigoteau, so I don't mean this as a personal criticism. I just want to set the record straight for people not familiar with the history. Art Ludwig |
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