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#361
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"Arny Krueger" wrote in message ... "Porky" wrote in message Bob, aren't they showing a 1KHz signal modulated with a 50 Hz signal? Yes, if you're referring to http://www.pcavtech.com/techtalk/doppler/ That isn't what happens with Doppler shift, the modulation frequency will depend on the velocity of the source, not its frequency of oscillation, The velocity of the source is given by the combination of the LF frequency and the amplitude of the motion of the speaker cone. and it sure won't be sidebands at +-50Hz!, Yes it will, check the referernces that you have been given! more like +-3 or 4Hz! The actual shift of the carrier is given by the vector sum of the sideband(s) and the carrier. From what I see the 4KHz signal shown is rock stable, no shift at all. The frequency of the low frequency signal has absolutely nothing to do with Doppler shift which is controlled solely by the velocity. The frequency of the Doppler modulated signal would vary by a few Hz shifting back and forth above and below the actual frequency at a 50Hz rate, but with a time/frequency analysis the 4KHz signal would simply show as being smeared, or if one were to sample at 50Hz synchronized to the 50 Hz of the cone there would be spikes at the plus or minus few Hz of the actual frequency shift, not at +-50Hz. |
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#362
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"Arny Krueger" wrote in message news  "Porky" wrote in message It still has nothing to do with Doppler shift, the spike is clean and sharp, indicating no small frequency shift as would occur with Doppler. Still havent read my references, I take it. http://contact.tm.agilent.com/Agilen...0-1/index.html http://contact.tm.agilent.com/Agilen...eFM_popup.html On this demo, if you just vary the frequency of the signal, as would happen with Doppler shift, you'll see that the center spike indicating the actual frequency moves back and forth, and if you could vary it fast enough over time, (say, 50 Hz) you'd note that the spike would be spread or blurred. This would indicate that Doppler shift is occurring in the speaker, and this is the only thing that could be attributed solely to Doppler shift. Since nothing i've seen so far indicates any evidence of this, I must conclude that Doppler shift in a speaker doesn't exist. |
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#363
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"Porky" wrote in message
"Arny Krueger" wrote in message ... "Porky" wrote in message Bob, aren't they showing a 1KHz signal modulated with a 50 Hz signal? Yes, if you're referring to http://www.pcavtech.com/techtalk/doppler/ That isn't what happens with Doppler shift, the modulation frequency will depend on the velocity of the source, not its frequency of oscillation, The velocity of the source is given by the combination of the LF frequency and the amplitude of the motion of the speaker cone. and it sure won't be sidebands at +-50Hz!, Yes it will, check the referernces that you have been given! more like +-3 or 4Hz! The actual shift of the carrier is given by the vector sum of the sideband(s) and the carrier. From what I see the 4KHz signal shown is rock stable, no shift at all. That's the carrier, and it will never change in frequency. However, as the modulation increases, energy will be taken from it to build the sidebands. The frequency of the low frequency signal has absolutely nothing to do with Doppler shift which is controlled solely by the velocity. In this case the velocity is itself a continuous signal with a known amplitude and frequency. The frequency of the Doppler modulated signal would vary by a few Hz shifting back and forth above and below the actual frequency at a 50Hz rate, Agreed. but with a time/frequency analysis the 4KHz signal would simply show as being smeared, This depends on what you analyze. These analysis were taken over a large number of cycles of the modulating signal. Things are highly averaged, and you see the true continuous spectrum of the wave. I can't believe that you've looked at the 3 references you've been offered before pursuing this line of reasoning any further. They show a very similar sideband structure. or if one were to sample at 50Hz synchronized to the 50 Hz of the cone there would be spikes at the plus or minus few Hz of the actual frequency shift, not at +-50Hz. In any case, the sidebands must be displaced from the carrier in increments of 50 Hz, which is the modulating frequency. The actual modulated wave is the vector sum of the carrier and the sidebands. Small deviations of the carrier are represented by smaller sidebands. I've written this at least 4 times this morning, and I'm running out of patience with a lack of appropriate dilligence on the part of the questioner. If you don't want to take time to read the references, some of which are pretty short, just FO! If you read them and questions remain, different story - I can work with those who drink, not those that won't even try to go to the water. |
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#364
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Arny Krueger wrote:
"ruffrecords" wrote in message Arny Krueger wrote: "ruffrecords" wrote in message Jim Carr wrote: Here's a link that you tech folks can argue about: http://www.pcavtech.com/techtalk/doppler/ Looks good to me. Plenty of evidence of harmonic distortion (casued by non-linearites Agreed. and no evidence of 50Hz sidebands arounf the 4KHz signal. What do you call the spikes around the 4,025 Hz carrier in, for example http://www.pcavtech.com/techtalk/dop...1-1-1+10dB.gif That pic was not in the OP I replied to - where did it come from? It was definately on the web site as of the date of the post I was responding to which was posted on the 17th. I haven't changed anything on the site for a week, since the 10th. Check the date stamp at the bottom of the page - its updated automatically when I update the site. I don't touch it! Ah, on the web site. I just looked at what the OP linked to. Is your pic a close up of the one he posted? Ian |
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#365
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"Porky" wrote in message
"Arny Krueger" wrote in message news "Porky" wrote in message It still has nothing to do with Doppler shift, the spike is clean and sharp, indicating no small frequency shift as would occur with Doppler. Still havent read my references, I take it. http://contact.tm.agilent.com/Agilen...0-1/index.html http://contact.tm.agilent.com/Agilen...eFM_popup.html On this demo, if you just vary the frequency of the signal, Carrier or modulation? I take it you are talking about the carrier. as would happen with Doppler shift, you'll see that the center spike indicating the actual frequency moves back and forth, and if you could vary it fast enough over time, (say, 50 Hz) you'd note that the spike would be spread or blurred. The model does not do that though, does it? That's because when you move the carrier and analyze the spectrum over a number of cycles, you get what they say. The concept of a modulated signal having a carrier and discrete sidebands has an assmuption, being that you're talking about a continuous process. If you look at a spectrum analyzer running in real time at a radio or TV station, you see the spread or blurring. This is due to the complex, changing modulation. But, if you stabilize the carrier and use a simple modulating signal, and measure across a number of cycles of the complex wave, you get the line spectrum shown at http://contact.tm.agilent.com/Agilen...eFM_popup.html This would indicate that Doppler shift is occurring in the speaker, and this is the only thing that could be attributed solely to Doppler shift. Since nothing i've seen so far indicates any evidence of this, I must conclude that Doppler shift in a speaker doesn't exist. You're still not getting modulation theory. Please read hp-am-fm.zip AKA http://contact.tm.agilent.com/data/s...1/hp-am-fm.zip . It takes much longer attention span than playing with the knobs on an interactive model, but it has the straight stuff. |
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#366
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"Arny Krueger" wrote in message ... "Porky" wrote in message "Arny Krueger" wrote in message news "Porky" wrote in message It still has nothing to do with Doppler shift, the spike is clean and sharp, indicating no small frequency shift as would occur with Doppler. Still havent read my references, I take it. http://contact.tm.agilent.com/Agilen...0-1/index.html http://contact.tm.agilent.com/Agilen...eFM_popup.html On this demo, if you just vary the frequency of the signal, Carrier or modulation? I take it you are talking about the carrier. as would happen with Doppler shift, you'll see that the center spike indicating the actual frequency moves back and forth, and if you could vary it fast enough over time, (say, 50 Hz) you'd note that the spike would be spread or blurred. The model does not do that though, does it? That's because when you move the carrier and analyze the spectrum over a number of cycles, you get what they say. The concept of a modulated signal having a carrier and discrete sidebands has an assmuption, being that you're talking about a continuous process. If you look at a spectrum analyzer running in real time at a radio or TV station, you see the spread or blurring. This is due to the complex, changing modulation. But, if you stabilize the carrier and use a simple modulating signal, and measure across a number of cycles of the complex wave, you get the line spectrum shown at http://contact.tm.agilent.com/Agilen...eFM_popup.html This would indicate that Doppler shift is occurring in the speaker, and this is the only thing that could be attributed solely to Doppler shift. Since nothing i've seen so far indicates any evidence of this, I must conclude that Doppler shift in a speaker doesn't exist. You're still not getting modulation theory. Please read hp-am-fm.zip AKA http://contact.tm.agilent.com/data/s...1/hp-am-fm.zip . It takes much longer attention span than playing with the knobs on an interactive model, but it has the straight stuff. Arny, with all due respect, I do understand modulation theory quite well. and I'm aware of what happens when you frequency modulate one signal with another. However this is NOT what we are doing here, we are dealing with a special case of frequency shift, Doppler shift, and we most certainly are not modulating the carrier (HF tone) with a LF tone. We are supposedly Doppler shifting the carrier frequency by some amount, depending on the cone velocity of the speaker. If you look at a freqiency alanysis of the train/whistle, you will see the carrier spike (the whistle) shifting to the right as the train approaches and then back to the left as it passes, sidebands have nothing to do with it. If the speaker were generating Doppler distortion, you would see exactly the same thing, but more rapidly, at 50Hz, the Spike would probably become a smear and it doesn't. Standard FM theory has little to do with the special case of Doppler shift. If you don't believe me, this can very easily be proven by using your train/whistle, or it's supposed equivalent by generating one half cycle of the gigantic speaker with huge excursion. Simply generate a ten second tone that starts at 500 Hz smoothly ascending to 600 Hz, this would simulate the train/whistle or gigantic speaker moving toward you, then scan it, what you get is a tabletop waveform from 500 Hz to 600 Hz (note that since you're simulating the Doppler shift by actually shifting the frequency, no LF Tone is necessary, in effect you're using a slowly varying DC voltage which would be moving the gigantic speaker cone, if the half cycle thing worries you, just create a wave that goes from 500 to 600 Hz and then back to 500 Hz again to simulate a full cycle, the results will be the same.). If it happens in your train/whistle or gigantic speaker analogy, then it would happen in the realworld, the only difference beign that the table top wave would extend from Carrier fequency minus shift to carrier frequency plus shift. Please note that if you find a train with whistle approaching and receding and scan it, you will get exactly the same thing, but you won't see in from a speaker. This should end the Doppler distortion thing once and for all! BTW, I tested this empirically with Cool Edit 96 and the results were exactly as I said, a nice flat tabletop. The whole problem stems from trying to use one audible waveform to modulate another, if you reduce it to its simplest form it becomes perfectly clear. |
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#367
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"Porky" wrote in message
"Arny Krueger" wrote in message ... You're still not getting modulation theory. Please read hp-am-fm.zip AKA http://contact.tm.agilent.com/data/s...1/hp-am-fm.zip . It takes much longer attention span than playing with the knobs on an interactive model, but it has the straight stuff. Arny, with all due respect, I do understand modulation theory quite well. Nope, and the evidence is squarely before me. You've missed the point of the interactive demo because you started playing with it before you got into the zip file text. and I'm aware of what happens when you frequency modulate one signal with another. You get sidebands, whether its AM or FM. However this is NOT what we are doing here, we are dealing with a special case of frequency shift, Doppler shift, and we most certainly are not modulating the carrier (HF tone) with a LF tone. Totally wrong. We are exactly modulating the HF tone with a LF tone, but its a bit confusing becasue there is both AM and FM and you clearly don't get FM. We are supposedly Doppler shifting the carrier frequency by some amount, depending on the cone velocity of the speaker. Right, which produces discrete sidebands that differ from the carrier by multiples of the modulating frequency. If you look at a freqiency alanysis of the train/whistle, you will see the carrier spike (the whistle) shifting to the right as the train approaches and then back to the left as it passes, sidebands have nothing to do with it. That's because the train is not moving back and forth in front of us, in a periodic sine-shaped pattern. If the speaker were generating Doppler distortion, you would see exactly the same thing, but more rapidly, at 50Hz, the Spike would probably become a smear and it doesn't. As I've pointed out about six times, whether you get a smear or a spike moving back and forth or what, depends on how you measure. Standard FM theory has little to do with the special case of Doppler shift. It does in the case of a speaker that is moving in and out in front of us, in a periodic sine-shaped pattern If you don't believe me, this can very easily be proven by using your train/whistle, or it's supposed equivalent by generating one half cycle of the gigantic speaker with huge excursion. Simply generate a ten second tone that starts at 500 Hz smoothly ascending to 600 Hz, this would simulate the train/whistle or gigantic speaker moving toward you, then scan it, what you get is a tabletop waveform from 500 Hz to 600 Hz (note that since you're simulating the Doppler shift by actually shifting the frequency, no LF Tone is necessary, in effect you're using a slowly varying DC voltage which would be moving the gigantic speaker cone, if the half cycle thing worries you, just create a wave that goes from 500 to 600 Hz and then back to 500 Hz again to simulate a full cycle, the results will be the same.). If it happens in your train/whistle or gigantic speaker analogy, then it would happen in the realworld, the only difference beign that the table top wave would extend from Carrier fequency minus shift to carrier frequency plus shift. Please note that if you find a train with whistle approaching and receding and scan it, you will get exactly the same thing, but you won't see in from a speaker. That's because the train isn't usually moving in and out in front of us, in a periodic sine-shaped pattern. This should end the Doppler distortion thing once and for all! You can't end a generally accpeted sceintific fact that has been around for 40 years or moer and keeps showing up everytime it is competently measured. At least it is darn tough. BTW, I tested this empirically with Cool Edit 96 and the results were exactly as I said, a nice flat tabletop. But, you simulated a train, not a speaker. The whole problem stems from trying to use one audible waveform to modulate another, if you reduce it to its simplest form it becomes perfectly clear. shaking head |
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#368
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I think we needto clarify what each of us are trying to say.
We have a speaker cone vibrating at both 50 Hz and 4 kHz. QUESTION 1 Does the 50 Hz cone vibration produce Doppler shift to the 4 kHz? I believe yes it does and that is what is shown on the spectrum analyzer previously. QUESTION 2 Does the 50 Hz cone vibration produce Doppler shift to the 50 Hz? I believe no it does not, the 50 Hz cone vibration is producing the 50 Hz wave and as far as the 50 Hz wave is concerned there is no additional movement. In any case, this is a very small effect and not the question I believe most of us are discussing. I believe most of us are addressing question #1. I think it would be helpful in the discussion to clarify which of these two questions you are discussing. Mark |
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#369
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I think we need to clarify what each of us are trying to say.
We have a speaker cone vibrating at both 50 Hz and 4 kHz. QUESTION 1 Does the 50 Hz cone vibration produce Doppler shift to the 4 kHz? I believe yes it does and that is what is shown on the spectrum analyzer previously. QUESTION 2 Does the 50 Hz cone vibration produce Doppler shift to the 50 Hz? I believe no it does not, the 50 Hz cone vibration is producing the 50 Hz wave and as far as the 50 Hz wave is concerned there is no additional movement. In any case, this is a very small effect and not the question I believe most of us are discussing. I believe most of us are addressing question #1. I think it would be helpful in the discussion to clarify which of these two questions you are discussing. Mark |
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#370
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So, it's easy to find the formula that shows how Doppler works (train
example) without taking into consideration the movement of the plane of the source that is creating the sound. I've seen numerous sites talk about Doppler Distortion, but I've yet to uncover one that shows the actual mathematical representation. I understand what you're saying, and YOU'RE WRONG. The velocity of the sound source "adds to" the velocity of the sound it produces. It doesn't matter whether the sound source is attached to the moving object, or the moving object IS the sound source. It doesn't matter whether the speaker is moved bodily, or the cone only is moved. To quote Galileo -- "Nevertheless, it does move." The other way of looking at it is that, as the sound propagates from the moving source, the source either "catches up with" or "lags behind" the propagating sound, thus (from the point of hearing of the observer) reducing or increasing its wavelength. That is, you're "squeezing" more cycles into a given space, or spreading them over a longer distance. Exactly the same thing happens with a driver producing LF and HF sounds at the same time. I'm going to spell this out in excruciating detail so you can SEE what happens... Suppose we feed a 5kHz signal into our classic KLH driver. As the driver vibrates, it alternately compresses and rarefies the air. These compressions and rarefactions propagate from the driver at the speed of sound -- about 1100 feet per second. This is much faster than the driver itself moves. Now let's add a strong 100Hz signal to the 5kHz signal, enough to visibly "pump" the driver. We'll also imagine that our senses have been speeded up by a thousand times or more ("The New Accelerator," anyone?) so we can "see" what's happening in front of the driver. Let's pick a point on the 100Hz signal where the driver is all the way back and is moving forward. Superimposed on that forward motion is the 5kHz signal. In the time the driver moves from back to front, 25 cycles of the 5kHz signal are generated and move away from the driver. Because THE DRIVER IS MOVING, each 5kHz cycle is generated "closer" to the previous cycle than it would be if the driver weren't moving, thus "squeezing" the wavelength. This is exactly the mechanism that creates Doppler shift. The opposite effect occurs when the driver moves "backwards." What you're forgetting, Bob, is that the acoustic waves generated are not "attached" to the cone. Once created, they move independently. If the sound source is moving -- for WHATEVER reason -- successive waves will be "closer to" or "farther from" the preceding waves. Suppose we could get the front surface of THE LOCOMOTIVE ITSELF to produce a 1kHz sound. Are you telling me that you WOULDN'T hear a Doppler shift as the train passed? If you think you wouldn't, then you don't understand Doppler shift. |
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#371
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How come you folks can't see that the train/whistle model is something
totally different from a loudspeaker (or other single source) producing a complex waveform? It isn't. It's exactly the same thing. See my othe post. |
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#372
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"Arny Krueger".. "Porky" BTW, I tested this empirically with Cool Edit 96 and the results were exactly as I said, a nice flat tabletop. But, you simulated a train, not a speaker. shaking head ** Arny is trying to teach a pig to sing. And Arny is getting annoyed faster than the pig. ............ Phil |
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#373
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"Arny Krueger" wrote in message ... "Porky" wrote in message "Arny Krueger" wrote in message ... You're still not getting modulation theory. Please read hp-am-fm.zip AKA http://contact.tm.agilent.com/data/s...1/hp-am-fm.zip . It takes much longer attention span than playing with the knobs on an interactive model, but it has the straight stuff. Arny, with all due respect, I do understand modulation theory quite well. Nope, and the evidence is squarely before me. You've missed the point of the interactive demo because you started playing with it before you got into the zip file text. and I'm aware of what happens when you frequency modulate one signal with another. You get sidebands, whether its AM or FM. However this is NOT what we are doing here, we are dealing with a special case of frequency shift, Doppler shift, and we most certainly are not modulating the carrier (HF tone) with a LF tone. Totally wrong. We are exactly modulating the HF tone with a LF tone, but its a bit confusing becasue there is both AM and FM and you clearly don't get FM. Once again, we are NOT modulating the Hf tone with a LF tone, we are modulating the HF tone with the supposed Doppler shift caused by the VELOCITY of the cone moving toward us and away from us! Since Doppler shift is supposed to occur any time the source is moving toward or away from the listener, the rate of the cone's moving back and forth doesn't matter, it's the velocity of the cone supposedly causing the shift in frequency! We are supposedly Doppler shifting the carrier frequency by some amount, depending on the cone velocity of the speaker. Right, which produces discrete sidebands that differ from the carrier by multiples of the modulating frequency. Try my experiment and see what you get, you are still thinking in terms of the LF tone modulating the HF tone and that ISN"T what's happening with Doppler shift! The Doppler shift is strictly a matter of the velocity of a sound source in relation to a listener, that's how Doppler is defined, it has nothing to do with how often the source changes direction!!! If you look at a freqiency alanysis of the train/whistle, you will see the carrier spike (the whistle) shifting to the right as the train approaches and then back to the left as it passes, sidebands have nothing to do with it. That's because the train is not moving back and forth in front of us, in a periodic sine-shaped pattern. If the speaker were generating Doppler distortion, you would see exactly the same thing, but more rapidly, at 50Hz, the Spike would probably become a smear and it doesn't. As I've pointed out about six times, whether you get a smear or a spike moving back and forth or what, depends on how you measure. Standard FM theory has little to do with the special case of Doppler shift. It does in the case of a speaker that is moving in and out in front of us, in a periodic sine-shaped pattern If you don't believe me, this can very easily be proven by using your train/whistle, or it's supposed equivalent by generating one half cycle of the gigantic speaker with huge excursion. Simply generate a ten second tone that starts at 500 Hz smoothly ascending to 600 Hz, this would simulate the train/whistle or gigantic speaker moving toward you, then scan it, what you get is a tabletop waveform from 500 Hz to 600 Hz (note that since you're simulating the Doppler shift by actually shifting the frequency, no LF Tone is necessary, in effect you're using a slowly varying DC voltage which would be moving the gigantic speaker cone, if the half cycle thing worries you, just create a wave that goes from 500 to 600 Hz and then back to 500 Hz again to simulate a full cycle, the results will be the same.). If it happens in your train/whistle or gigantic speaker analogy, then it would happen in the realworld, the only difference beign that the table top wave would extend from Carrier fequency minus shift to carrier frequency plus shift. Please note that if you find a train with whistle approaching and receding and scan it, you will get exactly the same thing, but you won't see in from a speaker. That's because the train isn't usually moving in and out in front of us, in a periodic sine-shaped pattern. This should end the Doppler distortion thing once and for all! You can't end a generally accpeted sceintific fact that has been around for 40 years or moer and keeps showing up everytime it is competently measured. At least it is darn tough. BTW, I tested this empirically with Cool Edit 96 and the results were exactly as I said, a nice flat tabletop. But, you simulated a train, not a speaker. The whole problem stems from trying to use one audible waveform to modulate another, if you reduce it to its simplest form it becomes perfectly clear. shaking head If you go from 500 to 600Hz and then back to 500, then you are simulating one cycle of the speaker cone. Whether the motion is linear or sinusoid won't matter. What applies to one cycle will apply to every other cycle. If Doppler shift exists, the result will be a table top, if not Wc will be represented as a spike. Try it! Go ahead, try it, go back and forth from one frequency to the other as many times as you wish, the results will be the same! Try this, generate a 20 second waveform of 500 Hz being modulated by 100 hz at a modulation frequency of .1 hz. this represents a HF tone of 500 Hz being moved on a giant speaker moving at a velocity high enough to result in a Doppler shift of +-20% with a LF component of .1 Hz. for two full p-p cycles. Note that you will see the same tabletop at higher frequencies if your resolution is fine enough. (I used an FFT of 65536 Blackman-Harris) That is as far as I can take this, but it seems adequate to me. The issue of Doppler distortion is related to the relative velocity of the listener and source, nothing in Doppler theory suggests that the frequency of change of direction will have anything to do with the amount of frequency shift. If the change of direction is such that the velocity graph forms a sine wave, the amount of shift will vary with the changing velocity from instant to instant, but the result will still be a varying amount of shift directly related to the instantaneous velocity at the time of measurement, and the result, assuming high enough resolution of the analyzer will be some form of tabletop waveform that varies from Wc minus shift to Wc plus shift. If you see this tabletop wave when recording a speaker producing a complex waveform consisting of a HF and a LF component, Doppler distortion exists in a speaker, if you don't, it doesn't. I tried it with my speakers and I didn't see it on the scan (no side bands either) or hear it on playback. As I said, that's as far as I can go with this, those with advanced education can debate further if they desire, but I've reached my limit and this time I'm going to respect that.:-) |
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#374
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"Phil Allison" wrote in message ... "Arny Krueger".. "Porky" BTW, I tested this empirically with Cool Edit 96 and the results were exactly as I said, a nice flat tabletop. But, you simulated a train, not a speaker. shaking head ** Arny is trying to teach a pig to sing. And Arny is getting annoyed faster than the pig. And Phil is an annoying ass... |
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#375
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"Porky" wrote in message
"Arny Krueger" wrote in message ... "Porky" wrote in message "Arny Krueger" wrote in message ... You're still not getting modulation theory. Please read hp-am-fm.zip AKA http://contact.tm.agilent.com/data/s...1/hp-am-fm.zip . It takes much longer attention span than playing with the knobs on an interactive model, but it has the straight stuff. Arny, with all due respect, I do understand modulation theory quite well. Nope, and the evidence is squarely before me. You've missed the point of the interactive demo because you started playing with it before you got into the zip file text. and I'm aware of what happens when you frequency modulate one signal with another. You get sidebands, whether its AM or FM. However this is NOT what we are doing here, we are dealing with a special case of frequency shift, Doppler shift, and we most certainly are not modulating the carrier (HF tone) with a LF tone. Totally wrong. We are exactly modulating the HF tone with a LF tone, but its a bit confusing becasue there is both AM and FM and you clearly don't get FM. Once again, we are NOT modulating the Hf tone with a LF tone, we are modulating the HF tone with the supposed Doppler shift caused by the VELOCITY of the cone moving toward us and away from us! But the toward/away motion is exactly due to the LF tone. Since Doppler shift is supposed to occur any time the source is moving toward or away from the listener, the rate of the cone's moving back and forth doesn't matter, it's the velocity of the cone supposedly causing the shift in frequency! The velocity of the cone and the rate of the cone's motion are closely tied together. We are supposedly Doppler shifting the carrier frequency by some amount, depending on the cone velocity of the speaker. Right, which produces discrete sidebands that differ from the carrier by multiples of the modulating frequency. Try my experiment and see what you get, you are still thinking in terms of the LF tone modulating the HF tone and that ISN"T what's happening with Doppler shift! I've both done and debunked your experiment. Next! The Doppler shift is strictly a matter of the velocity of a sound source in relation to a listener, that's how Doppler is defined, it has nothing to do with how often the source changes direction!!! But it does. If the source doesn't change direction often enough, and the peak velocity is high enough, the cone comes sailing out of the woofer frame, destroying the woofer. The cyclic changing direction creates the sidebands we observe. snip If you go from 500 to 600Hz and then back to 500, then you are simulating one cycle of the speaker cone. Whether the motion is linear or sinusoid won't matter. Oh but it does. A linear motion is characteristic of a sawtooth modulating wave, and that gives a different collection of sidebands. You convolve the spectrum of the carrier with the spectrum of the modulating signal.... What applies to one cycle will apply to every other cycle. Nope, the shape of the cycle matters. If Doppler shift exists, the result will be a table top, if not Wc will be represented as a spike. Try it! Were talking about a speaker cone that moves in a periodic cycle, not a train on a tabletop. Go ahead, try it, go back and forth from one frequency to the other as many times as you wish, the results will be the same! Trust me, I've tried a ton of things. Try this, generate a 20 second waveform of 500 Hz being modulated by 100 hz at a modulation frequency of .1 hz. this represents a HF tone of 500 Hz being moved on a giant speaker moving at a velocity high enough to result in a Doppler shift of +-20% with a LF component of .1 Hz. for two full p-p cycles. OK, its on the screen before me in Audition/CE Note that you will see the same tabletop at higher frequencies if your resolution is fine enough. Nope. I see a spike at whatever frequency the carrrier is at, where my analysis intersects it. I wish that people would do their own home work. (I used an FFT of 65536 Blackman-Harris) That is what I use. BTW, Audition/CE does not animate the spectral display when you play a file. That is as far as I can take this, but it seems adequate to me. I don't think you understand the relevance of the time period over which the analysis is performed. The issue of Doppler distortion is related to the relative velocity of the listener and source, So far so good. nothing in Doppler theory suggests that the frequency of change of direction will have anything to do with the amount of frequency shift. True, but the frequency of shift and the frequency of the sidebands are two vastly different things. If the change of direction is such that the velocity graph forms a sine wave, the amount of shift will vary with the changing velocity from instant to instant, but the result will still be a varying amount of shift directly related to the instantaneous velocity at the time of measurement, and the result, assuming high enough resolution of the analyzer will be some form of tabletop waveform that varies from Wc minus shift to Wc plus shift. OK, I think I follow this. If you see this tabletop wave when recording a speaker producing a complex waveform consisting of a HF and a LF component, Doppler distortion exists in a speaker, if you don't, it doesn't. In fact, an ideal frequency domain analysis will show a moving spike. Your tabletop is an artifact of the length of your sample, which is in turn determined by the sample size. If you analyze a wave during a time period when it has components at a range of frequencies (this time because the frequency is slowly changing) then you get your tabletop. A 65k point FFT is inappropriate for analyzing a wave with a modulation frequency of 0.1 Hz. 0.1 Hz corresponds to 10 seconds of data. Normally, we would use a FFT that covered several times that much data to reveal the sideband structure. A 44100 Hz you need a FFT that would be appropriate would have like 4 million points or more. I tried it with my speakers and I didn't see it on the scan (no side bands either) or hear it on playback. You picked a demonstration that is inapprorpiate for the data at hand. Futhermore its not representaive of speakers, because it is based on a 1/10 th Hz modulating frequency. As I said, that's as far as I can go with this, those with advanced education can debate further if they desire, but I've reached my limit and this time I'm going to respect that.:-) You've wasted both of our time with a very poorly-formed experiment. ;-( |
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"ruffrecords" wrote in message
... Suppose you have a tank of compressed air fed through a valve that can be modulated. This can be used to create compression waves from a source at rest. You mean the compressed air doesn't move? |
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#378
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In alt.music.home-studio,rec.audio.tech,rec.audio.pro, Bob Cain
wrote: Randy Yates wrote: ... How can I refute what you have refused to clearly restate? All right, I'll accept that you haven't read the thread. I've found a new way to state it anyway, that I think makes it clearer. I'm saying that the instantaneous velocity of the piston is transfered to the wave in the right position, since it is moving in step with it, to propegate that velocity out as an acoustic wave. It is _in_ the acoustic wave it is creating and it is at the right place at all times to impart the correct velocity _because_ it is in it. It doesn't matter what signals might have been mixed upstream to get the signal that controls the velocity of that piston. It will be moving in lock step with the wave defined by that signal and will always be in the right place to deliver the right velocity to the outgoing wave. If, on the other hand the piston is moving with a constant velocity superimposed on the signal velocity, it has no way to transfer that constant velocity to the air because contant velocity doesn't create a wave. At f=0 it runs out of punch. It ceases abruptly to transduce at all. In that case, the piston will always be in the wrong position to correctly impart the desired velocity signal and that error is Doppler shift. Suppose instead of this CONSTANT velocity, the piston is just barely speeding up or slowing down (accelerating/decellerating). Does the doppler effect stop? Is there an amount of acceleration above which the doppler effect stops? What if the acceleration is sinusoidal (thus the position is a cosine in relation)? In your explanation, there has to be some point where doppler shift goes away, as the velocity begins to change. Surely this could be measured? If the above is truly crazy, then I'll agree that I'm over the edge. But I don't think so. What, exactly, is wrong with it? I don't want to talk about trains and whistles. Bob ----- http://mindspring.com/~benbradley |
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#379
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"Phil Allison" :
I do not have mental disabilities. LOL! /Jonas |
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#380
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In alt.music.home-studio,rec.audio.tech,rec.audio.pro"Arny Krueger"
wrote: "Porky" wrote in message { Arny and Porky are arguing about FM sidebands... } ... and it sure won't be sidebands at +-50Hz!, Yes it will, check the referernces that you have been given! more like +-3 or 4Hz! The actual shift of the carrier is given by the vector sum of the sideband(s) and the carrier. From what I see the 4KHz signal shown is rock stable, no shift at all. That's the carrier, and it will never change in frequency. However, as the modulation increases, energy will be taken from it to build the sidebands. The frequency of the low frequency signal has absolutely nothing to do with Doppler shift which is controlled solely by the velocity. In this case the velocity is itself a continuous signal with a known amplitude and frequency. The frequency of the Doppler modulated signal would vary by a few Hz shifting back and forth above and below the actual frequency at a 50Hz rate, Agreed. but with a time/frequency analysis the 4KHz signal would simply show as being smeared, This depends on what you analyze. These analysis were taken over a large number of cycles of the modulating signal. Things are highly averaged, and you see the true continuous spectrum of the wave. I can't believe that you've looked at the 3 references you've been offered before pursuing this line of reasoning any further. They show a very similar sideband structure. or if one were to sample at 50Hz synchronized to the 50 Hz of the cone there would be spikes at the plus or minus few Hz of the actual frequency shift, not at +-50Hz. In any case, the sidebands must be displaced from the carrier in increments of 50 Hz, which is the modulating frequency. The actual modulated wave is the vector sum of the carrier and the sidebands. Small deviations of the carrier are represented by smaller sidebands. I've written this at least 4 times this morning, and I'm running out of patience with a lack of appropriate dilligence on the part of the questioner. If you don't want to take time to read the references, some of which are pretty short, just FO! If you read them and questions remain, different story - I can work with those who drink, not those that won't even try to go to the water. Arny, I can see a reluctance to even look, as the spectrum of an FM signal is SO counterintuitive (at least to me) that it's actually hard to believe (and thus the 'flattop' comment we saw earlier). I do 'believe' the sidebands are as you say they are, but can't imagine how FM results in those sidebands. Perhaps those with knowledge of higher math (more than a couple quarters of calculus) can visualize it. ----- http://mindspring.com/~benbradley |
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#381
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On Wed, 18 Aug 2004 06:26:11 -0500, "Porky" wrote:
"Arny Krueger" wrote in message news "Porky" wrote in message It still has nothing to do with Doppler shift, the spike is clean and sharp, indicating no small frequency shift as would occur with Doppler. Still havent read my references, I take it. http://contact.tm.agilent.com/Agilen...0-1/index.html http://contact.tm.agilent.com/Agilen...eFM_popup.html On this demo, if you just vary the frequency of the signal, as would happen with Doppler shift, you'll see that the center spike indicating the actual frequency moves back and forth, and if you could vary it fast enough over time, (say, 50 Hz) you'd note that the spike would be spread or blurred. Are you saying this page by HP (er, Agilent) is wrong? Have you written to the webmaster about it? This would indicate that Doppler shift is occurring in the speaker, and this is the only thing that could be attributed solely to Doppler shift. Since nothing i've seen so far indicates any evidence of this, I must conclude that Doppler shift in a speaker doesn't exist. ----- http://mindspring.com/~benbradley |
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#382
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Jim Carr wrote:
"ruffrecords" wrote in message ... Suppose you have a tank of compressed air fed through a valve that can be modulated. This can be used to create compression waves from a source at rest. You mean the compressed air doesn't move? No, but the device creating the sound does not move in the direction of the sound wave. Imagine it is like an iris opening and closing. Ian |
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#383
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"Ben Bradley" wrote in message
Arny, I can see a reluctance to even look, as the spectrum of an FM signal is SO counterintuitive (at least to me) that it's actually hard to believe (and thus the 'flattop' comment we saw earlier). Really? It's fairly intuitive to me. However, not everybody spent three years trying to grok Doppler radars that had LAYERS of AM & FM modulators in both the transmitter and receiver. That was a long time ago, but it seems to have stuck, at least a little. I do 'believe' the sidebands are as you say they are, but can't imagine how FM results in those sidebands. The 50 Hz part I understand. The phase inversion of the upper sidebands I understand. The multiple sidebands I take based on more-or-less faith, based on knowlege of Bessel functions and many hours spent with spectrum analyzers. Perhaps those with knowledge of higher math (more than a couple quarters of calculus) can visualize it. In the end, I think it is experience. The ultimate proof is when it works as predicted, theory and empirical knowlege tell the same story. |
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On Tue, 17 Aug 2004 22:34:06 -0700, "Jim Carr"
wrote: "Ben Bradley" wrote in message .. . So, it's easy to find the formula that shows how Doppler works (train example) without taking into consideration the movement of the plane of the source that is creating the sound. I've seen numerous sites talk about Doppler Distortion, but I've yet to uncover one that shows the actual mathematical representation. Somebody post this formula so we can all go to bed. I think that would be fo = fs . (v - vo) / (v - vs). This uneducated rube thinks you're wrong. Assume vo is zero. The source, a speaker, is producing a pure 50Hz signal and nothing else. It's stationary. I don't hear the Doppler effect, right? But according to you, I must take into account the velocity of the diaphragm creating the sound. That means I *will* hear the Doppler effect (or at least I could measure it). . It just doesn't make sense to me. The diaphragm is creating waves at the exact same point each cycle - compress-rarefy, compress-rarefy. It happens 50 times per second. The wavelength remains unchanged. So where is the Doppler shift? Short explanation, at the end of the compression, the cone is closer to the receiver than at the end of the rarification, so due to the speed of sound, it gets to the receiver sooner. See my other posts on this (in the last 16 hours or so). They way I've seen Doppler distortion explained is that higher frequencies are "riding" on the lower (or lowest, I don't know) frequency and it is *that* movement which creates the Doppler effect. Obviously your formula doesn't take that into account. Granted, I used to think algebra was the plural form of algae, but I'm pretty sure I understand the basic Doppler formual enough to know it doesn't factor in the movement of the source as it creates the wave. ----- http://mindspring.com/~benbradley |
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"Ben Bradley" wrote in message
On Wed, 18 Aug 2004 06:26:11 -0500, "Porky" wrote: "Arny Krueger" wrote in message news "Porky" wrote in message It still has nothing to do with Doppler shift, the spike is clean and sharp, indicating no small frequency shift as would occur with Doppler. Still havent read my references, I take it. http://contact.tm.agilent.com/Agilen...0-1/index.html Are you saying this page by HP (er, Agilent) is wrong? Have you written to the webmaster about it? FWIW, I take it as gospel. The simulation is good as gold. But, you have to *get* the text file. |
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#386
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Ben Bradley wrote: Suppose instead of this CONSTANT velocity, the piston is just barely speeding up or slowing down (accelerating/decellerating). Does the doppler effect stop? Is there an amount of acceleration above which the doppler effect stops? What if the acceleration is sinusoidal (thus the position is a cosine in relation)? In your explanation, there has to be some point where doppler shift goes away, as the velocity begins to change. Surely this could be measured? It is present if there is any constant component to the piston's velocity relative to the air. Otherwise it is not. It is what happens at the point where the air is incapable of carrying the signal contained in the piston's velocity function. The part it can carry is the component that changes with time. That which it can't has to appear in some way in the result and that way is the Doppler shift. Bob -- "Things should be described as simply as possible, but no simpler." A. Einstein |
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#387
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Ben Bradley wrote: Short explanation, at the end of the compression, the cone is closer to the receiver than at the end of the rarification, so due to the speed of sound, it gets to the receiver sooner. See my other posts on this (in the last 16 hours or so). At the end of the compression, the cone is exactly where it is supposed to be within the wave it is creating. It's position is that within the wave where the imparted velocity belongs. It is in the wave. Where else could it be? It could be in the wrong place if and only if there is a constant component to the velocity because it cannot impart that component to the air. The air cannot support it. Bob -- "Things should be described as simply as possible, but no simpler." A. Einstein |
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Porky wrote: thing and start from scratch, I think you'll see what Bob and I see. Remember two things, an oscilloscope pattern is at best a two dimensional representation of a three dimensional occurance and doesn't necessarily represent what is actually happening, and the speaker cone doesn't actually generate the sound wave (velocity too low and p-p cone travel too short for the wavelength), the air it moves does. No, the cone is actually moving with the right velocity for the wave, and it is in the right position to impart that to the air because its derivative gives the position in the wave where that velocity applies and that's just where it happens to be. Unless it's a constant component and then it can't impart that to the air so it's position is in error. That error is Doppler shift. Bob -- "Things should be described as simply as possible, but no simpler." A. Einstein |
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Porky wrote: Bob, aren't they showing a 1KHz signal modulated with a 50 Hz signal? That isn't what happens with Doppler shift, the modulation frequency will depend on the velocity of the source, not its frequency of oscillation, and it sure won't be sidebands at +-50Hz!, more like +-3 or 4Hz! If I'm right, this is an example of the fundamental mis-assumptions the pro-Doppler group is making, just like basing their logic on an analogy that doesn't meet the necessary criteria.. Something quite like that is what my inuition says too but without a deeper analysis and a better understanding of modulation theory I don't trust my intuition on this one. Bob -- "Things should be described as simply as possible, but no simpler." A. Einstein |
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"William Sommerwerck" wrote in message ... So, it's easy to find the formula that shows how Doppler works (train example) without taking into consideration the movement of the plane of the source that is creating the sound. I've seen numerous sites talk about Doppler Distortion, but I've yet to uncover one that shows the actual mathematical representation. I understand what you're saying, and YOU'RE WRONG. I'm wrong? Which part, exactly? The velocity of the sound source "adds to" the velocity of the sound it produces. The speed of sound is constant within a given medium, so I have to disagree with the above statement since your quotes around "adds to" means nothing to me. The other way of looking at it is that, as the sound propagates from the moving source, the source either "catches up with" or "lags behind" the propagating sound, thus (from the point of hearing of the observer) reducing or increasing its wavelength. That is, you're "squeezing" more cycles into a given space, or spreading them over a longer distance. That's one way of phrasing it. Suppose we could get the front surface of THE LOCOMOTIVE ITSELF to produce a 1kHz sound. Are you telling me that you WOULDN'T hear a Doppler shift as the train passed? If you think you wouldn't, then you don't understand Doppler shift. You just explained Doppler shift. Nobody is arguing that point. Here's a simple request. fo = fs . (v - vo) / (v - vs) describes the apparent frequency shift of a single frequency. So let's concentrate on a speaker creating one tone. If the source is sitting on a platform and that platforms moves towards (or away) from the observer, we agree that we will witness a Doppler shift. Suppose now that this platform is stationary. We experience no Doppler shift, right? So, we add the second tone. You argue that the first tone experiences no Doppler shift, but the second tone does (assuming it's a higher frequency). Right? So, what's the mathematical representation of a two-tone system where the lower tone remains unshifted and the higher tone shifts? Don't get me wrong. I fully and completely understand your mind experiment and it makes perfect sense. However, I have not seen a formula to predict it. I would think that experts such as yourself would be able to enlighten a rube like myself with some algebra. |
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"Ben Bradley" wrote in message
... It just doesn't make sense to me. The diaphragm is creating waves at the exact same point each cycle - compress-rarefy, compress-rarefy. It happens 50 times per second. The wavelength remains unchanged. So where is the Doppler shift? Short explanation, at the end of the compression, the cone is closer to the receiver than at the end of the rarification, so due to the speed of sound, it gets to the receiver sooner. See my other posts on this (in the last 16 hours or so). So, what's the shift? After all, EVERY source of sound requires movement of something to compress the air molecules, right? Okay, maybe some magnetic force can do it, but that's beyond me. Are you saying, for example, that a stationary speaker pumping at 50Hz may result in me hearing something like (numbers out of my ass) 51Hz and 49Hz since each compression and rarefication has a Doppler shift? If so, then the standard formula for computing Doppler shift is wrong since all sources have a compression and rarefication cycle of some distance, most likely very minute, but movement nonetheless. It would mean we never hear anything exactly as it sounds unless we managed to move our heads at exactly the same frequency. |
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You just explained Doppler shift. Nobody is arguing that point.
Here's a simple request. fo = fs . (v - vo) / (v - vs) describes the apparent frequency shift of a single frequency. So let's concentrate on a speaker creating one tone. If the source is sitting on a platform and that platforms moves towards (or away) from the observer, we agree that we will witness a Doppler shift. Suppose now that this platform is stationary. We experience no Doppler shift, right? So, we add the second tone. You argue that the first tone experiences no Doppler shift, but the second tone does (assuming it's a higher frequency). Right? So, what's the mathematical representation of a two-tone system where the lower tone remains unshifted and the higher tone shifts? Don't get me wrong. I fully and completely understand your mind experiment and it makes perfect sense. However, I have not seen a formula to predict it. I would think that experts such as yourself would be able to enlighten a rube like myself with some algebra. The behavior of the system is FULLY predicted by the Doppler formula, where THE VELOCITY OF THE CONE MOTION (as produced by the lower tone) IS THE SOURCE VELOCITY. That's all there is to it. It's no more complex than that. All you have to do is plug and grind. |
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Jim Carr wrote:
"William Sommerwerck" wrote in message ... So, it's easy to find the formula that shows how Doppler works (train example) without taking into consideration the movement of the plane of the source that is creating the sound. I've seen numerous sites talk about Doppler Distortion, but I've yet to uncover one that shows the actual mathematical representation. I understand what you're saying, and YOU'RE WRONG. I'm wrong? Which part, exactly? The velocity of the sound source "adds to" the velocity of the sound it produces. The speed of sound is constant within a given medium, so I have to disagree with the above statement since your quotes around "adds to" means nothing to me. The other way of looking at it is that, as the sound propagates from the moving source, the source either "catches up with" or "lags behind" the propagating sound, thus (from the point of hearing of the observer) reducing or increasing its wavelength. That is, you're "squeezing" more cycles into a given space, or spreading them over a longer distance. That's one way of phrasing it. Suppose we could get the front surface of THE LOCOMOTIVE ITSELF to produce a 1kHz sound. Are you telling me that you WOULDN'T hear a Doppler shift as the train passed? If you think you wouldn't, then you don't understand Doppler shift. You just explained Doppler shift. Nobody is arguing that point. Here's a simple request. fo = fs . (v - vo) / (v - vs) describes the apparent frequency shift of a single frequency. So let's concentrate on a speaker creating one tone. If the source is sitting on a platform and that platforms moves towards (or away) from the observer, we agree that we will witness a Doppler shift. Suppose now that this platform is stationary. We experience no Doppler shift, right? So, we add the second tone. You argue that the first tone experiences no Doppler shift, but the second tone does (assuming it's a higher frequency). Right? So, what's the mathematical representation of a two-tone system where the lower tone remains unshifted and the higher tone shifts? Don't get me wrong. I fully and completely understand your mind experiment and it makes perfect sense. However, I have not seen a formula to predict it. I would think that experts such as yourself would be able to enlighten a rube like myself with some algebra. That's what I was saying, that any tone would warp its own waveshape due to Doppler shift. |
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Jim Carr wrote:
"Ben Bradley" wrote in message ... It just doesn't make sense to me. The diaphragm is creating waves at the exact same point each cycle - compress-rarefy, compress-rarefy. It happens 50 times per second. The wavelength remains unchanged. So where is the Doppler shift? Short explanation, at the end of the compression, the cone is closer to the receiver than at the end of the rarification, so due to the speed of sound, it gets to the receiver sooner. See my other posts on this (in the last 16 hours or so). So, what's the shift? After all, EVERY source of sound requires movement of something to compress the air molecules, right? Okay, maybe some magnetic force can do it, but that's beyond me. Are you saying, for example, that a stationary speaker pumping at 50Hz may result in me hearing something like (numbers out of my ass) 51Hz and 49Hz since each compression and rarefication has a Doppler shift? If so, then the standard formula for computing Doppler shift is wrong since all sources have a compression and rarefication cycle of some distance, most likely very minute, but movement nonetheless. It would mean we never hear anything exactly as it sounds unless we managed to move our heads at exactly the same frequency. Like Heisenberg only different. |
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That's what I was saying, that any tone would warp
its own waveshape due to Doppler shift. It doesn't, though, because there's only one motion. Doppler requires two motions -- the wave, and the movement of the surface producing it. |
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#396
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William Sommerwerck wrote:
That's what I was saying, that any tone would warp its own waveshape due to Doppler shift. It doesn't, though, because there's only one motion. Doppler requires two motions -- the wave, and the movement of the surface producing it. LOL |
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#397
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"Jim Carr" wrote in message news:LIzUc.9970$yh.371@fed1read05...
Bob can correct me here, but from where I sit, Bob does not deny the Doppler shift. He completely accepts the formula fo = fs . (v - vo) / (v - vs). What he does not accept is that you can insert the speaker diaphragm's movement as vs. He accepts that you can put the speaker on the train and predict the Doppler shift with the above formula as does everyone he's arguing with. But here's the kicker: Either the above formula is wrong, there is no Doppler Distortion, or somebody left out an assumption somewhere. Why? Because the formula does not take into consideration the movement of the diaphragm. If the source and observer are not moving, there is no Doppler shift, right? But according to the proponents of Doppler Distortion, there *is* a shift. Everything that creates sound moves in some way. So why hasn't someone updated that formula? The problem isn't that somebody left out an assumption somewhere. The problem is that nobody involved in the discussion of this topic thus far seems to have asked the question "upon what assumption is the derivation of the formula based" Consequently those like Bob Cain, who foolishly deny the existence of demonstrated reality, "assume" that the equation that is cited for the constant-velocity case does not also apply to the dynamic case. On the other hand, those with a bit more common technical sense than Bob Cain similarly "assume" that the equation applies to the dynamic case as well as to the constant velocity case. The problem is that, for whatever reason, no one debating the issue seems to actually know which of the two mutually exclusive assumptions is correct. Those who are interested in knowing can find the answer as well as the derivation from fundamental principles of the formula for instantaneous Doppler shift in Allan Pierce's book "Acoustics, An Introduction to Its Physical Principles and Applications." |
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#398
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"Jim Carr" wrote in message news:LIzUc.9970$yh.371@fed1read05...
Bob can correct me here, but from where I sit, Bob does not deny the Doppler shift. He completely accepts the formula fo = fs . (v - vo) / (v - vs). What he does not accept is that you can insert the speaker diaphragm's movement as vs. He accepts that you can put the speaker on the train and predict the Doppler shift with the above formula as does everyone he's arguing with. But here's the kicker: Either the above formula is wrong, there is no Doppler Distortion, or somebody left out an assumption somewhere. Why? Because the formula does not take into consideration the movement of the diaphragm. If the source and observer are not moving, there is no Doppler shift, right? But according to the proponents of Doppler Distortion, there *is* a shift. Everything that creates sound moves in some way. So why hasn't someone updated that formula? The problem isn't that somebody left out an assumption somewhere. The problem is that nobody involved in the discussion of this topic thus far seems to have asked the question "upon what assumption is the derivation of the formula based" Consequently those like Bob Cain, who foolishly deny the existence of demonstrated reality, "assume" that the equation that is cited for the constant-velocity case does not also apply to the dynamic case. On the other hand, those with a bit more common technical sense than Bob Cain similarly "assume" that the equation applies to the dynamic case as well as to the constant velocity case. The problem is that, for whatever reason, no one debating the issue seems to actually know which of the two mutually exclusive assumptions is correct. Those who are interested in knowing can find the answer as well as the derivation from fundamental principles of the formula for instantaneous Doppler shift in Allan Pierce's book "Acoustics, An Introduction to Its Physical Principles and Applications." |
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#399
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The problem isn't that somebody left out an assumption somewhere. The
problem is that nobody involved in the discussion of this topic thus far seems to have asked the question "upon what assumption is the derivation of the formula based" Consequently those like Bob Cain, who foolishly deny the existence of demonstrated reality, "assume" that the equation that is cited for the constant-velocity case does not also apply to the dynamic case. On the other hand, those with a bit more common technical sense than Bob Cain similarly "assume" that the equation applies to the dynamic case as well as to the constant velocity case. The problem is that, for whatever reason, no one debating the issue seems to actually know which of the two mutually exclusive assumptions is correct. Those who are interested in knowing can find the answer as well as the derivation from fundamental principles of the formula for instantaneous Doppler shift in Allan Pierce's book "Acoustics, An Introduction to Its Physical Principles and Applications." In terms of the question Bob Cain initially asked, the formula doesn't matter, because Doppler shift exists regardless of whether you're working in constant-velocity medium ("the luminiferous ether") or a non-constant-velocity medium (air). The formulas are different, but that doesn't have anything to do with the nature of the argument. |
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#400
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Linear Mode
