Microphone in vaccuum


by Nick89
Tags: microphone, vaccuum
Nick89
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#1
Jun5-08, 03:36 PM
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Hi,

I was recently wondering this, might be a stupid question but hey, here to learn :)...

Since the sound pressure level is defined as:
[tex]L_p = 20 log_{10} \left( \frac{p}{p_{ref}} \right)[/tex]

What would happen if you put a microphone in a perfect vaccuum? Obviously there can be no sound pressure [itex]p[/itex], so would the microphone measure a value of [itex]-\infty[/itex] dB? ([itex]lim_{x \rightarrow 0} log(x) = -\infty[/itex], right?)

Intuitively I would say no since with most microphones if you are in a completely silent room (except for thermal noise), they would measure 0 dB, right? And this would probably be the lowest spl the microphone can measure, any lower pressure will not move the vibrating coil?

Meh, just thought I'd ask.
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berkeman
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Jun5-08, 03:42 PM
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Quote Quote by Nick89 View Post
Hi,

I was recently wondering this, might be a stupid question but hey, here to learn :)...

Since the sound pressure level is defined as:
[tex]L_p = 20 log_{10} \left( \frac{p}{p_{ref}} \right)[/tex]

What would happen if you put a microphone in a perfect vaccuum? Obviously there can be no sound pressure [itex]p[/itex], so would the microphone measure a value of [itex]-\infty[/itex] dB? ([itex]lim_{x \rightarrow 0} log(x) = -\infty[/itex], right?)

Intuitively I would say no since with most microphones if you are in a completely silent room (except for thermal noise), they would measure 0 dB, right? And this would probably be the lowest spl the microphone can measure, any lower pressure will not move the vibrating coil?

Meh, just thought I'd ask.
No, 0dB means that the pressure is equal to the reference pressure. Just like 0dBm means that the power is equal to 1mW (dBm is dB above 1mW into 50 Ohms).

So just take the log of 0 and see what you get. Yep, negative infinity.
Nick89
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#3
Jun5-08, 03:51 PM
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I know 0 dB means the pressure is the reference pressure. But was my 'problem' is, is that it doesn't make sense for a microphone to display -infinity dB... Does it? Probably just because there can be no perfect vaccuum.

BUT. A microphone is obviously never perfect either. If I'm right in thinking that a microphone works by a coil that is vibrated due to the sound waves, it will ofcourse take some minimal amount of force (pressure) to get the coil moving (due to friction etc). If the sound pressure is below this minimal pressure, the microphone should display -infinity, since the coil is not moving and the micrphone will detect no sound pressure, right? (That is ofcourse if the microphone is displaying the soundlevel instead of the sound pressure, but afaik most microphones do?)

russ_watters
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Jun5-08, 03:54 PM
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Microphone in vaccuum


Well, in a vacuum, Pref is 0, isn't it...?

So what's log(0/0)?
berkeman
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Jun5-08, 03:56 PM
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Quote Quote by russ_watters View Post
Well, in a vacuum, Pref is 0, isn't it...?

So what's log(0/0)?
No, I think Pref is some reference value, independent of the environment. Just like 1mW is used as the reference for dBm, and like 1uV is used as the reference for dBuv.
berkeman
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Jun5-08, 04:01 PM
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Quote Quote by Nick89 View Post
I know 0 dB means the pressure is the reference pressure. But was my 'problem' is, is that it doesn't make sense for a microphone to display -infinity dB... Does it? Probably just because there can be no perfect vaccuum.
To get the most sensitive measurement, you would connect the microphone to an audio spectrum analyzer. When the microphone is not picking anything up, the spectrum analyzer would display its own noise floor (it would see a very small level of white noise), probably around -70dB for a good analyzer.
Redbelly98
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Jun5-08, 05:25 PM
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Microphones don't measure dB directly. The microphone output will be a voltage signal, and that voltage would be 0 in a perfect vacuum (ignoring electronic noise that is unrelated to the sound).

dB is just something we humans calculate using the output voltage signal.
GTrax
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Jun5-08, 06:35 PM
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Just because you eliminated the energy contribution from sound pressure coupling to the diaphragm by sucking out the air, you do not have zero signal yet! There would still be electrical noise output at the terminals. Some may come from other mechanical coupling to the microphone, some from thermoelectric effects at the connections, some from local electromagnetic couplings from a.c. power distribution and more.

Even if all these are somehow eliminated, the noise of molecules shaking about will not subside until you resort to cryogenic cooling. The minimum signal you can discern, as ratio relative to some reference, expressed as dB=10*log(ratio) will not be infinite unless you can truly achieve absolute zero in a totally frozen, yet still somehow functional microphone..
( I think )
Nick89
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Jun11-08, 04:54 PM
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Sorry for waiting so long to respond, for some reason I did not get an email that someone replied...

Redbelly, is the voltage 0 because the sound pressure is 0? If so then, if you would convert that to dB's, you would get -infinity dB's...

GTrax, good point. I was using a vaccuum in my 'thought experiment' to cancel out the thermal noise of the air. But obviously the microphone itself contains molecules moving about so you can never actually meassure an absolute zero sound pressure (you might be able to acchieve that in a perfect vaccuum, but you cannot measure it with a mechanical microphone).
Redbelly98
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Jun11-08, 06:05 PM
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Quote Quote by Nick89 View Post
Redbelly, is the voltage 0 because the sound pressure is 0?
Yes. Or more accurately, because the amplitude of the sound pressure is zero.

If so then, if you would convert that to dB's, you would get -infinity dB's...
Yes. But my point was that it becomes -infinity only when you do the math conversion. The microphone itself is not "outputting" -infinity, so no physical laws are being violated.
lightarrow
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#11
Jun12-08, 08:50 AM
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Quote Quote by russ_watters View Post
Well, in a vacuum, Pref is 0, isn't it...?

So what's log(0/0)?
Pref is 20 micropascals (20 μPa) by definition.


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