Whaaa... I need help comparing two acoustic sensors
ElectricAye
Posts: 4,561
Hi all,
Through my tears, I'm trying to compare two acoustic emission sensors. The first sensor has a specification that says its sensitivity is −150 dBre 1V/μPa. The other sensor says its sensitivity is -70 dB ref 1V/μbar.
Wikipedia tells me that 1 Pa = 10-5 bar. So I'm guessing 0.1Pa = 1μbar .
Okay, so it's been a long time since I used dB for anything. Frankly, using dB to compare things has never made any sense to me. And with that negative thrown in there, I'm really confused. Are they telling me that the sensor provides negative voltage output, or is it something that gets plugged into the "20 log rule". (When I plug that negative into the 20 log rule, I get insanely small numbers. )
For "20 log rule", I'm looking at this on Wikipedia: en.wikipedia.org/wiki/Decibel
If I ignore the negative and just plug and chug, I get something like this for the first sensor: Pressure needed to get 1 Volt = 10150/20 = 31.6 Pa
and this for the second sensor: Pressure needed to get 1 Volt = 1070/20 = 3162 μbar = 316 Pa
They seem to be off by an order of magnitude, but ... am I doing this correctly?
thanks,
Mark
(EDIT) Well, now I'm fairly certain my approach doesn't work... I noticed other sensors, which happen to have integral pre-amps, have a sensitivity rated at -24 dB ref 1V/μbar. So what to do??? Anybody understand how this sort of specification is used? I'm trying to figure out how much voltage I can expect the sensors to output vs. some input of pressure.
Post Edited (ElectricAye) : 6/6/2009 6:16:57 PM GMT
Through my tears, I'm trying to compare two acoustic emission sensors. The first sensor has a specification that says its sensitivity is −150 dBre 1V/μPa. The other sensor says its sensitivity is -70 dB ref 1V/μbar.
Wikipedia tells me that 1 Pa = 10-5 bar. So I'm guessing 0.1Pa = 1μbar .
Okay, so it's been a long time since I used dB for anything. Frankly, using dB to compare things has never made any sense to me. And with that negative thrown in there, I'm really confused. Are they telling me that the sensor provides negative voltage output, or is it something that gets plugged into the "20 log rule". (When I plug that negative into the 20 log rule, I get insanely small numbers. )
For "20 log rule", I'm looking at this on Wikipedia: en.wikipedia.org/wiki/Decibel
If I ignore the negative and just plug and chug, I get something like this for the first sensor: Pressure needed to get 1 Volt = 10150/20 = 31.6 Pa
and this for the second sensor: Pressure needed to get 1 Volt = 1070/20 = 3162 μbar = 316 Pa
They seem to be off by an order of magnitude, but ... am I doing this correctly?
thanks,
Mark
(EDIT) Well, now I'm fairly certain my approach doesn't work... I noticed other sensors, which happen to have integral pre-amps, have a sensitivity rated at -24 dB ref 1V/μbar. So what to do??? Anybody understand how this sort of specification is used? I'm trying to figure out how much voltage I can expect the sensors to output vs. some input of pressure.
Post Edited (ElectricAye) : 6/6/2009 6:16:57 PM GMT
Comments
I'd reason like this to convert the second formula (givent in units of V/uBar) into units of V/uPa like the first one:
1 Pascal = 10 microbar. 1 microbar = 10E5 microPascal -70dB = 20 * Log V(1 uBar) / 1Volt ; voltage produced at 1 uBar ref 1 volt = 20 * Log V(10E5 uPa / 1Volt ; unit substitution = 20 * Log 10E5 * V(1 uPa) / 1Volt ; linearity = 20 * (Log 10E5 + Log V(1 uPa)) / 1Volt ; log of product is sum of logs = 100 + 20 * Log V(1 uPa) / 1Volt ; resolve 20 * Log 10E5 -170dB = 20 * Log V(1 uPa) / 1Volt ; rearrange, now commensurate unitsThat makes the two sensitivities 20dB apart, a factor of 10 in pressure ratio, in agreement with your reasoning.
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Tracy Allen
www.emesystems.com
I just saw your post. I need to think more about this. I appreciate your inputs. I have this mental block about dB that I've never been able to overcome.
thanks,
Mark