Solstace Posted September 4, 2012 Share Posted September 4, 2012 Hi everyone Here's an interesting thing to ponder for those who have a background of Acoustics training - looking for a sanity check I think? I've been working on an enclosure project to reduce noise in our venue from a particular projector. As part of this I'm in the middle of interpreting both theoretical modelling of the enclosure I designed and the real measurement data I've collected after the build. The levels are relatively low, especially in higher frequencies, and in some cases real-world measurements were not possible, leaving "gaps" in the measured data. For the purposes of the project I think I've got enough of a handle on things to draw sensible conclusions, but some finer details have been troubling me and, well, I need to share them so I can sleep at night! 1) When converting, say, 1/3-octave readings to whole-octave values... if those first three values happen to be 0dB, then (logarithmically) adding those three 0dB SPL values together becomes something like: 10*log((10^(0/10))+((10^(0/10))+((10^(0/10))), giving the final answer of 4.771dB SPL. This makes sense to me given that the "0dB" is a reference to an actual energy value, eg, the lowest amount of energy (or pressure difference) considered perceivable by human ears. That there is some energy at 0dB means, to me, that there is something tangible to add, and the result of the example logarithmic addition makes sense.2) When dealing with weighted figures, we could well end up dealing with negative dB values. For example, in a 1/3-octave RTA situation, we may have measured or calculated 34.629dB SPL (unweighted) at 31.5Hz. To get an A-weighted value, we arithmetically subtract the relevant value for that frequency from the measured value: 34.629-39.4 = -4.771dB. Now, if our measured unweighted 1/3-octave values happened to give A-weighted values of -4.771dB, -4.771dB and -4.771dB respectively, then I would expect (for finding their equivalent A-weighted whole-octave-band value) to add those together logarithmically and get 0dB - again, there is surely some real and tangible energy to add together, but it just happens to be below the lowest threshold of hearing.3) Is there any practical way that we could express true "zero" energy using the logarithmic figures? I'm guessing in practise this might be expressed on paper as something like -∞dB (SPL)... does that work?4) Has anyone found a way to express this kind of value in an Excel spreadsheet? Or do I simply need to fudge it by using a silly-small value like -192dB (SPL) so that the figure is so small it doesn't adversely affect calculations, but fills the cell value with *something* so that, say, averaging functions work correctly? In some of my sheets where negative values end up being created I'm encountering a number of #NUM and #DIV/0 errors in some cells that I'd like to make proper sense of. I realise that some of those finer points are probably comically insignificant for most, if not all real-world measurements, but I'm checking my understanding, both for the pleasure of knowing I (might) actually understand something, and more pragmatically so that I can error-check the results of my own or others' calculations and not end up looking/feeling stupid for believing values that might actually be utter nonsense! Thoughts? Link to comment Share on other sites More sharing options...
Brian Posted September 4, 2012 Share Posted September 4, 2012 Being pedantic but '0dB' is almost meaningless as it's a dimensionless number, it's just a ratio. Without saying what it's relative to you can't really do anything with it. Link to comment Share on other sites More sharing options...
Shez Posted September 4, 2012 Share Posted September 4, 2012 Could you use Pascals (or maybe micro Pascals) throughout the spreadsheet and just turn back to dBspl at the end? The range of numbers is bigger but I'm sure Excel's precision will be able to cope with them. That way, you'll always have a real 0 reference that you can't drop below. Link to comment Share on other sites More sharing options...
Simon Lewis Posted September 4, 2012 Share Posted September 4, 2012 0dB isn't zero energy.... (it's a pressure of course!) 2x10^-5 Pa. It's OK to show a figure as -dB (A) SPL since that simply says that it is a lower pressure than 20uPa? Link to comment Share on other sites More sharing options...
Simon Lewis Posted September 4, 2012 Share Posted September 4, 2012 Just to add... if the pressure was 2 microPascals, then the SPL would be: 20*LOG((2*10^-6)/(2*10^-5)) = -20dB If you were doing your disco underwater, the reference pressure is 1uPa, not 20uPa... Link to comment Share on other sites More sharing options...
Dmills Posted September 8, 2012 Share Posted September 8, 2012 It is generally best to convert to a linear amplitude before doing the summing,but remember that for uncorrelated noise the average level should be the square root of the sum of the squares of the average levels where for correlated signals it is simply the sum.... Converting from multiple per band figures to a single wideband number is not trivial, especially when you consider that peak levels rise faster then average ones when summing noise (But that some of the components will have correlations). The major gotcha with water is that it has a much higher characteristic acoustic impedance then air (1.5M Rayls IIRC), so the relationship between power and dbSPL is rather ah different (200dB spl in water is not that big a deal!). Regards, Dan. Link to comment Share on other sites More sharing options...
rolinger Posted September 10, 2012 Share Posted September 10, 2012 The major gotcha with water is that it has a much higher characteristic acoustic impedance then air (1.5M Rayls IIRC), so the relationship between power and dbSPL is rather ah different (200dB spl in water is not that big a deal!). 200dB re 1uPa in water is quite a big deal. In shallow water, it will be cavitating at that point (i.e. pressure getting so low that it boils and forms bubbles), which is a problem. But yes, you have to get used to a different range of 'normal' dBs! Link to comment Share on other sites More sharing options...
Dmills Posted September 11, 2012 Share Posted September 11, 2012 I can get 200dB re 1uPa @ 1M in water with a fairly simple piezo ceramic ring with about 800V RMS across it at 15Khz (Nearly 1KVA input according to the GB plot), granted it is 5M down, which limits cavitation, but that transducer is not even all that efficient as these things go, this in a fairly simple noise maker. Doing the same thing at low frequency is harder of course, and you can end up playing with large numbers of very large tonpils for things like sub bottom work at a Khz or so. I got out of theatre for a living in the end, guess what the new job is? Regards, Dan. Link to comment Share on other sites More sharing options...
dbuckley Posted September 11, 2012 Share Posted September 11, 2012 Sonar? Link to comment Share on other sites More sharing options...
Dmills Posted September 11, 2012 Share Posted September 11, 2012 Well, underwater transducer electronics anyway, I have yet to do an imaging project. I even go to the odd production as a punter again, it had got to the point where just walking into a theatre felt like work (When you start feeling that way, you HAVE to go do something unrelated), I may be back in the game one day, but for now I am seeking a amateur company that does something more interesting then endless badly done Priestly, Miller and Ayckbourn to keep my hand in. Regards, Dan. Link to comment Share on other sites More sharing options...
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