I Summing and averaging RMS pressure (amplitude) of sound waves

AI Thread Summary
To sum uncorrelated sound pressures, convert dB levels to pressure ratios and sum the square roots of their squares, which aligns with RMS pressure calculations. For averaging, take the RMS of the two RMS pressures, ensuring the correct conversion back to dB using the 20 log(ratio) equation. The discussion emphasizes that adding logarithmic values directly does not yield the correct result for sound signals; instead, linear scales should be used before converting back to logarithmic form. It is noted that when dealing with uncorrelated signals, the effective sum can be approximated by adding their powers. Understanding these principles is crucial for accurate sound pressure level calculations.
ngn
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TL;DR Summary
Want to check my understanding of summing and averaging RMS pressure (or RMS amplitude) of sound waves
Hello,
I've been trying to wrap my head around why, if given the sound pressure levels (dB1 and dB2) of two uncorrelated sounds, if you want to sum them together, you sum their intensities using the equation: 10 x log10(10^dB1/10 + 10^dB2/10). Likewise, if you want to average them, you average their intensities with the equation: 10 x log10([10^dB1/10 + 10^dB2/10] / 2). I know that the dB is a measure of the relative difference in power, but I've been struggling with how the summing and averaging would work if you converted the dB1 and dB2 back to pressure ratios and summed there and then used the 20xlog10(ratio) equation. I think I understand now, but I wanted to check.

SPL is calculated on RMS pressure. So, here are my two questions:

1. If I converted dB1 and dB2 back to their pressure ratios, in order to sum them, I would have to sum the square root of their squares? In other words, I would sum the square root of p1^2 + p2^2? Is that the right way to sum RMS pressures together (rather than just a straight sum of the pressures). When done that way, I can use the 20 x log(sum) and it converts back into dB correctly in line with the intensity equation above.

2. In order to average two RMS pressures together, I would take the RMS of those two RMS pressures? In other words it would be RMS(p1, p2). If that is my average, then when I use the 20 log(ratio) equation to convert that RMS value back into dB, it comes out right.

So, am I correct with my thinking? RMS pressures are summed by summing the square root of their squares. And RMS pressures are averaged by taking the RMS of these RMS values?

Thank you! This board has been a great help so far and I appreciate all of the feedback!
 
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1. To find the mean intensity you should add the two RMS pressures and then use 10 log (ratio).
2. I am not sure why you want to average two RMS pressures. If you imagine the two waves arriving at the ear drum, the two are added.
 
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ngn said:
TL;DR Summary: Want to check my understanding of summing and averaging RMS pressure (or RMS amplitude) of sound waves

So, am I correct with my thinking? RMS pressures are summed by summing the square root of their squares.
Ultimately you would like to add values of the pressures as they vary over time and then find their net RMS (resultant pressure). That info is probably not available but when the two sources are uncorrelated you are unlikely to have too many coincident peaks (random noise signals, for instance) so you can find the effective sum by just adding the powers. If you add two sine waves of equal amplitude, the resultant power will be four times. So you can only get an approximate answer with real signals that have any correlation between them

But you have got the message ok that adding logs is never the way to add signals - adding logs gives you the log of a product :headbang: . You convert to linear scales first and back to log scales at the end.👍
 
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