Summing and averaging RMS pressure (amplitude) of sound waves

In summary, summing and averaging RMS (Root Mean Square) pressure of sound waves involves calculating the effective amplitude of sound by taking the square root of the average of the squares of pressure values over time. This method provides a more accurate representation of perceived loudness compared to simple arithmetic averaging, as it accounts for the energy of the sound waves. RMS pressure is crucial in various applications, including audio engineering and acoustics, to quantify sound intensity and ensure consistent sound quality.
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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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  • #2
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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  • #3
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.👍
 

FAQ: Summing and averaging RMS pressure (amplitude) of sound waves

What is RMS pressure in the context of sound waves?

RMS (Root Mean Square) pressure is a statistical measure of the magnitude of a varying quantity and is used to quantify the amplitude of sound waves. It provides a single value that represents the effective pressure of the sound wave over time, which is useful for comparing different sounds and their potential impact on the environment or human hearing.

How do you calculate the RMS pressure of a sound wave?

To calculate the RMS pressure of a sound wave, you take the square root of the mean of the squares of the instantaneous pressure values over a given time period. Mathematically, it is expressed as:\[ P_{RMS} = \sqrt{\frac{1}{T} \int_{0}^{T} p(t)^2 \, dt} \]where \( p(t) \) is the instantaneous pressure at time \( t \), and \( T \) is the total time period over which the measurement is taken.

Why is RMS pressure used instead of peak pressure in sound measurements?

RMS pressure is used instead of peak pressure because it provides a more accurate representation of the sound wave's overall energy and its potential impact. Peak pressure only measures the maximum instantaneous pressure, which can be misleading as it does not account for the duration and distribution of the pressure variations. RMS pressure, on the other hand, averages these variations, giving a better sense of the sound wave's true power.

How do you sum RMS pressures from multiple sound sources?

Summing RMS pressures from multiple sound sources is not a straightforward addition because RMS values represent energy. Instead, you sum the squared pressures (which are proportional to the sound power) and then take the square root of the total. If \( P_{RMS1} \) and \( P_{RMS2} \) are the RMS pressures of two sound sources, the combined RMS pressure \( P_{RMS_{total}} \) is given by:\[ P_{RMS_{total}} = \sqrt{P_{RMS1}^2 + P_{RMS2}^2} \]

What is the difference between summing and averaging RMS pressures?

Summing RMS pressures involves combining the energy contributions from multiple sound sources, as described previously. Averaging RMS pressures, on the other hand, involves calculating the mean RMS pressure over multiple measurements or time periods. For example, if you have \( n \) RMS pressure values, the average RMS pressure \( \overline{P_{RMS}} \) is given by:\[ \overline{P_{RMS}} = \frac{1}{n} \sum_{i=1}^{n} P_{RMS_i} \]Averaging is useful for

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