Adding & averaging decibel values

Adding or averaging decibel values seems like a closed book. But the calculation with these hacks shown here is very easy to do with a calculator or in your head.
In words:
Addtion of 2 decibel values:
* Divide the decibel values by 10.
* Then calculate 10 to the power of this number - then add them together.
Now back to the logarithmic system:
* Calculate the log of the sum. Now we have "Bel"
* Multiply the result by 10 and we get the deci-Bel value .
Average of decibel values: Minute: 4:44
The same as Addition, but before you the log you first have to divide the summation of all summands by the number of summands. (This is the way you calculate an average in any case)
This is part 3 of a decibel explaining series.
Links to the other tutorials of this series:
Part 1 - Why decibels in acoustics - Acoustics Basics (Recommended)
• Why decibels in acoust...
Part 2 - Handling decibel values with ease
• Handling decibel value...
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Part 4 Working with decibels - Analysis Software
• Working with decibels ...

[@mhediseltana]

Very clear and instructive. Thanks for sharing. Hope you make a video about different DB scale.

[@nasturt2]

I had absolutely no idea Thomas Müller had a side hustle as an acoustic engineer 😄

[@HEADacousticsInternational]

That is where his passion lies. Football is just about earning some extra money.

[@nasturt2]

@@HEADacousticsInternational 🤣 great content in any case!

[@supperKingofworld]

This was really fun to watch. Like you saying come on do it in your head! that got me going, you couldve showed the paper for a few seconds longer and wrote a little neater but this really helped and wasnt boring to watch

[@HEADacousticsInternational]

Thanks for your detailed feedback. Feel invited to watch the other three videos in this series as well. Then you will have a good overview of the handling of acoustic measurement and analysis methods. You can find the links in the description box (More...)

[@dayekh]

thank you this was very helpful

[@HEADacousticsInternational]

You are welcome.

[@efti06]

thank you so much! this helped me so much

[@HEADacousticsInternational]

We like to read that. You are welcome.

[@brokenpromises3056]

what a great video thanks a lot 💖

[@HEADacousticsInternational]

Thanks. You are welcome!

[@paulgundo4755]

Thank you so much, I was having alot of difficulties as it was my first time doing this kind of calculations consisting of dB thanks alot

[@HEADacousticsInternational]

We are happy to help with any questions about NVH and acoustics

[@paulgundo4755]

I would like to ask why do we multiply by half when we are calculating the average between two decibels, can you please help me understand

[@HEADacousticsInternational]

A mean value is calculated by adding up the values of the individual summands and dividing the total sum by the number of summands. In the example calculation shown here, the mean value of two values is calculated. Therefore we divide here the sum by the number 2. With three values we would divide the sum afterwards by three. After that, the value is converted back to the logarithmic scaling. Please also have a look at the preceding video "Handling decibel values with ease".

[@paulgundo4755]

@@HEADacousticsInternational thank you so much and I will watch the following vid

[@OkanTandogan]

Thank you very much for these useful videos. Just a quick question, usually when it is said like "addition of dB" it is perceived as addition of Sound Pressure Levels. I think that, here we are not reffering to the addition of two sound pressure waves, but addition of two sound power, right (as the sound source)? So consequently, the created sound pressure level from this sound source at a given location r will also be increased as 3dB once the another source is added; is that right?

[@HEADacousticsInternational]

Hello Okan - long time - no see 😀. When talking about „addition of dB”, we usually want to know the resulting Sound Pressure Level (SPL) in dB, when two sound sources (with individual SPL) are summed. So we are in fact interested in the SPL of the sum of the sound pressure waves. However, the sound pressure waves should not be confused with the root mean square (RMS) sound pressure, which is used for calculation of sound pressure. To avoid confusion let’s use some variables: s1(t) is the sound pressure wave of the first sound, p1 is the RMS sound pressure and L1 is the SPL in dB. s2(t) is the sound pressure wave of the second sound, p2 is the RMS sound pressure and L2 is the SPL in dB. Note that L1 is calculated from p1 as L1 = 20*log10(p1/p0) (p0 is the reference sound pressure) Lets call this the equation for SPL using sound pressure. This is mathematically equivalent to calculating L1 = 10*log10(p1^2/p0^2). Lets call this the equation for SPL using sound power. We are now interested in the SPL L3 of the signal s3(t) = s1(t)+s(2). So this is the addition of the sound pressure waves. However, it is not correct to assume that this means that the RMS sound pressure p3 can be calculated as sum of p1 and p2. Usually we assume incoherent sources. In this case we can add the sound powers which means we can calculate p3^2 = p1^2+p2^2. With this equation we can easily calculate L3 = 10*log10(p3^2/p0^2). By substituting p3^2 by p1^2+p2^2 and solving the equation for SPL using sound power for p1^2/p0^2, we get L3 = 10*log10(10^(L1/10)+ 10^(L2/10)). This is the equation which is used in the video. So the short answer to the question: In the video we are referring to the addition of two sound pressure waves, but not addition of the RMS sound pressure of the waves. Instead we are in fact using the addition of sound power. This means, that we are assuming incoherent sources, which migth should have been mentioned in the video 🤔. This assumption is mostly valid in acoustics, if the two sounds are not emitted by the same source. The second question “So consequently, the created sound pressure level from this sound source at a given location r will also be increased as 3dB once the another source is added; is that right?” can not be answered without knowledge of the spatial positions of the sound sources and the definition of the position of r. If the two sources are close to each other and we are looking at a distance r in far field (where the two sources can be interpreted as one point source), the assumption is correct.

[@user-ey2jy7ok3p]

Thanks for the very informative and well explained video. Just one question though: why at 9'40, when measuring level vs time, no speaking equals to 50 dB? Is it an average over the duration of the measurement (with the loud speaking that reached 70-80 dB during the few seconds, then no noise for another few seconds would be half of that maximum) and eventually over time no speaking would end up dropping to 0 dB? Or is it that 0 dB should have been calibrated beforehand? Or is it something else?...

[@HEADacousticsInternational]

The device in this setup displays the current measured value - not an average value. The reason for the 50 dB is that ambient noise such as the ventilation system is humming quietly at a low level. People quickly get used to this and block it out in their perception. The decibel value does not do this. The dB(A) weighting is often used for such quiet background noises. This would reduce the level in the frequency range by a good 35 dB. If we set the device to dB(A), it would display 20 dB(A) as the current quiet level. Measuring 0 dB is hardly achievable in practice. We recommend using the loudness and many irritations are gone. Please have a look at the following video part 4. There these effects are explained in detail.

[@severinlutz953]

the fact that 0 + 0 makes 3dB makes me so mad

[@HEADacousticsInternational]

Yes, that's weird. That's why we did the math again with the calculator 😀. 1+1=4 is also surprising