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 Post subject: Addition of logarithms
PostPosted: Wed, 8 Dec 2010 17:22:30 UTC 
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Dear All,

Could someone out there explain what the mechanics of the addition of logs (log10) are - I use logs (as in the decibel scale) for acoustic assessments in relation to planning applications (ie a consultant submits a report to show a new development would not cause any problems related to noise).

I recently recieved a report from an acoustic consultant who had taken the noise from a piece of equipment and predicted it's noise level at a specific distance and attenuated the noise for distance, barrier effects etc.. After the attenuations to the noise source, at the specific distance the predicted noise was a negative log. In virtually every circumstance, no human can hear anything below 0dB, however it is possible for some noises to be below this (as the 0dB relates to human perception of noise).

Their argument for leaving in the negative values was spurious and simply stated "...this would be to falsely represent the level at the receptor due to this equipment." I know this is wrong and tried to argue the point with them but would have loved to back it up with mathematical reasoning.

In particular, when we add noise levels together (as they are not simply additive) we use:

Lp = 10*log10(10^(L1/10) + 10^(Ln/10))

where L is the noise source in decibels

However, using this it is possible to add -1dB and -1dB and get 2dB - which in terms of acoustics is nonsense as (and I said to the consultant) you cannot add two inaudible sounds and get and audible one.

Can someone explain to me (in layman's terms) whether this is a particular quirk of logs and or this method of addition?

It was actually much worse than this as there were serveral pieces of equipment, some with positive and negative dBs, so effectively he was using the negative dBs to "pull" the overall noise rating up (although mathematically this only changed the final answer by 0.1dB).

Many thanks

Gary


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 Post subject: Re: Addition of logarithms
PostPosted: Wed, 8 Dec 2010 17:49:19 UTC 
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Gary Park wrote:
Lp = 10*log10(10^(L1/10) + 10^(Ln/10))


You're certain this is the equation?

That translates to:

10\log_{10}(10^{L_1\over 10}+10^{L_n\over 10})

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 Post subject: re: Addition of logarithms
PostPosted: Mon, 13 Dec 2010 14:53:21 UTC 
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Yes that's the equation - as shown at:

http://www.sengpielaudio.com/calculator-spl.htm

This is the "standard" method for addition of noise sources.

Cheers

Gary


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 Post subject: Re: re: Addition of logarithms
PostPosted: Mon, 13 Dec 2010 20:00:59 UTC 
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Gary Park wrote:
Yes that's the equation - as shown at:

http://www.sengpielaudio.com/calculator-spl.htm

This is the "standard" method for addition of noise sources.

Cheers

Gary


You're almost right. If something is 0 decibels then the first term is a 1, which means no matter how soft the second sound is, you'll end up with the log of a number >1, which is automatically a positive number.

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