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Adding multiple decibels?

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  • Adding multiple decibels?


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Answer #1 | 14/01 2014 09:47
Well, if they were all in phase, then you would have the addition of 10 million sourcesof sound. Now recall that sound intensity in dB is given by: I(dB) = 10*log(p/p0) where p = pressure of sound wave and p0 = ambient pressure so I in decimal notation (Id) is: Id = 10^(I(dB)/10) If there are 10 million sources of sound, each with Id of intensity, the sound you get is: It = 10^7 * Id (10^7 = 10,000,000) going back to dB It(dB) = 10 log(It) = 10*log(10^7*Id) = 70 + 10*log(Id) = 70 + I(dB) = 135 dB using I(dB) = 65 dB
Positive: 50 %
Answer #2 | 14/01 2014 13:57
The sound sources will never be in phase to each other. All noise sources are incoherent. Given: Number of equal loud sound sources n = 10,000,000 (ten millions). Formulas: Δ L = 10 × log n or n = 10^(ΔL/10) Δ L = level difference; n = number of equal loud sound sources. n = 2 equally loud non-coherent sound sources result in a higher level of 10 × log (2) = +3.01 dB compared to the case that only one source is available. n = 3 equally loud non-coherent sound sources result in a higher level of 10 × log (3) = +4.77 dB compared to the case that only one source is available. n = 4 equally loud non-coherent sound sources result in a higher level of 10 × log (4) = +6.02 dB compared to the case that only one source is available. n =10 equally loud non-coherent sound sources result in a higher level of 10 × log (10) = +10.00 dB compared to the case that only one source is available n =10,000,000 equally loud non-coherent sound sources result in a higher level of 10 × log (100000000) = +70.00 dB compared to the case that only one source is available. If one source has a level of 65 dB then the total noise sum of the 10 million sources is 65 dB + 70 dB = 135 dB. "Total level adding of incoherent sound sources": http://www.sengpielaudio.com/calculator-leveladding.htm Cheers ebs
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