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[Uncle Freddie]

Hi,

I know enough audio to know that "twice as loud" is a 6 dB increase. If you add 6 dB on a fader, you have made that sound 2x as loud.

Can someone give me the right formula for calculating things like "X" as loud? Where X can be anything.

For example, a formula to calculate how many dB to add (or subtract) to make something 1.5x as loud. Or -1.7x as loud. Or any factor.


Thanks,
Fred

[wireman]

The general formula relating to your first sentence is:

dB = 20 log10(ratio)

where log10 is logarithm to base 10 (can't work out how to write that properly in the forum)

[Hugh Robjohns]

I know enough audio to know that "twice as loud" is a 6 dB increase. If you add 6 dB on a fader, you have made that sound 2x as loud.

Kind of.... You've made it twice as big... but you may not perceive that as twice as loud!

The fader calibrations refer to the electrical signal voltage inside the analogue mixing console, and a 6dB increase is a doubling of the signal voltage -- NOT the audio volume we perceive subjectively through our ears.

So while +6dB is twice the signal voltage, it won't necessarily sound twice as loud. To achieve that you'd need to raise the signal voltage by about 10dB (although the exact value depends on the frequency as well as personal subjectivity).

Can someone give me the right formula for calculating things like "X" as loud? Where X can be anything. For example, a formula to calculate how many dB to add (or subtract) to make something 1.5x as loud. Or -1.7x as loud. Or any factor.

[/quote]

Not really, no, but I can give you a formula to convert signal voltage ratios into decibels.

Decibels = 20 x log (ratio) where 'ratio' is the amount you want to increase or decrease the signal voltage. (And the log is a standard base10 logarithm)

So if you want double the voltage it's log 2 times 20 which is 6dB, and if you want half the voltage it's log 0.5 times 20 which is -6dB.

If you were working with signal powers rather than voltages, the formula would be:

Decibels = 10 x log (ratio). ... Which means a doubling of power is +3dB...

But as I say, it's important to understand that these formulae relate to signal voltages (or powers), and not how we perceive acoustic sound volumes through our ears, subjectively.

... And Wireman got there before me!

H

[wireman]

... And Wireman got there before me!

H

But I ran away from any discussion of the conventions in using dB or of Loudness, knowing that someone else would come along...

[Uncle Freddie]

Thanks everyone. Using this info, I wrote some quick shell scripts that I can use at the command line.


The first script takes the x-factor as its input and outputs the corresponding dB change:

$ dbxchange 1
1 x increase = 0 dB

$ dbxchange 1.5
1.5 x increase = 3.52182518111362 dB

$ dbxchange 2
2 x increase = 6.02059991327962 dB

$ dbxchange 1.125
1.125 x increase = 1.02305044894763 dB

$ dbxchange 0.5
0.5 x decrease = -6.02059991327962 dB

$ dbxchange 0.1
0.1 x decrease = -20 dB


The second script takes the dB as its input and outputs the corresponding x-factor change:

$ dbchangex 0
0 dB increase = 1 x

$ dbchangex 1
1 dB increase = 1.12201845430196 x

$ dbchangex -1
-1 dB decrease = 0.891250938133746 x

$ dbchangex 3
3 dB increase = 1.41253754462275 x

$ dbchangex 6
6 dB increase = 1.99526231496888 x

$ dbchangex 9
9 dB increase = 2.81838293126445 x

[Hugh Robjohns]

There's a handy calculator here too:

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

H

[Jadoube]

but it goes to eleven...