# Mathmatics of dbFS

trying to understand how to calculate the difference between two sounds. for example, if the first sound is -.1dbFS and 2nd sound is -.01dBFS how much louder is the second sound.

The second track is 0.09 dB higher than the first track.

To calculate the difference, just subtract.

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sorry I meant how much louder is it on a linear scale not in dbFS

Itâs simple subtraction - The difference is 0.09dB. (An unnoticeable difference.)

According to my handy-dandy spreadsheet, itâs an amplitude factor (voltage or a digital value) of about 1.01, or about 1% higher.

dB = 20 x log(A/Aref), where A is amplitude (a voltage or a digital value).
Or, dB = 10 x log(P/Pref), where P is power (wattage) (1)

Voltage ratio = 10 to the power of (dB/20)
Power ratio = 10 to the power of (dB/10)

(1) There are different formulas for power because power is calculated as Voltage X Current. And current is proportional to voltage. For example, if you double the voltage you also double the current for 4 times the power. The decibels are the same, itâs just a different formula.

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Was there a specific job that needed to know this?

Are you designing a microphone preamplifier? They deal with values down there.

Koz

P.S.

Some more numbers for youâŚ

With integer audio, 0dBFS is the highest you can âcount toâ with given number of bits. 16-bits can hold values between â32,768 and +32,767. (1)

A wave (or single sample) that hits those values on the negative & positive peaks has a peak of 0dBFS.

Back to playing around with my spreadsheet, and considering only 16-bit positive valuesâŚ

0dB = 32,767
-0.09dB = 32,429
-0.1dB = 32,392

Internally, Audacity uses floating point where 1.0 represents 0dBFS.

You can print the sample values to a text file with Tools â Sample Data Export. But remember that you are (usually) looking at a waveform with (typically) 44,100 samples-per-second so itâs too much data to visualize or comprehend.

(1) Negative numbers can count one-higher because in the binary format there is no negative zero.

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no Im not designing a pre amplifier

thank you for the in depth information its a lot to take in but I think I can figure it out. Would you be able to send the spread sheet you speak of?

Thanks again,

I canât attach the spreadsheet but here are the formulas. Same as above but in âspreadsheet formatâ:

=20*LOG(A4) Gives dB where cell A4 contains the voltage (or digital amplitude) gain, factor or ratio.
Enter 2 and you should get about +6dB. Enter 0.5 and you should get about -6dB.

=10*LOG(D4) Gives dB where D4 contains the power factor or ratio.
Enter 2 and you should about +3dB. Enter 0.5 and you should get about -3dB.

=POWER(10,A8/20) Gives the voltage gain or ratio where A8 contains the dB (or dB difference from the 0dB reference).
Enter +6 and you should get about 2. Enter -6 and you should get about 0.5.

=POWER(10,A11/10) Gives the power ratio where A11 contains dB.
Enter +3 and you should get about 2. Enter -3 and you should gat about 0.5

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