I asked myself how to compute dBFS (dB full scale) from a value of sample between 1 and -1?
and in general?
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It is really quite simple for values between 1 and -1:
valueDBFS = 20*log10(abs(value))
If you have values between another range, for example 16bit, then it's:
valueDBFS = 20*log10(abs(value)/32768)
(because 16bit signed has values between -32768 and +32767, e.g. 2^15=32768)
And so the general formula is:
valueDBFS = 20*log10(abs(value)/maxValue)
The code examples are given in MATLAB code.
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2$\begingroup$ This is technically correct, but most dBFS meters actually display headroom, so you will often want to display the negative of these values. $\endgroup$ – Bjorn Roche Apr 22 '13 at 14:41
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2$\begingroup$ According to the wikipedia article on full-scale, when using integers the positive value of the range is used for the max (i.e. 32767) and the maximum negative value (-32768) actually exceeds full-scale, which would mean the formula here should be
20*log10(abs(value)/32767)$\endgroup$ – paul Aug 27 '20 at 18:11
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All the standards define dBFS as an RMS measurement, relative to the RMS level of a full-scale sine wave, so the calculation is:
value_dBFS = 20*log10(rms(signal) * sqrt(2)) = 20*log10(rms(signal)) + 3.0103
- A full-scale sine wave is 0 dBFS
- A full-scale square wave is +3 dBFS
The similar unit dBov is defined in relation to power ratios (so it's also an RMS measurement), such that full-scale DC or square wave is 0 dBov, so that calculation is:
value_dBov = 20*log10(rms(signal))
- A full-scale sine wave is −3 dBov
- A full-scale square wave is 0 dBov
(This definition of dBFS is explicitly designed such that the dBFS value of a full-scale sine wave equals 0 (and in consequence, that of a full-scale square wave is +3 dBFS). Since the RMS of the full-scale sine wave is 1/sqrt(2), multiplying rms(signal) by sqrt(2) ensures that the formula evaluates to 0 for the full scale sine wave: 20*log10(rms(signal) * sqrt(2)) = 20*log10((1/sqrt(2)) * sqrt(2)) = 20*log10(1) = 0.)
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Relative to the RMS level of a full-scale sine wave, so the calculation is:
value_dBFS = 20*log10(rms(signal) * sqrt(2)) = 20*log10(rms(signal)) + 3.0103
Need some clarity about - > value_dBFS = 20*log10(rms(signal) * sqrt(2))
why in sin wave multiplying by sqrt(2).
Please share your valuable information.
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The reason for the sqrt(2) in the definition of dBFS given by
value_dBFS = 20*log10(rms(signal) * sqrt(2)) = 20*log10(rms(signal)) + 3.0103
is that this definition is explicitly designed such that the dBFS value of a full-scale sine wave equals 0. Since the RMS of the full-scale sine wave is 1/sqrt(2), multiplying rms(signal) by sqrt(2) ensures that the formula evaluates to 0 when signal is a full-scale sine wave:
20*log10(rms(signal) * sqrt(2)) = 20*log10((1/sqrt(2)) * sqrt(2))
= 20*log10(1)
= 0
In the more general case where the signal is in the range [-v, v], the corresponding full-scale sine wave would be the standard sine wave multiplied by v. The RMS of this sine wave is v/sqrt(2), which implies the correction factor should be sqrt(2)/v:
value_dBFS = 20*log10(rms(signal) * sqrt(2) / v)
