7
$\begingroup$

I wonder why several RMS meters I'm using for music production show RMS values above 0.

For example, when using an input signal with a constant DC of 1 (0 dbFS), I get these results, although Wikipedia says that in such cases, the RMS is equal to the amplitude of the direct current.

Samplitude Pro X (RMS: 2.9)

enter image description here

TT Dynamic Range Meter (RMS 3.0)

enter image description here

Whereas other meters show the values I expect:

Voxengo SPAN (RMS: 0.0)

enter image description here

Why is it that meters come to such different results?

$\endgroup$
9
$\begingroup$

0 dBFS refers to a full-scale sine wave or more rarely a full-scale square wave.

ITU-T telephony Recs. P.381 and P.382 and reportedly the paywalled digital audio device measurement standards AES17-1998 and IEC 61606 define that full-scale sine wave is 0 dBFS. ITU-T Rec. P.10/G.100 (11/2017) does not make the distinction and defines dBfs (sic) as "relative power level expressed in decibels, referred to the maximum possible digital level (full scale)".

Full scale DC has the same RMS value as a full-scale square wave, but is 3.010299956 dBFS RMS if the 0 dBFS reference is a full-scale sine wave. Looks like the different programs use a different reference (square or sine). Samplitude Pro X might use some approximation of RMS that gives in this case a 0.1 dB unit error.

The calculations:

$$\int_0^{2\pi}\frac{\sin(x)^2dx}{2\pi} = \frac{1}{2}$$ $$10\log_{10}\left(\frac{1}{2}\right) \approx -3.010299956$$

$\endgroup$
3
  • 3
    $\begingroup$ All the standards I've read agree that FS sine is 0 dBFS, and that FS square is +3 dBFS, so it seems it's always an RMS measurement (never peak) and there's no ambiguity. dBov is similar but 3 dB lower, so FS sine is -3 dBov and FS square is 0 dBov. $\endgroup$
    – endolith
    Aug 1 '18 at 16:29
  • 3
    $\begingroup$ @endolith Adobe Audition 3 has RMS Settings: 0 dB = FS Sine Wave or 0 dB = FS Square Wave. They don't directly call the measure dBFS. $\endgroup$ Aug 4 '18 at 7:37
  • $\begingroup$ Yes, Audition was my first introduction to this concept $\endgroup$
    – endolith
    Aug 4 '18 at 13:53

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.