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Is it accurate to say in digital audio that, when a fader is down, then its value is "-$\infty$"?

What has confused me is, whether it's really $-\infty$ or some lower bound of the amplitude. it could be that the $-\infty$ is just a "joke" by the developers and it doesn't have mathematical validity.

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    $\begingroup$ I would associate that position with 0 rather than minus infinity. A fader denotes amplification, or in mathematical terms, multiplication (by the value of the fader). Multiplication by minus infinity would involve reversal of phase and infinite multiplication at which point the console is blowing the fuse... I wonder if this is the fader of a specific parameter though. Where did you see it? BTW, this is a rare occurence but this sounds more like something we could have answered in chat rather than here where it feels a bit off-topic $\endgroup$ – A_A Aug 9 '18 at 13:38
  • $\begingroup$ @A_A It's the full scale. 0dB corresponds to full amplitude and $\min_{a} -a$ is the lowest value representable in the used bit depth. $\endgroup$ – mavavilj Aug 9 '18 at 13:39
  • $\begingroup$ @mavavilj 0 dB corresponds to unity gain. A fader can amplify or attenuate, so they typically go from −∞ dB to +12 dB or so, with a mark at 0 dB. $\endgroup$ – endolith Aug 9 '18 at 17:49
  • $\begingroup$ @A_A The minus infinity is in decibels, so you're not multiplying by infinity; you're multiplying by 0. Negative decibels are attenuation, not polarity flip. (and positive decibels are gain.) Polarity flip cannot be represented in decibels. $\endgroup$ – endolith Aug 9 '18 at 17:51
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    $\begingroup$ @endolith Thanks, you are right and so is Jojek. I wasn't thinking about dB at all at that point. $\endgroup$ – A_A Aug 9 '18 at 18:41
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If by "its" you mean the value of the fader, then yes - it's absolutely correct. Fader defines the attenuation of the signal with respect to the reference level. The units are in logarithmic scale.

For example, no amplification (or in other words, multiplication of the signal by 1) corresponds to $0\; \mathrm{dB}$ since $10\log_{10} 1 = 0$.

If you want to completely attenuate your signal, then you have to multiply everything by $0$, hence $10\log_{10} 0 = -\infty$.

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  • $\begingroup$ or, stated differently: $$ 10^{\frac{-\infty}{10}}=0$$ so the gain coefficient is $0$. and when you multiply any sample by $0$, you get $0$ as a result. and $0$ for a slew of audio samples sound like silence. $\endgroup$ – robert bristow-johnson Aug 10 '18 at 16:58
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Maybe a table of decibels vs multiplication factors will help:

\begin{array} {|r|r|r|} \hline +12\ \textrm{dB} & 4× & \textrm{amplification}\\ \hline +6\ \textrm{dB} & 2× & \textrm{amplification}\\ \hline 0\ \textrm{dB} & 1× & \textrm{unity gain, no change}\\ \hline -6\ \textrm{dB} & \frac 1 2× & \textrm{attenuation}\\ \hline -20\ \textrm{dB} & \frac 1 {10}× & \textrm{attenuation}\\ \hline -\infty\ \textrm{dB} & 0× & \textrm{silence}\\ \hline \end{array}

So if you set the fader to +6 dB, you're doubling the height of the waves. If you set it to −6 dB you're halving the height of the waves, etc.

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