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![]() Lord Hasenpfeffer wrote: If the WAV of a sine wave was normalized (i.e. maximum peak just below 0dB, no clipping) and the "level" of that WAV was -12dBFS, Here's were what dB is relative to begins to come into the picture. A sin wave that is normalized is defined as 0 dBFS (usually.) To clarify (or perhaps confuse) things, if you consider full digital scale to be +/- 1.0 and do the RMS calculation I mentioned on that normalized sin you will get a value of .707 (sqrt(2)/2). This means that dBFS is the average RMS value of the signal relative to .707. I.e., considering samples to have the range +/- 1, square all their values, take the average and then take the square root of that. Call that number N. Then the level of the file in dBFS would be: 20*log10(N/.707) If a signal is -12 dBFS (averaged over its whole length) then you can go backwards to the to its average RMS level as: (10^(-12/20))*.707 - .1776 dB is always 20 times log base 10 of the ratio of two things. The thing on the bottom is what it is relative to. dBu for example is voltage relative to .775 V (for historical reasons). dBV is voltage relative to 1 V. dBA is sound pressure relative to 2E-5 Pascals. would the bisector (i.e. the average level) of the sine wave be located 12 decibels below 0dB? The average level of a sin is zero and zero is -inf dB. Now you are thinking about instantaneous rather than average RMS values. An instantaneous level that is 12 dB below another is 1/4 of the other. Confused yet? :-) Bob -- "Things should be described as simply as possible, but no simpler." A. Einstein |
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