Re: Amplitude Decibel conversion problem

jungledmnc wrote:
> Hi,
> this question might be very trivial, but I haven't found good
> explanation anywhere else.
>
> I have an audio signal - some sine. Use a wavelabl to look at the
> waveform, and its maximum is at about 0.6. When I change the display
> into dB it shows about -5dB at that level.
> But this does not make sense - the formula is 10*log10(amplitude). So
> it should be -2.5dB. I have found, that this is because we use
> squares for comparison :
> 10 * log10(amp1^2 / amp2^2) = 20 * log10(amp1 / amp2)
>
> They say it is often used in electronics to compare voltages, but why
> is it used in digital audio? Yeah I know that the digital values are
> equivalent to resulting electrical power, but we are not comparing
> anything to anything. So why should we square?
>
> And if so, then when should we NOT square?
>
> Thanks a lot.
> dmnc

I checked your numbers.  You need to pay close attention to what is being 
compared when you talk about dB.  Because, dB is always a comparison of 
amplitudes.

You say "its maximum is about 0.6" but don't give any dimensions like 
"volts" for example.
Then, you don't give the reference which is usually 1 volt.... but that is

*usually* understood to mean 1vrms.  So, 1vrms would be 0dB.

From 20log(0.6) I get -4.4dB ... which really means
20log(0.6) - 20log(1.0) so the dimensions are likely mixed:
Peak voltage for the 0.6 and rms voltage for the 1.0.

Converting 0.6v peak to rms is 0.6 x 0.707 = 0.42 vrms
20log(0.42) = -7.4dB

So, I'm not sure where you are on all this with your wavelab1.