Re: Amplitude Decibel conversion problem

On Oct 7, 2:52=A0pm, "jungledmnc" <jungled...@[EMAIL PROTECTED]
> 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) =3D 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
anythi=
ng
> to anything. So why should we square?
>
> And if so, then when should we NOT square?
>
> Thanks a lot.
> dmnc


Hello dmnc,


dB =3D 10*log(power/power_reference)

now since power is pro****tional to amplitude squared then

dB =3D 20*log(amplitude/amplitude_reference)

With dB we are really always talking about power relative to a
reference power.

Even in acoustics, we have 100 dB is 0.946 watts per square meter.

I hope this helps.

Clay