Re: Amplitude Decibel conversion problem

On 7 Okt, 20:52, "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) = 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?

Fred Marshall already answered this, so you have seen t before,
but the amp2 in your formula above is a reference level. The dB
numbers are expressed relative to that number.

Lots of people are not aware of that fact, so one often hears
that 'some sonar is so loud it is like a jet fighter at full
blast.' Well, once you dip your head under the sea surface
you experience 130-140 dB of sound.

Such numbers occur because the air measurements are relative
to 20 micropascal wheras the water numbers are relative
to 1 micropascal.

> And if so, then when should we NOT square?

Look at the reference unit. If it is a linear unit, don't
square. If it's a quadratic unit, square.

Rune