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

dB's always compare one power to another power by the formula that you
have written down.

In electronics we often do not directly measure power, but rather
voltage.  Most of our measuring equipment measures things in voltage.

So we are stuck in a position where we want to compare one power level
to another power level, but our measuring equipment measures voltages.

(Also in DSP most of the time the A/D converter is getting a voltage
equivelent input)

To properly compare one power level when we made the measurement in
voltage you have to first convert the measured voltage level into a
measured power level, and the reference voltage level into a reference
power level.

Here is where the shortcut comes in, but remember, dB's always compare
power levels.  It is the definition.

But when you are comparing voltage levels, it turns out you get to use
the trick of multiplying by 20*log(ratio) instead of 10*log(ratio),
becuase power pro****tional to the voltage squared.  so 10*log(V1^2/
V2^2) is the same as

20*log(V1/V2).  This is a shortcut or trick. Always go back to dB's
compare power per the definition formula 10*log(P1/P2).

The 20 log thing comes into play becuase in the real world you rarely
measure power directly.  You measure voltages directly, and then
apply the 20* trick to convert the voltages to power.