Pat Farrell wrote:
> Technically:
> Say you have a signal with 100dB of signal and 30dB of noise.
> SNR is 20 log10(S/n)
> where S is signal and n is noise, in volts, or volume.
> SNR == 20 log10( 100/30)
> == 20 log10 (3.33)
> == 20 * 0.5228
> == 10.45
>
> Crank the sound down to 50% of what it was so 100dB becomes 94 dB.
> SNR == 20 log10( 94/30)
> == 20 log10 (3.133)
> == 20 * 0.4960
> == 9.920
>
> It is interesting to note that cutting the voltage (which roughly
> approximates loudness) in half, reduces the power needed by a factor of
> four.
Actually, SNR is the power ratio of the signal to the noise, i.e.
SNR = ( Ps / Pn )
Where Ps and Pn are the power levels of the signal and noise respectively.
Since this is a power ratio, the decibel equivalent is:
SNR = 10 log10 ( Ps / Pn )
In this case, you don't know the power levels since you only have the
values in dB. Fortunately, that's not a problem since you don't need to
take the log when the values are in dB already, assuming both signal and
noise levels are equivalent, i.e. measured relative to the same
reference power level. i.e.
Ls = 10 log10 ( Ps / Pr )
Ln = 10 log10 ( Pn / Pr )
Where Ls and Ln are the signal levels in dB, and Pr is the reference
power level.
Doing some re-arranging:
Ps / Pr = 10^(Ls/10)
Pn / Pr = 10^(Ln/10)
SNR = Ps / Pn = 10^(Ls/10) / 10^(Ln/10)
In DB,
SNR = 10 log10 (Ps / Pn) = Ls - Ln
So, with 100dB of signal and 30dB of noise the SNR is 100-30 = 70dB.
Reduce the signal to 94dB and the SNr becomes 94-30 = 64dB
R.
_______________________________________________
audiophiles mailing list
[email protected]
http://lists.slimdevices.com/lists/listinfo/audiophiles
