How to calculate +/-db as a percentage of the source signal. - diyAudio
 How to calculate +/-db as a percentage of the source signal.
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 7th October 2003, 09:58 PM #1 diyAudio Member   Join Date: Mar 2002 Location: Montreal, Quebec, Canada How to calculate +/-db as a percentage of the source signal. When attenuating a source signal, I understand that 100% of the signal = -0db, 0% of the signal = - infinity db. I wish to calculate exactly much % of the source signal is –1db, or –2db, or, -50db... What’s the formula? __________________ _______ Brian
 7th October 2003, 10:13 PM #2 diyAudio Member     Join Date: Mar 2003 Location: London, Ontario, Canada __________________ Those who claim to be making history are often the same ones repeating it.
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Location: Croatia
Re: How to calculate +/-db as a percentage of the source signal.

Quote:
 Originally posted by Brian Guralnick [B]When attenuating a source signal, I understand that 100% of the signal = -0db, 0% of the signal = - infinity db. I wish to calculate exactly much % of the source signal is –1db, or –2db, or, -50db...
Hi,

20 log x/y where y is reference level (voltage, current, ..)

10 log x/y for power

Look here
http://www.rane.com/par-d.html
"decibel" and "0dBV,u,m"

for example for voltage

-40dBV=20 log x/1V = x=0,01V or 10mV

then you can express this value in %

regards

 7th October 2003, 10:25 PM #4 diyAudio Member   Join Date: Dec 2001 Location: Germany Working alot with these values it´s nice to know some and you can easily calculate the "rest". For example +6dB means ~twice the signal, -6dB ~half the signal. So +12dB is ~4 times (2*2) the original and +18dB ~8 times(2*2*2) +20dB is 10 times and so +40dB 100 times (10*10). It´s that easy. Cheers Jens
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Join Date: Mar 2002
Re: Re: How to calculate +/-db as a percentage of the source signal.

Quote:
 Originally posted by moamps Hi, 20 log x/y where y is reference level (voltage, current, ..) 10 log x/y for power regards
Thanks a million.

So, If im reading this correctly, calculating for my linear 256 step attenuator:
------------------------------------------------------------------
a volume of 0 = ' 20*log( 0/255) ' = -(infinity) db.
a volume of 1 = ' 20*log( 1/255) ' = -48 db.
a volume of 2 = ' 20*log( 2/255) ' = -42 db.

...

a volume of 253 = ' 20*log(253/255) ' = -0.06 db.
a volume of 254 = ' 20*log(254/255) ' = -0.03 db.
a volume of 255 = ' 20*log(255/255) ' = -0 db.
------------------------------------------------------------------
Did I calculate this corectly?

I see the importance of a log pot. Arrrrrrg, I need to change the resistor wireings on my circuit...
__________________
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Brian

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