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7th October 2003, 10:58 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?
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7th October 2003, 11:13 PM  #2 
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Join Date: Mar 2003
Location: London, Ontario, Canada

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7th October 2003, 11:14 PM  #3  
diyAudio Member
Join Date: Oct 2002
Location: Croatia

Re: How to calculate +/db as a percentage of the source signal.
Quote:
20 log x/y where y is reference level (voltage, current, ..) 10 log x/y for power Look here http://www.rane.com/pard.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, 11:25 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 
8th October 2003, 12:18 AM  #5  
diyAudio Member
Join Date: Mar 2002
Location: Montreal, Quebec, Canada

Re: Re: How to calculate +/db as a percentage of the source signal.
Quote:
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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