
Home  Forums  Rules  Articles  diyAudio Store  Gallery  Wiki  Blogs  Register  Donations  FAQ  Calendar  Search  Today's Posts  Mark Forums Read  Search 
Pass Labs This forum is dedicated to Pass Labs discussion. 

Please consider donating to help us continue to serve you.
Ads on/off / Custom Title / More PMs / More album space / Advanced printing & mass image saving 

Thread Tools  Search this Thread 
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. 
7th October 2003, 10:14 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, 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 
7th October 2003, 11:18 PM  #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...
__________________
_______ Brian 

Thread Tools  Search this Thread 


Similar Threads  
Thread  Thread Starter  Forum  Replies  Last Post 
Signal Detecting Source Selector  error401  Digital Source  12  16th October 2007 02:42 PM 
what is the percentage of current across Ground ?  Chris Daly  Power Supplies  3  24th April 2007 06:33 PM 
twisted pair signal, source selector and ground buss???  cbutterworth  Tubes / Valves  2  30th March 2007 07:39 AM 
Creating a differential signal from a singleended signal for bridged amps  Peter Daniel  Chip Amps  45  8th November 2004 12:47 PM 
Noise figure vs signal source resistances in JFETS datasheet  rlim  Solid State  25  19th December 2003 07:11 PM 
New To Site?  Need Help? 