XPost: rec.audio.pro, comp.dsp   
   From: spam@spam.com   
      
   On Mon, 22 Nov 2010 12:41:10 +0000,   
   adrian@poppyrecords.invalid.invalid (Adrian Tuddenham) wrote:   
      
   >davew  wrote:   
   >   
   >> On Nov 22, 7:14 am, PStamler  wrote:   
   >[...]   
   >> > 1V into 1k, at the input, means 1mA, so the power is 1mW. 10V into   
   >> > 100k at the output means 0.1mA, so the power is 1mW again. There's 0dB   
   >> > power gain, but there's voltage gain of 10x, which is coded +20dB in   
   >> > the voltage-gain realm of decibel calculation. This deviates from the   
   >> > "real" standard of what decibels are, by divorcing the voltage and   
   >> > power gains, but using the same unit for them, dB.   
   >> But, if I were to double the input voltage I would see a 6dB increase   
   >> in power output would I not?  In context, the use of dB for voltage   
   >> gain is entirely correct and acceptable.   
   >   
   >It can be used that way as long as the two voltages being compared are   
   >across the same impedance and both relate to the power in the same way.   
   >This was the case in your example.   
   >   
   >You can compare voltages at two different places as long as the   
   >impedances at those places are the same.  So an amplifier working with   
   >600-ohm input and output terminations could have its power gain   
   >expressed in dB by just measuring the voltage gain.   
   >   
   >What you must not do (which is all-too frequently done) is to use the   
   >same calculation to compare voltages in different impedances.  For that,   
   >you must use the voltages to calculate the power at each point and then   
   >compare the powers to get a result in dB.  If you want to use a   
   >logarithmic scale to compare voltages, call it something different (dBu,   
   >dBV) because it is not "dB".   
   >   
   >The convenient shortcut dB formula relating voltages is often taught   
   >with insufficient emphasis being placed on the conditions for which it   
   >is valid.  After a while, people begin to believe that the voltage   
   >formula is the true representation of dB until a situation arises where   
   >it generates nonsense results - then they get confused and sometimes   
   >adamant.  At least one well-known textbook has got it wrong.   
      
   The full equation is   
      
   20 log(v2/v1 *sqrt(z1/z2))   
      
   That gets the power gain right for any impedance. (V1 is across   
   impedance z1, and v2 is across z2).   
      
   d   
      
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