#25




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According properties of logarithm, division is equivalent to subtraction. We see this because dividing two exponential of the same base is equal to subtracting the exponents, that is (C^A)/(C^B)=C^(AB). Thus my math IS correct: 20*log10((P_out/P_ref) / (V_out/V_ref)) = 20*log10(P_out/P_ref)  20*log10(V_out/V_ref) = X dB  Y dB = output dB. This is also why I said i was off by a "factor of 2" because i'm used to working with power as milliwatts and converting it to dBm. To convert X milliwatts to dBm it's 10*log10(mW_power/1mW). Often when dealing with voltages it's assumed that the reference has the same impedance and the measurement that was taken so it's converted to power: P = V^2/Z (As it's been said on this thread may times) 10*log((V_out^2/Z) / (V_ref^2/Z)) we see that the impedance cancels out 10*log((V_out^2) / (V_ref^2)) the squares can be pushed out: 10*log((V_out/V_ref)^2) exponential in logs becomes multiplications and tadum: 20*log(V_out/V_ref) So to clarify my point of view on "dB/V" further, I did not know if dB/V was a relationship of sound pressure in dB to volts or if it was a relationship of sound pressure in pascals to volts. I assumed that it was sound pressure in dB to volts which because I read the units "dB/V" as "dB per Volt" which is incorrect because it's actually describing dB of Pascals per 1V. Last edited by ninjackn; 10262009 at 02:30 AM. Reason: Clarifying that P is pressure and not power. 
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amplifier, how to, testing 
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