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dB From Voltages
Although dB is technically a ratio of power levels, it is commonly used with voltages by assuming that both voltages are driving a constant resistive load, such that:
power = voltage^2 / resistance
So if we have 2 different voltages V1 and V0 and resistance R, the R drops out:
dB = 10 * log10((V1^2 / R) / (V0^2 / R)) = 10 * log10(V1^2 / V0^2) = 10 * log10(V1 / V0)^2
Then using the property that the square of a log is equal to twice the log:
dB = 10 * 2 * log10(V1 / V0) = 20 * log10(V1 / V0)
We can rearrange the dB formula to get the voltage ratio from a specified dB value:
V1 / V0 = 10^(dB / 20)
This assumption of constant resistance may not always be true, such as when the voltage is driving a complex load like a loudspeaker whose impedance changes with frequency. But it is still proper, for example, for a manufacturer to assume constant resistance when discussing sound card amplifier performance, since there are no standards for loudspeaker impedance versus frequency... and there are a lot of different loudspeakers on the market.
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