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When converting a power ratio into dB, what is the correct expression? I've seen references to: 20 * log (y2/y1) = y dBand I've also seen: 10 * log (y2/y1) = y dBWhich one is correct, and why the discrepancy? Dave, Connecticut
The Doctor Responds: dB = 10 log (P2/P1)Now, some people try to derive dB indirectly from voltage measurements. This is fine, as long as the impedance across which they're being measured is constant. In that very special case, power ratio equals voltage ratio squared. So it is mathematically correct to say that: dB = 10 log (V2/V1)2but only if impedance is constant. Finally, since the logarithm of a number is simply the power to which its base must be raised to equal that number, you can move an exponent of a log out in front. In other words, for example: log (x)y = y log (x)So we can take the "voltage dB" equation, and move the exponent 2 up front, giving us: dB = 10 log (V2/V1)2which explains the origin of the confusing "20 log" dB equation. This is a special case, that only works when you're measuring voltages, in a constant-impedance situation. And, though mathematically correct, it is conceptually flawed, and confusing, because it gets the formula too far away from its roots of "deci" and "Bel." But, it's still frequently published, and does work out mathematically. I'm going to recommend that you stick to the first formula cited above. If you have voltages, convert them to power ratio first, and then use the "official" dB equation. This will keep you out of trouble. For a more detailed treatment, see my trilogy of articles from QST, circa 1986: |
