Mike Hyder -N4NT- wrote:
Since we measure signal strength in volts and power in watts and since watts
vary with the square of the voltage, are your 6dB per S-unit dB of voltage
difference or of power difference? In other words, if I double my power am
I quadrupling my volts? Inquiring minds want to know.
Mike,
If the impedance remains the same doubling voltage quadruples power.
Either way it is a 6 dB increase. That explains why there are two
formulas for dB change, one for power ratios and one for voltage ratios:
10 log P1/P2 and 20 log V1/V2. We measure signal strength in a lot of
different ways. Probably a better way would be dBm, which is decibles
relative to 1 milliwatt. This is they way receiver sensitivities are
usually expressed, these days. Sometimes you will see specifications in
microvolts and dBm. Even when we are using 50 uV as our point of
reference (when S9 = 50 uV) the ratios relative to that reference point
are still in dB or S units, from which either power or voltage ratios
can easily be derived. S units are generally agreed to equal 6 dB. A one
S unit change would be a power ratio of 4 (or 1/4) and that would be a
voltage ratio of 2 (or 1/2). Either way it is 6 dB.
DE N6KB
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