You're correct, of course. In this instance, a given level
was used as a reference, and an attenuator was used to drop
the signal to the various readings. Voltage or power doesn't
matter, because a 10 dB attenuator will attenuate voltage 10 dB
and power 10 dB.
de KL7HF
>Oh, oh, here we go again. When I was in engineering school, the
>definition of dB was that it is 10 times the log(base 10) of the
>power ratio. For voltage change ratio in the circuit, it is
>20 times the log(base 10) of the voltage ratio, and the answer is
>the same numeric number of dB's. Because, by definition a dB is
>a power ratio change with respect to whereever P1 is measured.
>
>If so, then the above statement that 25 dB power ratio is the
>same (in the same circuit) as a 50 dB voltage ratio is not true.
>A dB is a dB is a dB (as is a rose a rose).
>
>You can, of course, express voltage ratios in logarithmic terms,
>to make big ratios smaller, and small ratios bigger which is part of
>the point of logarithmic arithmetic( to eliminate lots of zeros both before
>or after the decimal point), but you must specify Dbv, not just dB. And
>the arithmetic of dBv must also be specified, that is dBv= what times
>the log of the voltage ratio?
>
>At least, I think that is what I learned! Please help me if I
>am wrong.
>
>73, Jim, KH7M
>
>
>
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