On Fri, 10 Feb 2006 18:46:13 -0500, "Robert11"
wrote:

Hello:

When a table gives the attenuation (at some freq.) for a length of a
particular coax in db,
are they referring to db in voltage or db in power ?

Is is correct to ask a question as simply as the following:

If the attenuation is given as, e.g., 2 db, what Percentage therefore of a
received signal is "lost"
going thru the coax length ?

Decibels are a means of expressing a power ratio when used properly.
Of course, the power ratio is implied by a voltage or current ratio
when the impedance is the same in the two cases being compared. For
that reason, you can say that the power ratio in db is 10*log(P1/P2)
or 20*log(V1/V2) provided Z remains constant.

Nevertheless, voltage or current ratios are often compared using dB
where the Z is not the same, and stricly speaking, the power ratio
rquires an adjustment for the changed impedance.

If the loss is 2dB, the the ratio of power out to power in is
10^(-2/10) or 63%. If the impedance is the same, the ratio of the
voltages will be the square root of 63% or 79%.

Owen
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