Bob,

To more directly answer your question...

One way to look at it is, that it is power which does the deed. To be
fair, however, you can't have power without the voltage and visa-versa.
This is when we are talking about receivers not OP amps. When a receiver
receives a signal, there is a voltage present and current flowing at its
input. They both occur. They both have to. Together these represent some
amount of power. The receiver's input stage needs some RF energy to tickle
its input so it can amplify it for the next stage. It is this power which
is the energy. Because both of these quantities are there, we can use
either to talk about the sensitivity of said receiver. Although not
"normal" we could just as easily use current as a metric for receiver
sensitivity.

Voltage is directly measurable so it is easy to use as a measure. It has
been used over the centuries (not really) as a standard way of measuring
what is happening in a circuit.

Power, on the other hand, cannot be directly measured it has to be
calculated. Power, we know, is Volts times Amps or E x I. There are other
ways to figure out power which I won't go into here, but this is how we must
be satisfied doing it.

Why, you may ask then, do we use these power numbers instead of just voltage
AND to make matters worse, why do we express them in some obscure thing like
dBm? Good question! So we Engineers can keep you Hams in a steady state of
confusion and therefore keep us in high esteem and power... Not really, but
I'll stop here to see if this helped any.

73, Steve, K9DCi




You can't have
"Robert11" wrote in message
...
Hi Roy,

Thanks for reply re my older post on db question. Very clear explanations
by all.
Will put this as a new thread, though.

New at this, and relize I'm not thinking about this the correct way,
probably.

Relative to a receiving antenna's signal: what does the receiver actually

respond to; power or voltage at its input ?

Bob

--------------------------------------------------------------


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


100 * (1 - 10^(-dB/10)) ~ 37% is the fraction of power lost.

100 * (1 - 10^(-dB/20)) ~ 21% is the fraction of voltage lost.

These assume that the coax is terminated with its characteristic
impedance.

And you don't need to put "lost" in quotation marks. It is truly lost as
a signal, having been turned into heat.

Roy Lewallen, W7EL