In message
,
Cecil Moore writes
On Sep 1, 10:29 am, Ian Jackson
wrote:
Oh, indeed. But I have to confess that I've realised that I'm suddenly
unsure of where, on a long, lossy, feeder with a mismatched load, most
of the absolute power is actually lost. I'm assuming that it is at the
TX end, where the power is greatest - even though that's where the SWR
is best. Please advise!

It seems logical that the highest I^2*R losses would be where the
standing-wave current is maximum and the highest dielectric losses
wold be where the the standing-wave voltage is maximum. As Owen is
fond of pointing out, the locations of those points are very
important. If a current maximum point exists at the load and a voltage
maximum point exists at the source, the losses at the load are
probably higher than the losses at the source on HF. If a voltage
maximum point exists at the load and a current maximum point exists at
the source, the losses at the load are probably lower than the losses
at the source on HF (assuming that losses due to SWR are mostly I^2*R
losses on HF).
--
The SWR on a feeder will always be greatest at the load end. However,
because of the feeder losses, the power will also be lowest at the load
end.

So, although the power loss per unit length (in dB) due to SWR will
steadily increase as you approach the load end, because of the feeder
loss, there will be less absolute power (in watts) to lose as you
approach the load end.

The question is, at which end of the feeder is the most absolute power
lost per unit length?
--
Ian