In message , Jeff writes

Is that good? If the cable length is 1/4 wavelength (as it is in the
info provided for this case) the reflected power is nearly 180 degrees
out of phase with the initial power at the antenna. I think reducing
this through cable losses would not be so bad, or better to dump it in
the ATU?

It is good in as much as some of the re-reflected power is radiated (and
some re-re-reflected) since the ATU causes the phase of the re-reflected
wave to be 'in-phase' at the antenna.

It is bad in as much as the reflected power suffers 2 times the cable
loss, and dissipates that in heat, on each return trip, up and down the
coax.

With a high VSWR at the antenna there will be many return trips before
the re-reflected power drops to a negligible level.

No power is 'dumped' in the ATU; although there will be losses, but that
is another story.

If not the ATU, then the transmitter. I'm sure not all of the power is
reflected back from the ATU. Exactly what is the phase of the reflected
power from the ATU? I haven't seen an actual circuit for the ATU in
question. For that matter, what is the phase of the power reflected
from the antenna? I'm pretty confident we are not looking at the return
of the reflected wave in phase with the incident wave.


If the ATU is adjusted so that the Tx sees a 1:1 match then no power is
reflected to the transmitter. A 1:1 match means that there is no power
reflected to the Tx by definition (and can be proved by measurement).

By conservation of energy then all of the power must be radiated or
lost as heat, (mostly in the coax).

When the ATU gives a conjugate match, (ie the Tx sees a 1:1 vswr) the
phase of the re-reflected wave at the antenna is 0, ie it is in-phase
with the original forward wave and so any power that is not
re-re-reflected again adds to the power supplied to the antenna.

If the coax were lossless, and there were no losses in-the ATU, then
ALL of the power that was supplied by the tx would be radiated
regardless of the mismatch at coax to antenna interface.
However, in reality even small coax losses add up to a significant loss
when the mismatch at the coax to antenna interface is high due to the
number of times that the power bounces up & down the coax suffering
loss on each trip..


A concise explanation.

What is the easiest way of calculating the power loss (say, assuming
it's all in coax, and none in the ATU)? Is it simply a case of adding up
a large number of diminishing losses as the signal repeatedly rattles up
and down the coax (until it becomes so small that the losses can be
ignored), or are there some more-elegant (and accurate) methods?
--
Ian