lu6etj wrote in
:

....
PS: Owen do not be upset with me :) most of my available newsgroup
time it is spent in translate to english without flaws that may induce
to misinterpretations (all of you are very demanding with precise
wording and exact definitions), one mistake and I will need three or
four painly translations more to clarify :D

Today I tested your interesting formula with a Half wave 50 ohms TL,
loaded with 100 and 25 ohms respectively. Vs=100 Vrms, Rs=50 ohm give
2:1 VSWR. In both cases then Pf 50 W, Pr=5.5 W, Pnet=44.4 W. Giving
Vf=50 V and Vr=16,6 V aprox. for both loads.

I posted a correction to the formula to properly account for the fact
that Vr is a complex quantity.

The corrected expression is Prs=|(Vs/2-Vr)|^2/Rs . Apologies if you
missed it. Of course, the V quantities are RMS, it is a bit of a botch of
RMS with phase as we often do in thinking... but I see you using the same
shorthand in your calcs. Although we are using RMS values, don't overlook
that where they add (eg Vf+Vr) you must properly account for the phase.

For your cases:

For the 25+j0 load, Vr=-16.67+j0, so Prs=|50--16.67|^2/50=88.9W.

For the 100+j0 load, Vr=16.67+j0, so Prs=|50-16.67|^2/50=22.2W.

Of course, for a 50+j0 load, Vr=0, so Prs=|50-0|^2/50=50.0W.

As you can see, the first two cases have the same |Vr| (though different
phase), the same 'reflected power', and yet Prs is very different.

Consider an extreme case, Zl=1e6, VSWR is extreme, almost infinity, Vref=
50+j0, so Prs=|50-50|^2/50=0.0W. Here, your 'reflected power' is as large
as it gets, but the power dissipated in the source is zero.

The notion that reflected power is simply and always absorbed in the real
source resistance is quite wrong. Sure you can build special cases where
that might happen, but there is more to it. Thinking of the reflected
wave as 'reflected power' leads to some of the misconception.

Owen