Roy Lewallen wrote:
*Sigh*

The same misconceptions keep coming up, as they have countless times on
this newsgroup and I'm sure they will for decades or perhaps centuries
to come. After one of the many previous discussions, I wrote a little
tutorial on the topic. Originally in the form of plain text files, I've
combined it into a pdf file for easier viewing. You can find it at
http://eznec.com/misc/Food_for_thought.pdf.

On page 8 you'll find the statement "THE REVERSE POWER IS NOT DISSIPATED
IN OR ABSORBED BY THE SOURCE RESISTANCE". Above it is a chart of several
examples which clearly show that there's no relationship between the
"reverse power" and the source dissipation. The remainder of the
tutorial explains why.

Any theory about "forward" and "reverse" power, what they do, and their
interaction with the source, will have to explain the values in the
example chart on page 8. Does yours?

Mine does. All of your values can be understood by looking
at the destructive and constructive interference and applying
the irradiance (power density) equations from the field of
optics. You see, optical engineers and physicists don't have
the luxury of measuring voltage and current in their EM waves.
All they can measure is power density and interference and
thus their entire body of knowledge of EM waves rests upon
measurements of those quantities. Those power density and
interference theories and equations are directly applicable
and 100% compatible with RF theories and equations. Any
analysis based on power density and interference will yield
identical results to the ones you reported in your "food for
thought" article which includes the following false statement:

"While the nature of the voltage and current waves when
encountering an impedance discontinuity is well understood,
we're lacking a model of what happens to this "reverse power"
we've calculated."

We are not lacking a model of what happens to this
"reverse power" we've calculated. The model is explained
fully in "Optics", by Hecht. When one has standing waves of
light in free space, it is hard to hide the details under
the transmission line rug.

In general, it is just as easy, and sometimes easier, to deal
with the energy values and then calculate voltage and current
as it is to start with voltage and current and then calculate
the power.

All this is explained in my WorldRadio article at:

http://www.w5dxp.com/energy.htm

The great majority of amateur antenna systems are Z0-matched.
For such systems, an energy analysis is definitely easier to
perform than a voltage analysis. Here's an example:

100W------50 ohm---+---Z050 ohms-----load
Pfor1=100W-- Pfor2--
--Pref1=0W --Pref2

The power reflection coefficient is 0.5 at point '+'.
The power reflection coefficient is 0.5 at the load.

What are the values of Pfor2 and Pref2? What is the physics
equation governing what happens to Pref2 at point '+'?
--
73, Cecil http://www.w5dxp.com