It's true in the case of transmission lines that there are standing
waves and reflections but, unfortunately, this concept has somehow
come to dominate and confuse the concept of matching a tranmitter to
an antenna -- a generator to a load. Like in many areas of science,
mathematicians and scientists often find convenient ways to
mathematically describe and predict physical phenomenon that hinders,
even misleads, the understanding of how it actually works.

If you leave out the complex part of impedences for the moment and
think of 100 volt generator that has a 50 ohm internal impedance
driving a 50 ohm load, current is 1 amp and the power dissipated by
the load is 50 watts. There is also 50 watts dissipated by the
generator's internal impedance, for a total of 100 watts dissipated by
the entire system. Therefore, the "available" power for this generator
is 50 watts.

Maximum available I^2*R power only occurs when the load impedance is
equal to the generator's characteristic impedance, 50 ohms (do the
math). Any load impedance higher or lower, ALWAYS produces less
"available" power.

Herein lies one of the big problems with the "reflection" definition,
conceptually. The generator (transmitter) is not a constant-power
device. When a manufacturer says that it's XYZ transmitter produces
100 watts, it only produces (has available) 100 watts (after internal
dissipation) into a 50 ohm load. Any other load *always* produces less
available power, due to simple I^2*R laws. It has nothing to do with
reflections or standing waves, although, mathmatically, reflection
formulas accurately describe it.

A couple of examples using a 100 volt constant voltage generator and
an internal impedance (RG) of 50 ohms:

1) Load (RL) = 50 ohms. Current = 100 / (50 + 50) = 1 amp
Power dissipated in RL (PL) = (1)^2*50 = 50 watts
Power dissipated in RG (PG) = (1)^2*50 = 50 watts
SWR = RL/RG = PL/PG = 1:1

2) Load (RL) = 100 ohms. Current = 100 / (50 + 100) = .667 amp
Power dissipated in RL (PL) = (.667)^2*100 = 44.5 watts
Power dissipated in RG (PG) = (.667)^2*50 = 22.25 watts
SWR = RL/RG = PL/PG = 2:1

3) Load (RL) = 25 ohms. Current = 100 / (50 + 25) = 1.34 amp
Power dissipated in RL (PL) = (1.34)^2*25 = 44.9 watts
Power dissipated in RG (PG) = (1.34)^2*50 = 89.8 watts
SWR = RG/RL = PG/PL = 2:1

Notice that the total power dissipated in all three examples is
different. The transmitter is NOT a constant-power source, but it's
also not a unlimited power source and has operational limits.

Therefore, what is commonly called "reflected power" is power that
never leaves the transmitter and is dissipated as heat by the
transmitter's internal 50 ohm impedance (if the transmitter's design
doesn't prematurely shut down first).

Al


"Henry Kolesnik" wrote in message ...
I know that any power not dissipated by an antenna is reflected back to the
transmitter. Then the transmitter "reflects" this reflection back to
antenna, ad nauseum until its all gone. I also know that a short or an open
is required to reflect power and I'm searching for which it is, an open or a
short. I'm inclined to think it's a virtual open but I'm at a loss to
understand that and I wonder if someone has a good explanation or analogy
and some math wouldn't hurt.
tnx
Hank WD5JFR