Keith Dysart wrote:

On Mar 21, 1:32 pm, Dan Bloomquist wrote:


Belief has nothing to do with it. Observations suffice. If you have any
doubt that EM radiation can be reflected, just look in a mirror. Did you
need to 'believe' you saw your reflection to confirm your observation?


I have yet to question the reflection of EM radiation, just the
existence of
"reverse power" in transmission lines.

Use of a TDR makes for a valid observation on the line.

Red Herring. There is no 'failure' in the accounting of reflected power.
The observations and numbers work out quite nicely.

A simple example that I can never make add up is a 50 Watt generator
with a 50 ohm output impedance, driving a 50 ohm line which is open at
the end. Using the "reverse power" explanation, 50 W of "forward
power"
from the generator is reflected at the open end, providing 50 W of
"reverse
power". Since the generator is matched to the line there is no
reflection
when this "reverse power" reaches the generator so it disappears into
the generator....

I explained this last week, albeit for a different reason. I'll paste:
***
Hi Richard,
He says it in the last sentence. But here is an example. Take a 50 ohm
thevinin source. Power off, it looks like 50 ohms back into it. Take a
second thevinin source to represent a reflection and drive 5 volts into
the first source. Now set your first source 180 degrees to the
reflection and drive forward 5 volts.

(s)-----/\/\/\--------(c)-----/\/\/\--------(r)

(s)source (c)connection (r)reflection. With (s) 180 degrees out of phase
from (r), (r) will see a short at (c). It is because of the power
generated at the source that the impedance into it can look purely
reactive. And, you can use 5 ohms with 1 volt at the source, (c) will
still look like a short to (r). The source resistance doesn't matter as
long as a 'match' is made.

And for the same reason, why the 50 ohm line doesn't look like 50 ohms
is because of reflected power. Drive an open quarter wave line and it
looks like a short because the reflected voltage is 180 degrees out from
the source.
***

If this is truly power, it must go somewhere else, be
dissipated, transformed into some other form or stored (based on the
conservation of energy principle). Where did it go?

The energy is sitting on the line. It didn't disappear. See all the
posts I made last week.


Most correspondents agree that what happens depends on the design
of generator;

Actually, it doesn't. The exception is minor and pointed out in the case
of a non linear source.

dissipation either increases, decreases or stays the
same (compared to when the line was terminated in 50 Ohms and the
power going down the line is dissipated in the termination). This
does
not make an easy explanation for where that supposedly real power
goes. Of course, if it is not real power, then there is no issue,
which
leads one back to looking for explanations other than "reverse power".

The mistake is in assuming the energy must disappear. It doesn't.

...Keith

Best, Dan.