On Mar 21, 4:25 pm, Dan Bloomquist wrote:
Keith Dysart wrote:
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.
***

So you don't like my example? Well I can use yours then.

While you use the word 'power', the real analysis in your example is
all
done with volts. This is excellent and helps demonstrate my point that
'reverse power' is not needed as an explanation. We can carry on from
the analysis you have done and compute some powers. The real power at
(c)
is 0 Watts (the voltage is 0 at all times so using P=VI, the power
must
be zero). Assuming that when you say "drive 5 volts", you mean that
the
voltage source in the Thevenin equivalent generator is set to 10 V,
the 'forward power' at (c) is 0.5 W and the 'reverse power' is 0.5 W.
When subtracted, these produce the expected result of 0 W which agrees
with the actual computed power. All is well. (And yes, the Bird works
for determining this result).

Now consider someone who believes in the reality of 'forward' and
'reflected power'. There is 0.5 W of 'forward power' which reaches
the generator at the right. Since the impedance of this generator is
the
same as the characteristic impedance of the line (or the left
generator
in this example because there is no line), there is no reflection
so the 'power' must go into the generator. Similarly with no
reflection,
the 'reverse power' goes into the generator. Where does this power go?
If 'reverse power' is real, then it needs to be accounted. Since there
are many examples for which this accounting can not be done
it leads to the inescapable conclusion that 'reverse power' does not
really exist. (Please though, this does not mean that forward and
reverse voltages and currents do not exist. The fundamental error
being committed is that it is not valid to multiply the voltages and
currents computed using superposition to produce powers that actually
represent flowing energy).

Any reader is invited to prove me wrong by providing an accounting for
the 'reverse power' when it reaches the generator whose output
impedance
matches the characteristic impedance of the line (i.e. no reflection).

....Keith