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Old March 23rd 07, 03:05 PM posted to rec.radio.amateur.antenna
Dan Bloomquist Dan Bloomquist is offline
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First recorded activity by RadioBanter: Mar 2007
Posts: 23
Default Revisiting the Power Explanation

Keith Dysart wrote:

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....


So you don't like my example?


Your example assumes that the reflected power will see the 50 ohms of
the generator. And I had shown you a condition where it will see a short
no mater what the 'output' impedance of the generator.

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).


It is a short. It is just that simple. If you have any kind of lab, take
a long piece of coax and drive it with a pulse and watch with a scope.
Find out for yourself that energy will reflect off a short.

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.


No. (c) is a short. It is just that simple. The power does not flow past
it 'into the other generator'. If you have any doubt that power is
reflected from a short you have not made the observation. And if you
have not made an observation that contradicts this, you can not make the
claim of a contradiction.

snip assumptions based on a false premise

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).


If you stick to the premise that the reflected power sees the generator
impedance it can't be 'proven wrong'. I can say that gravity doesn't
exist all day long but that won't make it so. The observation will
'prove' me wrong.