On Mar 17, 10:05*am, Cecil Moore wrote:
Keith Dysart wrote:
Regardless, if you use the snippet above to support
your claim, you have effectively modified your claim.

False. My claim is what it has always been which is:
An amateur radio antenna system obeys the conservation
of energy principle and abides by the principles of
superposition (including interference) and the wave
reflection model.

Well who could argue with that. But don't all systems
obey? And why limit it to amateur?

Everything I have claimed falls out from those principles.

But the question then becomes "Have the prinicples been
correctly applied?"

Your claims, however, are in direct violation of the
principles of superposition and of the wave reflection
model, e.g. waves smart enough to decide to be reflected
when the physical reflection coefficient is 0.0. Your
claims even violate the principles of AC circuit theory,
e.g. a reactance doesn't store energy and deliver it back
to the system at a later time in the same cycle.

An intriguing set of assertions. It would be good if you
could point out the equations that are in violation.

Is there an error in either
Ps(t) = Prs(t) + Pg(t)
or
Pg(t) = Pf.g(t) + Pr.g(t)

Both by derivation and by example, these seem to be true.

Do you disagree?

You are now saying that the reflected energy is
dissipated in the source resistor only at particular
times, such as when the source voltage is 0, and that
you are not interested in what happens during the rest
of the cycle.

You haven't read my article yet, have you? Here's a quote:
"For this *special case*, it is obvious that the reflected energy
from the load is flowing through the source resistor, RS, and
is being dissipated there.

I would suggest that it is not 'obvious'. Some times the
energy from the reflected wave is being absorbed in the
source.

'Obvious' is often an excuse for an absence of rigour.

But remember, we chose a special
case (resistive RL and 1/8 wavelength feedline) in order to
make that statement true and it is *usually not true* in the
general case."

If there is one case where your assertion is wrong, then
your assertion is false. I found that special case when
the source voltage is zero that makes your assertions
false.

Which assertion is wrong?
Ps(t) = Prs(t) + Pg(t) ?
Pg(t) = Pf.g(t) + Pr.g(t) ?

For example, if you were to do the same analysis,
except do it for t such that wt equals 100 degrees,
instead of 90, you would find that you need more terms
than just Pr.g to make the energy flows balance.

Yes, you have realized that destructive and constructive
interference energy must be accounted for to balance the
energy equations. I have been telling you that for weeks.

You do use the words, but do not offer any equations that
describe the behaviour. The claim is thus quite weak.

I repeat: The *only time* that reflected energy is 100%
dissipated in the source resistor is when the two component
voltages satisfy the condition: (V1^2 + V2^2) = (V1 + V2)^2.
None of your examples have satisfied that necessary condition.
All it takes is one case to prove the following assertion false:

"Reflected energy is *always* re-reflected from the
source and redistributed back toward the load."

I have never made that claim. So if that is your concern, I agree
completely that it is false.

My claim is that "the energy in the reflected wave can not usually be
accounted for in the source resistor dissipation, and that this is
especially so for the example you have offerred".

You appear to think that if you can find many cases where
an assertion is true, then you can simply ignore the cases
where it is not true. I have presented some special cases
where it is not true. It may be true for 99.9% of cases,
but that nagging 0.1% makes the statement false overall.

Having never claimed the truth of the statement, I have
never attempted to prove it.

Equally false is the assertion: "Reflected energy is
always dissipated in the source resistor." The amount
of reflected energy dissipated in the source resistor
can vary from 0% to 100% depending upon network
conditions. That statement has been in my article from
the beginning.

But you have claimed special cases where the energy
*is* dissipated in the source resistor.

Ps(t) = Prs(t) + Pg(t)
and
Pg(t) = Pf.g(t) + Pr.g(t)
demonstrate this to be false for the example you have offerred.

So are you now agreeing that it is not the energy
in the reflected wave that accounts for the
difference in the heating of the source resistor
but, rather, the energy stored and returned from
the line, i.e. Pg(t)?

I have already presented a case where there is *zero*
power dissipated in the source resistor in the presence
of reflected energy so your statement is obviously just
false innuendo, something I have come to expect from
you when you lose an argument.

Since it was a question, you were being invited to clarify
your position. But it begs the question:
When zero energy is being dissipated in the source resistor
in the presence of reflected energy, where does that reflected
energy go?

Recalling that by substitution
Ps(t) = Prs(t) + Pf.g(t) + Pr.g(t)
so that when the resistor dissipation is 0 for certain values of t
Ps(t) = 0 + Pf.g(t) + Pr.g(t)
= Pf.g(t) + Pr.g(t)
Pr.g(t) = Ps(t) - Pf.g(t)

Please provide equations that describe how the reflected energy is
split between the source and the forward wave.
For completeness, these equations should generalize to describe
the splitting of the energy for all values of t, not just those
where the dissipation in the source resistor is 0.

...Keith