On Mar 20, 10:02*am, Cecil Moore wrote:
Keith Dysart wrote:
Sentence one says "no limitations". Sentence two specifies a
limitation.

Semantic games.

The whole question here revolves around the meaning of the
limitations on your claim. That is semantics. Not games.
And it is key to the discussion.

There are no limitations within the
stated boundary conditions just like any number of
other concepts.

But you need to clearly state your limitations and stop
flip flopping.

Could you clarify whether your claim for the circuit of
Fig 1-1 applies to all time, or just to those instances
when the source voltage is 0.

There are no claims regarding instantaneous power in
Fig 1-1 or anywhere else in my web article. All the
claims in my web article refer to *average* powers and
the article states exactly that. For the purpose and
subject of the web article, the subject of instantaneous
power is just an irrelevant diversion. No other author
on the subject has ever mentioned instantaneous power.

I am surprised, this being 2008, that I could actually be
offering a new way to study the question, but if you insist,
I accept the accolade.

Given average values, time doesn't even appear in any
of their equations. Apparently, those authors agree
with Eugene Hecht that instantaneous power is "of
limited utility".

Here's my claim made in the article: When the *average*
interference at the source resistor is zero, the *average*
reflected power is 100% dissipated in the source resistor.
I gave enough examples to prove that claim to be true.
Since the instantaneous interference averages out to
zero, this claim about *average* power is valid.

Without prejudice to the accuracy of the above, let us
explore a bit.

We know that conservation of energy requires that the
energy flows balance at all times, which means that at
any instance, the flows must account for all the energy.

Analysis has shown that when examined with fine granularity,
that for the circuit of Fig 1-1, the energy in the reflected
wave is not always dissipated in the source resistor. For
example, some of the time it is absorbed in the source.

Now when the instantaneous flows are averaged, it is true
that the increase in dissipation is numerically equal to
the average power for the reflected wave. But this does
not mean that the energy in the reflecte wave is dissipated
in the source resistor, merely that the averages are equal.

Now you qualify your claim with the term "*average* power".
You say "the *average* reflected power is 100% dissipated in
the source resistor."

But the actual energy in the reflected wave is not dissipated
in the source resistor. So what does it mean to say that the
*average* is?

When Tom, K7ITM, asserted that the same concepts work
for instantaneous power, I took a look and realized that
he was right. One can make the same claim about instantaneous
power although I do not make that claim in my web article.

When the instantaneous interference at the source resistor
is zero, the instantaneous reflected power is 100% dissipated
in the source resistor.

Perhaps this has clarified. So you are only claiming that
reflected energy is dissipated in the source resistor at
those instances when the source voltage is zero. Good.

I am claiming no such thing. Please cease and desist with
the mind fornication, Keith. You cannot win the argument
by being unethical.

Unfortunately you struck the sentence which I paraphrased
and then failed to explain which parts of it I may have
misinterpreted. That does not help. It would have been
more valuable for you to rewrite the original sentence
to increase its clarity.

Mostly to prove that my analysis has an error.

I have pointed out your error multiple times before, Keith,
and you simply ignore what I say. Why should I waste any
more time on someone who refuses to listen?

One more time:
Over and over,

I only have two expression involving power.

you use the equation Ptot = P1 + P2 even
though every sophomore EE student knows the equation is
(usually) invalid. The valid method for adding AC powers is:

Ptot = P1 + P2 + 2*SQRT(P1*P2)cos(A)

Which of the two needs the 'cos' term?

Ps(t) = Prs(t) + Pg(t)
or
Pg(t) = Pf.g(t) + Pr.g(t)

In fact neither do. They both stand quite well on their own.
Both are so correct, that they apply for any voltage function
provided by the source. What would you propose to use for cos(A)
when the source voltage function is aperiodic pulses? Fortunately,
the 'cos' term is not needed so the question is completely moot.

Non-the-less do feel free to offer corrected expression that include
the 'cos(A)' term.

The last term is called the interference term which you
have completely ignored in your analysis. Therefore, your
analysis is obviously in error. When you redo your math
to include the interference term, your conceptual blunders
will disappear. Until then, you are just blowing smoke.

The math holds as it is. But I invite you to offer an alternative
analysis that includes cos(A) terms. We can see how it holds up.

If you can't, then you should definitely reconsider who is
'blowing smoke'.

...Keith