Keith Dysart wrote:
I then restated your claim as applying only at those particular
times and not at other points in the cycle, but you were
unhappy with that limitation.

There are no limitations. If zero interference exists,
then 100% of the reflected energy is dissipated in
the source resistor.
--
73, Cecil http://www.w5dxp.com

I have consolidated three replies below...

On Mar 20, 12:29*am, Cecil Moore wrote:
Keith Dysart wrote:
I then restated your claim as applying only at those particular
times and not at other points in the cycle, but you were
unhappy with that limitation.

There are no limitations. If zero interference exists,
then 100% of the reflected energy is dissipated in
the source resistor.

Sentence one says "no limitations". Sentence two specifies a
limitation. But your paper did provide that limitation and
indicated that circuit (Fig 1-1) with a 45 degree line was
an example which satisfied that limitation.

But in subsequent discussion you have waffled about whether,
for the circuit in Fig 1-1, "100% of the reflected energy is
dissipated in the source resistor" is applicable for all
time or only for those instances when the source voltage is
equal to 0.

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.

On Mar 20, 12:34 am, Cecil Moore wrote:
Keith Dysart wrote:
If you wish to have your equalities apply at only selected
points within the cycle, that works for me.

Not only at selected points within the cycle but
also for average values. If zero average interference
exists then 100% of the average reflected energy is
dissipated in the source resistor which is the subject
of my Part 1 article. If the instantaneous interference
is zero, 100% of the instantaneous reflected power is
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.

Now as to averages: Averaging is a mathematical operation
applied to the signal which reduces information. I do agree
that the increase in the average dissipation in the source
resistor is numerically equal to the average value of the
reflected power. But this is just numerical equivalency.
It does not prove that the energy in the reflected wave
is dissipated in the source resistor. To prove the latter,
one must show that the energy in the reflected wave, on
an instance by instance basis is dissipated in the source
resistor because conservation of energy applies at the
instantaneous level.

And I have shown in an evaluation of the instantaneous
energy flows that the energy dissipated in the source
resistor is not the energy from the reflected wave.

I repeat: When zero interference exists, 100% of the
reflected energy is dissipated in the source resistor.

But only at those instances where the source voltage is
zero.

On Mar 20, 12:50 am, Cecil Moore wrote:
Keith Dysart wrote:
But I notice that you have not yet indicated which
energy equation I may have written that was unbalanced.

Why should I waste my time finding your conservation of
energy violations?

Mostly to prove that my analysis has an error.

I repeat: When there exists zero interference, 100%
of the reflected energy is dissipated in the source
resistor. Since you think you provided an example where
that statement is not true, your example violates the
conservation of energy principle.

But if there is no error in my analysis (and you have not
found one), then perhaps you should examine whether the
clause "When there exists zero interference, 100% of the
reflected energy is dissipated in the source resistor"
is in error. Sometimes one has to re-examine one's deeply
held beliefs in the light of new evidence. It is the only
rational thing to do.

...Keith

Keith Dysart wrote:
Sentence one says "no limitations". Sentence two specifies a
limitation.

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

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.
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.

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.

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, 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)

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.
--
73, Cecil http://www.w5dxp.com

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

Keith Dysart wrote:
But you need to clearly state your limitations and stop
flip flopping.

What you are calling "flip flopping" is me correcting
my errors. Once I correct an error, I don't flip-flop
back. My error was in assuming that the power-density
(irradiance) equation only works on average powers.
K7ITM convinced me that it works on instantaneous powers
also.

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.

I'm sure you are not the first, just the first to think
there is anything valid to be learned by considering
instantaneous power to be important. Everyone except
you discarded that notion a long time ago.

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.

Yes, yes, yes, now you are starting to get it. When
interference is present, the energy in the reflected
wave is NOT dissipated in the source resistor. Those
facts will be covered in Part 2 & 3 of my web article.

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.

For instantaneous values of voltage, the phase angle is
either 0 or 180 degrees so the cosine term is either +1
or -1. The math is perfectly consistent. That you don't
recognize the sign of the instantaneous interference term
as the cosine of 0 or 180 is amazing.

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

I did and you ignored it. There is no negative sign in the
power equation yet you come up with negative signs. That
you don't recognize your negative sign as cos(180) is
unbelievable.

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.

It it unfortunate that you don't comprehend that the cos(0) is
+1 and the cos(180) is -1. The sign of the instantaneous
interference term is the cosine term. The math certainly does
hold as it is - you are just ignorant of the "is" part.
--
73, Cecil http://www.w5dxp.com

On Mar 21, 12:37*pm, Cecil Moore wrote:
Keith Dysart wrote:
But you need to clearly state your limitations and stop
flip flopping.

What you are calling "flip flopping" is me correcting
my errors. Once I correct an error, I don't flip-flop
back.

Actually the flip-flopping I was referring to was the
constant changes in your view of the limitations that
apply to your claim.

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.

I'm sure you are not the first, just the first to think
there is anything valid to be learned by considering
instantaneous power to be important. Everyone except
you discarded that notion a long time ago.

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.

Yes, yes, yes, now you are starting to get it.

Is this a flop or a flip? Are you now agreeing that the
energy in the reflected wave is only dissipated in the
source resistor for those instances when Vs is 0?

When
interference is present, the energy in the reflected
wave is NOT dissipated in the source resistor. Those
facts will be covered in Part 2 & 3 of my web article.

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.

For instantaneous values of voltage, the phase angle is
either 0 or 180 degrees so the cosine term is either +1
or -1. The math is perfectly consistent.
[gratuitous insult snipped]

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

I did and you ignored it.

I could not find them in the archive. Could you kindly provide
them again, showing where the 'cos(A)' term fits in the
equations:
Ps(t) = Prs(t) + Pg(t)
and
Pg(t) = Pf.g(t) + Pr.g(t)

There is no negative sign in the
power equation yet you come up with negative signs.
[gratuitous insult snipped]

Negative signs also arise when one rearranges equations.

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.
[gratuitous insult snipped]

...Keith

Keith Dysart wrote:
Actually the flip-flopping I was referring to was the
constant changes in your view of the limitations that
apply to your claim.

As I said, I corrected an error in my thinking. You
are free to consider that to be a flip-flop, I consider
it to be a step forward.

Are you now agreeing that the
energy in the reflected wave is only dissipated in the
source resistor for those instances when Vs is 0?

Yes, for instantaneous reflected energy exactly as I
previously stated, but not true for average reflected
energy. 100% of the average reflected energy is
dissipated in the source resistor when the transmission
line is 45 degrees long. That intra-cycle interference
exists, thus delaying the dissipation by 90 degrees,
is irrelevant to where the net energy winds up going.

Your conservation of power principle would have you
demanding that the power sourced by a battery charger
must be instantaneously dissipated. Everyone except
you seems to realize that is an invalid concept. The
dissipation of energy can be delayed by a battery
or a reactance. In the present example, the dissipation
of the instantaneous reflected energy is delayed by
90 degrees by the reactance.

I could not find them in the archive. Could you kindly provide
them again, showing where the 'cos(A)' term fits in the
equations:

Go back and read the part where I said cos(0)=+1 and
cos(180)=-1. There is no such thing as conservation of
power. Your equations assume a conservation of power
principle that doesn't exist in reality. The forward
power is positive power. The reflected power is positive
power. The only negative power is destructive interference
which must be offset by an equal magnitude of constructive
interference. When the instantaneous interference power is
negative, the two voltages are 180 degrees out of phase and
that is your cos(180)=-1. When the instantaneous interference
power is positive, the two voltages are in phase and that
is your cos(0)=+1. Why don't you already know all this
elementary stuff?
--
73, Cecil http://www.w5dxp.com