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On Mon, 17 Mar 2008 03:00:10 -0700 (PDT)
Keith Dysart wrote:

On Mar 17, 1:06*am, Cecil Moore wrote:
Keith Dysart wrote:
Pr.g(t) = -50 + 50 *cos(2wt)
* * * * = -50 -50
* * * * = -100
which would appear to be the 100 watts needed to
heat the source resistor.


Thanks for a comprehensive analysis Keith. It was helpful to me to see how you step by step analyzed the circuit.

--
73, Roger, W7WKB

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Roger Sparks wrote:
Keith Dysart wrote:
Pr.g(t) = -50 + 50 cos(2wt)
= -50 -50
= -100
which would appear to be the 100 watts needed to
heat the source resistor.

Thanks for a comprehensive analysis Keith. It was helpful to me to see how you step by step analyzed the circuit.

Note that his analysis proves that destructive interference
energy is indeed stored in the reactance of the network and
dissipated later in the cycle as constructive interference
in the source resistor at a time when the instantaneous
source power equals zero.
--
73, Cecil http://www.w5dxp.com

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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

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On Mar 18, 10:58*am, Roger Sparks wrote:
Thanks for a comprehensive analysis Keith. *It was helpful to me to see how you step by step analyzed the circuit.

You are most welcome.

...Keith

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On Mar 18, 12:16*pm, Cecil Moore wrote:
Roger Sparks wrote:
Keith Dysart wrote:
Pr.g(t) = -50 + 50 *cos(2wt)
* * * * = -50 -50
* * * * = -100
which would appear to be the 100 watts needed to
heat the source resistor.

Thanks for a comprehensive analysis Keith. *It was helpful to me to see how you step by step analyzed the circuit.

Note that his analysis proves that destructive interference
energy is indeed stored in the reactance of the network and
dissipated later in the cycle as constructive interference
in the source resistor at a time when the instantaneous
source power equals zero.

I've made no claims about destructive or other interference
so I am intrigued by what you say my analysis proves.

Could you expand on how my analysis "proves" that
"destructive interference energy is indeed stored in the
reactance of the network and dissipated later in the cycle
as constructive interference in the source resistor at a
time when the instantaneous source power equals zero."

I am not sure why you focus on the infinitesimally small
times when "the instantaneous source power equals zero".
Are you saying that at other times, an explanation other
than interference is needed?
Or does interference explain the flows of the energy
in the reflected waves for all times?

It would be good if you could provide the equations that
describe these flows, especially the ones describing the
storage of some of the energy from the reflected wave in
the reactance of the network.

...Keith

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Keith Dysart wrote:
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".

The instantaneous example to which you are referring does
not meet my special case requirement of *ZERO INTERFERENCE*
so your above statement is just another straw man and is
thus irrelevant to my claim. But you already know that.

I agree with you that "the energy in the reflected wave can
not usually be accounted for in the source resistor dissipation."
I have stated over and over that the reflected energy
dissipation in the source resistor can range from 0% to 100%.

My claim is that when the special case of *ZERO INTERFERENCE*
exists between the two voltages superposed at the source resistor,
then 100% of the reflected energy is dissipated in the source
resistor. You have not provided a single example which proves
my claim false.

The test for *ZERO INTERFERENCE* is when, given the two
voltages, V1 and V2, superposed at the source resistor,

(V1^2 + V2^2) = (V1 + V2)^2

None of your examples have satisfied that special case
condition. My average reflected power example does NOT
satisfy that condition for instantaneous power!

Here's the procedure for proving my claim to be false.
Write the equations for V1(t) and V2(t), the two instantaneous
voltages superposed at the source resistor. Find the time
when [V1(t)^2 + V2(t)^2] = [V1(t) + V2(t)]^2. Calculate
the instantaneous powers *at that time*. You will find that,
just as I have asserted, 100% of the instantaneous reflected
energy is dissipated in the source resistor.

Until you satisfy my previously stated special case condition
of *ZERO INTERFERENCE*, you cannot prove my assertions to be
false.
--
73, Cecil http://www.w5dxp.com

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Keith Dysart wrote:
I am not sure why you focus on the infinitesimally small
times when "the instantaneous source power equals zero".

:-) :-) That was my exact reaction when you wanted to
focus on the infinitesimally small times associated
with instantaneous powers. :-) :-) You are to blame
for that focus, not I.

It only takes one example to prove your claims wrong.
I didn't want to focus on instantaneous power at all
but you insisted. Now that your own techniques are
being used to prove you wrong, you are objecting.

At the time when the instantaneous source power is zero,
the only other sources of energy in the entire network
is reflected energy and interference energy. That's
a fact of physics.

You have gotten caught superposing powers, something
that every sophomore EE knows not to do. I repeat:
The only time that power can be directly added is
when there is *ZERO INTERFERENCE*. The test for
zero interference is: (V1^2 + V2^2) = (V1 + V2)^2
--
73, Cecil http://www.w5dxp.com

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Keith Dysart wrote:
On Mar 17, 10:05 am, Cecil Moore wrote:
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.

Well, of course, you do when you argue with me. For
instance, you believe that reflections can occur when
the reflections see a reflection coefficient of 0.0,
i.e. a source resistance equal to the characteristic
impedance of the transmission line. This obviously
flies in the face of the wave reflection model. When
you cannot balance the energy equations, you are
arguing with the conservation of energy principle.
--
73, Cecil http://www.w5dxp.com

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On Mar 19, 11:47*am, Cecil Moore wrote:
Keith Dysart wrote:
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".

The instantaneous example to which you are referring does
not meet my special case requirement of *ZERO INTERFERENCE*
so your above statement is just another straw man and is
thus irrelevant to my claim. But you already know that.

I agree with you that "the energy in the reflected wave can
not usually be accounted for in the source resistor dissipation."
I have stated over and over that the reflected energy
dissipation in the source resistor can range from 0% to 100%.

My claim is that when the special case of *ZERO INTERFERENCE*
exists between the two voltages superposed at the source resistor,
then 100% of the reflected energy is dissipated in the source
resistor. You have not provided a single example which proves
my claim false.

The test for *ZERO INTERFERENCE* is when, given the two
voltages, V1 and V2, superposed at the source resistor,

(V1^2 + V2^2) = (V1 + V2)^2

None of your examples have satisfied that special case
condition. My average reflected power example does NOT
satisfy that condition for instantaneous power!

Here's the procedure for proving my claim to be false.
Write the equations for V1(t) and V2(t), the two instantaneous
voltages superposed at the source resistor. Find the time
when [V1(t)^2 + V2(t)^2] = [V1(t) + V2(t)]^2. Calculate
the instantaneous powers *at that time*. You will find that,
just as I have asserted, 100% of the instantaneous reflected
energy is dissipated in the source resistor.

Until you satisfy my previously stated special case condition
of *ZERO INTERFERENCE*, you cannot prove my assertions to be
false.
--
73, Cecil *http://www.w5dxp.com

In a previous iteration, I had accepted that your claim only
applied when there was zero interference. I converted that to
the particular values of wt for which your claim aaplied,
e.g. 90 degrees.

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.

So you have to choose...
Does it only apply at those particular points in the cycle
where wt is equal to 90 degrees or appropriate multiples?
(A restatement of only applying when there is no interference.)
or
Does it apply at all times in the cycle?

...Keith

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On Mar 19, 12:00*pm, Cecil Moore wrote:
Keith Dysart wrote:
I am not sure why you focus on the infinitesimally small
times when "the instantaneous source power equals zero".

:-) :-) That was my exact reaction when you wanted to
focus on the infinitesimally small times associated
with instantaneous powers. :-) :-) You are to blame
for that focus, not I.

It only takes one example to prove your claims wrong.

Could you state the claim that you think you are proving
wrong?

I didn't want to focus on instantaneous power at all
but you insisted. Now that your own techniques are
being used to prove you wrong, you are objecting.

If you wish to have your equalities apply at only selected
points within the cycle, that works for me.

It only fails when you make the assertion that
for the circuit under consideration, the energy in
the reflected wave is dissipated in the source
resistor.

At the time when the instantaneous source power is zero,
the only other sources of energy in the entire network
is reflected energy and interference energy. That's
a fact of physics.

You have gotten caught superposing powers, something
that every sophomore EE knows not to do.

Could you kindly show me where?

The two expressions in which I have summed power are
Ps(t) = Prs(t) + Pg(t)
and
Pg(t) = Pf.g(t) + Pr.g(t)

You have not previously indicated that either of these
are incorrect. Please take this opportunity.

I repeat:
The only time that power can be directly added is
when there is *ZERO INTERFERENCE*. The test for
zero interference is: (V1^2 + V2^2) = (V1 + V2)^2

Surely you overstate. Powers can be directly added
whenever you have a system with ports adding and
removing energy from the system. Conservation of
energy says that the sum of the energy being stored
in the system is equal to the sums of all the ins
and outs at the ports. There are no conditions on
the ports adding or removing energy.

...Keith