The Rest of the Story
 

On Mon, 31 Mar 2008 13:08:59 -0700 (PDT)
Keith Dysart wrote:

On Mar 31, 2:22*pm, Roger Sparks wrote:
On Mon, 31 Mar 2008 10:03:52 -0700

Roger Sparks wrote:
On Sun, 30 Mar 2008 07:43:59 -0700 (PDT)
Keith Dysart wrote:

On Mar 29, 7:18 pm, Roger Sparks wrote:
On Sat, 29 Mar 2008 12:45:48 -0700 (PDT)

Keith Dysart wrote:
On Mar 27, 2:06 am, Roger Sparks wrote:
Cecil Moore wrote:
Roger Sparks wrote:
You need to take a look at the spreadsheets.
clip
http://www.fairpoint.net/~rsparks/Sm...Reflection.pdf
clip

I doubt that this will satisfy your power location concerns because the spread sheet shows more power being delivered to the resistor than is present in the voltage. *This is because the impedance of the power equation has changed due to the contribution of the current component. Consider that for columns B and C, the same current flows whether the voltage in B is applied or the voltage in C is applied. *This can only happen if the impedance seen by each respective voltage is different. *This is interference at work *
--
73, Roger, W7WKB

After posting previosly, I got to thinking that interference here is wrecking the analysis of Column D. *The traveling wave analysis is correct (Column H). *Only one current is flowing through Rs, and the current is not enough to supply the power suggested in column D. *While it is logical to add the voltages from Column B and Column C, the two voltages are often in opposition so they are not "seen" by Rs. *As a result, we must have a reflection from Rs that I am not taking into account. *

Column B is correct; this being the voltage produced by the source
divided by two.
It is also the forward voltage on the line.
Vrs.source(t) = Vf(t) = 70.7 sin(wt)

Column C is the reflected voltage (not the reflected voltage impressed
across the
source resistor). The reflection coefficient is -1, and the delay is
90 degrees
so the reflected voltage at the generator is
Vr(t) = -1 * Vf(t - 90 degrees)
= - 70.7 sin(wt-90)
= 70.7 sin(wt+90)

But Vr is impressed across the resistor in the opposite direction to
that of
Vrs.source, so the equation for total Vrs is
Vrs.total(t) = Vrs.source(t) - Vr(t)
thus column D should be B31-C31.

Alternatively,
Vrs.reflect(t) = -Vr(t)
and then
Vrs.total(t) = Vrs.source(t) + Vrs.reflect(t)

Column E is correctly computing the instantaneous power from Column D
since
P(t) = V(t) * I(t)
= V(t) * V(t) / R
= V(t) * V(t) / 50 (in this example)
but has the wrong data because of the error in Column D.

Column F is integrating the power to yield either the energy in a
cycle or
the average power per cycle (though presented in unusual units).

I agree G is erroneous and I am not sure what H is computing.

...Keith

I made the change in Column D and the trend is more believable in Column E. I think the math here is noncommutative in the sense that time must rotate forward. I think this change does that even though the average power results stay the same.
--
73, Roger, W7WKB

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Cecil Moore wrote:
Roger Sparks wrote:
While it is logical to add the voltages from Column B and Column C,
the two voltages are often in opposition so they are not "seen" by
Rs. As a result, we must have a reflection from Rs that I am not
taking into account.

It's not a "reflection" of a single wave, Roger, it is
a "redistribution" of energy caused by superposition
of two waves accompanied by interference. Eugene Hecht
explains it all in Chapter 9: Interference in "Optics".

For anyone who thinks he is already omniscient
about EM waves, I would highly recommend reading
Hecht's chapter on interference.

Here's an even better idea.

Dump Hecht and read about interference in Born and Wolf. Chapter VII (in
the 7th edition) is one of the main contributions. I particularly like
Section 7.6. In this section the authors derive a general set of
equations that deal with all sorts of reflection configurations. There
is no need to worry about constructive or destructive interference. The
equations smoothly transition from one to the other as appropriate. The
equations don't fall apart as the reflection goes to zero. No
"redistribution" is needed.

This is the way physics typically works; it is not necessary to separate
superposition from interference or separate constructive from
destructive. If the analysis and the equations are correct they will
work for a wide range of parameters. Equations that must be fine-tuned
for every possible change in reflection coefficient or other parameters
are very limited and most troublesome.

73,
Gene
W4SZ

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On Mar 31, 8:04*am, Cecil Moore wrote:
Keith Dysart wrote:
You state that your hypothesis is that for this specific
circuit, "the energy in the reflected wave is dissipated in
the source resistor".

First, let's correct your out-of-context quotation.
Here is what you should have quoted: "When zero
interference exists at the source resistor, the
energy in the reflected wave is dissipated in the
source resistor."

This is actually a fact for both average powers
and instantaneous powers. Since all of your examples
are associated with a non-zero level of interference,
they are irrelevant to the stated conditions.

Since my example is *your* example (q.v. your Fig 1-1), your
example has non-zero interference (as you state above), so
you have just said that your example violates the stated
conditions.

Or are you going to say that the circuit exhibits interference
when an instantaneous analysis is performed, but knows that
it should refrain from doing so when only an average analysis
is done?

Here is a quote from that article:
"Please note that any power referred to in this paper is an AVERAGE
POWER. Instantaneous power is irrelevant to the following discussion."
The word "average" is implied in every statement I make.

Yes, "implied" is the word. Why not clearly state that, while the
average energy appears to be dissipated in the source resistor, the
actual energy is not.

Or *is* the intent to deceive?

This claim is amenable to analysis
using instantaneous energy flows. When so analyzed, the
hypothesis fails.

No, it doesn't fail. You have simply failed to satisfy
the zero interference precondition.

If the precondition fails for the circuit, then it fails for
the circuit.

If you wish to narrow your hypothesis to "the average energy
in the reflected wave is simply numerically equal to the
increase in the average dissipation in the source
resistor" I will not object since that hypothesis would
be completely accurate and not misleading.

That is, in fact, the only hypothesis presented in
my Part 1 article. Since my hypothesis never applied
to instantaneous power, I don't have to narrow the
hypothesis. My article stands as written. Please
cease and desist with the unfair innuendo.

You insist that the narrowing is "implied", but then
refuse to explicitly state such to make it clear to
the reader. Why?

Not a waste at all.

Obviously, your opinion differs from mine. To the best
of my knowledge, you are the first person to spend any
mental effort on instantaneous power. If that's what
you want to do, be my guest. I consider it to be little
more than mental masturbation, "of limited utility" as
Hecht said.

Yes. It does not support your hypothesis, so it is wise
to ignore it.

In fact, I proved my assertion was true even at the
instantaneous power level when the "zero interference"
precondition is met.

Ah, yes. X**2 + Y**2 = (X+Y)**2 only when X or Y equals 0,
which for the example at hand applies at exactly 4 instances
per cycle.

The rest of the time the circuit exhibits interference.

Since you start with an unshakeable belief in the
existance of energy in the reflected wave, this would be
your natural conclusion.

Since you are incapable of producing an EM wave devoid
of energy (or an angel dancing on the head of a pin) both
concepts are unrelated to reality IMO.

Your challenge is the same as it has always been. Just
produce an EM wave containing zero energy and get it
over with.

Tis a problem isn't it. You won't let go of energy in the
reflected wave long enough to even explore the circuit to
discover the inconsistencies that result from the belief.

You can not find a reason why instantaneous analysis should
not work, but the conclusions are uncomfortable, so you
decide that Hecht has told you not to bother, and you stop.
Without knowing why.

...Keith

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On Mar 31, 8:43*am, Cecil Moore wrote:
Keith Dysart wrote:
But the meaning of the disclaimer is not clear to the
reader. You really need to restate your hypothesis to
remove the possibility of misleading the reader.

What is it about "Please note that any power referred to
in this paper is an AVERAGE POWER. Instantaneous power
is irrelevant to the following discussion." that you
do not understand?

After many posts and back and forth, I understand. But the
poor first reader will miss the implications: that the
imputed energy in the reflected wave is not dissipated
in the source resistor.

Why not save the reader the challenge and just state it
clearly?

I would suggest ...

I would suggest that you write your own article.
Mine stands as written in the *stated context*
of zero interference and average powers. I am
not interested in attempting a unified theory
of everything.

Except that you have now indicated that there is
interference in the circuit of Fig 1-1.

I personally don't think
that anyone else cares about instantaneous powers.

I am sure some do not. But anyone interested in a full
understanding does.

Anyone interested in a *full* understanding would
take the discussion down to the quantum level which,
interestingly enough, you have chosen to ignore.

Yes. I have stopped at the level that disproves that the
imputed energy in the reflected wave is dissipated in
the source resistor. That is sufficient for me.

I do not think that deeper analysis will show this to
be wrong, but you are invited to do so.

On the other hand, average analysis can be shown to
produce misleading results by applying instantaneous
analysis. You should be interested because it disproves
that the imputed energy in the reflected wave is
dissipated in the source resistor.

It is convenient when you just ignore the analysis
that disproves your hypothesis. But it does not make
the hypothesis more correct.

If you think your unethical innuendo, out-of-context
quotes, and straw man arguments disprove anything,
I feel sorry for you.

Once again, the context of my Part 1 assertions is
*ZERO INTERFERENCE* and *AVERAGE POWERS*. You have
disproved nothing so far. You were even taken aback
when it was true at the instantaneous level in the
context of zero instantaneous interference.

I was? If so, I have now moved beyond. Especially since
you now assert that the circuit does exhibit interference,
the hypothesis becomes moot.

...Keith

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On Mar 31, 9:01*am, Cecil Moore wrote:
Keith Dysart wrote:
Perhaps, but it is highly improbable that it falls apart
in a manner that ends up supporting the original failed
hypothesis.

Since the original hypothesis is in the context of
zero interference (and average powers) it has not
failed. So far, I have made no assertions about
conditions when interference is present as it is
in all of your examples. None of your observations
are relevant to my Part 1 article because they are
all outside the stated context of the article.

The challenge for you is to present a zero interference
example for which my hypothesis is false. So far, you
have failed to do so.

We are talking about the same circuit, which you now
claim exhibits interference, rendering your hypothesis
moot.

I have asserted, "If zero interference exists, then 'A'
is true". You have said 'A' is not true when interference
exists. I actually agree with you but it is irrelevant
to the stated 'if' portion of my premise. Where did
you study logic?

Good that you agree. And now that you state that the
circuit exhibits interference, it might be best to
withdraw your example.

...Keith

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Keith Dysart wrote:
Since my example is *your* example (q.v. your Fig 1-1), your
example has non-zero interference (as you state above), so
you have just said that your example violates the stated
conditions.

Keith, if I send you $100, would you use it to buy
yourself some ethics?
--
73, Cecil http://www.w5dxp.com

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Keith Dysart wrote:
After many posts and back and forth, I understand.

Do you understand that you need to go out and buy
some ethics?
--
73, Cecil http://www.w5dxp.com

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Keith Dysart wrote:
We are talking about the same circuit, which you now
claim exhibits interference, rendering your hypothesis
moot.

If the average interference is zero, the average
reflected power is dissipated in the source resistor.
All of your unethical lies, innuendo, and hand-waving
will not change that fact of physics.
--
73, Cecil http://www.w5dxp.com

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Roger Sparks wrote:
The storage of the redistributed energy must be close to the
wires of the circuit so we should be able to describe it
mathmatically, if I just knew how.

The math is pretty easy, Roger. Keith seems to believe
that an inductor stores power which is, of course, a
ridiculous concept. As soon as I can see a character
that is smaller than 2 inches tall, I will respond.
--
73, Cecil http://www.w5dxp.com

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Gene Fuller wrote:
Dump Hecht ...

Dump Gene Fuller. :-)
--
73, Cecil http://www.w5dxp.com

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Keith Dysart wrote:
After many posts and back and forth, I understand. But the
poor first reader will miss the implications: that the
imputed energy in the reflected wave is not dissipated
in the source resistor.

You have yet to provide an example of zero interference
where the reflected power is not dissipated in the source
resistor. Until you do that, you are just waving your hands.

Examples containing interference will be covered in Parts
2 & 3 but the poor first reader will not get to read them
until you cease your present unethical behavior.
--
73, Cecil http://www.w5dxp.com

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On Apr 1, 12:06*am, Cecil Moore wrote:
Keith Dysart wrote:
We are talking about the same circuit, which you now
claim exhibits interference, rendering your hypothesis
moot.

If the average interference is zero, the average
reflected power is dissipated in the source resistor.
All of your unethical lies, innuendo, and hand-waving
will not change that fact of physics.

More precisely, the average value of the imputed reflected
power is numerically equal to the increase in the average
power dissipated in the source resistor.

It is not obvious why you reject this more precise, less
misleading description. Is there an intent to mislead the
reader?

...Keith

The Rest of the Story
 

On Apr 1, 12:17*am, Cecil Moore wrote:
Roger Sparks wrote:
The storage of the redistributed energy must be close to the
wires of the circuit so we should be able to describe it
mathmatically, if I just knew how.

The math is pretty easy, Roger. Keith seems to believe
that an inductor stores power which is, of course, a
ridiculous concept.

You must have misunderstood. The power measured at the
terminals of an inductor is the rate of change with respect
to time of the energy stored within the inductor.

In calculus terms, the power is the derivative of the energy
within the inductor.

This is equivalent to the rate of flow of energy into
the inductor (or out of, if you choose a different convention
for the sign of the value).

...Keith

The Rest of the Story
 

On Apr 1, 12:39*am, Cecil Moore wrote:
Keith Dysart wrote:
After many posts and back and forth, I understand. But the
poor first reader will miss the implications: that the
imputed energy in the reflected wave is not dissipated
in the source resistor.

You have yet to provide an example of zero interference
where the reflected power is not dissipated in the source
resistor. Until you do that, you are just waving your hands.

You misunderstand. I am not attempting to do that. Though
somewhat bizzarre, I have, for the purposes of this
discussion, accepted your definition of interference.

And using your definition, that there is no interference
when (V1**2 + V2**2) = (V1+V2)**2, it can be seen that
for the circuit at hand, your Fig 1-1, there is zero
interference in the terms you wish to add, four times
in each cycle. From this one might conclude that the
imputed reflected power is dissipated in the source
resistor at four instances during the cycle. For the
remainder of the cycle, again using your definition of
interference, there is interference and hence the
imputed reflected power is not all dissipated in the
source resistor.

Thus any unqualified assertion that the imputed reflected
power is dissipated in the source resistor is somewhat
disingenuous.

Examples containing interference will be covered in Parts
2 & 3 but the poor first reader will not get to read them
until you cease your present unethical behavior.

But you have been claiming that the circuit of Part 1 already
exhibits interference.

...Keith

The Rest of the Story
 

Keith Dysart wrote:
Cecil Moore wrote:
If the average interference is zero, the average
reflected power is dissipated in the source resistor.
All of your unethical lies, innuendo, and hand-waving
will not change that fact of physics.

More precisely, the average value of the imputed reflected
power is numerically equal to the increase in the average
power dissipated in the source resistor.

More precisely, at the *exact* time of arrival of the
reflected wave, the average power dissipated in the
source resistor increases by *exactly* the magnitude
of the average reflected power. Do you think that is
just a coincidence?

It is not obvious why you reject this more precise, less
misleading description.

Why do you use such unfair ill-willed debating techniques
based on innuendo and not on facts in evidence?

IMO, our two statements above say the same thing
with mine being the more precise and detailed.
I don't reject yours - I just prefer mine.

If the source of the increased dissipation in the
source resistor is not the reflected energy, exactly
where did that "extra" energy come from at the exact
time of arrival of the reflected wave?

Hint: Since the only other source of energy in the
entire system is the reflected wave, any additional
source would violate the conservation of energy
principle.
--
73, Cecil http://www.w5dxp.com

The Rest of the Story
 

Keith Dysart wrote:
And using your definition, that there is no interference
when (V1**2 + V2**2) = (V1+V2)**2, it can be seen that
for the circuit at hand, your Fig 1-1, there is zero
interference in the terms you wish to add, four times
in each cycle.

Correction for omitted word above: And using my
definition, that there is no *average* interference
when (V1**2 + V2**2) = (V1+V2)**2,"
Those are average (RMS) values of voltage.

The test for zero *instantaneous* interference is:
[V1(t)^2 + V2(t)^2] NOT= [V1(t)^2+V2(t)^2]
Those are instantaneous values of voltage.

Please correct your confusion about what I have said.
It is also clear that you don't understand when
interference exists and when it doesn't.

The instantaneous destructive interference equals
the instantaneous constructive interference 90
degrees later. That's why the interference averages
out to zero.

I believe, although I have not taken the time to
prove it, that the instantaneous interference is
zero only at the zero-crossings of the source
voltage and reflected voltage.

Again, the existence and magnitude of the
instantaneous interference is irrelevant to
the assertions in my Part 1 article. It is
obvious that the interference averages out
to zero over each cycle for the example
presented.
--
73, Cecil http://www.w5dxp.com

The Rest of the Story
 

On Apr 1, 8:08 am, Cecil Moore wrote:
Keith Dysart wrote:
Cecil Moore wrote:
If the average interference is zero, the average
reflected power is dissipated in the source resistor.
All of your unethical lies, innuendo, and hand-waving
will not change that fact of physics.

More precisely, the average value of the imputed reflected
power is numerically equal to the increase in the average
power dissipated in the source resistor.

More precisely, at the *exact* time of arrival of the
reflected wave, the average power dissipated in the
source resistor increases by *exactly* the magnitude
of the average reflected power. Do you think that is
just a coincidence?

Not quite, but close. And averages can not change instantly.

But your phraseology suggests another way to approach
the problem. Let us consider what happens just after
the arrival of the reflected wave. For computational
convenience, we'll use a frequency of 1/360 hertz so
that 1 degree of the wave is one second long.

The reflected wave arrives at 90 seconds (or 90 degrees)
after the source is turned on. If the line is
terminated in 50 ohms, then no reflection arrives and
during the one second between degree 90 and degree 91,
the source resistor dissipates 0.01523 J of energy. The
line also receives 0.01523 J of energy and the source
provides 0.03046 J.
Esource.50[90..91] = 0.03046 J
Ers.50[90..91] = 0.01523 J
Eline.50[90..91] = 0.01523 J
Esource = Ers + Eline
as expected.
Efor = Eline
since Eref is 0.

Now let us examine the case with a 12.5 ohm load and
a non-zero reflected wave.
During the one second between degree 90 and degree 91,
the source resistor dissipation with no reflection
is still 0.01523 J and the imputed reflected wave
provides 99.98477 J for a total of 100.00000 J.

But the source resistor actually absorbs 98.25503 J
in this interval.

Ooooopppps. It does not add up. So the dissipation
in the source resistor went up, but not enough to
account for all of the imputed energy in the
reflected wave. This does not satisfy conservation
of energy, which should be sufficient to kill the
hypothesis.

But Esource = Ers + Eline as expected.
Esource.12.5[90..91] = -1.71451 J
Ers.12.5[90..91] = 98.25503 J
Eline.12.5[90..91] = -99.96954 J
Note that in the interval 90 to 91 degrees, the source
is absorbing energy. This is quite different than what
happens when there is no reflected wave.

You said earlier "Do you think that is just a coincidence?"
It is to be expected that the dissipation in the source
resistor changed; after all, the load conditions changed.
But it is mere happenstance that the average of the
increase in the dissipation is the same as the average
power imputed to the reflected wave; an ideosyncratic
effect of the selection of component values.

It is not obvious why you reject this more precise, less
misleading description.

Why do you use such unfair ill-willed debating techniques
based on innuendo and not on facts in evidence?

IMO, our two statements above say the same thing
with mine being the more precise and detailed.
I don't reject yours - I just prefer mine.

Authors often do have difficulties detecting when
their words mislead. Excellent authors, and there
are not many, use the feedback from their readers
to adjust their wording to eliminate misleading
prose.

If the source of the increased dissipation in the
source resistor is not the reflected energy, exactly
where did that "extra" energy come from at the exact
time of arrival of the reflected wave?

Exactly. This is what calls into question the
notion that the reflected wave is transporting
energy. This imputed energy can not be accounted for.

Now the energy that can be accounted for is the
energy that flows in or out of the line. This
energy, along with the energy dissipated in the
source resistor will *always*, no matter how
you slice and dice it, be equal to the energy
being delivered by the source; completely satisfying
the requirements of conservation of energy.

The best that can be said for the imputed power in
the reflected wave is that when it is subtracted
from the imputed power in the forward wave, the
result will be the actual energy flow in the line.
But this is just a tautology; a result from the
very definition of Vforward and Vreflected.

Hint: Since the only other source of energy in the
entire system is the reflected wave, any additional
source would violate the conservation of energy
principle.

Alternatively, since it turns out that trying to
use this imputed power to calculate the power
dissipated in the source resistor results in a
violation of conservation of energy, this
energy flow imputed to the reflected wave is
a figment. The only thing that is real is the
total energy flow.

Now you have asked repeatedly about the reflection
from the mirror because you are sure that this
is proof of the energy in the reflected wave.

The energy in the light entering your eye is
the total energy; it is not imputed energy of a
wave that is a partial contributor to the total.
If the eye were also a source, such that there
was a Pfor to go along with Pref, then your
question would align with the situation under
discussion. But when the energy flow is only
in one direction, that flow is a total flow
and it definitely contains energy.

To recap, it is when a total flow is
broken into multiple non-zero constituent flows
that energy flow imputed to the constituents
is a dubious concept.

....Keith

The Rest of the Story
 

On Wed, 2 Apr 2008 02:41:24 -0700 (PDT)
Keith Dysart wrote:

clip
But your phraseology suggests another way to approach
the problem. Let us consider what happens just after
the arrival of the reflected wave. For computational
convenience, we'll use a frequency of 1/360 hertz so
that 1 degree of the wave is one second long.

The reflected wave arrives at 90 seconds (or 90 degrees)
after the source is turned on. If the line is
terminated in 50 ohms, then no reflection arrives and
during the one second between degree 90 and degree 91,
the source resistor dissipates 0.01523 J of energy. The
line also receives 0.01523 J of energy and the source
provides 0.03046 J.
Esource.50[90..91] = 0.03046 J
Ers.50[90..91] = 0.01523 J
Eline.50[90..91] = 0.01523 J
Esource = Ers + Eline
as expected.
Efor = Eline
since Eref is 0.
I come up with 141.4v across 50 plus 50 ohms. The current should be 1.414a. Power to each 50 ohm resistor would be 50*1.4142^2 = 200w. For the 1 degree interval of 1 second, that would be 200 joules.

Right? Peak current flows at 90 degrees?
--
73, Roger, W7WKB

The Rest of the Story
 

On Apr 2, 9:17*am, Roger Sparks wrote:
On Wed, 2 Apr 2008 02:41:24 -0700 (PDT)

Keith Dysart wrote:

clip



But your phraseology suggests another way to approach
the problem. Let us consider what happens just after
the arrival of the reflected wave. For computational
convenience, we'll use a frequency of 1/360 hertz so
that 1 degree of the wave is one second long.

The reflected wave arrives at 90 seconds (or 90 degrees)
after the source is turned on. If the line is
terminated in 50 ohms, then no reflection arrives and
during the one second between degree 90 and degree 91,
the source resistor dissipates 0.01523 J of energy. The
line also receives 0.01523 J of energy and the source
provides 0.03046 J.
* Esource.50[90..91] = 0.03046 J
* Ers.50[90..91] * * = 0.01523 J
* Eline.50[90..91] * = 0.01523 J
* Esource = Ers + Eline
as expected.
* Efor = Eline
since Eref is 0.

I come up with 141.4v across 50 plus 50 ohms. *The current should be 1.414a. *Power to each 50 ohm resistor would be *50*1.4142^2 = 200w. *For the 1 degree interval of 1 second, that would be 200 joules.

Right? *Peak current flows at 90 degrees?
--
73, Roger, W7WKB- Hide quoted text -

- Show quoted text -

We may be using different sources. My Vs is 141.4cos(wt)
so that between 90 degrees and 91 degrees, my source
voltage is going from 0 to -2.468 V.

And an ooopppps. I actually did the calculations for
a shorted load rather than 12.5 ohms as stated.
With the reflection coefficient of -1 and a 90 degree
delay, the reflected voltage between 90 and 91 degrees
changes from -70.711 V to -70.700 V.

Hoping these details resolve the disparity,

...Keith

The Rest of the Story
 

Keith Dysart wrote:
And averages can not change instantly.

Now there is an assertion that should make you
famous. :-)

I have just had an operation on my eyes. My
prescription will be changing for a couple
of weeks and until then, I will be without
glasses. I have my newsreader characters set
to about 1/2 inch in height so I can see them.
I will be quite handicapped for a couple of
weeks.

This does not satisfy conservation
of energy, which should be sufficient to kill the
hypothesis.

What hypothesis? The scope of my hypothesis is
limited to average power. You have not presented even
one example where my average power hypothesis is
incorrect. I have no hypothesis about instantaneous
power. Any failed hypothesis about instantaneous
power must have been presented by someone else -
it sure wasn't presented by me. I say my GMC
pickup is white. You say it is not white because
the tires are black. Your diversionary argument
is obviously a straw man because you cannot win
the main argument.

Instantaneous power, as Hecht says, is "of limited
utility". IMO, it is irrelevant and certainly far
beyond the scope of my Part 1 article. Please feel
free to write your own article and publish it. Such
an article would, IMO, be a waste of time.

To recap, it is when a total flow is
broken into multiple non-zero constituent flows
that energy flow imputed to the constituents
is a dubious concept.

Since you have not offered a single average power
example that disagrees with my average power
hypothesis, I guess we will just have to agree to
disagree about that.

You have completely ignored the fact that instantaneous
destructive interference energy is stored for part of
the cycle and then released back into the network as
constructive interference energy 90 degrees later.

I will save your posting and digest it better when
I get my eyesight back along with new glasses.
--
73, Cecil http://www.w5dxp.com

The Rest of the Story
 

On Wed, 2 Apr 2008 06:33:46 -0700 (PDT)
Keith Dysart wrote:

On Apr 2, 9:17*am, Roger Sparks wrote:
On Wed, 2 Apr 2008 02:41:24 -0700 (PDT)

Keith Dysart wrote:

clip



But your phraseology suggests another way to approach
the problem. Let us consider what happens just after
the arrival of the reflected wave. For computational
convenience, we'll use a frequency of 1/360 hertz so
that 1 degree of the wave is one second long.

The reflected wave arrives at 90 seconds (or 90 degrees)
after the source is turned on. If the line is
terminated in 50 ohms, then no reflection arrives and
during the one second between degree 90 and degree 91,
the source resistor dissipates 0.01523 J of energy. The
line also receives 0.01523 J of energy and the source
provides 0.03046 J.
* Esource.50[90..91] = 0.03046 J
* Ers.50[90..91] * * = 0.01523 J
* Eline.50[90..91] * = 0.01523 J
* Esource = Ers + Eline
as expected.
* Efor = Eline
since Eref is 0.

I come up with 141.4v across 50 plus 50 ohms. *The current should be 1.414a. *Power to each 50 ohm resistor would be *50*1.4142^2 = 200w. *For the 1 degree interval of 1 second, that would be 200 joules.

Right? *Peak current flows at 90 degrees?
--
73, Roger, W7WKB- Hide quoted text -

- Show quoted text -

We may be using different sources. My Vs is 141.4cos(wt)
so that between 90 degrees and 91 degrees, my source
voltage is going from 0 to -2.468 V.

And an ooopppps. I actually did the calculations for
a shorted load rather than 12.5 ohms as stated.
With the reflection coefficient of -1 and a 90 degree
delay, the reflected voltage between 90 and 91 degrees
changes from -70.711 V to -70.700 V.

Hoping these details resolve the disparity,

...Keith

I am sorry Keith, but I still can not duplicate your work. I wonder if the difference is in how we each apply the recharge voltage to the transmission line.

I redid my table showing the power in the source and reflected waves. I now have the power to resistor Rs coming from voltage agreeing with the expected power calculated from current.

The table can be found at http://www.fairpoint.net/~rsparks/Sm...Reflection.pdf.

-*/---------
--
73, Roger, W7WKB

The Rest of the Story
 

On Apr 3, 9:47*am, Roger Sparks wrote:
On Wed, 2 Apr 2008 06:33:46 -0700 (PDT)
I am sorry Keith, but I still can not duplicate your work. *I wonder if the difference is in how we each apply the recharge voltage to the transmission line.

I do not know. The spreadsheet that did the work is now available from
this page:
http://keith.dysart.googlepages.com/radio6

Perhaps with the extra detail, you can find the discrepancy.

I redid my table showing the *power in the source and reflected waves. *I now have the power to resistor Rs coming from voltage agreeing with the expected power calculated from current. *

My sheet agrees with yours for the samples I checked.

...Keith

The Rest of the Story
 

Keith Dysart wrote:
My sheet agrees with yours for the samples I checked.

Here's an example that changes nothing but allows
isolation of the component energy flows.

50 ohms
+---1WL-50-ohm---\/\/\/\---1WL-50-ohm---+
| |
Source 141.4*cos(wt) volts Load +j50
| |
+---------------------------------------+
GND GND

It appears to me that the confusion arises from
the *series* resistor.
--
73, Cecil http://www.w5dxp.com

The Rest of the Story
 

On Thu, 3 Apr 2008 10:35:17 -0700 (PDT)
Keith Dysart wrote:

On Apr 3, 9:47*am, Roger Sparks wrote:
On Wed, 2 Apr 2008 06:33:46 -0700 (PDT)
I am sorry Keith, but I still can not duplicate your work. *I wonder if the difference is in how we each apply the recharge voltage to the transmission line.

I do not know. The spreadsheet that did the work is now available from
this page:
http://keith.dysart.googlepages.com/radio6

Perhaps with the extra detail, you can find the discrepancy.

I redid my table showing the *power in the source and reflected waves.. *I now have the power to resistor Rs coming from voltage agreeing with the expected power calculated from current. *

My sheet agrees with yours for the samples I checked.

...Keith

I think that at 91 degrees, the sign of the returning wave should be positive, not negative. Thus, it should be +1.234v. My reasoning is that the voltage from the line side of the resistor adds to the voltage from the source side of the resistor, but in this case it takes a little longer to get around the loop. The 90 degrees is timing, the lead edge of the wave is not reversed by turning the corner at what we call the short.

Another way of thinking about it is that if the transmission line was shorter, the added voltage to the resistor from the line side would arrive even earlier, and add. If the line were zero length, there would be no question but that the voltage would add.

Lastly, your current peaks at 135 degrees, not 45 degrees, which should be the delay.

It is interesting to notice that the current is delayed 45 degrees, but the voltage is delayed 90 degrees. It is as if the current does not take the entire trip around the loop, but the voltage does. The current apparently travels an average of half the distance around the loop.

I know that the convention for calculating the reflection coefficient is to make it negative for a shorted transmission line. The negative sign makes sense because the voltages at the short cancel so we have a zero voltage across the short. That convention sure messes up our spread sheet calculations however.

--
73, Roger, W7WKB

The Rest of the Story
 

On Thu, 03 Apr 2008 15:49:15 -0500
Cecil Moore wrote:

Keith Dysart wrote:
My sheet agrees with yours for the samples I checked.

Here's an example that changes nothing but allows
isolation of the component energy flows.

50 ohms
+---1WL-50-ohm---\/\/\/\---1WL-50-ohm---+
| |
Source 141.4*cos(wt) volts Load +j50
| |
+---------------------------------------+
GND GND

It appears to me that the confusion arises from
the *series* resistor.
--
73, Cecil http://www.w5dxp.com

This circuit has a reflection at the 50 ohm resistor because the impedance transitions from 50 ohms to 100 ohms, then back to 50 ohms.

--
73, Roger, W7WKB

The Rest of the Story
 

On Apr 3, 6:47*pm, Roger Sparks wrote:
On Thu, 3 Apr 2008 10:35:17 -0700 (PDT)

Keith Dysart wrote:
On Apr 3, 9:47*am, Roger Sparks wrote:
On Wed, 2 Apr 2008 06:33:46 -0700 (PDT)
I am sorry Keith, but I still can not duplicate your work. *I wonder if the difference is in how we each apply the recharge voltage to the transmission line.

I do not know. The spreadsheet that did the work is now available from
this page:
http://keith.dysart.googlepages.com/radio6

Perhaps with the extra detail, you can find the discrepancy.

I redid my table showing the *power in the source and reflected waves. *I now have the power to resistor Rs coming from voltage agreeing with the expected power calculated from current. *

My sheet agrees with yours for the samples I checked.

...Keith

I think that at 91 degrees, the sign of the returning wave should be positive, not negative. *Thus, it should be +1.234v. *My reasoning is that the voltage from the line side of the resistor adds to the voltage from the source side of the resistor, *but in this case it takes a little longer to get around the loop. *The 90 degrees is timing, the lead edge of the wave is not reversed by turning the corner at what we call the short. *

I do believe that -1.234 is correct. The reflection coefficient is -1,
so at the
short, Vr(t) is always equal to -Vf(t), which, when summed, give a
Vtot(t) of 0
which is what is expected at a short. Vr.g(t) is delayed by 90 degrees
from
Vf.g(t) so at 91 degrees, we should expect Vr.g(91) to be (-1 *
Vf.g(1)) or
-1.234. This is also how the spreadsheet computes it.

Another way of thinking about it is that if the transmission line was shorter, the added voltage to the resistor from the line side would arrive even earlier, and add. *If the line were zero length, there would be no question but that the voltage would add.

Vg(t) is the voltage across the line, so Vg(t) is the sum of Vf.g(t)
and Vr.g(t),
recalling that Vtot = Vf + Vr.

And
Vrs(t) = Vs(t) - Vg(t)
Since Vf.g(t) = Vs(t)/2, and Vg(t) = Vf.g(t) + Vr.g(t) we have, by
substitution,
Vrs(t) = 2 * Vf.g(t) - Vf - Vr.g(t)
= Vf.g(t) - Vr.g(t)

Lastly, your current peaks at 135 degrees, not 45 degrees, which should be the delay. *

It is interesting to notice that the current is delayed 45 degrees, but the voltage is delayed 90 degrees. *It is as if the current does not take the entire trip around the loop, but the voltage does. *The current apparently travels an average of half the distance around the loop. *

Be careful if you are comparing things before and after the 90 degree
point. At 90
degrees, the circuit conditions change dramatically.

At the entrance to the line, Ig is delayed 90 degrees from Vg which is
to be expected
since the line is behaving as a pure inductor from any time after 90
degrees. Before
that the line is behaving as a resistance.

At the resistor, the offset is only half this because the resistor and
the line have
the same magnitude of impedance.

I know that the convention for calculating the reflection coefficient is to make it negative for a shorted transmission line. *The negative sign makes sense because the voltages at the short cancel so we have a zero voltage across the short. *That convention sure messes up our spread sheet calculations however.

One does have to be careful to treat the signs consistently. In this
sheet, the convention
is that flows (current and energy) to the right (assuming the load is
to the right) are
positive. Some conventions use positive to mean flows to the right for
the forward wave
and flows to the left for the reflected wave.

On this sheet, Vtot = Vf+Vr, Itot=If+Ir and Ptot=Pf+Pr.
I hope no gremlins snuck in.

...Keith

The Rest of the Story
 

On Apr 3, 4:49*pm, Cecil Moore wrote:
Keith Dysart wrote:
My sheet agrees with yours for the samples I checked.

Here's an example that changes nothing but allows
isolation of the component energy flows.

* * * * * * * * * * *50 ohms
* * +---1WL-50-ohm---\/\/\/\---1WL-50-ohm---+
* * | * * * * * * * * * * * * * * * * * * * |
Source 141.4*cos(wt) volts * * * * * * * Load +j50
* * | * * * * * * * * * * * * * * * * * * * |
* * +---------------------------------------+
* *GND * * * * * * * * * * * * * * * * * * *GND

It appears to me that the confusion arises from
the *series* resistor.

It seems to me that this circuit is quite different.
There is a reflection where the left TL connects
to the 50 ohm resistor. There is a reflection when
the reflected wave on the left TL arrives back at the
source and there is a reflection when the reflected
wave on the right TL hits the 50 ohm resistor.

This circuit takes an infinitely long time to settle.
The previous circuit only had a reflection at the load
and settled after one round trip.

Quite different.

...Keith

The Rest of the Story
 

Roger Sparks wrote:
This circuit has a reflection at the 50 ohm resistor because the impedance transitions from 50 ohms to 100 ohms, then back to 50 ohms.

Yet conditions are identical to the earlier example.
That should tell us something.
--
73, Cecil http://www.w5dxp.com

The Rest of the Story
 

Keith Dysart wrote:
It seems to me that this circuit is quite different.

Yet, steady-state conditions are identical. That
should tell us something about what is wrong
with the analysis so far. Wave reflection theory
works if the transmission line is a multiple of
one wavelength even if the multiple is zero.
--
73, Cecil http://www.w5dxp.com

The Rest of the Story
 

On Thu, 3 Apr 2008 17:50:05 -0700 (PDT)
Keith Dysart wrote:

On Apr 3, 6:47*pm, Roger Sparks wrote:
On Thu, 3 Apr 2008 10:35:17 -0700 (PDT)

Keith Dysart wrote:
On Apr 3, 9:47*am, Roger Sparks wrote:
On Wed, 2 Apr 2008 06:33:46 -0700 (PDT)
I am sorry Keith, but I still can not duplicate your work. *I wonder if the difference is in how we each apply the recharge voltage to the transmission line.

I do not know. The spreadsheet that did the work is now available from
this page:
http://keith.dysart.googlepages.com/radio6

Perhaps with the extra detail, you can find the discrepancy.

I redid my table showing the *power in the source and reflected waves. *I now have the power to resistor Rs coming from voltage agreeing with the expected power calculated from current. *

My sheet agrees with yours for the samples I checked.

...Keith

I think that at 91 degrees, the sign of the returning wave should be positive, not negative. *Thus, it should be +1.234v. *My reasoning is that the voltage from the line side of the resistor adds to the voltage from the source side of the resistor, *but in this case it takes a little longer to get around the loop. *The 90 degrees is timing, the lead edge of the wave is not reversed by turning the corner at what we call the short. *

I do believe that -1.234 is correct. The reflection coefficient is -1,
so at the
short, Vr(t) is always equal to -Vf(t), which, when summed, give a
Vtot(t) of 0
which is what is expected at a short. Vr.g(t) is delayed by 90 degrees
from
Vf.g(t) so at 91 degrees, we should expect Vr.g(91) to be (-1 *
Vf.g(1)) or
-1.234. This is also how the spreadsheet computes it.

OK, I see now. You reversed the sign to calculate Vrs so once I get to the Vrs column, we are the same. You used the -1 reversal to calculate Vg, but plus one to calculate Vrs.

clip

On this sheet, Vtot = Vf+Vr, Itot=If+Ir and Ptot=Pf+Pr.
I hope no gremlins snuck in.

...Keith

I will go back and look some more at the power relationships from your earlier posting.


--
73, Roger, W7WKB

The Rest of the Story
 

On Wed, 2 Apr 2008 02:41:24 -0700 (PDT)
Keith Dysart wrote:

On Apr 1, 8:08 am, Cecil Moore wrote:
Keith Dysart wrote:
Cecil Moore wrote:
If the average interference is zero, the average
reflected power is dissipated in the source resistor.
All of your unethical lies, innuendo, and hand-waving
will not change that fact of physics.

More precisely, the average value of the imputed reflected
power is numerically equal to the increase in the average
power dissipated in the source resistor.

More precisely, at the *exact* time of arrival of the
reflected wave, the average power dissipated in the
source resistor increases by *exactly* the magnitude
of the average reflected power. Do you think that is
just a coincidence?

Not quite, but close. And averages can not change instantly.

But your phraseology suggests another way to approach
the problem. Let us consider what happens just after
the arrival of the reflected wave. For computational
convenience, we'll use a frequency of 1/360 hertz so
that 1 degree of the wave is one second long.

The reflected wave arrives at 90 seconds (or 90 degrees)
after the source is turned on. If the line is
terminated in 50 ohms, then no reflection arrives and
during the one second between degree 90 and degree 91,
the source resistor dissipates 0.01523 J of energy. The
line also receives 0.01523 J of energy and the source
provides 0.03046 J.
Esource.50[90..91] = 0.03046 J
Ers.50[90..91] = 0.01523 J
Eline.50[90..91] = 0.01523 J
Esource = Ers + Eline
as expected.
Efor = Eline
since Eref is 0.

OK, Power at the source is found from (V^2)/50 = 0.03046w. Another way to figure the power to the source would be by using the voltage and current through the source. Using current and voltage, the power at time 91 degrees of the reflected wave is 1.234v*1.43868a= 1.775w. Over one second integrated, the energy should be 1.775 J.

Now let us examine the case with a 12.5 ohm load and
a non-zero reflected wave.
During the one second between degree 90 and degree 91,
the source resistor dissipation with no reflection
is still 0.01523 J and the imputed reflected wave
provides 99.98477 J for a total of 100.00000 J.

But the source resistor actually absorbs 98.25503 J
in this interval.

100 - 1.775 = 98.225 J

The apparent impedance of the reflect wave has changed from 50 ohms to 0.8577 ohms.

There is still a confusion between what power comes from where at time 91 degrees, at least partly coming from how Vg is used. Or maybe the confusion is because we are mixing a circuit with a shorted line with calculations from a line with a 12.5 ohm load.


Ooooopppps. It does not add up. So the dissipation
in the source resistor went up, but not enough to
account for all of the imputed energy in the
reflected wave. This does not satisfy conservation
of energy, which should be sufficient to kill the
hypothesis.

But Esource = Ers + Eline as expected.
Esource.12.5[90..91] = -1.71451 J
Ers.12.5[90..91] = 98.25503 J
Eline.12.5[90..91] = -99.96954 J
Note that in the interval 90 to 91 degrees, the source
is absorbing energy. This is quite different than what
happens when there is no reflected wave.

clip
To recap, it is when a total flow is
broken into multiple non-zero constituent flows
that energy flow imputed to the constituents
is a dubious concept.

...Keith

The power seems pretty well accounted for degree by degree so far as I can see right now.

I think your spread sheet would be easier to follow if the reflected power beginning at 91 degrees was positive, like adding batteries in series. That is what happens and there is no reversal of voltage when examined from the perspective of the resistor. There is only a delay in time of connection to the voltage source on the line side of the resistor.

--
73, Roger, W7WKB

The Rest of the Story
 

On Apr 4, 1:29*am, Roger Sparks wrote:
On Wed, 2 Apr 2008 02:41:24 -0700 (PDT)

Keith Dysart wrote:
On Apr 1, 8:08 am, Cecil Moore wrote:
Keith Dysart wrote:
Cecil Moore wrote:
If the average interference is zero, the average
reflected power is dissipated in the source resistor.
All of your unethical lies, innuendo, and hand-waving
will not change that fact of physics.

More precisely, the average value of the imputed reflected
power is numerically equal to the increase in the average
power dissipated in the source resistor.

More precisely, at the *exact* time of arrival of the
reflected wave, the average power dissipated in the
source resistor increases by *exactly* the magnitude
of the average reflected power. Do you think that is
just a coincidence?

Not quite, but close. And averages can not change instantly.

But your phraseology suggests another way to approach
the problem. Let us consider what happens just after
the arrival of the reflected wave. For computational
convenience, we'll use a frequency of 1/360 hertz so
that 1 degree of the wave is one second long.

The reflected wave arrives at 90 seconds (or 90 degrees)
after the source is turned on. If the line is
terminated in 50 ohms, then no reflection arrives and
during the one second between degree 90 and degree 91,
the source resistor dissipates 0.01523 J of energy. The
line also receives 0.01523 J of energy and the source
provides 0.03046 J.
* Esource.50[90..91] = 0.03046 J
* Ers.50[90..91] * * = 0.01523 J
* Eline.50[90..91] * = 0.01523 J
* Esource = Ers + Eline
as expected.
* Efor = Eline
since Eref is 0.

OK, Power at the source is found from (V^2)/50 = 0.03046w. *

Not quite.

Another way to figure the power to the source would be by using the voltage and current through the source. *

This is how I did it.

Taking Esource.50[90..91] = 0.03046 J as an example ...

Psource.50[90] = V * I
= 0.000000 * 0.000000
= 0 W
Psource.50[91] = -2.468143 * -0.024681
= 0.060917 W

Using the trapezoid rule for numerical integration,
Esource.50[90..91] = ((Psource.50[90]+Psource.50[91])/2) * interval
= ((0+0.060917)/2)*1
= 0.030459 J

The other powers and energies in the spreadsheet are computed
similarly.

There was an error in the computation of the 'delta's; the sign was
wrong.
The spreadsheet available at http://keith.dysart.googlepages.com/radio6
has
now been corrected. (And, for Cecil, this spreadsheet no longer has
macros
so it may be downloadable.)

To use this spreadsheet to compute my numbers above, set the formulae
for Vr.g and Ir.g in the rows for degrees 90 and 91 to zero.

Using current and voltage, the power at time 91 degrees of the reflected wave is 1.234v*1.43868a= 1.775w. *Over one second integrated, the energy should be 1.775 J.

I could not match to the above data to any rows, so I can't comment.
But
perhaps the explanation above will correct the discrepancy.

Now let us examine the case with a 12.5 ohm load and
a non-zero reflected wave.

This was an ooooopppps. The spreadsheet actually calculates
for a shorted load.

During the one second between degree 90 and degree 91,
the source resistor dissipation with no reflection
is still 0.01523 J and the imputed reflected wave
provides 99.98477 J for a total of 100.00000 J.

But the source resistor actually absorbs 98.25503 J
in this interval.

100 - 1.775 = 98.225 J

I computed 98.25503 J from applying the trapezoid rule for
numerical integration of the power in the source resistor
at 90 degrees and 91 degrees.

I am not sure where you obtained 1.775 J.

The apparent impedance of the reflect wave has changed from 50 ohms to 0.8577 ohms.

There is still a confusion between what power comes from where at time 91 degrees, at least partly coming from how Vg is used. *Or maybe the confusion is because we are mixing a circuit with a shorted line with calculations from a line with a 12.5 ohm load.

I do not think the latter is happening. If you want to see
the results for a 12.5 ohm load, set the Reflection Coefficient
cell in row 1 to -0.6.

Ooooopppps. It does not add up. So the dissipation
in the source resistor went up, but not enough to
account for all of the imputed energy in the
reflected wave. This does not satisfy conservation
of energy, which should be sufficient to kill the
hypothesis.

But Esource = Ers + Eline as expected.
* Esource.12.5[90..91] = -1.71451 J
* Ers.12.5[90..91] * * = 98.25503 J
* Eline.12.5[90..91] * = -99.96954 J
Note that in the interval 90 to 91 degrees, the source
is absorbing energy. This is quite different than what
happens when there is no reflected wave.

clip
To recap, it is when a total flow is
broken into multiple non-zero constituent flows
that energy flow imputed to the constituents
is a dubious concept.

...Keith

The power seems pretty well accounted for degree by degree so far as I can see right now. *

I think your spread sheet would be easier to follow if the reflected power beginning at 91 degrees was positive, like adding batteries in series. *That is what happens and there is no reversal of voltage when examined from the perspective of the resistor. *There is only a delay in time of connection to the voltage source on the line side of the resistor.

When I wrote my original spreadsheet I had to decide which convention
to
use and settled on positive flow meant towards the load. Care is
certainly
needed in some of the computations. Choosing the other convention
would
have meant that care would be needed in a different set of
computations.
So I am not sure it would be less confusing, though it would be
different.

When using superposition, one has to pick a reference voltage and
current
direction for each component and then add or subtract the contributing
voltage or current depending on whether it is in the reference
direction
or against the reference direction. Mistakes and confusion with
relation
to the signs are not uncommon. Careful choice of reference directions
will
sometimes help, but mostly it simply leads to some other voltage or
current that needs to be subtracted instead of added.

...Keith

The Rest of the Story
 

On Fri, 4 Apr 2008 06:30:08 -0700 (PDT)
Keith Dysart wrote:

On Apr 4, 1:29*am, Roger Sparks wrote:
On Wed, 2 Apr 2008 02:41:24 -0700 (PDT)

Keith Dysart wrote:
On Apr 1, 8:08 am, Cecil Moore wrote:
Keith Dysart wrote:
Cecil Moore wrote:
If the average interference is zero, the average
reflected power is dissipated in the source resistor.
All of your unethical lies, innuendo, and hand-waving
will not change that fact of physics.

More precisely, the average value of the imputed reflected
power is numerically equal to the increase in the average
power dissipated in the source resistor.

More precisely, at the *exact* time of arrival of the
reflected wave, the average power dissipated in the
source resistor increases by *exactly* the magnitude
of the average reflected power. Do you think that is
just a coincidence?

Not quite, but close. And averages can not change instantly.

But your phraseology suggests another way to approach
the problem. Let us consider what happens just after
the arrival of the reflected wave. For computational
convenience, we'll use a frequency of 1/360 hertz so
that 1 degree of the wave is one second long.

The reflected wave arrives at 90 seconds (or 90 degrees)
after the source is turned on. If the line is
terminated in 50 ohms, then no reflection arrives and
during the one second between degree 90 and degree 91,
the source resistor dissipates 0.01523 J of energy. The
line also receives 0.01523 J of energy and the source
provides 0.03046 J.
* Esource.50[90..91] = 0.03046 J
* Ers.50[90..91] * * = 0.01523 J
* Eline.50[90..91] * = 0.01523 J
* Esource = Ers + Eline
as expected.
* Efor = Eline
since Eref is 0.

OK, Power at the source is found from (V^2)/50 = 0.03046w. *

Not quite.

Another way to figure the power to the source would be by using the voltage and current through the source. *

This is how I did it.

Taking Esource.50[90..91] = 0.03046 J as an example ...

Psource.50[90] = V * I
= 0.000000 * 0.000000
= 0 W
Psource.50[91] = -2.468143 * -0.024681
= 0.060917 W

Using the trapezoid rule for numerical integration,
Esource.50[90..91] = ((Psource.50[90]+Psource.50[91])/2) * interval
= ((0+0.060917)/2)*1
= 0.030459 J

The other powers and energies in the spreadsheet are computed
similarly.

There was an error in the computation of the 'delta's; the sign was
wrong.
The spreadsheet available at http://keith.dysart.googlepages.com/radio6
has
now been corrected. (And, for Cecil, this spreadsheet no longer has
macros
so it may be downloadable.)

To use this spreadsheet to compute my numbers above, set the formulae
for Vr.g and Ir.g in the rows for degrees 90 and 91 to zero.

Using current and voltage, the power at time 91 degrees of the reflected wave is 1.234v*1.43868a= 1.775w. *Over one second integrated, the energy should be 1.775 J.

I could not match to the above data to any rows, so I can't comment.
But
perhaps the explanation above will correct the discrepancy.

I took a look at your revised spreadsheet entitled "Reflected45degrees-1.xls". Using the numbers from row 94 (91 degrees), the voltage developed at the source would be Vs = -2.468143v. The current folowing through the source would be found from Ig which is 1.389317a in todays version of the spreadsheet (it was 1.439a previously). The power flowing INTO the source is 2.468143* 1.389317 = 3.429032w. This is the power Ps found in Column 11.

This returning power is all from the reflected wave. The source is acting like a resistor with an impedance of 2.468143/1.389317 = 1.776 ohms. As a result, the returning reflection does not truely see 50 ohms but sees 50 + 1.776 = 51.776 ohms. Of course the result is a another reflection. Is this the idea you were trying to communicate Cecil?

To me, this is destructive interference at work, so all the power in the reflected wave does not simply disappear into the resistor Rs on the instant basis.

clip
When using superposition, one has to pick a reference voltage and
current
direction for each component and then add or subtract the contributing
voltage or current depending on whether it is in the reference
direction
or against the reference direction. Mistakes and confusion with
relation
to the signs are not uncommon. Careful choice of reference directions
will
sometimes help, but mostly it simply leads to some other voltage or
current that needs to be subtracted instead of added.

...Keith

Yes. I don't want to force my conventions onto you, so I am trying to understand yours while insisting that the answers from each convention must be the same. I think we both agree that the reflections are carrying power now.
--
73, Roger, W7WKB

The Rest of the Story
 

Roger Sparks wrote:
Is this the idea you were trying to communicate Cecil?

What I am trying to communicate is that the distributed
network model is closer to Maxwell's equations that is
the lumped circuit model. If the lumped circuit model
disagrees with the distributed network model, then it
is wrong.

Steady-state conditions are identical whether the
ideal transmission line is zero wavelength or one
wavelength. If adding one wavelength of ideal
transmission between the source voltage and the
source resistance changes steady-state conditions
in Keith's mind, then there is something wrong in
Keith's mind.

To me, this is destructive interference at work, so all the

power in the reflected wave does not simply disappear into the

resistor Rs on the instant basis.

90 degrees later, an exactly equal magnitude of
constructive interference exists so it is obvious
that the constructive interference energy has been
delayed by 90 degrees from the destructive interference
energy.

One advantage of moving the source voltage one wavelength
away from the source resistor is that it is impossible for
the source to respond instantaneously to the requirements
of the source resistor thus making the energy flow easier
to track. (I'll be glad when I get my sight back so I can
read my calculator.)
--
73, Cecil http://www.w5dxp.com

The Rest of the Story
 

On Apr 4, 1:54 pm, Cecil Moore wrote:
Roger Sparks wrote:
Is this the idea you were trying to communicate Cecil?

What I am trying to communicate is that the distributed
network model is closer to Maxwell's equations that is
the lumped circuit model. If the lumped circuit model
disagrees with the distributed network model, then it
is wrong.

snip
(I'll be glad when I get my sight back so I can
read my calculator.)
--
73, Cecil http://www.w5dxp.com

Yes Cecil, if the above is a sample of your auguement on this subject
I suggest you stop posting until you get your spectacles back
Regards
Art

The Rest of the Story
 

On Apr 4, 12:01*am, Cecil Moore wrote:
Keith Dysart wrote:
It seems to me that this circuit is quite different.

Yet, steady-state conditions are identical. That
should tell us something about what is wrong
with the analysis so far.

Please expand on what it tells us "about what is wrong
with the analysis so far".

...Keith

The Rest of the Story
 

On Apr 4, 11:41*am, Roger Sparks wrote:
On Fri, 4 Apr 2008 06:30:08 -0700 (PDT)

Keith Dysart wrote:
On Apr 4, 1:29*am, Roger Sparks wrote:
[snip]
Another way to figure the power to the source would be by using the voltage and current through the source. *

This is how I did it.

Taking Esource.50[90..91] = 0.03046 J as an example ...

Psource.50[90] = V * I
* * * * * * * *= 0.000000 * 0.000000
* * * * * * * *= 0 W
Psource.50[91] = -2.468143 * -0.024681
* * * * * * * *= 0.060917 W

Using the trapezoid rule for numerical integration,
Esource.50[90..91] = ((Psource.50[90]+Psource.50[91])/2) * interval
* * * * * * * * * *= ((0+0.060917)/2)*1
* * * * * * * * * *= 0.030459 J

The other powers and energies in the spreadsheet are computed
similarly.

There was an error in the computation of the 'delta's; the sign was
wrong.
The spreadsheet available athttp://keith.dysart.googlepages.com/radio6
has
now been corrected. (And, for Cecil, this spreadsheet no longer has
macros
so it may be downloadable.)

To use this spreadsheet to compute my numbers above, set the formulae
for Vr.g and Ir.g in the rows for degrees 90 and 91 to zero.

Using current and voltage, the power at time 91 degrees of the reflected wave is 1.234v*1.43868a= 1.775w. *Over one second integrated, the energy should be 1.775 J.

I could not match to the above data to any rows, so I can't comment.
But
perhaps the explanation above will correct the discrepancy.

I took a look at your revised spreadsheet entitled "Reflected45degrees-1.xls". Using the numbers from row 94 (91 degrees), the voltage developed at the source would be Vs = -2.468143v. *The current folowing through the source would be found from Ig which is 1.389317a in todays version of the spreadsheet (it was 1.439a previously). *The power flowing INTO the source is 2..468143* 1.389317 = 3.429032w. *This is the power Ps found in Column 11. *

This returning power is all from the reflected wave. *

I would not say this. The power *is* from the line, but this is Pg,
and it satisfies
the equation
Ps(t) = Prs(t) + Pg(t)

The imputed power in the reflected wave is Pr.g(t) and is equal to
-99.969541 W, at
91 degrees. This can not be accounted for in any combination of Ps(91)
(-3.429023 W)
and Prs(91) (96.510050 W). And recall that expressing Cecil's claim
using instantaneous
powers requires that the imputed reflected power be accounted for in
the source
resistor, and not the source. This is column 26 and would require that
Prs(91) equal
100 W (which it does not).

The source is acting like a resistor with an impedance of 2.468143/1.389317 = 1.776 ohms. *

This is not a good way to describe the source. The ratio of the
voltage to the current
is 1.776 but this is not a resistor since if circuit conditions were
to change, the
voltage would stay the same while the current could take on any value;
this being
the definition of a voltage source. Since the voltage does not change
when the current
does, deltaV/deltaI is always 0 so the voltage source is more properly
described as
having an impedance of 0.

As a result, the returning reflection does not truely see 50 ohms but sees 50 + 1.776 = 51.776 ohms. *

The returning reflection is affectively a change in the circuit
conditions. Using
the source impedance of 0 plus the 50 ohm resistor means the
reflection sees 50
ohms, so there is no reflection.

Using your approach of computing a resistance from the instantaneous
voltage and
current yeilds a constantly changing resistance. The reflection would
alter this
computed resistance. This change in resistance would then alter the
reflection
which would change the resistance. Would the answer converge?

The only approach that works is to use the conventional approach of
considering
that a voltage source has an impedance of 0.

Of course the result is a another reflection. *Is this the idea you were trying to communicate Cecil?

To me, this is destructive interference at work, so all the power in the reflected wave does not simply disappear into the resistor Rs on the instant basis. *

I agree with latter, but not for the reason expressed. Rather, because
the imputed
power of the reflected wave is a dubious concept. This being because
it is impossible
to account for this power.

clip

When using superposition, one has to pick a reference voltage and
current
direction for each component and then add or subtract the contributing
voltage or current depending on whether it is in the reference
direction
or against the reference direction. Mistakes and confusion with
relation
to the signs are not uncommon. Careful choice of reference directions
will
sometimes help, but mostly it simply leads to some other voltage or
current that needs to be subtracted instead of added.

...Keith

Yes. *I don't want to force my conventions onto you, so I am trying to understand yours while insisting that the answers from each convention must be the same. *

I think we both agree that the reflections are carrying power now.

Not I. Not until the imputed power can be accounted for.

...Keith

The Rest of the Story
 

On Apr 4, 2:54*pm, Cecil Moore wrote:
Roger Sparks wrote:
Is this the idea you were trying to communicate Cecil?

What I am trying to communicate is that the distributed
network model is closer to Maxwell's equations that is
the lumped circuit model. If the lumped circuit model
disagrees with the distributed network model, then it
is wrong.

The joys of motherhood statements.

Steady-state conditions are identical whether the
ideal transmission line is zero wavelength or one
wavelength. If adding one wavelength of ideal
transmission between the source voltage and the
source resistance changes steady-state conditions
in Keith's mind, then there is something wrong in
Keith's mind.

There was an 'if' there, wasn't there? Do you think
the 'if' is satisfied? Or not? The rest is useless
without knowing.

To me, this is destructive interference at work, so all the
power in the reflected wave does not simply disappear into the
resistor Rs on the instant basis.

90 degrees later, an exactly equal magnitude of
constructive interference exists so it is obvious
that the constructive interference energy has been
delayed by 90 degrees from the destructive interference
energy.

You still have to explain where this destructive energy is stored
for those 90 degrees. Please identify the element and its energy
flow as a function of time.

One advantage of moving the source voltage one wavelength
away from the source resistor is that it is impossible for
the source to respond instantaneously

You have previously claimed that the steady-state conditions
are the same (which I agree), but now you have moved to discussing
transients, for which the behaviour is quite different.

If you want to claim similarity, then you need to allow the
circuit to settle to steady state after any change. Instantaneous
response is not required if the analysis is only steady-state.

If you wish to study transient responses, then the circuits do
not behave similarly.

...Keith

The Rest of the Story
 

Keith Dysart wrote:
Please expand on what it tells us "about what is wrong
with the analysis so far".

You have not been able to tell where the instantaneous
reflected energy goes. This new example should help
you solve your problem. If your steady-state equations
for the new example are not identical to the steady-
state equations for the previous example, then something
is wrong with your previous analysis. IMO, something
is obviously wrong with your previous analysis since
it requires reflected waves to contain something other
than ExH joules/sec, a violation of the laws of EM
wave physics.

feedline1 feedline2
source---1WL 50 ohm---Rs---1WL 50 ohm---+j50

The beauty of this example is that conditions at
the source resistor, Rs, are isolated from any
source of energy other than the ExH forward wave
energy in feedline1 and the ExH reflected wave
energy in feedline2. That's all the energy there
is available at Rs and that energy cannot be denied.
--
73, Cecil http://www.w5dxp.com

The Rest of the Story
 

Keith Dysart wrote:
And recall that expressing Cecil's claim
using instantaneous
powers requires that the imputed reflected power be accounted for in
the source
resistor, and not the source.

Keith, I hope that Roger knows you are uttering falsehoods
about what I have said.

I. My Part 1 claim applies *ONLY* to a zero interference
precondition. Your example contains interference. Therefore,
my claim does NOT apply to your example. Examples containing
interference are yet to be covered in Parts 2 and 3 of
my series of articles. I am going to state my claims once
again.

1. If zero interference exists at the source resistor, all of
the reflected energy is dissipated in the source resistor.
Your example does NOT contain zero interference.

2. If interference exists at the source resistor, the energy
associated with the interference flows to/from the source
and/or to/from the load. This claim covers the present
example under discussion.

II. I have said many times that the source can adjust
its output to compensate for the destructive interference
and constructive interference in the system. It is ONLY
under zero interference conditions that all of the
reflected energy is dissipated in the source resistor.
You have failed to offer an example where that assertion
is not true GIVEN THE ZERO INTERFERENCE PRECONDITION.

The example under discussion is not covered by any of
my Part 1 claims.
--
73, Cecil http://www.w5dxp.com