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

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

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

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

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

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

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

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

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

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