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