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

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

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

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

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

Under the laws of physics governing transmission lines
inserting an ideal 1WL line does not change the steady-
state conditions. If you think it does, you have invented
some new laws of physics.

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.

Your request is beyond the scope of my Part 1 article.
If interference exists at the source resistor, the energy
associated with the interference flows to/from the source
and/or to/from the load. That condition is NOT covered in
my Part 1 article. Please stand by for Part 2 which will
explain destructive interference and Part 3 which will
explain constructive interference.

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),

Glad you agree so there is nothing stopping you from an
analysis of the following example:

source---1WL 50 ohm---Rs---1WL 50 ohm---+j50
Pfor1-- Pfor2--
--Pref1 --Pref2

Make Rs a 4-terminal network and a standard s-parameter
analysis is possible.

but now you have moved to discussing
transients, for which the behaviour is quite different.

Nope, you are confused. I am saying absolutely nothing about
transients. Why do you think an instantaneous power analysis
during steady-state is not possible?

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.

Are you saying that an analysis of instantaneous power
does not apply during steady-state? If that is true,
then all of your earlier analysis involving transmission
lines is bogus.
--
73, Cecil http://www.w5dxp.com

On Apr 5, 10:02*am, Cecil Moore wrote:
Keith Dysart wrote:
There was an 'if' there, wasn't there? Do you think
the 'if' is satisfied? Or not? The rest is useless
without knowing.

Under the laws of physics governing transmission lines
inserting an ideal 1WL line does not change the steady-
state conditions. If you think it does, you have invented
some new laws of physics.

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.

Your request is beyond the scope of my Part 1 article.
If interference exists at the source resistor, the energy
associated with the interference flows to/from the source
and/or to/from the load. That condition is NOT covered in
my Part 1 article. Please stand by for Part 2 which will
explain destructive interference and Part 3 which will
explain constructive interference.

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),

Glad you agree so there is nothing stopping you from an
analysis of the following example:

True, but as you say, the results will be the same.

source---1WL 50 ohm---Rs---1WL 50 ohm---+j50
* * * * * *Pfor1-- * * * * *Pfor2--
* * * * * *--Pref1 * * * * *--Pref2

Make Rs a 4-terminal network and a standard s-parameter
analysis is possible.

Yes, but that would be an average analysis and we have already
seen how averages mislead.

but now you have moved to discussing
transients, for which the behaviour is quite different.

Nope, you are confused. I am saying absolutely nothing about
transients.

You did say: "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."

The words "respond instantaneously" suggested transient, rather
than waiting for the system to settle.

Why do you think an instantaneous power analysis
during steady-state is not possible?

Haven't said that. In fact, I think that is what I have been
doing.

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.

...Keith

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

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

On Sat, 5 Apr 2008 03:06:18 -0700 (PDT)
Keith Dysart wrote:

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).

Taking these numbers and adding the Psource.50[91] = -2.468143 * -0.024681*= 0.060917 W found previously above,
we have Total = 0.060917 + 3.429023 + 96.510050 = 99.99999. Very close to 100w, but I am not sure of exactly what I am adding here.

I think you are right, that we should always be able to add the power on the line to 100w if we truely account for the power at any instant. This results from the trig identity that sin(x) + cos(x) = 1. Sin(x+90)= cos(x) so sin(x)+sin(x+90)= 1. This explains why the sum of the forward power and reflected power should always equal 100w.

Pg(t) is the result of a standing wave, containing power from Pf(x) and Pr(x+90). Only the power from Pr(x+90) is available to at a later time Prs(t+90+delta). Power from Pf(x+delta) is found in the transmission line.


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.

The overriding issue is to account for all the power, which we are having a hard time doing. The current is flowing the wrong way for the source voltage at Ps(91) so the source is absorbing power. That negates the idea that the source has an impedance of zero when we also assign the source a voltage.

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.

Does it help to notice that we are applying the power through two wires/paths. We then divide the power in half and say that the power to the resistor from the close half flows seamlessly from source to resistor. On the other hand, the second half of the source power flows through the transmission line and we say that it may never get to the resistor. Something is wrong with this logic.

If we can't account for the power, it is because we are doing the accounting incorrectly.


clip


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

I am convinced that accounting for the power can be done on an instant basis, but neither of us has done it on a mutually acceptable basis yet. Maybe if we pursue the trig identity
sin(x) + sin(x+90) = 1, we can have a constant reference and avoid mixing forward and reflected powers?
--
73, Roger, W7WKB

correction
On Sat, 5 Apr 2008 07:06:22 -0700
Roger Sparks wrote:

On Sat, 5 Apr 2008 03:06:18 -0700 (PDT)
Keith Dysart wrote:

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).

Taking these numbers and adding the Psource.50[91] = -2.468143 * -0.024681*= 0.060917 W found previously above,
we have Total = 0.060917 + 3.429023 + 96.510050 = 99.99999. Very close to 100w, but I am not sure of exactly what I am adding here.

I think you are right, that we should always be able to add the power on the line to 100w if we truely account for the power at any instant. This results from the trig identity that sin(x) + cos(x) = 1. Sin(x+90)= cos(x) so sin(x)+sin(x+90)= 1. This explains why the sum of the forward power and reflected power should always equal 100w.

Pg(t) is the result of a standing wave, containing power from Pf(x) and Pr(x+90). Only the power from Pr(x+90) is available to Rs at a later time Prs(t+90+delta). Power from Pf(x+delta) is found in the transmission line.
I omitted the Rs in "available to Rs at a later time Prs(t+90+delta)" Sorry!


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.

The overriding issue is to account for all the power, which we are having a hard time doing. The current is flowing the wrong way for the source voltage at Ps(91) so the source is absorbing power. That negates the idea that the source has an impedance of zero when we also assign the source a voltage.

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.

Does it help to notice that we are applying the power through two wires/paths. We then divide the power in half and say that the power to the resistor from the close half flows seamlessly from source to resistor. On the other hand, the second half of the source power flows through the transmission line and we say that it may never get to the resistor. Something is wrong with this logic.

If we can't account for the power, it is because we are doing the accounting incorrectly.


clip


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

I am convinced that accounting for the power can be done on an instant basis, but neither of us has done it on a mutually acceptable basis yet. Maybe if we pursue the trig identity
sin(x) + sin(x+90) = 1, we can have a constant reference and avoid mixing forward and reflected powers?
--
73, Roger, W7WKB


--
73, Roger, W7WKB