The Rest of the Story
 

After discovering the error on Roy's web page at:

http://eznec.com/misc/Food_for_thought.pdf

I have begun a series of articles that convey "The Rest
of the Story" (Apologies to Paul Harvey). Part 1 of
these articles can be found at:

http://www.w5dxp.com/nointfr.htm

Stand by for the other three articles.
--
73, Cecil http://www.w5dxp.com

The Rest of the Story
 

On Mar 4, 3:36*pm, Cecil Moore wrote:
After discovering the error on Roy's web page at:

http://eznec.com/misc/Food_for_thought.pdf

I have begun a series of articles that convey "The Rest
of the Story" (Apologies to Paul Harvey). Part 1 of
these articles can be found at:

http://www.w5dxp.com/nointfr.htm

Looks good. And well presented. There is only one small problem
with the analysis.

When the instantaneous energy flows are examined it can be seen
that Prs is not equal to 50 W plus Pref.

Taking just the second example (12.5 ohm load) for illustrative
purposes...

The power dissipated in Rs before the reflection arrives is
Prs.before = 50 + 50cos(2wt) watts

The reflected power at the source is
Pref.s = 18 + 18cos(2wt) watts

But the power dissipated in Rs after the reflection arrives is
Prs.after = 68 + 68cos(2wt-61.9degrees) watts

Prs.after is not Prs.before + Pref, though the averages do sum.

And since the energy flows must be accounted for on a moment by
moment basis (or we violate conservation of energy), it is the
instantaneous energy flows that provide the most detail and
allow us to conclude with certainty that Prs.after is not
Prs.before + Pref.

The same inequality holds for all the examples except those
with Pref equal to 0.

Thus these examples do not demonstrate that the reflected power
is dissipated in the source resistor.

...Keith

The Rest of the Story
 

On Mar 4, 8:00*pm, Keith Dysart wrote:
On Mar 4, 3:36*pm, Cecil Moore wrote:

After discovering the error on Roy's web page at:

http://eznec.com/misc/Food_for_thought.pdf

I have begun a series of articles that convey "The Rest
of the Story" (Apologies to Paul Harvey). Part 1 of
these articles can be found at:

http://www.w5dxp.com/nointfr.htm

Looks good. And well presented. There is only one small problem
with the analysis.

When the instantaneous energy flows are examined it can be seen
that Prs is not equal to 50 W plus Pref.

Taking just the second example (12.5 ohm load) for illustrative
purposes...

The power dissipated in Rs before the reflection arrives is
Prs.before = 50 + 50cos(2wt) watts

The reflected power at the source is
Pref.s = 18 + 18cos(2wt) watts

But the power dissipated in Rs after the reflection arrives is
Prs.after = 68 + 68cos(2wt-61.9degrees) watts

Prs.after is not Prs.before + Pref, though the averages do sum.

And since the energy flows must be accounted for on a moment by
moment basis (or we violate conservation of energy), it is the
instantaneous energy flows that provide the most detail and
allow us to conclude with certainty that Prs.after is not
Prs.before + Pref.

The same inequality holds for all the examples except those
with Pref equal to 0.

Thus these examples do not demonstrate that the reflected power
is dissipated in the source resistor.

...Keith

I should have mentioned that there are some energy flows that do
add, as expected.

The energy delivered by the generator to the line (or equivalently,
the energy flowing in the line at the generator end) is the sum
of the forward and reverse energy flows...

Pf.g = 50 + 50cos(2wt)
Pr.g = -18 + 18cos(2wt)
Pline.g = 32 + 68cos(2wt)

And the energy delivered by the source is always equal to the
energy being dissipated in the resistor plus the energy being
delived to the line...

Prs = 68 + 68cos(2wt-61.9degrees)
Rline.g = 32 + 68cos(2wt)

Psource = 100 + 116.6cos(2wt-30.96degrees)

Psource is equal to Prs + Pline.g

So the energy flows that should add up, do add up.

...Keith

The Rest of the Story
 

Keith Dysart wrote:
When the instantaneous energy flows are examined it can be seen
that Prs is not equal to 50 W plus Pref.

That is irrelevant. The power (irradiance) model
doesn't apply to instantaneous energy and power.
Hecht says as much in "Optics".

Nobody has ever claimed that the energy/power
analysis applies to instantaneous values. The
energy/power values are all based on RMS voltages
and currents. There is no such thing as an
instantaneous RMS value.
--
73, Cecil http://www.w5dxp.com

The Rest of the Story
 

Keith Dysart wrote:
Thus these examples do not demonstrate that the reflected power
is dissipated in the source resistor.

There are no reflections at the source so the reflected
energy flows through the source resistor. There is no
interference to redistribute any energy. There is no
other place for the reflected energy to go.
--
73, Cecil http://www.w5dxp.com

The Rest of the Story
 

Cecil Moore wrote:
Keith Dysart wrote:
When the instantaneous energy flows are examined it can be seen
that Prs is not equal to 50 W plus Pref.

That is irrelevant. The power (irradiance) model
doesn't apply to instantaneous energy and power.
Hecht says as much in "Optics".

Nobody has ever claimed that the energy/power
analysis applies to instantaneous values. The
energy/power values are all based on RMS voltages
and currents. There is no such thing as an
instantaneous RMS value.

Interesting.

Do you also use only the RMS phase and RMS interference to come up with
your RMS answers?

8-)

73,
Gene
W4SZ

The Rest of the Story
 

On Mar 4, 10:25*pm, Cecil Moore wrote:
Keith Dysart wrote:
When the instantaneous energy flows are examined it can be seen
that Prs is not equal to 50 W plus Pref.

That is irrelevant. The power (irradiance) model
doesn't apply to instantaneous energy and power.
Hecht says as much in "Optics".

Nobody has ever claimed that the energy/power
analysis applies to instantaneous values.

Are you saying that conservation of energy does NOT apply to
instantaneous values?

The
energy/power values are all based on RMS voltages
and currents. There is no such thing as an
instantaneous RMS value.

I understand the analysis technique you are proposing. That
it leads to the wrong conclusion can be easily seen when the
instantaneous energy flows are studied. This merely demonstrates
that the analysis technique has its limitations.

The bottom line remains that the reflected energy is not
dissipated in the source resistor, even for the special cases
under discussion.

...Keith

The Rest of the Story
 

On Mar 4, 10:27*pm, Cecil Moore wrote:
Keith Dysart wrote:
Thus these examples do not demonstrate that the reflected power
is dissipated in the source resistor.

There are no reflections at the source so the reflected
energy flows through the source resistor. There is no
interference to redistribute any energy. There is no
other place for the reflected energy to go.

That is the conundrum, isn't it?

And yet the analysis of instantaneous energy flows definitely
shows that the reflected energy is not the energy being dissipated
in the source resistor.

Encountering this conundrum, and not wanting to give up on
conservation of energy, is what helped me form my views on
the nature of reflected energy.

...Keith

The Rest of the Story
 

Keith Dysart wrote:
....
The bottom line remains that the reflected energy is not
dissipated in the source resistor, even for the special cases
under discussion.

...Keith

What if the source resistor is of finite length :)
Alan

The Rest of the Story
 

On Tue, 4 Mar 2008 17:00:31 -0800 (PST)
Keith Dysart wrote:

On Mar 4, 3:36*pm, Cecil Moore wrote:
After discovering the error on Roy's web page at:

http://eznec.com/misc/Food_for_thought.pdf

I have begun a series of articles that convey "The Rest
of the Story" (Apologies to Paul Harvey). Part 1 of
these articles can be found at:

http://www.w5dxp.com/nointfr.htm

Looks good. And well presented. There is only one small problem
with the analysis.

When the instantaneous energy flows are examined it can be seen
that Prs is not equal to 50 W plus Pref.

Taking just the second example (12.5 ohm load) for illustrative
purposes...

The power dissipated in Rs before the reflection arrives is
Prs.before = 50 + 50cos(2wt) watts

How do you justify this conclusion? It seems to me that until the reflection arrives, the source will have a load of Rs = 50 ohms plus line = 50 ohms for a total of 100 ohms. Prs.before = 50 watts.

There would be no reflected power at the source until the reflection returns, making the following statements incorrect.


The reflected power at the source is
Pref.s = 18 + 18cos(2wt) watts

But the power dissipated in Rs after the reflection arrives is
Prs.after = 68 + 68cos(2wt-61.9degrees) watts

Prs.after is not Prs.before + Pref, though the averages do sum.

And since the energy flows must be accounted for on a moment by
moment basis (or we violate conservation of energy), it is the
instantaneous energy flows that provide the most detail and
allow us to conclude with certainty that Prs.after is not
Prs.before + Pref.

The same inequality holds for all the examples except those
with Pref equal to 0.

Thus these examples do not demonstrate that the reflected power
is dissipated in the source resistor.

...Keith

Rs is shown as a resistance on only one side of the line. It would simplify and focus the discussion if Rs were broken into two resistors, each placed on one side of the source to make the circuit balanced.
--
73, Roger, W7WKB

The Rest of the Story
 

On Mar 5, 8:12*am, Roger Sparks wrote:
On Tue, 4 Mar 2008 17:00:31 -0800 (PST)

Keith Dysart wrote:
On Mar 4, 3:36*pm, Cecil Moore wrote:
After discovering the error on Roy's web page at:

http://eznec.com/misc/Food_for_thought.pdf

I have begun a series of articles that convey "The Rest
of the Story" (Apologies to Paul Harvey). Part 1 of
these articles can be found at:

http://www.w5dxp.com/nointfr.htm

Looks good. And well presented. There is only one small problem
with the analysis.

When the instantaneous energy flows are examined it can be seen
that Prs is not equal to 50 W plus Pref.

Taking just the second example (12.5 ohm load) for illustrative
purposes...

The power dissipated in Rs before the reflection arrives is
Prs.before = 50 + 50cos(2wt) watts

How do you justify this conclusion? *It seems to me that until the reflection arrives, the source will have a load of Rs = 50 ohms plus line = 50 ohms for a total of 100 ohms. *Prs.before = 50 watts.

The voltage on Rs, before the reflection returns is
Vrs(t) = 70.7cos(wt)
Prs(t) = Vrs(t)**2/50
= 50 + 50cos(2wt)

Prs.before.average = average(Prs(t))
= 50
since the average of cos is 0.

My apologies for leaving out the "(t)" everywhere which would have
made it clearer.

There would be no reflected power at the source until the reflection returns, making the following statements incorrect.

I should have been more clear. The below applies after the reflected
wave returns.
And I should have included the "(t)" for greater clarity.

The reflected power at the source is
Pref.s = 18 + 18cos(2wt) watts

But the power dissipated in Rs after the reflection arrives is
Prs.after = 68 + 68cos(2wt-61.9degrees) watts

Prs.after is not Prs.before + Pref, though the averages do sum.

And since the energy flows must be accounted for on a moment by
moment basis (or we violate conservation of energy), it is the
instantaneous energy flows that provide the most detail and
allow us to conclude with certainty that Prs.after is not
Prs.before + Pref.

The same inequality holds for all the examples except those
with Pref equal to 0.

Thus these examples do not demonstrate that the reflected power
is dissipated in the source resistor.

...Keith

Rs is shown as a resistance on only one side of the line. *It would simplify and focus the discussion if Rs were broken into two resistors, each placed on one side of the source to make the circuit balanced. *

I am not sure how this would help. It would make the arithmetic
somewhat more
complex.

...Keith

The Rest of the Story
 

Gene Fuller wrote:
Cecil Moore wrote:
Keith Dysart wrote:
When the instantaneous energy flows are examined it can be seen
that Prs is not equal to 50 W plus Pref.

That is irrelevant. The power (irradiance) model
doesn't apply to instantaneous energy and power.
Hecht says as much in "Optics".

Nobody has ever claimed that the energy/power
analysis applies to instantaneous values. The
energy/power values are all based on RMS voltages
and currents. There is no such thing as an
instantaneous RMS value.

Interesting.

What is interesting is that in the formula for power
dissipated in the source resistor, the 50 watts is
an average power. It is an invalid procedure to try
to add instantaneous power to an average power.

Do you also use only the RMS phase and RMS interference to come up with
your RMS answers?

I didn't say anything about "RMS phase and RMS interference".
The phase angle used in Hecht's irradiance equation is the
phase between the electric fields of the two waves. The
magnitude of the interference is an average magnitude based
on the RMS values of voltage and current.

I do exactly what Eugene Hecht did in "Optics". He said:

"Furthermore, since the power
arriving cannot be measured instantaneously, the detector
must integrate the energy flux over some finite time, 'T'.
If the quantity to be measured is the net energy per unit
area received, it depends on 'T' and is therefore of limited
utility. If however, the 'T' is now divided out, a highly
practical quantity results, one that corresponds to the
average energy per unit area per unit time, namely 'I'."
'I' is the irradiance (*AVERAGE* power density).

i.e. The irradiance/interference equation does not work
for instantaneous powers which are "of limited utility".
--
73, Cecil http://www.w5dxp.com

The Rest of the Story
 

Keith Dysart wrote:
On Mar 4, 10:25 pm, Cecil Moore wrote:
Nobody has ever claimed that the energy/power
analysis applies to instantaneous values.

Are you saying that conservation of energy does NOT apply to
instantaneous values?

Of course not. I am saying that the 50 watts in the
source resistor power equation is an average power.
It is invalid to try to add instantaneous power to
an average power, as you tried to do.

The
energy/power values are all based on RMS voltages
and currents. There is no such thing as an
instantaneous RMS value.

I understand the analysis technique you are proposing. That
it leads to the wrong conclusion can be easily seen when the
instantaneous energy flows are studied. This merely demonstrates
that the analysis technique has its limitations.

The proposed analysis technique is a tool. Trying to apply
that tool to instantaneous powers is like trying to use a
DC ohm-meter to measure the feedpoint impedance of an antenna.
Only a fool would attempt such a thing.

The bottom line remains that the reflected energy is not
dissipated in the source resistor, even for the special cases
under discussion.

Saying it doesn't make it so, Keith. There is nothing to
keep the reflected energy from being dissipated in the
source resistor.

If the reflected energy is not dissipated in the source
resistor, where does it go? Please be specific because it
is obvious to me that there is nowhere else for it to go
in the special zero interference case presented.

Here are some of your choices:

1. Reflected energy flows through the resistor and into
the ground without being dissipated.

2. The 50 ohm resistor re-reflects the reflected energy
back toward the load. (Please explain how a 50 ohm load
on 50 ohm coax can cause a reflection.)

3. There's no such thing as reflected energy.

4. Reflected waves exist without energy.

5. ______________________________________________.
--
73, Cecil http://www.w5dxp.com

The Rest of the Story
 

Keith Dysart wrote:
On Mar 4, 10:27 pm, Cecil Moore wrote:
Keith Dysart wrote:
Thus these examples do not demonstrate that the reflected power
is dissipated in the source resistor.
There are no reflections at the source so the reflected
energy flows through the source resistor. There is no
interference to redistribute any energy. There is no
other place for the reflected energy to go.

That is the conundrum, isn't it?

And yet the analysis of instantaneous energy flows definitely
shows that the reflected energy is not the energy being dissipated
in the source resistor.

Your analysis seems to be flawed. You are adding average
power terms to instantaneous power terms which is mixing
apples and oranges.
--
73, Cecil http://www.w5dxp.com

The Rest of the Story
 

Cecil Moore wrote:
Keith Dysart wrote:
On Mar 4, 10:25 pm, Cecil Moore wrote:
Nobody has ever claimed that the energy/power
analysis applies to instantaneous values.

Are you saying that conservation of energy does NOT apply to
instantaneous values?

Of course not. I am saying that the 50 watts in the
source resistor power equation is an average power.
It is invalid to try to add instantaneous power to
an average power, as you tried to do.

The
energy/power values are all based on RMS voltages
and currents. There is no such thing as an
instantaneous RMS value.

I understand the analysis technique you are proposing. That
it leads to the wrong conclusion can be easily seen when the
instantaneous energy flows are studied. This merely demonstrates
that the analysis technique has its limitations.

The proposed analysis technique is a tool. Trying to apply
that tool to instantaneous powers is like trying to use a
DC ohm-meter to measure the feedpoint impedance of an antenna.
Only a fool would attempt such a thing.

The bottom line remains that the reflected energy is not
dissipated in the source resistor, even for the special cases
under discussion.

Saying it doesn't make it so, Keith. There is nothing to
keep the reflected energy from being dissipated in the
source resistor.

If the reflected energy is not dissipated in the source
resistor, where does it go? Please be specific because it
is obvious to me that there is nowhere else for it to go
in the special zero interference case presented.

Here are some of your choices:

1. Reflected energy flows through the resistor and into
the ground without being dissipated.

2. The 50 ohm resistor re-reflects the reflected energy
back toward the load. (Please explain how a 50 ohm load
on 50 ohm coax can cause a reflection.)

3. There's no such thing as reflected energy.

4. Reflected waves exist without energy.

5. ______________________________________________.


Gentlemen;

What I was taught long ago was that the phenomenon that we call
reflected energy is radiated on the return. That portion of energy that
is not radiated is re-reflected and is radiated. This continues until
the level of energy no longer supports radiation. Resistance also
dissipates a portion of the transmitted energy. That resistance includes
the resistor under discussion above and that resistance found in the
coax conductors.

Of course my Elmer could have been wrong.


Dave WD9BDZ

The Rest of the Story
 

David G. Nagel wrote:
What I was taught long ago was that the phenomenon that we call
reflected energy is radiated on the return. That portion of energy that
is not radiated is re-reflected and is radiated. This continues until
the level of energy no longer supports radiation. Resistance also
dissipates a portion of the transmitted energy. That resistance includes
the resistor under discussion above and that resistance found in the
coax conductors.

Of course my Elmer could have been wrong.

Your Elmer was parroting the party line which is:
Any reflected energy dissipated in the source was
never sourced in the first place. Therefore,
at the source (by convention and by definition):

Sourced power = forward power - reflected power

If that is true, it follows that all reflected power
must necessarily be re-reflected back toward the load
(even if, in the process, it violates the laws of physics
governing the reflection model). Since contradictions
don't exist in reality, there must be another explanation.

An antenna tuner which achieves a Z0-match allows no
reflected energy to be incident upon the source so, for
that most common configuration, all is well and your
Elmer was right about those Z0-matched systems.

However, when reflected energy is allowed to reach
the source, it is naive to think that none of that
reflected energy is ever dissipated in the source
resistance when the source resistance is dissipative
as it is in the example under discussion here.

Both of the following assertions are false:
1. Reflected energy is never dissipated in the source.
2. Reflected energy is always dissipated in the source.
Most assertions containing the words "always" and "never"
are false.

There will be three more parts on this topic published
on my web page. The top page will be on the subject
of interference which will explain how reflected energy
can be redistributed back toward the load after not
being re-reflected.
--
73, Cecil http://www.w5dxp.com

The Rest of the Story
 

On Mar 5, 11:11*am, Cecil Moore wrote:
Keith Dysart wrote:
On Mar 4, 10:27 pm, Cecil Moore wrote:
Keith Dysart wrote:
Thus these examples do not demonstrate that the reflected power
is dissipated in the source resistor.
There are no reflections at the source so the reflected
energy flows through the source resistor. There is no
interference to redistribute any energy. There is no
other place for the reflected energy to go.

That is the conundrum, isn't it?

And yet the analysis of instantaneous energy flows definitely
shows that the reflected energy is not the energy being dissipated
in the source resistor.

Your analysis seems to be flawed. You are adding average
power terms to instantaneous power terms which is mixing
apples and oranges.

I do not think that is the case.

The expression for instantaneous power in Rs before the reflection
(or, if you prefer, when a 50 ohm load is used), is

Prs(t) = 50 + 50cos(2wt)

It is trivial to compute the average of this since the average
of a sine wave is 0, but that does not make the expression the
sum of an average and an instantaneous power.

As an exercise, compute the power in a 50 ohm resistor that has
a 100 volt sine wave across at, that is
V(t) = 100 cos(wt)

You will find the result is of the form shown above.

So when you add the instantaneous power in Rs before the reflection
arrives with the instantaneous power from the reflection it will
not sum to the instantaneous power dissipated in Rs after the
reflection returns.

Thus conveniently showing that for this example, the reflected
power is not dissipated in Rs.

...Keith

The Rest of the Story
 

On Mar 5, 11:06*am, Cecil Moore wrote:
Keith Dysart wrote:
On Mar 4, 10:25 pm, Cecil Moore wrote:
Nobody has ever claimed that the energy/power
analysis applies to instantaneous values.

Are you saying that conservation of energy does NOT apply to
instantaneous values?

Of course not. I am saying that the 50 watts in the
source resistor power equation is an average power.
It is invalid to try to add instantaneous power to
an average power, as you tried to do.

The
energy/power values are all based on RMS voltages
and currents. There is no such thing as an
instantaneous RMS value.

I understand the analysis technique you are proposing. That
it leads to the wrong conclusion can be easily seen when the
instantaneous energy flows are studied. This merely demonstrates
that the analysis technique has its limitations.

The proposed analysis technique is a tool. Trying to apply
that tool to instantaneous powers is like trying to use a
DC ohm-meter to measure the feedpoint impedance of an antenna.
Only a fool would attempt such a thing.

You used your tool to attempt to show that the reflected power
is dissipated in Rs.

I did a finer grained analysis using instantaneous power to show
that it is not.

The use of averages in analysis can be misleading.

The bottom line remains that the reflected energy is not
dissipated in the source resistor, even for the special cases
under discussion.

Saying it doesn't make it so, Keith. There is nothing to
keep the reflected energy from being dissipated in the
source resistor.

If the reflected energy is not dissipated in the source
resistor, where does it go? Please be specific because it
is obvious to me that there is nowhere else for it to go
in the special zero interference case presented.

Here are some of your choices:

1. Reflected energy flows through the resistor and into
the ground without being dissipated.

2. The 50 ohm resistor re-reflects the reflected energy
back toward the load. (Please explain how a 50 ohm load
on 50 ohm coax can cause a reflection.)

3. There's no such thing as reflected energy.

4. Reflected waves exist without energy.

5. ______________________________________________.

Now you have got the issue. Since the reflected power is
not dissipated in Rs, the answer must be one of 1 to 5.

5. is probably the best choice.

And that is why it became necessary to rethink the nature
of energy in reflected waves.

...Keith

The Rest of the Story
 

Keith Dysart wrote:
Thus conveniently showing that for this example, the reflected
power is not dissipated in Rs.

The *average* reflected power is certainly dissipated in Rs
because there is nowhere else for it to go. Your instantaneous
power, according to Eugene Hecht, is "of limited utility" which
you have proved with your straw man assertion above.

I have made no assertions about instantaneous power. All of
my assertions have been about average power and you have proved
none of my assertions about average power to be false.

Here is what you are doing:
Cecil: My GMC pickup is white.
Keith: No, your GMC pickup has black tires.

Your diversions are obvious. Instantaneous power is irrelevant
to my assertions.
--
73, Cecil http://www.w5dxp.com

The Rest of the Story
 

Keith Dysart wrote:
You used your tool to attempt to show that the reflected power
is dissipated in Rs.

The tool proves that the reflected average power is dissipated
in Rs because it has no where else to go. If instantaneous
reflected power were relevant, why don't we read about it in
any of the technical textbooks under the wave reflection model?

I did a finer grained analysis using instantaneous power to show
that it is not.

My tool is known not to work for instantaneous power and
was never intended to work for instantaneous power. So your
argument is just a straw man diversion. Eugene Hecht explains
why average power density (irradiance) must used instead of
instantaneous power.

"Furthermore, since the power
arriving cannot be measured instantaneously, the detector
must integrate the energy flux over some finite time, 'T'.
If the quantity to be measured is the net energy per unit
area received, it depends on 'T' and is therefore of limited
utility. If however, the 'T' is now divided out, a highly
practical quantity results, one that corresponds to the
average energy per unit area per unit time, namely 'I'."
'I' is the irradiance (*AVERAGE* power density).

The use of averages in analysis can be misleading.

The misuse of a tool, designed to be used only with
averages, can be even more misleading. When you measure
an open-circuit using a DC ohm-meter on a dipole, are
you really going to argue that the DC ohm-meter is not
working properly? That's exactly what you are arguing here.
When one misuses a tool, as you are doing, one will get
invalid results. There's no mystery about that at all.

You are saying that the energy model, designed to be
used with average powers, does not work for instantaneous
values. When you try to use it for instantaneous values,
you are committing a well understood error. Why do you
insist on committing that error?

5. ______________________________________________.

Now you have got the issue. Since the reflected power is
not dissipated in Rs, the answer must be one of 1 to 5.

5. is probably the best choice.

Until you fill in the blank for number 5, you are just
firing blanks. :-) Exactly what laws of physics are you
intending to violate with your explanation?

And that is why it became necessary to rethink the nature
of energy in reflected waves.

Nope, it's not. Reflected waves obey the laws of superposition
and reflection physics. That's all you need to understand.
Now new laws of physics or logical diversions required.
--
73, Cecil http://www.w5dxp.com

The Rest of the Story
 

On Mar 5, 1:58*pm, Cecil Moore wrote:
Keith Dysart wrote:
Thus conveniently showing that for this example, the reflected
power is not dissipated in Rs.

The *average* reflected power is certainly dissipated in Rs
because there is nowhere else for it to go. Your instantaneous
power, according to Eugene Hecht, is "of limited utility" which
you have proved with your straw man assertion above.

Hecht seems to have sufficient reputation that I trust that he made
this statement in the context of optics and not in the context of
electrical circuits.

I have made no assertions about instantaneous power. All of
my assertions have been about average power and you have proved
none of my assertions about average power to be false.

True. But analysis using instantaneous power reveals a different
answer.
Which is more likely to be correct?

Here is what you are doing:
Cecil: My GMC pickup is white.
Keith: No, your GMC pickup has black tires.

Your diversions are obvious. Instantaneous power is irrelevant
to my assertions.

Only if you give up on conservation of energy in instantaneous flows.

...Keith

The Rest of the Story
 

Keith Dysart wrote:
Hecht seems to have sufficient reputation that I trust that he made
this statement in the context of optics and not in the context of
electrical circuits.

EM waves are EM waves, Keith, no matter what the frequency.
EM waves all obey the laws of reflection physics, superposition,
and conservation of energy principle. If you want to prove those
laws to be invalid and replace them with ones of your own design,
be our guest.

But analysis using instantaneous power reveals a different
answer. Which is more likely to be correct?

Analysis using an MFJ-259B on an antenna system reveals a
different impedance than is indicated by a DC ohm-meter.
So what? You used an average power tool, known to be
invalid for instantaneous powers, to incorrectly analyze
instantaneous powers. You are the one who made the error
- not the model. Your error was the (deliberate?) misuse
of the tool in order to try to create your straw man.

Using the power equation, derived from RMS values of
voltage, on instantaneous powers is an invalid thing
to do and will give known erroneous results which are
not the fault of the average power model. The fault is
in the *misuse* of the average power model.

We have already laid your straw man argument to rest
when we discussed the power in standing waves.

1. There is non-zero instantaneous power in standing waves.
2. There is zero average power in standing waves.

Does statement 1 contradict statement 2? Of course not.
They are both true. The same holds true for the present
discussion.

I have a probable explanation for your calculations. I
set the example up such that the average interference
is zero inside the source. It is entirely possible that
localized interference exists within each cycle such
that there is destructive interference for part of the
cycle and constructive interference in another part of
the cycle. In fact, based on the conservation of energy
principle, I am willing to state that is a fact and the
destructive interference magnitude exactly equals the
constructive interference magnitude for each cycle.
--
73, Cecil http://www.w5dxp.com

The Rest of the Story
 

On Mar 5, 8:06 am, Cecil Moore wrote:
....blah, blah...

So consider the case of a section of lossless uniform transmission
line of characteristic impedance R0, which I write as R instead of Z
since it of course must be real-valued, connected between two sources
S1 at end 1 and S2 at end 2. These sources each have source impedance
R0: they are perfectly matched to the characteristic impedance of the
line. The line is long enough that we can observe any standing waves
that may be on it. (For believers in directional couplers, that can
be short indeed, but it does not need to be short.) Source S1 is set
to output a sinusoidal signal of amplitude A1 into a matched load, on
frequency f1. Similarly S2 outputs a sinusoidal signal A2 into a
matched load at frequency f2, which is distinct from f1.

It is easy to show mathematically, and to measure in practice, that
the amplitude of the frequency f1 is constant along the line, and
similarly that the amplitude of the frequency f2 is constant along the
line. That is to say, there is no standing wave at either frequency.
Energy at f1 travels on the line only in the direction from S1 to S2,
and vice-versa for f2.

That says to me that the energy on the line at f1 is absorbed entirely
by source S2, and the energy at f2 is absorbed entirely by S1, with no
reflection at the boundaries between S1 and the line, and the line and
S2.

At this point, I leave it as an exercise for the reader to interpret
or explain exactly what is meant by "absorbed by." This may involve
understanding that in a Thevenin or Norton simple model of each
source, the energy delivered by the voltage or current source at any
moment in time may not equal that which it would deliver into a
matched load at the same point in the cycle...

Cheers,
Tom

The Rest of the Story
 

K7ITM wrote:
So consider the case of a section of lossless uniform transmission
line of characteristic impedance R0, which I write as R instead of Z
since it of course must be real-valued, connected between two sources
S1 at end 1 and S2 at end 2. These sources each have source impedance
R0: they are perfectly matched to the characteristic impedance of the
line. The line is long enough that we can observe any standing waves
that may be on it. (For believers in directional couplers, that can
be short indeed, but it does not need to be short.) Source S1 is set
to output a sinusoidal signal of amplitude A1 into a matched load, on
frequency f1. Similarly S2 outputs a sinusoidal signal A2 into a
matched load at frequency f2, which is distinct from f1.

What you have described is a system with two sources which
are incapable of interfering with each other because they
are not coherent. Note that this example bears zero resemblance
to a system where the sources are coherent, i.e. frequency-
locked and phase-locked and therefore, capable of interference.

It is easy to show mathematically, and to measure in practice, that
the amplitude of the frequency f1 is constant along the line, and
similarly that the amplitude of the frequency f2 is constant along the
line. That is to say, there is no standing wave at either frequency.
Energy at f1 travels on the line only in the direction from S1 to S2,
and vice-versa for f2.

Obviously true for non-coherent sources.

That says to me that the energy on the line at f1 is absorbed entirely
by source S2, and the energy at f2 is absorbed entirely by S1, with no
reflection at the boundaries between S1 and the line, and the line and
S2.

Obviously true for non-coherent sources.

Unfortunately, "non-coherent sources" is not the subject of
this discussion. The rules change between non-coherent, non-inter-
fering sources and coherent, interfering sources. I suggest you
reference the "Interference" chapter in "Optics", by Hecht.
--
73, Cecil http://www.w5dxp.com

The Rest of the Story
 

Cecil Moore wrote:
Keith Dysart wrote:
Hecht seems to have sufficient reputation that I trust that he made
this statement in the context of optics and not in the context of
electrical circuits.

EM waves are EM waves, Keith, no matter what the frequency.
EM waves all obey the laws of reflection physics, superposition,
and conservation of energy principle. If you want to prove those
laws to be invalid and replace them with ones of your own design,
be our guest.

Cecil,

It is likely that all of these interference-related items you like to
quote from Hecht are cast in an environment of lossless optical
components. The characteristic impedance is set by the index of
refraction of the various layers, but none of the optical layers have
any absorption.

Soooo, how does any of this optical stuff extend to making arguments
about the absorption or re-reflection of energy in the source resistor
for the HF case?

The laws of physics may be inviolate, but it is not quite so clear that
your derived and extended models share the same characteristic.

73,
Gene
W4SZ

The Rest of the Story
 

On Mar 5, 1:27 pm, Cecil Moore wrote:
The rules change between non-coherent,
non-interfering sources and coherent, interfering sources.

And exactly which part of "linear system" do you fail to understand?

The Rest of the Story
 

Gene Fuller wrote:
Soooo, how does any of this optical stuff extend to making arguments
about the absorption or re-reflection of energy in the source resistor
for the HF case?

What happens at a flat black interface in optics?
--
73, Cecil http://www.w5dxp.com

The Rest of the Story
 

K7ITM wrote:
On Mar 5, 1:27 pm, Cecil Moore wrote:
The rules change between non-coherent,
non-interfering sources and coherent, interfering sources.

And exactly which part of "linear system" do you fail to understand?

When two equal amplitude coherent signals are superposed,
the resulting power density can be four times the power
density of one of the single waves due to constructive
interference.

That is not true for two equal amplitude non-coherent
waves of different frequencies. The interference term
averages out to zero so there are no bright rings and
dark rings.

I'm surprised you don't know that.
--
73, Cecil http://www.w5dxp.com

The Rest of the Story
 

On Mar 5, 5:25 pm, Cecil Moore wrote:
K7ITM wrote:
On Mar 5, 1:27 pm, Cecil Moore wrote:
The rules change between non-coherent,
non-interfering sources and coherent, interfering sources.

And exactly which part of "linear system" do you fail to understand?

When two equal amplitude coherent signals are superposed,
the resulting power density can be four times the power
density of one of the single waves due to constructive
interference.

That is not true for two equal amplitude non-coherent
waves of different frequencies. The interference term
averages out to zero so there are no bright rings and
dark rings.


Yes, yes, you've posted all that a billion times before in this NG.
Now, exactly what part of "linear system" do you fail to understand?

(I might also ask why you're going to so much trouble to be
disagreeable with something that agrees with what you were
posting...but I think I already know the answer to that one.)

The Rest of the Story
 

Cecil Moore wrote:
Gene Fuller wrote:
Soooo, how does any of this optical stuff extend to making arguments
about the absorption or re-reflection of energy in the source resistor
for the HF case?

What happens at a flat black interface in optics?

And that would be relevant in what manner? Does Hecht discuss
interference at flat black interfaces?

Nobody is questioning the laws of physics or Hecht's writings. Many are
questioning your extensions to your own models. Name-dropping and
invoking the sacred laws of physics do not automatically validate your
models.

Did you ever wonder why all of the basic phenomena, both optical and RF,
were known to the "ancients", yet you are the first one to pull
everything together in this miraculous new version of a reflection model?

73,
Gene
W4SZ

The Rest of the Story
 

K7ITM wrote:
Now, exactly what part of "linear system" do you fail to understand?

Have you stopped beating your wife?
--
73, Cecil http://www.w5dxp.com

The Rest of the Story
 

Gene Fuller wrote:
Did you ever wonder why all of the basic phenomena, both optical and RF,
were known to the "ancients", yet you are the first one to pull
everything together in this miraculous new version of a reflection model?

I am the one quoting the wisdom of those ancients
which seems to have somehow fallen by the wayside
and been replaced by some pseudo scientific religion.
What is it about the conservation of energy principle
that you disagree with? What is it about the wave
reflection model that you disagree with? What is it
about the principle of superposition that you disagree
with?
--
73, Cecil http://www.w5dxp.com

The Rest of the Story
 

On Wed, 5 Mar 2008 06:06:04 -0800 (PST)
Keith Dysart wrote:

On Mar 5, 8:12*am, Roger Sparks wrote:
On Tue, 4 Mar 2008 17:00:31 -0800 (PST)

Keith Dysart wrote:
On Mar 4, 3:36*pm, Cecil Moore wrote:
After discovering the error on Roy's web page at:

http://eznec.com/misc/Food_for_thought.pdf

I have begun a series of articles that convey "The Rest
of the Story" (Apologies to Paul Harvey). Part 1 of
these articles can be found at:

http://www.w5dxp.com/nointfr.htm

Looks good. And well presented. There is only one small problem
with the analysis.

When the instantaneous energy flows are examined it can be seen
that Prs is not equal to 50 W plus Pref.

Taking just the second example (12.5 ohm load) for illustrative
purposes...

The power dissipated in Rs before the reflection arrives is
Prs.before = 50 + 50cos(2wt) watts

How do you justify this conclusion? *It seems to me that until the reflection arrives, the source will have a load of Rs = 50 ohms plus line = 50 ohms for a total of 100 ohms. *Prs.before = 50 watts.

The voltage on Rs, before the reflection returns is
Vrs(t) = 70.7cos(wt)
Prs(t) = Vrs(t)**2/50
= 50 + 50cos(2wt)

Prs.before.average = average(Prs(t))
= 50
since the average of cos is 0.

For the equation Vrs(t) = 70.7cos(wt), are you finding the 70.7 from sqrt(50^2 + 50^2)?

If so, how do you justify that proceedure before the reflection returns? I think the voltage across Rs is 50v until the reflection returns. I also think the current would be 1 amp, for power of 50w, until the reflection returns.


My apologies for leaving out the "(t)" everywhere which would have
made it clearer.

There would be no reflected power at the source until the reflection returns, making the following statements incorrect.

I should have been more clear. The below applies after the reflected
wave returns.
And I should have included the "(t)" for greater clarity.

The reflected power at the source is
Pref.s = 18 + 18cos(2wt) watts

But the power dissipated in Rs after the reflection arrives is
Prs.after = 68 + 68cos(2wt-61.9degrees) watts

Prs.after is not Prs.before + Pref, though the averages do sum.

And since the energy flows must be accounted for on a moment by
moment basis (or we violate conservation of energy), it is the
instantaneous energy flows that provide the most detail and
allow us to conclude with certainty that Prs.after is not
Prs.before + Pref.

The same inequality holds for all the examples except those
with Pref equal to 0.

Thus these examples do not demonstrate that the reflected power
is dissipated in the source resistor.

...Keith

Rs is shown as a resistance on only one side of the line. *It would simplify and focus the discussion if Rs were broken into two resistors, each placed on one side of the source to make the circuit balanced. *

I am not sure how this would help. It would make the arithmetic
somewhat more
complex.

...Keith

Well, as drawn, the circuit is unbalanced. Reflections are about time and distance traveled during that time, so we really don't know where the far center point is located for finding instantaneous values. We are making an assumption that the length of the source and resistor rs is zero which is OK, but we need to be aware that we are making that assumption. By splitting the resistor into two parts, and then adding them together to make the calculations, we can see that a balanced circuit is intended, but the source resistors are combined for ease of calculation.

--
73, Roger, W7WKB

The Rest of the Story
 

Cecil Moore wrote:
K7ITM wrote:
Now, exactly what part of "linear system" do you fail to understand?

Have you stopped beating your wife?

(I might also ask why you're going to so much trouble to be
disagreeable with something that agrees with what you were
posting...but I think I already know the answer to that one.)

Tom, we were getting along quite well before you asked
your leading question, obviously designed to elicit
anger, not just once but twice.
--
73, Cecil http://www.w5dxp.com

The Rest of the Story
 

Keith Dysart wrote:
So the energy flows that should add up, do add up.

How did you take care of the fact that the forward wave
and reflected wave are flowing in opposite directions
through the resistor?

How did you take care of the 90 degree phase difference
between the forward wave and the reflected wave through
the resistor?
--
73, Cecil http://www.w5dxp.com

The Rest of the Story
 

On Mar 6, 12:04*am, Roger Sparks wrote:
On Wed, 5 Mar 2008 06:06:04 -0800 (PST)

Keith Dysart wrote:
On Mar 5, 8:12*am, Roger Sparks wrote:
On Tue, 4 Mar 2008 17:00:31 -0800 (PST)

Keith Dysart wrote:
On Mar 4, 3:36*pm, Cecil Moore wrote:
After discovering the error on Roy's web page at:

http://eznec.com/misc/Food_for_thought.pdf

I have begun a series of articles that convey "The Rest
of the Story" (Apologies to Paul Harvey). Part 1 of
these articles can be found at:

http://www.w5dxp.com/nointfr.htm

Looks good. And well presented. There is only one small problem
with the analysis.

When the instantaneous energy flows are examined it can be seen
that Prs is not equal to 50 W plus Pref.

Taking just the second example (12.5 ohm load) for illustrative
purposes...

The power dissipated in Rs before the reflection arrives is
Prs.before = 50 + 50cos(2wt) watts

How do you justify this conclusion? *It seems to me that until the reflection arrives, the source will have a load of Rs = 50 ohms plus line = 50 ohms for a total of 100 ohms. *Prs.before = 50 watts.

The voltage on Rs, before the reflection returns is
Vrs(t) = 70.7cos(wt)
Prs(t) = Vrs(t)**2/50
* * * *= 50 + 50cos(2wt)

Prs.before.average = average(Prs(t))
* * * * * * * * * *= 50
since the average of cos is 0.

For the equation Vrs(t) = 70.7cos(wt), are you finding the 70.7 from sqrt(50^2 + 50^2)?

No. Cecil's circuit has a 100 V RMS source. For this source

Vs(t) = 141.4 cos(wt)

Before the reflection returns, the line acts as 50 ohms. This, in
series with
the 50 source resistor means that Vs(t)/2 is across Rs and Vs(t)/2
appears
across the line.

Vrs(t) = Vf.g(t) = Vs(t)/2 = 70.7 cos(wt)

The power dissipated in the resistor is

Prs(t) = Vrs(t)**2 / 50
= ((70.7 * 70.7) / 50) * cos(wt) * cos(wt)
= 100 * 0.5 * (cos(wt+wt) + cos(wt-wt))
= 50 (cos(2wt) + 1)
= 50 cos(2wt) + 50

If so, how do you justify that proceedure before the reflection returns? *I think the voltage across Rs is 50v until the reflection returns. *I also think the current would be 1 amp, for power of 50w, until the reflection returns.

That is the average power. But since the voltage is a sinusoid, the
instantaneous power as a function of time is
Prs(t) = 50 + 50 cos(2wt)
which does average to 50 W, thus agreeing with your calculations.


My apologies for leaving out the "(t)" everywhere which would have
made it clearer.

There would be no reflected power at the source until the reflection returns, making the following statements incorrect.

I should have been more clear. The below applies after the reflected
wave returns.
And I should have included the "(t)" for greater clarity.

The reflected power at the source is
Pref.s = 18 + 18cos(2wt) watts

But the power dissipated in Rs after the reflection arrives is
Prs.after = 68 + 68cos(2wt-61.9degrees) watts

Prs.after is not Prs.before + Pref, though the averages do sum.

And since the energy flows must be accounted for on a moment by
moment basis (or we violate conservation of energy), it is the
instantaneous energy flows that provide the most detail and
allow us to conclude with certainty that Prs.after is not
Prs.before + Pref.

The same inequality holds for all the examples except those
with Pref equal to 0.

Thus these examples do not demonstrate that the reflected power
is dissipated in the source resistor.

...Keith

Rs is shown as a resistance on only one side of the line. *It would simplify and focus the discussion if Rs were broken into two resistors, each placed on one side of the source to make the circuit balanced. *

I am not sure how this would help. It would make the arithmetic
somewhat more
complex.

...Keith

Well, as drawn, the circuit is unbalanced. *Reflections are about time and distance traveled during that time, so we really don't know where the far center point is located for finding instantaneous values. *We are making an assumption that the length of the source and resistor rs is zero which is OK, but we need to be aware that we are making that assumption. *

I find it valuable to understand how such an ideal circuit would
operate, then
extend the solution if necessary.

In practice, many circuits are small compared to the wavelength and
these assumptions
do not materially affect the answer.

Of course, it is important to know when they do, at which time more
complex analysis
becomes necessary.

By splitting the resistor into two parts, and then adding them together to make the calculations, we can see that a balanced circuit is intended, but the source resistors are combined for ease of calculation.

...Keith

The Rest of the Story
 

On Wed, 5 Mar 2008 21:37:06 -0800 (PST)
Keith Dysart wrote:

On Mar 6, 12:04*am, Roger Sparks wrote:
On Wed, 5 Mar 2008 06:06:04 -0800 (PST)

Keith Dysart wrote:
On Mar 5, 8:12*am, Roger Sparks wrote:
On Tue, 4 Mar 2008 17:00:31 -0800 (PST)

Keith Dysart wrote:
On Mar 4, 3:36*pm, Cecil Moore wrote:
After discovering the error on Roy's web page at:

http://eznec.com/misc/Food_for_thought.pdf

I have begun a series of articles that convey "The Rest
of the Story" (Apologies to Paul Harvey). Part 1 of
these articles can be found at:

http://www.w5dxp.com/nointfr.htm

Looks good. And well presented. There is only one small problem
with the analysis.

Thanks Keith. I see what you are doing now, although I still don't understand your logic in faulting Cecil on the instantaneous values. I agree with you that the instantaneous values can be tracked, but don't see a fault in Cecil's presentation.

Roy and I went around a few times on whether the source reflects in a case like this. The source reflection controls whether the 50 ohm source resistor acts like 50 ohms to the reflected wave, or acts like a short circuit in parallel with the 50 ohm source resistor.

--
73, Roger, W7WKB

The Rest of the Story
 

Roger Sparks wrote:
Thanks Keith. I see what you are doing now, although I still
don't understand your logic in faulting Cecil on the instantaneous
values. I agree with you that the instantaneous values can be
tracked, but don't see a fault in Cecil's presentation.

I made no assertions about instantaneous power at all and
have added a note in my article to that effect. My assertions
about instantaneous power cannot possibly be wrong because
I didn't make any. :-) Nothing in my article applies to or
is concerned with instantaneous power.

For the power density/irradiance/interference equation to
be applicable, certain conditions must be met. One of those
conditions is that all component powers must be *average*
powers resembling power density/irradiance from optics.
Another condition is that the phase angle between the two
waves being superposed must be constant and therefore the
two associated waves must be coherent (phase-locked) with
each other.

Roy and I went around a few times on whether the source reflects
in a case like this. The source reflection controls whether the
50 ohm source resistor acts like 50 ohms to the reflected wave,
or acts like a short circuit in parallel with the 50 ohm source
resistor.

What Roy (and others) are missing is that there is more than
one mechanism in physics that can cause a redistribution of
reflected energy back toward the load. An ordinary reflection
is not the only cause. In a one-dimensional environment,
e.g. a transmission line, there is an additional mechanism
present that can redirect and redistribute the reflected
energy back toward the load.

1. Reflection - what happens when a *single wave* encounters
an impedance discontinuity. Some (or all) of the reflected
energy reverses direction.

2. Wave interaction - what happens when *two waves* superpose,
interact, AND effect a redistribution of their energy components
as described on the FSU web page at:

http://micro.magnet.fsu.edu/primer/j...ons/index.html

"... when two waves of equal amplitude and wavelength that
are 180-degrees ... out of phase with each other meet, they
are not actually annihilated, ... All of the photon energy
present in these waves must somehow be recovered or
redistributed in a new direction, according to the law of
energy conservation ... Instead, upon meeting, the photons
are redistributed to regions that permit constructive
interference, so the effect should be considered as a
*redistribution of light waves and photon energy* ..."

In the simple ideal voltage source described in my article,
there are no reflections because the source resistance equals
the characteristic impedance of the transmission line.

In Part 1 of the article, there is also no wave interaction
because the forward wave and reflected wave are 90 degrees
out of phase. So for that special case, none of the reflected
energy is redistributed back toward the load. Therefore, for
that special case, all of the reflected energy is dissipated
in the source resistor because all conditions for a redistribution
of the reflected energy have been eliminated.

In the special case described in Part 1, because of the 90
degree phase difference, the forward wave and reflected
wave are completely independent of each other almost as if
they were not coherent. The result in that special case is
that the power components can simply be added because in
that special case, (V1^2 + V2^2) = (V1 + V2)^2, something
that is obviously NOT true in the general case.

Part 2 of the article will describe what happens when
the forward wave and reflected wave interact at the source
resistor and effect a redistribution of reflected energy
back toward the load *even when there are no reflections*.
This is the key concept, understood for many decades in the
field of optical physics, that most RF people seem to be
missing.
--
73, Cecil http://www.w5dxp.com

The Rest of the Story
 

On Mar 6, 1:17*am, Roger Sparks wrote:
On Wed, 5 Mar 2008 21:37:06 -0800 (PST)
[snip]
Thanks Keith. *I see what you are doing now, although I still don't understand your logic in faulting Cecil on the instantaneous values. *I agree with you that the instantaneous values can be tracked, but don't see a fault in Cecil's presentation. *

For the special situation described in
http://www.w5dxp.com/nointfr.htm
Cecil is attempting to show that the reflected energy is dissipated in
the source resistor.

The logic he employs is:
- before the reflection arrives back at the generator, the source
resistor
is dissipating X watts.
- the reflected wave has an energy flow of Y watts.
- after the reflection arrives back at the generator, the source
resistor
is dissipating Z watts.
- since Z is equal to X + Y, the energy in the reflected wave is being
dissipated in the source resistor.

In other words, since the dissipation in the source resistor increases
by the same amount as the power in the reflected wave, the energy in
the reflected wave must be being dissipated in the source resistor.

Cecil analyzes the circuit for a number of load resistances and
suggests
that the equality holds for any load resistance.

For example, with a load resistance of 12.5 ohms, the original
dissipation
in the source resistor is 50 W which increases to 68 W when the 18 W
reflected wave arrives back at the generator. That is, X = 50, Y = 18
and Z = 68, so Z is equal to X + Y.

Cecil does all of this analysis using average powers.

But we know that the power dissipation varies as a function of time
and
that the power in the reflected wave is a function of time. It is my
contention that if it is the energy in the reflected wave that is
increasing the dissipation in the source resistor, the dissipation in
source resistor should occur at the same time that the reflected wave
delivers the energy.

In other words, not only should Z.average = X.average + Y.average,
but Z.instantaneous should equal X.instantaneous + Y.instantaneous
for if the dissipation in the source resistor is not tracking the
energy in the reflected wave, it can not be the energy in the
reflected wave that is heating the resistor.

So using the same 12.5 ohm example,
X.inst = 50 + 50 cos(2wt)
Y.inst = 18 + 18 cos(2wt)
X.inst + Y.inst = 68 + 68 cos(2wt)
but
Z.inst = 68 + 68 cos(2wt - 61.9degrees)

So Z.inst is not equal to Y.inst + X.inst.

This means that the dissipation in the resistor is not happening
at the same time as the energy is being delivered by the reflected
wave, which must mean that it is not the energy from the reflected
wave that is heating the source resistor.

So while analyzing average power dissipations suggests that the
energy from the reflected wave is dissipated in the source resistor,
analysis of the instantaneous power shows that it is not.

...Keith

The Rest of the Story
 

Keith Dysart wrote:
Cecil does all of this analysis using average powers.

That's because the power density/irradiance/interference
equation only works for average powers. One condition
for valid use of that equation is that the powers must
have been averaged over one complete cycle and usually
over many, many cycles. How many cycles does it take
for a bright interference ring to register on the human
retina?

My example is set up such that the average interference
is zero over each complete cycle. I cannot think of a
way to eliminate interference internal to a cycle. Both
destructive and constructive interference occur during a
cycle, which is what you are seeing, but interference
averages out to zero over each complete cycle.

In other words, not only should Z.average = X.average + Y.average,
but Z.instantaneous should equal X.instantaneous + Y.instantaneous
for if the dissipation in the source resistor is not tracking the
energy in the reflected wave, it can not be the energy in the
reflected wave that is heating the resistor.

Destructive interference from one part of the cycle is
being delivered as constructive interference in another
part of the cycle. That is completely normal.

The same thing happens with an ideal standing wave. The
average power in a standing wave equals zero although
some instantaneous power can be calculated as existing
in the standing wave. You have discovered why Eugene
Hecht says that instantaneous power is "of limited utility".

In fact, what we are dealing with in the example is a
standing wave because the forward wave and reflected
wave are flowing in opposite directions through the
source resistor. I hope your math took that into account.

This means that the dissipation in the resistor is not happening
at the same time as the energy is being delivered by the reflected
wave, which must mean that it is not the energy from the reflected
wave that is heating the source resistor.

All it means is that there must be some localized interference
during each cycle for which you have failed to account.
In other words, you are superposing powers which is a
known invalid thing to do when interference is present.
--
73, Cecil http://www.w5dxp.com