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

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

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

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

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

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

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

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

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

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