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

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

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

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, 11:06*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:
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

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

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

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?