On Jun 8, 5:57*am, Keith Dysart wrote:
As soon as one assigns tangible energy to the reflected wave, it
becomes reasonable to ask for an accounting of this energy and
the model is incapable of properly accounting for the energy.

Then one is using the wrong model which is typical of many RF
engineers. Optical physicists have been accounting for reflected
energy for decades. That one is ignorant of where reflected energy
goes is not a good reason to abandon the law of physics that says that
*ALL* EM waves contain ExH energy, including reflected EM waves. EM
waves cannot exist without ExH energy so you might as well say that
reflected waves do not exist at all because that is the logical
conclusion based on your false premises.

Optical physicists have been aware of the power-density/interference
equation for a long, long time. It is covered in any good reference
book on optics, e.g. by Hecht and by Born and Wolf. I have posted the
details - why do choose to remain ignorant? At least take time to
comprehend the technical information available from the field of
optics and report back to us why many decades of that EM wave
knowledge is wrong. Here is the equation that explains where the
reflected energy goes. All you have to to is track the energy back in
time.

Ptot = P1 + P2 + 2*SQRT(P1*P2)*cos(A)

where A is the angle between the two voltages. The last term is known
as the "interference term".

When two coherent, collimated, signals interfere while traveling in
the same direction in a transmission line: If they are 90 degrees
apart, there is zero interference and the I and Q components are
easily recovered.

If they are less than 90 degrees apart, the interference is
constructive and the phasor superposition of the two waves results in
*more power in the total superposed wave* than in the arithmetic sum
of the two powers. The extra energy has to come from somewhere. In the
absence of a local source, the extra energy has to come from
destructive interference in the opposite direction, e.g. at a Z0-
match.

If they are between 90 degrees and 180 degrees apart, the interference
is destructive and the phasor superposition of the two waves results
in *less power in the total superposed wave* than in the arithmetic
sum of the two waves. The "left-over" energy has to go somewhere so it
has to be delivered in the opposite direction to the area that
supports constructive interference.

Here is what Hecht said: "Briefly then, optical (EM wave) interference
corresponds to the interaction of two or more (EM) lightwaves yielding
a resultant irradiance (power density) that deviates from the sum of
the component irradiances."

Hecht's statement hints to where the reflected energy goes. In the
case of wave cancellation at a Z0-match that eliminates ExH reflected
energy flowing toward the load, all of the ExH reflected energy is
recovered and redistributed back toward the load.

Following this weakness back through the model, the root cause
is the attempt to assign tangible energy to the reflected wave.
Think of it as a reflected voltage or current wave and all will
be well, but assign power to it and eventually incorrect
conclusions will be drawn.

Only if the model is inadequate. Optical physicists assign tangible
energy to reflected waves all the time. Many are probably reading
these postings and laughing at the collective ignorance about EM
waves.

Take a look at this web page:

http://www.teachspin.com/instruments...eriments.shtml

Scroll down to, "Using Dielectric Beamsplitters to find the "missing
energy" in destructive interference". There it is, all spelled out for
you - where the reflected energy goes. Reflected energy is never
actually missing - what are missing are a few of the brain cells that
need to be used to think about reflected energy. :-)

For those who understand this, and know that "where did the
reflected energy go?" is an invalid question, using the power
model within its limits will not cause difficulties. But for
those who are not careful, great difficulties arise and a lot
of fancy dancing is offered to work around the difficulties,
unsuccessfully.

Yes, throughout history, ignorant people have brought great
difficulties upon themselves because they refuse to alleviate their
ignorance. Those who refuse to learn from their mistakes are destined
to repeat those same mistakes. And you were making these exact same
mistakes years ago. You can lead a horse to water ...

In his Nov/Dec 2001 QEX article, Dr. Best attempted to explain where
the reflected energy goes. Shackled by ignorance, he got many things
wrong e.g. phantom EM waves that exist without energy. But the article
alludes to the conceptual path that needs to be taken to alleviate
that ignorance. In "Wave Mechanics of Transmission Lines, Part 3", Dr.
Best published the above power-density/interference equation although
he ignorantly asserted on this newsgroup that interference did not
exist.

Interference resulting from superposition at an impedance
discontinuity is the key missing link in explaining everything that
happens to the energy in a transmission line including Roy's food-for-
thought article. Why is there so much reluctance to adopt a proven law
of EM wave physics from the field of optics?
--
73, Cecil, w5dxp.com

On Jun 8, 2:14*pm, Cecil Moore wrote:
On Jun 8, 5:57*am, Keith Dysart wrote:

As soon as one assigns tangible energy to the reflected wave, it
becomes reasonable to ask for an accounting of this energy and
the model is incapable of properly accounting for the energy.

Then one is using the wrong model which is typical of many RF
engineers. Optical physicists have been accounting for reflected
energy for decades. That one is ignorant of where reflected energy
goes is not a good reason to abandon the law of physics that says that
*ALL* EM waves contain ExH energy, including reflected EM waves. EM
waves cannot exist without ExH energy so you might as well say that
reflected waves do not exist at all because that is the logical
conclusion based on your false premises.

Optical physicists have been aware of the power-density/interference
equation for a long, long time. It is covered in any good reference
book on optics, e.g. by Hecht and by Born and Wolf. I have posted the
details - why do choose to remain ignorant? At least take time to
comprehend the technical information available from the field of
optics and report back to us why many decades of that EM wave
knowledge is wrong. Here is the equation that explains where the
reflected energy goes. All you have to to is track the energy back in
time.

Ptot = P1 + P2 + 2*SQRT(P1*P2)*cos(A)

where A is the angle between the two voltages. The last term is known
as the "interference term".

When two coherent, collimated, signals interfere while traveling in
the same direction in a transmission line: If they are 90 degrees
apart, there is zero interference and the I and Q components are
easily recovered.

If they are less than 90 degrees apart, the interference is
constructive and the phasor superposition of the two waves results in
*more power in the total superposed wave* than in the arithmetic sum
of the two powers. The extra energy has to come from somewhere. In the
absence of a local source, the extra energy has to come from
destructive interference in the opposite direction, e.g. at a Z0-
match.

If they are between 90 degrees and 180 degrees apart, the interference
is destructive and the phasor superposition of the two waves results
in *less power in the total superposed wave* than in the arithmetic
sum of the two waves. The "left-over" energy has to go somewhere so it
has to be delivered in the opposite direction to the area that
supports constructive interference.

Here is what Hecht said: "Briefly then, optical (EM wave) interference
corresponds to the interaction of two or more (EM) lightwaves yielding
a resultant irradiance (power density) that deviates from the sum of
the component irradiances."

Hecht's statement hints to where the reflected energy goes. In the
case of wave cancellation at a Z0-match that eliminates ExH reflected
energy flowing toward the load, all of the ExH reflected energy is
recovered and redistributed back toward the load.

Following this weakness back through the model, the root cause
is the attempt to assign tangible energy to the reflected wave.
Think of it as a reflected voltage or current wave and all will
be well, but assign power to it and eventually incorrect
conclusions will be drawn.

Only if the model is inadequate. Optical physicists assign tangible
energy to reflected waves all the time. Many are probably reading
these postings and laughing at the collective ignorance about EM
waves.

Take a look at this web page:

http://www.teachspin.com/instruments...eriments.shtml

Scroll down to, "Using Dielectric Beamsplitters to find the "missing
energy" in destructive interference". There it is, all spelled out for
you - where the reflected energy goes. Reflected energy is never
actually missing - what are missing are a few of the brain cells that
need to be used to think about reflected energy. :-)

For those who understand this, and know that "where did the
reflected energy go?" is an invalid question, using the power
model within its limits will not cause difficulties. But for
those who are not careful, great difficulties arise and a lot
of fancy dancing is offered to work around the difficulties,
unsuccessfully.

Yes, throughout history, ignorant people have brought great
difficulties upon themselves because they refuse to alleviate their
ignorance. Those who refuse to learn from their mistakes are destined
to repeat those same mistakes. And you were making these exact same
mistakes years ago. You can lead a horse to water ...

In his Nov/Dec 2001 QEX article, Dr. Best attempted to explain where
the reflected energy goes. Shackled by ignorance, he got many things
wrong e.g. phantom EM waves that exist without energy. But the article
alludes to the conceptual path that needs to be taken to alleviate
that ignorance. In "Wave Mechanics of Transmission Lines, Part 3", Dr.
Best published the above power-density/interference equation although
he ignorantly asserted on this newsgroup that interference did not
exist.

Interference resulting from superposition at an impedance
discontinuity is the key missing link in explaining everything that
happens to the energy in a transmission line including Roy's food-for-
thought article. Why is there so much reluctance to adopt a proven law
of EM wave physics from the field of optics?
--
73, Cecil, w5dxp.com

interference is nothing but what you observe after you superimpose two
or more waves. the general principle at work is superposition, giving
it more specific names like destructive or constructive interference
is just describing the result that you observe.

On Tue, 8 Jun 2010 03:57:44 -0700 (PDT), Keith Dysart
wrote:

But all of these papers have the appropriate
discipline

Hi Keith,

But? A pejorative tied to the phrase "appropriate discipline?" What
an interesting segue into denial. Quite original.

and do not ask the question "where does the reflected
energy go?"

OK, their having stated explicitly that, and having it in black and
white from competent authority is just as insufficient as Walt's
method and data. Bench experience to this issue makes a lot of
armchair theorists uncomfortable.

To your credit, you responded to the witness of evidence. It's like
the rest threw you under the bus.

73's
Richard Clark, KB7QHC

On Tue, 8 Jun 2010 03:57:44 -0700 (PDT), Keith Dysart
wrote:

Just for fun, here is a simple example.

-sigh-

No wonder traffic dies here. This stuff is a wheeze:

The source is no longer
providing energy,

If it had been a battery instead, then the battery would have as much
energy as it had the moment before - (minus) the few coulombs taken
during the short step interval.

Stick with the topic, draw from authorities (that are agreeable to all
parties) that demonstrate the concepts at the bench using the ordinary
tools of the trade (well, ordinary for a Metrologist, perhaps, but is
increasingly available at Ham fests) on actual equipment (what we
actually use in the Shack).

73's
Richard Clark, KB7QHC

On Jun 8, 10:57*am, Keith Dysart wrote:
On Jun 8, 3:47*am, Richard Clark wrote:



On Sun, 6 Jun 2010 14:21:16 -0700 (PDT), Keith Dysart

wrote:
The 100W forward and 50W reflected have no relation to actual powers

From MECA, makers of Isolators - their application:

The isolator is placed in the measurement path of a test bench between
a signal source and the device under test (DUT) so that any
reflections caused by any mismatches will end up at the termination of
the isolator and not back into the signal source. This example also
clearly illustrates the need to be certain that the termination at the
isolated port is sufficient to handle 100% of the reflected power
should the DUT be disconnected while the signal source is at full
power. If the termination is damaged due to excessive power levels,
the reflected signals will be directed back to the receiver because of
the circular signal flow.
...
MECA offers twenty-four models of isolators and circulators in both N
and SMA-female connectors with average power ratings from 2 - 250
watts.

Good day Richard,

You have located several examples from reputable vendors where the
behaviour of directional couplers is described in terms of power
in a forward and reflected wave. This model of behaviour works within
its limits and allows for convenient computation and prediction of
the behaviour. But all of these papers have the appropriate
discipline and do not ask the question "where does the reflected
energy go?" which is good, for this exceeds the limits of the
model.

As soon as one assigns tangible energy to the reflected wave, it
becomes reasonable to ask for an accounting of this energy and
the model is incapable of properly accounting for the energy.

Following this weakness back through the model, the root cause
is the attempt to assign tangible energy to the reflected wave.
Think of it as a reflected voltage or current wave and all will
be well, but assign power to it and eventually incorrect
conclusions will be drawn.

For those who understand this, and know that "where did the
reflected energy go?" is an invalid question, using the power
model within its limits will not cause difficulties. But for
those who are not careful, great difficulties arise and a lot
of fancy dancing is offered to work around the difficulties,
unsuccessfully.

Just for fun, here is a simple example.

100V DC source, connected to a 50 ohm source resistor,
connected to 50 ohm transmission line, connected to a 50 ohm
load resistor. Turn on the source. A voltage step propagates
down the line to the load. The impedances are matched, so
there are no reflections. The source provides 100W. 100W is
dissipated in the source resistor. 100W is dissipated in the
load resistor. Energy moves along the transmission line
from the source to the load at the rate of 50W.
All is well.

lets see, when steady state is reached you have 100v through 2 50ohm
resistors in series i assume since you say the source provides 100w...
that would mean the current is 1a and each resistor is dissipating
50w.


Disconnect the load. A voltage step propagates back along
the line from the load to the source. In front of this
step current continues to flow. Behind the step, the
current is 0. When the step reaches the source there is
no longer any current flowing. The source is no longer
providing energy, the source resistor is dissipating
nothing, and neither is the load resistor.

Proponensts of the power model claim that energy is
still flowing down the line, being reflected from the
open end and flowing back to the source. Since the
source is clearly no longer providing energy, great
machinations are required to explain why the reflected
'energy' is re-reflected to provide the forward 'energy'.

why would there be power flowing down the line, in this simple dc case
once the voltage in the line equals the source voltage, which in this
perfect case would be after one transient down the line, all is in
equilibrium... and art is happy.

HOWEVER!!! replace that dc source with an AC one and everything
changes, now there are forward and reflected waves continuously going
down the line and back. your DC analysis is no longer valid as it is
a very special case. now you have to go back to the general equations
and use the length of the line to transform the open circuit back to
the source based on the length of the line in wavelengths... for the
specific case of the line being 1/4 wave long the open now looks like
a short and the current in the source resistor doubles, now making it
dissipate 100w!


But really, does anyone believe that a length of
transmission line, charged to 100V voltage with zero
current flowing, is actually simultaneously transporting
energy in both directions?

For even more fun, replace the ideal conductors in the
transmission line with some lossy conductors. How much
of the reflected and re-reflected energy flowing up and
down the line will be dissipated in the conductors?
Remember that the current is zero, everywhere along
the line.

...Keith

On Jun 8, 7:39*pm, K1TTT wrote:
On Jun 8, 10:57*am, Keith Dysart wrote:





On Jun 8, 3:47*am, Richard Clark wrote:

On Sun, 6 Jun 2010 14:21:16 -0700 (PDT), Keith Dysart

wrote:
The 100W forward and 50W reflected have no relation to actual powers

From MECA, makers of Isolators - their application:

The isolator is placed in the measurement path of a test bench between
a signal source and the device under test (DUT) so that any
reflections caused by any mismatches will end up at the termination of
the isolator and not back into the signal source. This example also
clearly illustrates the need to be certain that the termination at the
isolated port is sufficient to handle 100% of the reflected power
should the DUT be disconnected while the signal source is at full
power. If the termination is damaged due to excessive power levels,
the reflected signals will be directed back to the receiver because of
the circular signal flow.
...
MECA offers twenty-four models of isolators and circulators in both N
and SMA-female connectors with average power ratings from 2 - 250
watts.

Good day Richard,

You have located several examples from reputable vendors where the
behaviour of directional couplers is described in terms of power
in a forward and reflected wave. This model of behaviour works within
its limits and allows for convenient computation and prediction of
the behaviour. But all of these papers have the appropriate
discipline and do not ask the question "where does the reflected
energy go?" which is good, for this exceeds the limits of the
model.

As soon as one assigns tangible energy to the reflected wave, it
becomes reasonable to ask for an accounting of this energy and
the model is incapable of properly accounting for the energy.

Following this weakness back through the model, the root cause
is the attempt to assign tangible energy to the reflected wave.
Think of it as a reflected voltage or current wave and all will
be well, but assign power to it and eventually incorrect
conclusions will be drawn.

For those who understand this, and know that "where did the
reflected energy go?" is an invalid question, using the power
model within its limits will not cause difficulties. But for
those who are not careful, great difficulties arise and a lot
of fancy dancing is offered to work around the difficulties,
unsuccessfully.

Just for fun, here is a simple example.

100V DC source, connected to a 50 ohm source resistor,
connected to 50 ohm transmission line, connected to a 50 ohm
load resistor. Turn on the source. A voltage step propagates
down the line to the load. The impedances are matched, so
there are no reflections. The source provides 100W. 100W is
dissipated in the source resistor. 100W is dissipated in the
load resistor. Energy moves along the transmission line
from the source to the load at the rate of 50W.
All is well.

lets see, when steady state is reached you have 100v through 2 50ohm
resistors in series i assume since you say the source provides 100w...
that would mean the current is 1a and each resistor is dissipating
50w.

Ouch. I can't even get the arithmetic right. You are, of course,
correct. Fortunately, the rest of the examples do not attempt
to compute actual values.

Disconnect the load. A voltage step propagates back along
the line from the load to the source. In front of this
step current continues to flow. Behind the step, the
current is 0. When the step reaches the source there is
no longer any current flowing. The source is no longer
providing energy, the source resistor is dissipating
nothing, and neither is the load resistor.

Proponensts of the power model claim that energy is
still flowing down the line, being reflected from the
open end and flowing back to the source. Since the
source is clearly no longer providing energy, great
machinations are required to explain why the reflected
'energy' is re-reflected to provide the forward 'energy'.

why would there be power flowing down the line,

Exactly. And yet, were you to attach a DC-couple directional
wattmeter to the line, it would indicate 50W forward and 50W
reflected (using your corrected numbers).

in this simple dc case
once the voltage in the line equals the source voltage, which in this
perfect case would be after one transient down the line, all is in
equilibrium... and art is happy.

HOWEVER!!! *replace that dc source with an AC one and everything
changes,

Not at all. The measurements and computations for forward and
reflected
power that work for AC, work perfectly well for DC, pulses, steps,
or pick your waveform. There is no magic related to AC.

now there are forward and reflected waves continuously going
down the line and back. *your DC analysis is no longer valid as it is
a very special case. *

Not so. And if it were, at exactly what very low frequency does
the AC analysis begin to fail?

now you have to go back to the general equations
and use the length of the line to transform the open circuit back to
the source based on the length of the line in wavelengths... for the
specific case of the line being 1/4 wave long the open

The transformation properties apply only to sinusoidal AC (note, for
example, that the line length is expressed in wavelengths), but
reflection coefficients and the the computation of forward and
reflected voltages, currents and powers are waveshape independant.

When analyzing in the time domain, use the delay of the line to decide
the values of the signals to superposed. Doing this for sinusoids
will lead to exactly the same results that are expected from, for
example, 1/4 wave lines.

....Keith

On 6/6/2010 9:22 AM, JC wrote:
Basic question (at least for me) for a very poor antenna matching :
-100 w reach the antenna and 50 w are radiated.
- 50 w are "reflected", what is their fate ?
Are they definitely lost for radiation and just heat the line, the final....
JC




It would be really really REALLY effing great if you dorks would take
this to a suitable forum. This has nothing to do with antennas.

And it is an unproductive infinite length, in case you bunch have missed
that point. But I'm guessing that given how intelligent you all are,
you already figured that out, and are ignoring it. No one is changing
their position, and they never will.

And yes, I have the thread set to "ignore". Which everyone should.

I will NOT see your responses.

tom
K0TAR

On 6/17/2010 9:59 PM, tom wrote:
On 6/6/2010 9:22 AM, JC wrote:
Basic question (at least for me) for a very poor antenna matching :
-100 w reach the antenna and 50 w are radiated.
- 50 w are "reflected", what is their fate ?
Are they definitely lost for radiation and just heat the line, the
final....
JC




It would be really really REALLY effing great if you dorks would take
this to a suitable forum. This has nothing to do with antennas.

And it is an unproductive infinite length, in case you bunch have missed
that point. But I'm guessing that given how intelligent you all are, you
already figured that out, and are ignoring it. No one is changing their
position, and they never will.

And yes, I have the thread set to "ignore". Which everyone should.

I will NOT see your responses.

tom
K0TAR


Actually, I'm following and enjoying the conversation. Since you set
your computer to ignore responses, then you should have no objection if
they continue. It makes more sense to ignore subjects which don't
interest you than to complain about them.

John
KD5YI

On Jun 17, 9:59*pm, tom wrote:
It would be really really REALLY effing great if you dorks would take
this to a suitable forum. *This has nothing to do with antennas.

Is there a rec.radio.amateur.feedlines group that I have missed?
--
73, Cecil, w5dxp.com