what happens to reflected energy ?
 

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.

73's
Richard Clark, KB7QHC

what happens to reflected energy ?
 

On 7 jun, 03:04, Richard Clark wrote:
On Sun, 6 Jun 2010 15:21:59 -0700 (PDT), Wimpie
wrote:

Try the following experiment:

-sigh-

Did you do the experiment (forward power versus mismatch)?

Hi Wimpie,

This is not what I asked for. *You said what the source was NOT, but
you cannot say what the source IS.

The source is not a bipedal reptile. *The source is not an elongated
diphthong. *The source is not the resurrection of a deity (some may
argue that more than I would care to follow). *The source is not....

There are many ways to say what the source is NOT, and that will never
inform us about the source. *I see many draw deuces to this question
and try to convince everyone that the card is a pair of winning aces.

Probably you consumed something wrong here.

Now make some mismatch (for example VSWR=2 at different phase) and
read the forward power. Did it change? If so, the output impedance is
no (longer) 50 Ohms.

What IS it now? *If you could measure it once, you should be able to
tell us what it is this time too. *I did this for years to methods set
by the National Bureau of Standards. *You have drawn a deuce, not two
aces.

Some values:
9 +/-1 Ohm (real impedance), reference impedance 16 Ohms, 8 MHz
amplifier (ISM). Exact value requried because of additional filtering,
based on several IRF110, push pull.

2 Ohms (small reactance present), 10V/1A driver (sinusoidal), 700
kHz, slightly saturating class C output stage based on single mosfet.
RF feedback present to guarantee value below 2 Ohms. also here,
certain maximum value was required for the application

|RC| 0.80 (actual value depending on type of additional filtering in
between), 5kW pulsed power amplfiier, non 50 Ohm application.
Actual value not important, but is a recent project, so I knew from
memory.

See also EL34 example in posting to Walt.



Measuring method used: change in resistive load, from voltage change
you can calculate the current change, hence the output impedance. One
note, except two, all where solid state.

Yes, that is called a load pull, but you offer no data - another
deuce. *I have done a lot of load pulls. *

All high efficiency designs (class E, D) that I did have output
impedance far from the expected load impedance. With "far" I mean
factor 2 or factor 0.5. I did not measure that (as it is not
important in virtually all cases), but know it from the overload
simulation/measurement and I did the design myself.

You didn't measure it, but you can state the value - interesting state
of guessing. *So, what did the simulation/non-measurement give you as
a value? *What IS the value? *Another deuce.

Please read again last part of sentence above your text.


The reason for not being 50 Ohms

Reasons abound. *BP is giving us reasons why the Gulf Coast shouldn't
worry. *Data has proven that reasons don't work and neither does their
well. *That is a Joker draw.

High efficiency CW amplifiers

The TS830S is not such an animal. *Another Deuce.

Not al hams use TS830S, See more recent reply of Walt to Tom's
posting.

If you can drop me a link to the discussion of the TS830s, I would
appreciate that.

Google "Plate Resistance" in this group.


73's
Richard Clark, KB7QHC

I hope, further replies from you will be constructive,


Wim
PA3DJS

what happens to reflected energy ?
 

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.

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'.

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

what happens to reflected energy ?
 

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

what happens to reflected energy ?
 

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

what happens to reflected energy ?
 

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

what happens to reflected energy ?
 

On Tue, 8 Jun 2010 02:50:17 -0700 (PDT), Wimpie
wrote:

There are many ways to say what the source is NOT, and that will never
inform us about the source. *I see many draw deuces to this question
and try to convince everyone that the card is a pair of winning aces.

Probably you consumed something wrong here.

Hi Wimpie,

Hmmm, an ad hominem usually reserved for Art, but I will let it pass
as your perceiving it as a response in kind. Fair enough.

Now make some mismatch (for example VSWR=2 at different phase) and
read the forward power. Did it change? If so, the output impedance is
no (longer) 50 Ohms.

What IS it now? *If you could measure it once, you should be able to
tell us what it is this time too. *I did this for years to methods set
by the National Bureau of Standards. *You have drawn a deuce, not two
aces.

Some values:
9 +/-1 Ohm (real impedance), reference impedance 16 Ohms, 8 MHz
amplifier (ISM).

Taking my statement and your response at face value, you are
describing how a source that formerly exhibited 50 Ohms was drawn down
to 9 +/-1 Ohm (real impedance). What "real impedance" is? is not
something that I will dwell on. Instead, I will ask how this shift
was induced. Or perhaps you are using a 16 Ohm system. If so, this
lacks too many details for consideration. When I had Walt's 300+ line
hypothesis, I also had access to his service manual for his equipment,
the final tube specification, and his measurement data. In short, I
am impressed only by data, not results.

Given the "some values" that you offer continues in disjointed
discussion as a sort of short-hand description for a project known
only to you, then these distractions draw this further away from a
conclusion. I also see your directions to wander the threads for
vague side bars of discussion. Supposedly, this is instruction for me
to draw together and connect the important details of your point for
you. No, thank you. Too many in this group contribute teasing
details in their coy writing and none of them have the power to
intrigue me with that.

Your "some values" looks vaguely interesting, but I would suggest you
put effort into describing your complete scenario with as much care as
Walt's 300+ line posting that laid bare his complete hypothesis. I
offered you a specifically new thread that I originated solely for one
issue where I see no discussion from you - quid pro quo?

And returning to the matter at hand - Up or Down:

"Does Walt's data support the evidence of a Conjugate Match?"

I cannot see pursuing your inquiry with your own example if you cannot
offer a fixed answer here. That would put me in a position of
embarking on a ponderous journey where, simultaneously, you and Walt
both DID AND DID NOT demonstrate your claims. Paradox is suitable
only for Operetta by Gilbert and Sullivan.

73's
Richard Clark, KB7QHC

what happens to reflected energy ?
 

Hello Richard,

You asked for what it IS, several times, so you virtually left me no
choice.

Especially the first one (Zout = 9 ohms where the load is 16 Ohms,
RCout = -0.28) is very well documented as it was designed to have this
value. Without the extra measures, the value was below 9 Ohms. The
amplifier is time limited fully short circuit proof, so it was not
difficult to determine Zout at various loads. I put myself into
problems when publishing the documentation over here.

When we go back to my first posting in this thread (the first reply)
where I stated that amplifiers do not show 50 Ohms in general, is not
that strange and doesn't violate Walt's conclusions.

When I understand it well, Walt assumes tuning for maximum output at a
given drive (not necessarily the maximum drive). As mentioned earlier,
I support his findings.

When you don't touch the plate and load capacitor and change from 50
Ohms to a load with VSWR = 2, you violate Walt's conditions and the
amplifier will no long behave as a 50 Ohms source, hence what happens
with the 50W reflected power from JC's posting becomes more
complicated then just a transmission line problem.

Walt made a statement in this thread that an amplifier that obeys ZL =
Zout can have higher then 50% efficiency. I fully support this
statement also (Example Class C amplifier with narrow conduction angle
at the edge of current/voltage saturation, with efficiency over
75%). However, change the load and you get a complete other
situation. You have to retune to get back to Walt's condition.

From my experience, many devices with conventional amplifiers (like
the valve with pi-filter, or single transistor stages) are now being
replaced by other topologies where ZL Zout under specified load (as
these types of amplifiers are not optimized for maximum gain, but for
maximum power added efficiency). I think Tom, KT7ITM also have this
experience.

The example given by JC (huge reflection) and my own experience with
conventional and high efficiency topologies resulted in my statement
that a PA is not a 50 Ohms source. Maybe I had to be more precise to
mention the exceptions: forward power control loop as mentioned by
Roy, -tuning to maximum output after each change of load, -specially
designed negative feedback, -adding an attenuator. The reader can now
determine the category for his PA.

In case of doubt use the forward power measurement technique, when it
changes under varying load, your amplifier doesn’t behave as a 50 Ohms
source.

Best Regards,


Wim
PA3DJS
www.tetech.nl
without abc, PM will reach me

what happens to reflected energy ?
 

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.

what happens to reflected energy ?
 

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