what happens to reflected energy ?
 

On Jun 18, 7:55*pm, K1TTT wrote:
On Jun 18, 11:26 pm, Keith Dysart wrote:

On Jun 18, 7:04 am, K1TTT wrote:

On Jun 18, 12:00 am, Keith Dysart wrote:
except i can measure If and Ir separately and can see they are both
zero.

How would you measure them?

you said you had a directional wattmeter, so do i. *mine reads zero,
what does yours read?

That would most likely be because your wattmeter is AC coupled.
Working
with DC, you need a DC-coupled meter, in which case it would read as
I predicted.

mine is dc coupled... still nothing. *please provide the schematic or
model number for the one you use to get such a reading.

Oh my. Could you kindly provide the schematic or a reference? Before
posting I spent an hour searching my library and the web for the
circuit
for a dc-coupled directional wattmeter and was unsuccessful.

Nothing prevents one from being constructed. That one is not readily
available should not prevent you from using the equations that they
implement and then predicting the DC behaviour of the line.

A highly instructive exercise is to examine the schematic for your
directional wattmeter and work out how it implements the expressions
* Vf(t)=(V(t)+R0*I(t))/2
* Vr(t)=(V(t)-R0*I(t))/2

* Pf=avg(Vf(t)*Vf(t)/R0)
* Pr=avg(Vr(t)*Vr(t)/R0)
or some equivalent variant using If and Ir.

You will find a voltage sampler, a current sampler, some scaling,
and an addition (or subtraction). If your meter is analog, the
squaring function is usually implemented in the non-linear meter
scale. And the averaging function might be peak-reading or rely
on meter inertia.

You can, though, have an arbitrary current as an intermediate
result in solving a problem. With the two battery example, reduce
the resistor to 0 and an infinitely large current is computed in
both directions, though we know in reality that no current is
actually flowing.

no it isn't. *in the limit at R-0 *0V/R still looks like 0 to me..

When computing the intermediate results using superposition, you
have the full battery voltage in to 0 ohms. First the battery on
the left, then the one on the right, and then you add the two
currents to get the total current. 0 is always the total, but
the intermediate results can go to infinity if you connect
the two batteries directly in parallel.

there is no battery on the right, unless you have changed the case
again. *i thought we were doing the open circuit coax with the dc
source on one end.

No. I was referring to the simple DC circuit with two batteries and
a resistor. I was pretty sure that was your referent when you used
"R-0".

I understand that you have not found any flaws in my exposition,
except that you are still uncomfortable with the results. It is
possible to overcome this.

And I re-iterate the value of studying your directional wattmeter's
schematic, especially if it is DC-coupled.

....Keith

what happens to reflected energy ?
 

On Jun 17, 8:38*pm, Cecil Moore wrote:
On Jun 17, 7:00*pm, Keith Dysart wrote:

I will caution again, do not assign too much reality to these
forward and reflected waves; very useful for solving problems,
but trouble if carried too far.

I'm afraid you are too late with that advice. Optical physicists have
assigned reality to forward and reflected waves inside an
interferometer and have tracked all of the energy including the energy
in canceled wavefronts.

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

Scroll down to, "Using Dielectric Beamsplitters to find the "missing
energy" in destructive interference" It says: "... when interference
is destructive at the standard output, it is constructive at the non-
standard output."

In RF terms, when interference is destructive at a Z0-match, it is
constructive in the direction of the load.
--
73, Cecil, w5dxp.com

You have fallen in to the same trap that you did with circulators.
When you introduce a circulator in to the transmission line, you
change the experiment and you get certain results. The circulator
provides some numerology that appears to support the hypothesis
of 'reflected energy' being real. The same with looking for
"missing energy" (note they even quote it) with a beamsplitter.
The beamsplitter reveals what is happening at the beamsplitter,
not what is happening at points between the two beamsplitters.

Just like a circulator.

....Keith

what happens to reflected energy ?
 

On Jun 19, 12:22*am, Keith Dysart wrote:
On Jun 17, 8:38*pm, Cecil Moore wrote:





On Jun 17, 7:00*pm, Keith Dysart wrote:

I will caution again, do not assign too much reality to these
forward and reflected waves; very useful for solving problems,
but trouble if carried too far.

I'm afraid you are too late with that advice. Optical physicists have
assigned reality to forward and reflected waves inside an
interferometer and have tracked all of the energy including the energy
in canceled wavefronts.

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

Scroll down to, "Using Dielectric Beamsplitters to find the "missing
energy" in destructive interference" It says: "... when interference
is destructive at the standard output, it is constructive at the non-
standard output."

In RF terms, when interference is destructive at a Z0-match, it is
constructive in the direction of the load.
--
73, Cecil, w5dxp.com

You have fallen in to the same trap that you did with circulators.
When you introduce a circulator in to the transmission line, you
change the experiment and you get certain results. The circulator
provides some numerology that appears to support the hypothesis
of 'reflected energy' being real. The same with looking for
"missing energy" (note they even quote it) with a beamsplitter.
The beamsplitter reveals what is happening at the beamsplitter,
not what is happening at points between the two beamsplitters.

Just like a circulator.

...Keith- Hide quoted text -

- Show quoted text -

he is even more hopeless than most, first he determines there are
forward and reflected waves when a dc source has charged the line,
then he rejects circulators and beamsplitters that obviously work
because of the waves. so dc waves exist, but rf and optical waves
don't. thats a unique viewpoint.

what happens to reflected energy ?
 

On Jun 18, 8:36*pm, K1TTT wrote:
On Jun 19, 12:22*am, Keith Dysart wrote:





On Jun 17, 8:38*pm, Cecil Moore wrote:

On Jun 17, 7:00*pm, Keith Dysart wrote:

I will caution again, do not assign too much reality to these
forward and reflected waves; very useful for solving problems,
but trouble if carried too far.

I'm afraid you are too late with that advice. Optical physicists have
assigned reality to forward and reflected waves inside an
interferometer and have tracked all of the energy including the energy
in canceled wavefronts.

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

Scroll down to, "Using Dielectric Beamsplitters to find the "missing
energy" in destructive interference" It says: "... when interference
is destructive at the standard output, it is constructive at the non-
standard output."

In RF terms, when interference is destructive at a Z0-match, it is
constructive in the direction of the load.
--
73, Cecil, w5dxp.com

You have fallen in to the same trap that you did with circulators.
When you introduce a circulator in to the transmission line, you
change the experiment and you get certain results. The circulator
provides some numerology that appears to support the hypothesis
of 'reflected energy' being real. The same with looking for
"missing energy" (note they even quote it) with a beamsplitter.
The beamsplitter reveals what is happening at the beamsplitter,
not what is happening at points between the two beamsplitters.

Just like a circulator.

...Keith- Hide quoted text -

- Show quoted text -

he is even more hopeless than most, first he determines there are
forward and reflected waves when a dc source has charged the line,
then he rejects circulators and beamsplitters that obviously work
because of the waves. *so dc waves exist, but rf and optical waves
don't. *thats a unique viewpoint.- Hide quoted text -

- Show quoted text -

You might consider the following experiment:

- a sinusoidal generator that will produce 50W in to 50 ohms
- attach one half wavelength of transmission line
- leave the far end of the transmission line open

After the line settles, a directional "wattmeter" anywhere on
the line will indicate 50W forward and 50W reflected.
The generator will not be putting any energy in to the line
since the line input appears as an open circuit at the
generator output.

Insert a circulator (with the reflected port terminated in
50 ohms) between the generator and the line. The circulator
termination will be dissipating 50W and the generator will
now be delivering 50W in to the circulator.

1. Please explain how inserting the circulator did not change
the circuit conditions. The generator went from delivering
0W to delivering 50W.

2. Where do you think the 50W being dissipated in the circulator
termination resistor is coming from? The line? Or the
generator (which is now outputting 50W)?

Just for fun, you may be interested in a DC coupled circulator
that works all the way down to 0 Hz:
http://www.techlib.com/files/RFDesign3.pdf

Answer question 2 for this circulator design.

....Keith

what happens to reflected energy ?
 

On Jun 18, 7:00*pm, Keith Dysart wrote:
Reflected waves heat finals, no they
don't, yes they do, ... but eventually I learned.

Hopefully, you learned that rail arguments are rarely valid and the
truth usually lies somewhere in between. Sometimes reflected waves
heat finals and sometimes they don't. It all depends on what kind and
how much interference exists between the forward wave and reflected
wave at the source impedance and (apparently) how much of that
impedance is dissipative or non-dissipative. Roy's food-for-thought
article doesn't take interference into account at all and therefore
arrives at magic conclusions divorced from reality.

Anyhow, I suggest you read my examples and find the flaws, for that
is the surest way to convince me that I am wrong.

I have pointed out many of the flaws in your examples. The way you
parse the energy violates the laws of physics. Maxwell's equations
only work on a function that contains classic EM traveling waves.
Standing waves and DC steady-state examples don't meet the necessary
boundary conditions. In fact, it can be proved that EM waves do not
existent during DC steady-state. The fact that an instrument designed
to detect EM waves malfunctions during DC steady-state is not
unexpected. Even an SWR meter calibrated for a different Z0 than the
one being used will indicate false results.

You really need to learn to recognize when your experiment and your
math model are contradictory, when your concepts violate the laws of
physics, and when you have chosen the wrong measuring equipment. You
remind me of myself when I was 14 years old and tried to measure the
screen voltage on a vacuum tube using a Simpson multimeter. It wasn't
anywhere near the voltage specified in the manual so I wasted my money
by buying a new tube the next time my family made a trip to Houston.

You would report those same measurement results as proof of a strange,
magical happening.
--
73, Cecil, w5dxp.com

what happens to reflected energy ?
 

On Jun 18, 7:22*pm, Keith Dysart wrote:
The beamsplitter reveals what is happening at the beamsplitter,
not what is happening at points between the two beamsplitters.

Oh yeah, according to you, the light beams are changing directions at
every voltage and current node. Sorry, I do not accept your
metaphysics, especially when they violate the accepted laws of
physics.
--
73, Cecil, w5dxp.com

what happens to reflected energy ?
 

On Jun 20, 8:03*pm, Keith Dysart wrote:
1. Please explain how inserting the circulator did not change
* *the circuit conditions. The generator went from delivering
* *0W to delivering 50W.

The generator went from experiencing total destructive interference to
experiencing zero interference. Of course, the entire purpose of
inserting the circulator is to cause the generator to see 50 ohms as a
load impedance.

2. Where do you think the 50W being dissipated in the circulator
* *termination resistor is coming from? The line? Or the
* *generator (which is now outputting 50W)?

TV ghosting experiments will verify that the signal being dissipated
in the circulator load resistor has made a round trip from the
generator to the end of the N(1/2WL) stub and back to the circulator.
Please don't insult our intelligence by arguing that is not a steady-
state condition.
--
73, Cecil, w5dxp.com

what happens to reflected energy ?
 

On Jun 21, 1:03*am, Keith Dysart wrote:
On Jun 18, 8:36*pm, K1TTT wrote:



On Jun 19, 12:22*am, Keith Dysart wrote:

On Jun 17, 8:38*pm, Cecil Moore wrote:

On Jun 17, 7:00*pm, Keith Dysart wrote:

I will caution again, do not assign too much reality to these
forward and reflected waves; very useful for solving problems,
but trouble if carried too far.

I'm afraid you are too late with that advice. Optical physicists have
assigned reality to forward and reflected waves inside an
interferometer and have tracked all of the energy including the energy
in canceled wavefronts.

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

Scroll down to, "Using Dielectric Beamsplitters to find the "missing
energy" in destructive interference" It says: "... when interference
is destructive at the standard output, it is constructive at the non-
standard output."

In RF terms, when interference is destructive at a Z0-match, it is
constructive in the direction of the load.
--
73, Cecil, w5dxp.com

You have fallen in to the same trap that you did with circulators.
When you introduce a circulator in to the transmission line, you
change the experiment and you get certain results. The circulator
provides some numerology that appears to support the hypothesis
of 'reflected energy' being real. The same with looking for
"missing energy" (note they even quote it) with a beamsplitter.
The beamsplitter reveals what is happening at the beamsplitter,
not what is happening at points between the two beamsplitters.

Just like a circulator.

...Keith- Hide quoted text -

- Show quoted text -

he is even more hopeless than most, first he determines there are
forward and reflected waves when a dc source has charged the line,
then he rejects circulators and beamsplitters that obviously work
because of the waves. *so dc waves exist, but rf and optical waves
don't. *thats a unique viewpoint.- Hide quoted text -

- Show quoted text -

You might consider the following experiment:

- a sinusoidal generator that will produce 50W in to 50 ohms
- attach one half wavelength of transmission line
- leave the far end of the transmission line open

After the line settles, a directional "wattmeter" anywhere on
the line will indicate 50W forward and 50W reflected.
The generator will not be putting any energy in to the line
since the line input appears as an open circuit at the
generator output.

Insert a circulator (with the reflected port terminated in
50 ohms) between the generator and the line. The circulator
termination will be dissipating 50W and the generator will
now be delivering 50W in to the circulator.

1. Please explain how inserting the circulator did not change
* *the circuit conditions. The generator went from delivering
* *0W to delivering 50W.

2. Where do you think the 50W being dissipated in the circulator
* *termination resistor is coming from? The line? Or the
* *generator (which is now outputting 50W)?

well, its coming from the generator via the far end of the line of
course.


Just for fun, you may be interested in a DC coupled circulator
that works all the way down to 0 Hz:http://www.techlib.com/files/RFDesign3.pdf

Answer question 2 for this circulator design.

...Keith

the result is no current, but 50 v, as would be expected of course.

what happens to reflected energy ?
 

On Jun 21, 9:04*am, Cecil Moore wrote:
On Jun 20, 8:03*pm, Keith Dysart wrote:

1. Please explain how inserting the circulator did not change
* *the circuit conditions. The generator went from delivering
* *0W to delivering 50W.

The generator went from experiencing total destructive interference to
experiencing zero interference. Of course, the entire purpose of
inserting the circulator is to cause the generator to see 50 ohms as a
load impedance.

Along with the other use for circulators: To present an impedance
match to the reflected wave so that it will not be re-reflected.

2. Where do you think the 50W being dissipated in the circulator
* *termination resistor is coming from? The line? Or the
* *generator (which is now outputting 50W)?

TV ghosting experiments will verify that the signal being dissipated
in the circulator load resistor has made a round trip from the
generator to the end of the N(1/2WL) stub and back to the circulator.

It is not clear why you think this verifies something.

Enhancing the details slightly of the generator in the original
example: The generator is constructed using the Thevenin model
of a source followed by a 50ohm resistor.

With the circulator, there is no re-reflection at the driven
end of the line, 50W is dissipated in the circulator
resistor and 50W is dissipated in the source resistor.

Without the circulator, there is no re-reflection and nothing is
dissipated in the source resistor (and there is no circulator
resistor).

The dissipation seems to correlate much more strongly with the
circuit design than it does with the magnitude of the "reflected
power".

Please don't insult our intelligence by arguing that is not a steady-
state condition.

Well you are right, it is not steady-state when there is modulation,
but modulation does not affect the examples I have provided.

....Keith

PS: For further understanding, substitute a Norton style generator,
then do it again for both kinds of generators but with the end
of the line shorted yielding more examples where the "reflected
power" does not correlate with the dissipation.

what happens to reflected energy ?
 

On Jun 21, 5:40 pm, K1TTT wrote:
On Jun 21, 1:03 am, Keith Dysart wrote:

You might consider the following experiment:

- a sinusoidal generator that will produce 50W in to 50 ohms
- attach one half wavelength of transmission line
- leave the far end of the transmission line open

After the line settles, a directional "wattmeter" anywhere on
the line will indicate 50W forward and 50W reflected.
The generator will not be putting any energy in to the line
since the line input appears as an open circuit at the
generator output.

Insert a circulator (with the reflected port terminated in
50 ohms) between the generator and the line. The circulator
termination will be dissipating 50W and the generator will
now be delivering 50W in to the circulator.

1. Please explain how inserting the circulator did not change
the circuit conditions. The generator went from delivering
0W to delivering 50W.

2. Where do you think the 50W being dissipated in the circulator
termination resistor is coming from? The line? Or the
generator (which is now outputting 50W)?

well, its coming from the generator via the far end of the line of
course.

That is the classic answer from the “reflected power” model, but
consider the following...

We construct two experiments similar to the above.
The first one:
- a sinusoidal generator that will produce 50W in to 50 ohms
- attach one half wavelength of transmission line
- leave the far end of the transmission line open
- the generator is constructed in the Thevenin style with a
voltage source and 50 ohm resistor

The second experiment:
- same as above except there is a circulator between the
generator and the line and the circulator is terminated
with 50 ohms

Examine experiment 1. After the line settles, a directional
wattmeter indicates 50W “forward power” and 50W “reflected
power”. The current in the voltage source and source resistor
is 0 so no energy is dissipated in the source resistor and
the voltage source is delivering no energy to the system.
The “reflected power” is not being dissipated in the source
resistor or voltage source so where is it going. Cecil will
offer some explanation where the “reflected energy”
is “redistributed” in the other direction as the “forward
energy”, which “must” be happening since it is not dissipated
in the source and it can not be reflected because there is
no impedance discontinuity.

Examine experiment 2. After the line settles, a directional
wattmeter indicates 50W “forward power” and 50W “reflected
power”. But with a circulator, the voltage source is
delivering 100W, the source resistor is dissipating 50W
and the circulator resistor is also dissipating 50W.
You offer that the 50W being dissipated in the circulator
is the 50W of “reflected power” coming from the line.
Presumably then, the 50W of “forward power” is coming from
the generator. This agrees numerically since the voltage
source is providing 100W and the source resistor is
dissipating 50W.

In both experiments, conditions on the line are identical.
There is 50W of “forward and reflected power” indicated.
There is no energy flowing past the end of the line since
it is open circuited.
Because the line is one-half wavelength long, the line
presents an open circuit to the generator (exp 1) or
circulator (exp 2).

In both experiments, the line is presented with a 50 ohm
source impedance, either from the generator or from the
circulator.

What puzzles me then, is how the “reflected power” knows
that in experiment 1, it should stay out of the generator
so that it is not dissipated in the source resistor
but in experiment 2, it should enter the circulator so
that it can be dissipated in the circulator load resistor.

Can you explain how the “reflected power” “knows”?

....Keith