what happens to reflected energy ?
 

On Jun 18, 12:00*am, Keith Dysart wrote:
On Jun 17, 5:53*pm, K1TTT wrote:



On Jun 17, 10:31*am, Keith Dysart wrote:

On Jun 14, 7:04 pm, K1TTT wrote:

On Jun 14, 12:09 pm, Keith Dysart wrote:
Choose an arbitrary point on the line and observe the voltage. The
observed voltage will be a function of time. Let us call it V(t).
At the same point on the line, observe the current and call it I(t).

if you can choose an arbitrary point, i can choose an arbitrary
time... call it a VERY long time after the load is disconnected.

That works for me, as long as the time is at least twice the
propagation
time down the line.

The energy flowing (i.e. power) at this point on the line will
be P(t)=V(t)I(t). If the voltage is measured in volts and the current
in amps, then the power will be in joules/s or watts.
We can also compute the average energy flow (power) over some
interval by integrating P(t) over the interval and dividing by
the length of the interval. Call this Pavg. For repetitive
signals it is convenient to use one period as the interval.

Note that V(t) and I(t) are arbitrary functions. The analysis
applies regardless of whether V(t) is DC, a sinusoid, a step,
a pulse, etc.; any arbitrary function.

To provide some examples, consider the following simple setup:
1) 50 ohm, Thevenin equivalent circuit provides 100 V in to an
* *open circuit.

this is the case we were discussing... the others are irrelevant at
this point

2) The generator is connected to a length of 50 ohm ideal
* *transmission line. We will stick with ideal lines for the
* *moment, because, until ideal lines are understood, there
* *is no hope of understanding more complex models.
3) The ideal line is terminated in a 50 ohm resistor.

snip the experiments you do not consider relevent

Now let us remove the terminating resistor. The line is
now open circuit.

OK, now we are back to where i had objections above.

Again, set the generator to DC. After one round trip
time, the observations at any point on the line will
be:
* V(t)=100V
* I(t)=0A
* P(t)=0W
* Pavg=0W

ok, p(t)=0w, i=0a, i connect my directional wattmeter at some very
long time after the load is disconnected so the transients can be
ignored and i measure 0w forever.

Not correct. See the equations for a directional wattmeter below.

there is no power flowing in the line.

I agree with this, though a directional wattmeter indicates
otherwise. See below.

Now it is time to introduce forward and reflected waves to
the discussion. Forward an reflected waves are simply a
mathematical transformation of V(t) and I(t) to another
representation which can make solving problems simpler.

The transformation begins with assuming that there is
a forward wave defined by Vf(t)=If(t)*Z0, a reflected
wave defined by Vr(t)=Ir(t)*Z0 and that at any point
on the line V(t)=Vf(t)+Vr(t) and I(t)=If(t)-Ir(t).

Since we are considering an ideal line, Z0 simplifies
to R0.

Simple algebra yields:
* Vf(t)=(V(t)+R0*I(t))/2
* Vr(t)=(V(t)-R0*I(t))/2
* If(t)=Vf(t)/R0
* Ir(t)=Vr(t)/R0

The above equations are used in a directional wattmeter.

Now these algebraically transformed expressions have all
the capabilities of the original so they work for DC,
60 HZ, RF, sinusoids, steps, pulses, you name it.

hmmm, they work for all waveforms you say... even dc?? *ok, open
circuit load, z0=50, dc source of 100v with 50ohm thevenin
resistance...

You have made some arithmetic errors below. I have re-arranged
your prose a bit to allow correction, then comment.

well, maybe i missed something... lets look at the voltages:

Vf(t)=(100+50*2)/2 = 100
Vr(t)=(100-50*2)/2 = 0

You made a substitution error here. On the transmission line
with a constant DC voltage:
* V(t)=100V
* I(t)=0A

substituting in to
* Vf(t)=(V(t)+R0*I(t))/2
* Vr(t)=(V(t)-R0*I(t))/2

yields
* Vf(t)=50V
* Vr(t)=50V

and
* If(t) = 50/50 = 1a
* Ir(t) = 50/50 = 1a

Recalling that the definition of forward and reflected is
* V(t)=Vf(t)+Vr(t)
* I(t)=If(t)-Ir(t)
we can check our arithmetic
* V(t)= 50+50 = 100V
* I(t)= 1-1 * = 0A
Exactly as measured.
If you are disagree with the results, please point out
my error in the definitions, derivations or arithmetic.

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?

The standard definition of forward
and reflected waves are as I provided above. I suppose you could
construct your own definitions, but such a new definition would
be confusing and unlikely to have the nice properties of the
standard definitions.

this is where your argument about putting a resistor between
two batteries falls apart... there is a length of cable with a finite
time delay in the picture here. *so it is possible to actually watch
the waves reflect back and forth and see what really happens. *in the
case where the line is open at the far end we can measure the voltage
at both ends of the line and see the constant 100v, so where is the
voltage difference needed to support the forward and reverse current
waves? *

Well, as I suggested, the forward and reflected wave are somewhat
fictitious. They are very useful as intermediate results in solving
problems but, as you note above, quite problematic if one starts
to assign too much reality to them.

fictitious?? i can measure them, watch them flow from here to there,
and see physical effects of them, doesn't sound very fictitious to
me. weren't you the one who teaches tdr use? what are you seeing in
the steps of the tdr if not for reflected waves coming back to your
scope?


That is why they are quite analogous to the two currents in the two
battery example.

no, that is why they aren't analogous. lumped components can't
represent the traveling waves.


you can't have current without a voltage difference. *

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.



plus i
can substitute in a lossy line and measure the line heating and
measure no i^2r loss that would exist if there were currents flowing
in the line, so there are none.

I certainly agree that there is no real current flowing, but the
fictititious If and Ir are indeed present (or pick a better word
if you like).

now you say they are fictitious, yet you have formulas for them and
can measure them... until you come to your senses and admit that
reflections are real there isn't much we can do for you so the rest of
your post is based on an incorrect premise.

snip

what happens to reflected energy ?
 

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

what happens to reflected energy ?
 

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

what happens to reflected energy ?
 

On Jun 18, 3:06*am, Roy Lewallen wrote:
Nothing I have ever
posted contradicts, or intends to contradict, the very well known and
established theory which has been found to be correct and useful for
over a hundred years.

On the contrary, what you have said indeed does contradict 100 years
of optical physics. All you have to do to alleviate your ignorance is
to grok that superposition (wave-cancellation/interference) can and
does redistribute energy even when there are no reflections present.
The ONLY time that a forward wave and a reflected wave don't interfere
with each other is when they are 90 degrees different in phase.

What I find silly is the idea that there are waves
of average power bouncing back and forth, needing to "go" somewhere, and
the specious arguments put forth in desperate attempts to explain where
these elusive waves "go" and how they supposedly behave.

Have you never stood in a hall of mirrors where the reflections seem
to go on out to infinity on both sides? You would have us believe that
those reflections are not bouncing back and forth, mirror to mirror,
and that those reflections that you can see with your own eyes have
zero energy??? Roy, you are just spouting guru-type metaphysics, and
should be ashamed of yourself for trying to spread your religion to
the unwashed masses.
--
73, Cecil, w5dxp.com

what happens to reflected energy ?
 

On Jun 18, 6:00*am, Keith Dysart wrote:
"Is this a situation
where the computed reflected power represents something real?"
because if it does not, the original question is moot.

It certainly depends upon your definition of "power" and that
definition is different between pure physics and RF engineering. One
of the accepted definitions of "power" in "The IEEE Dictionary"
contradicts the definition of "power" given in my college physics
book. "The IEEE Dictionary" says that flowing energy passing a fixed
measurement point is power, by definition.

If there is anything at all that can be measured, it must necessarily
contain energy. If the EM energy is moving, it is power, by IEEE
definition. If you disagree, take it up with The IEEE.

So the actual question that needs to be answered is: Does the
electromagnetic reflected wave contain energy traveling at the speed
of light in the medium? The answer is yes, being photonic in nature,
an EM wave must necessarily contain an ExH power density and,
consisting of photons, must necessarily be traveling at the speed of
light in the medium. The EM wave will continue to travel in the
direction of energy flow at the speed of light in the medium until it
encounters an impedance discontinuity which causes a reflection and/or
interference.

There is really no difference in a forward wave and a reflected wave
except for the direction of travel. They both contain an associated
ExH power density. One can set up identical signal generators at each
end of a transmission line to emulate forward and reflected waves.
Which generator yields forward waves and which generator yields
reflected waves? It doesn't really matter because direction is only a
convention.
--
73, Cecil, w5dxp.com

what happens to reflected energy ?
 

On 18 jun, 12:26, Cecil Moore wrote:
On Jun 18, 6:00*am, Keith Dysart wrote:

"Is this a situation
where the computed reflected power represents something real?"
because if it does not, the original question is moot.

It certainly depends upon your definition of "power" and that
definition is different between pure physics and RF engineering. One
of the accepted definitions of "power" in "The IEEE Dictionary"
contradicts the definition of "power" given in my college physics
book. "The IEEE Dictionary" says that flowing energy passing a fixed
measurement point is power, by definition.

If there is anything at all that can be measured, it must necessarily
contain energy. If the EM energy is moving, it is power, by IEEE
definition. If you disagree, take it up with The IEEE.

So the actual question that needs to be answered is: Does the
electromagnetic reflected wave contain energy traveling at the speed
of light in the medium? The answer is yes, being photonic in nature,
an EM wave must necessarily contain an ExH power density and,
consisting of photons, must necessarily be traveling at the speed of
light in the medium. The EM wave will continue to travel in the
direction of energy flow at the speed of light in the medium until it
encounters an impedance discontinuity which causes a reflection and/or
interference.

There is really no difference in a forward wave and a reflected wave
except for the direction of travel. They both contain an associated
ExH power density. One can set up identical signal generators at each
end of a transmission line to emulate forward and reflected waves.
Which generator yields forward waves and which generator yields
reflected waves? It doesn't really matter because direction is only a
convention.
--
73, Cecil, w5dxp.com

Hello friends. good day for you.

I am glad that I have not bothered with my joke, Keith! :) I was to
do some comments to your and Roy posts, but I prefer to ask questions
rather than giving personal opinions :)
.....
Please Cecil, friend, do not mention photons and optics (do not bother
me, don't worry), I believe you better play your game without such
things. I know is like fighting with one arm tied, but I think you
can, make an effort :)

Please Owen explain to me what is "negative power".

In the other thread you said = "The notion that reflected power is
simply and always absorbed in the real source resistance is quite
wrong. Sure you can build special cases where that might happen, but
there is more to it. Thinking of the reflected wave as 'reflected
power' leads to some of the misconception.
Then I asked; "remember me what reflected power definition are you
using here and expand the sentence idea".
I did not ask for the quoting, I ask (again) for your concept of
Reflected Power. I do not sure if you accept (or not) the very notion
of reflected power as legitimate. The sentence it is not clear enough
to me.

Please, tell us if you accept the idea of two very directive
electromagnetic waves from different sources flowing in opposite
directions as carring independent energy (with different frequencies
at first) and information, and the posibbility that energy be
dissipated and/or transmitted by de opposite system and the
information being recovered in both ends.
If your answer is yes, then tell us if you conceive such similar
system using a same wave guide to simultaneously vinculate both
system.

73 - Miguel - LU6ETJ

PS: Roy, friend, take it ease (is OK, or polite say "take it easy?),
references to the article were only to complement (bring close?) some
useful notions related, mentioning a possible common reference (my
english books are few, old and possibly useless as reference
nowadays)

what happens to reflected energy ?
 

On Jun 18, 1:01*pm, lu6etj wrote:
Please Cecil, friend, do not mention photons and optics (do not bother
me, don't worry), ...

What do I do about people who assert that RF waves possess the ability
to violate the known laws of EM wave physics? It doesn't matter what
the frequency of an EM wave is, it must obey the laws of physics, two
of which a

1. It cannot exist without a Poynting vector ExH power density.

2. It must necessarily move at the speed of light in the medium.
--
73, Cecil, w5dxp.com

what happens to reflected energy ?
 

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.

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.

I certainly agree that there is no real current flowing, but the
fictititious If and Ir are indeed present (or pick a better word
if you like).

now you say they are fictitious, yet you have formulas for them and
can measure them... until you come to your senses and admit that
reflections are real there isn't much we can do for you so the rest of
your post is based on an incorrect premise.

Then what are the equations for computing forward and reflected waves
and 'power'? That is where I started. Is that the premise you consider
to be incorrect?

Do follow through the arithmetic in the previously provided examples
and see if you can find any faults. Point me to any errors and I will
gladly reconsider my position.

If not, perhaps it is time for you to start questioning your
assumptions.

....Keith

what happens to reflected energy ?
 

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.


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.

but even if you have the case above and connect another 100v battery
on the open end of the line there will be no current flowing back as
there is no voltage difference to drive it.



I certainly agree that there is no real current flowing, but the
fictititious If and Ir are indeed present (or pick a better word
if you like).

now you say they are fictitious, yet you have formulas for them and
can measure them... until you come to your senses and admit that
reflections are real there isn't much we can do for you so the rest of
your post is based on an incorrect premise.

Then what are the equations for computing forward and reflected waves
and 'power'? That is where I started. Is that the premise you consider
to be incorrect?

yep, your 'waves' don't include the proper term to make them
propagating waves that satisfy maxwell's equations... so they can't be
real... so your initial premise is incorrect.


Do follow through the arithmetic in the previously provided examples
and see if you can find any faults. Point me to any errors and I will
gladly reconsider my position.

If not, perhaps it is time for you to start questioning your
assumptions.

...Keith

my assumptions are perfectly valid... it is yours that are flawed from
the very beginning.

what happens to reflected energy ?
 

On Jun 17, 9:12*am, Cecil Moore wrote:
On Jun 17, 5:38*am, Keith Dysart wrote:

You are an RF dude and there is nothing
that can possibly be learned from simple DC circuits.

We studied DC circuits in EE101. Later we were forced to expand our
knowledge into distributed networks because the DC circuit model
failed at RF frequencies.

One does, however, usually require that the more sophisticated model
continue to provide the correct answer for the simpler problems that
could be solved in another way.

And it does here as well, though the answers make some uncomfortable.

Still, I observe that you have discovered no errors in the exposition.

I have not wasted my time time trying to discover any errors. In a
nutshell, what new knowledge have you presented?

Well, at least you continue to read the posts, so perhaps there is
some slim hope, but I am not holding my breath.

As for me, I have learned a lot from your postings since I first
encountered them circa 1994. Back then, I was not so knowledgeable
about transmission lines and I was trying to decide who was right.
You posted well thought out, convincing arguments and I found myself
switching from side to side. Reflected waves heat finals, no they
don't, yes they do, ... but eventually I learned. Then I continued
to read your posts, visited your web site and studied your arguments.
The challenge was to discover where you had gone wrong. Occasionally
I would think up a suitable counter example and offer it to you, but,
even then, you refused to look at examples that might threaten your
beliefs. So I no longer post with any expectation that you might be
persuaded.

Mostly now I post to ensure that a view other than yours is voiced
to help prevent those who are not yet sure from being sucked in to
your vortex.

Anyhow, I suggest you read my examples and find the flaws, for that
is the surest way to convince me that I am wrong. And don't just
post other examples that support your position, for it only takes
one counter-example to disprove a hypothesis, no matter how many
examples are in agreement. And don't reject simple DC examples
because
they lead to uncomfortable answers for it is by examining the
examples
that do not support your hypotheses that you will learn, not by
sticking
just with the ones that do.

....Keith