On Jan 3, 1:56*pm, Cecil Moore wrote:
Keith Dysart wrote:
So there is NO reference that claims
that the output impedance can not be used to
compute the reflection coefficient.

If I say I am not going to look for a reference
to "creation" in The Bible, are you going to assert
there are no references to creation in The Bible?
Good luck on your ridiculous assertions.

You do seem to like to clip the important bits.

It was your sentence: "If there was a reference,
Mr. Maxwell or Dr. Bruene would have reported it
by now but their argument continues to rage" that
made it clear you did not expect to be able to
find a reference.

So that settles it, then.

And that was what settled it. No expectation
of a reference... Then no reason for you to
argue further.

...Keith

On Thu, 3 Jan 2008 07:55:48 -0800 (PST), Keith Dysart
wrote:

The presence of this poster providing misleading
information makes this group a rather unique learning
environment.

Hi Keith,

By a certain opaque style of writing, the tenor of questions offered,
and an aversion to to deliberating the evidence given in response; I
would say the crown of Trolling is being challenged and Cecil may slip
beneath a new prince's claim.

73's
Richard Clark, KB7QHC

On Jan 3, 2:15*pm, Cecil Moore wrote:
Keith Dysart wrote:
Good answers. Exactly as I expected. Now please
explain the applicability of EM waves to the
state of an open circuited line excited with a
step function, especially after it settles to a
constant voltage (where only an E field will be
present).

Before it settles to a constant voltage, there is
acceleration of electrons that results in an EM
photonic wave.

After it settles to a constant voltage, there is
no acceleration of electrons and the EM photonic
wave disappears.

So when the edge of the step is travelling towards
the right, is there an EM wave to the right of the
step, to left of the step, at the step, or all three?
Similar question for when the step is travelling
back to the generator?

When the line has settled, how do you add the forward
and reflected wave to compute the voltage on the line,
or does the disappearance of the wave mean this is
now impossible?

If only the step itself has an EM wave, how are
voltages computed using reflection coefficient after
the step has reflected from the open end?

...Keith

Keith Dysart wrote:

clip.....
I fully agree with the philosophy you express here Keith. But I can see
how you would doubt that I am practicing what I just agreed with.

You may have mis-interpreted my comments. I have NOT
seen evidenace of the behaviour I describe above in
your writings.

The comments mostly apply to a single poster who has
been posting on this group for many years, at least
since when I first started viewing this group in the
mid 90s and began to really gain an understanding of
transmission lines.

The presence of this poster providing misleading
information makes this group a rather unique learning
environment.

In most learning environments, the information is
neatly packaged and presented from a consistent
point of view with no challenge.

Here, a lot of chaff is mixed with the wheat. This
has the "benefit" of forcing the learner to
understand well enough to make decisions between
competing explanations. The learner who makes the
right choices comes out with a much more solid
understanding than one who has just been (spoon)
fed the story. On the other hand, some have
probably been lead seriously astray.

For sure, I have a better understanding than
I would have had without the challenging
misleading information.

So for sure it would be better for the poster
in question were he to let go of some of his
incorrect beliefs, it would also reduce some of
the opportunities for learning provided to
others lurking or partaking in the discussions.

...Keith


Thanks Keith. I am learning a tremendous amount here. As you say, the
interaction really helps focus, reason, and justify, and finally,
readjust thinking as understanding improves.

I certainly like your example of two opposite traveling pulses. I used
it again today in a posting responding to Roy.

73, Roger, W7WKB

Keith Dysart wrote:
Please describe the final state of the step
excited open circuited line using photons.

Photons are emitted and absorbed by the electrons
as the electrons lose/gain energy. Photons are not
conserved. Only the energy in photons is conserved.

In a DC system with no accelerating or decelerating
electrons, all of the photons have been absorbed
back into the electrons (or lost to radiation).
Of course, this describes an ideal system.
--
73, Cecil http://www.w5dxp.com

On Jan 3, 2:10*pm, Cecil Moore wrote:
Keith Dysart wrote:
Excellent. So there is NO reference that claims
that the output impedance can not be used to
compute the reflection coefficient.

That is probably a false statement. I just haven't
wasted my time looking for a reference that uses
those exact words.

There are many references that do.

I seriously doubt that they say what you are
asserting. Please produce those references.

One has been directly provided, though many
more are available using the google searches
previously suggested.

But that one is infinitely more than those
available supporting the opposite view.

In another thread, I proved your assertion wrong.

Asserting that you have proved an assertion wrong
is not the same as proving it wrong.

A Bird wattmeter placed at the output of your source
will read forward power = reflected power. The
reflection coefficient can be calculated from
that. rho = SQRT(Pref/Pfor) = plus or minus 1.0

Of course. With one side of the Bird wattmeter
left open, it will happily measure the reflection
coefficient of that open. This says nothing about
the reflection coefficient of the line connection
with the source.

...Keith

On Thu, 03 Jan 2008 12:25:59 -0600, Cecil Moore
wrote:

The presence of this poster providing misleading
information makes this group a rather unique learning
environment.

For the record: The only controversial assertion
that I have ever made is that coherent EM wave
cancellation can cause a redistribution of the
EM energy in the opposite direction in a transmission
line. No one has proved that assertion to be wrong.

What an ego to rush to slip into a TNT vest in the hope of being
associated with Nobel.

As usual, Cecil's arguments are so script driven, that I cannot pass
up this mocking opportunity:

I shall assert that coherent EM wave cancellation can not cause a
redistribution of the EM energy in the opposite direction in a
transmission line.

No one has proved that assertion to be wrong.

Does that misleading statement qualify me for Keith's anointed villain
of the group? Cecil certainly has described me as being scurrilous
enough to so qualify! ;-)

Besides, I think I look better in that vest than he does.

73's
Richard Clark, KB7QHC

Keith Dysart wrote:
It was your sentence: "If there was a reference,
Mr. Maxwell or Dr. Bruene would have reported it
by now but their argument continues to rage" that
made it clear you did not expect to be able to
find a reference.

Make that *easily* find a reference and you will
have it correct. Just because I am lazy is not
a proof that the reference doesn't exist.

And that was what settled it. No expectation
of a reference... Then no reason for you to
argue further.

Just a minute. What about the proof I offered
that the actual reflection coefficient is 1.0
based on Bird wattmeter readings?

The Bird tells us that at the source terminals,
the forward power equals the reflected power.
rho = SQRT(Pref/Pfor) = plus or minus 1.0
--
73, Cecil http://www.w5dxp.com

Cecil Moore wrote:
Mike Monett wrote:

The term "bounce" means they interact. Electromagnetic signals do
not interact. They superimpose. Each is completely unaware and
unaffected by the other.


Except when they are coherent, collinear in the same direction,
equal in magnitude and 180 degrees out of phase.

Fabricated nonsense.

Thensomething
permanent happens as signified by the s-parameter equation.

b1 = s11*a1 + s12*a2 = 0

s11*a1 and s12*a2 are coherent, collinear in the same
direction, equal in magnitude and 180 degrees out of
phase. Their combined energy components are redistributed
in the direction of b2 = s21*a1 + s22*a2

Squaring the above equation results in the power density
irradiance equation from the field of optics.

Preflected = (b1)^2 = (s11*a1)^2 + (s12*a2)^2 + 2*s11*a1*s12*a2 = 0

micro.magnet.fsu.edu/primer/java/scienceopticsu/interference/waveinteractions/index.html


"... when two waves of equal amplitude and wavelength that
are 180-degrees ... out of phase with each other meet, ...
All of the photon energy present in these waves must somehow be
recovered or redistributed in a new direction, according to the
law of energy conservation ... Instead, upon meeting, the photons
are redistributed to regions that permit constructive interference, ..."

That certainly sounds like an "interaction" to me.

But to truly be considered official proof of an interaction, it must
walk like one as well as sound like one, IIRC.

:-) ac6xg

On Thu, 3 Jan 2008 10:28:07 -0800 (PST), art
wrote:

When current gets to the top of a
fractional
wave antenna it just does not turn back. It has to wait until half a
period time has
elapsed

Guru Prior Art, sir,

Which is more rankling to your celebrity:
1. being ignored for such stupid remarks;
1. being criticised for such stupid remarks?

To reduce confusion, select 1 of the above in response.

73's
Richard Clark, KB7QHC