Keith Dysart wrote:
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?

You are confusing cause and effect. There is an EM
wave wherever there are photons being exchanged
among the electrons. Any speed-of-light movement
is evidence of the existence of photons.

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?

After all the photons have been absorbed or radiated,
there is no forward EM wave or reflected EM wave.
They simply cease to exist in the DC steady-state
where electrons are not being accelerated or
decelerated.

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?

If the step is reflected, the reflection consists
of photons. As long as anything is flowing at the
speed of light in the medium, photons exist.
--
73, Cecil http://www.w5dxp.com

Cecil Moore wrote:
Gene Fuller wrote:
I challenged you to find any case in AN 95-1 that supports your claim
of counter-traveling waves in a transmission line, with each wave
carrying its own energy that somehow nets out to zero.

I did exactly that earlier and you didn't comprehend
it then - but here it is again.

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

(b1)^2 is reflected power. It is only zero when
(s11*a1 + s12*a2)^2 = 0

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

Since a1 and a2 are phasors, their multiplication
involves cos(A) of the Angle between them.

Pref1 = P1 + P2 + 2*SQRT(P1*P2)cos(A)

Does that equation look familiar? Please reference the
s-parameter ap note, pages 16 & 17, for the meaning of
those squared terms. The power density equation can be
derived from the s-parameter equation.

http://www.ecs.umass.edu/ece/labs/an...parameters.pdf

Wow! You missed again! And I thought that you actually understood what
s-parameters are all about.

Get a clue. None of your rantings say anything about the behavior of
waves on the transmission line. As usual you keep ducking the question
by answering a different one.

73,
Gene
W4SZ

On Thu, 03 Jan 2008 19:44:31 GMT, Cecil Moore
wrote:

of a reference... Then no reason for you to
argue further.

Just a minute.

Reminds me of that Monty Python moment during the Black Plague and the
cart passing in the street.

Old man struggling with family against being put in:
"Wait! WAIT! I'm not dead yet!"

Family struggling harder:
"Oh YES YOU ARE!"

73's
Richard Clark, KB7QHC

Keith Dysart wrote:
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.

Any way you choose to look at the example, the same
amount of joules are flowing into the source as are
flowing out of the source during any particular time
period. That is a power reflection coefficient of 1.0
Take the square root to find the voltage reflection
coefficient of plus or minus 1.0
--
73, Cecil http://www.w5dxp.com

On Thu, 3 Jan 2008 11:19:26 -0800 (PST), Keith Dysart
wrote:

Hint: electrons cannot move at the speed of
light. EM waves move at the speed of light.

I love these built-in failures of argument. :-)

Shine the sun on a pie pan. How fast is light moving in getting
through it? How fast is an electron moving in getting through it?

Is light traveling at the speed of light? Would it travel faster than
an electron if we took out the pie? Would it travel faster than an
electron if we kept the pie and took out the pan?

73's
Richard Clark, KB7QHC

Richard Clark wrote:
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.

The Melles-Groit and FSU web pages certainly seem to
disagree with you. To the best of my knowledge, they
prove your assertion to be wrong and support my
contention of redistribution of energy after wave
cancellation.

http://www.mellesgriot.com/products/optics/oc_2_1.htm

"Clearly, if the wavelength of the incident light and
the thickness of the film are such that a phase difference
exists between reflections of p, then reflected wavefronts
interfere destructively, and overall reflected intensity is
a minimum. If the two reflections are of equal amplitude,
then this amplitude (and hence intensity) minimum will be
zero." (Referring to 1/4 wavelength thin films.)

"In the absence of absorption or scatter, the principle of
conservation of energy indicates all 'lost' reflected intensity
will appear as enhanced intensity in the transmitted beam.
The sum of the reflected and transmitted beam intensities is
always equal to the incident intensity. This important fact
has been confirmed experimentally."

http://micro.magnet.fsu.edu/primer/j...ons/index.html

"... when two waves of equal amplitude and wavelength that are
180-degrees ... out of phase with each other meet, they are not
actually annihilated, ... 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, so the effect should be considered as
a redistribution of light waves and photon energy rather than
the spontaneous construction or destruction of light."

In an RF transmission line, since there are only two possible
directions, the only "regions that permit constructive interference"
at an impedance discontinuity is the opposite direction from the
direction of destructive interference.
--
73, Cecil http://www.w5dxp.com

On 3 Jan, 11:50, Richard Clark wrote:
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

Fortunately idiots such as you are not able to debate civily
so what you think doesn't count.You may have been a good sailor
man but without a degree in engineering you will never be able
to understand first principals. Time and time again your questions
given but no technical answers provided other than taunts is
enought to show what a miserable man you are. For a person not
to understand that a time variable added to Gaussian law
equals Maxwell's lawa plus argueing about same with a man with
a doctorate who works for the space agency at MIT shows just how
much your ego has been inflated. You are just a large inflated
ballon looking for somebody who will place a prick in you.
Begone you miserable urchin.

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

Fabricated nonsense.

Coherent waves are fabricated nonsense?
Collinear, same direction waves are fabricated nonsense?
Equal magnitude waves are fabricated nonsense?
Opposite phase waves are fabricated nonsense?
--
73, Cecil http://www.w5dxp.com

On Thu, 03 Jan 2008 14:17:11 -0600, Cecil Moore
wrote:

No one has proved that assertion to be wrong.

The Melles-Groit and FSU web pages certainly seem to
disagree with you.

There is a vast gulf between seeming and proving.

My assertion stands unassailed! [except for a few pecks by a duck]

Art wrote:
"It is obvious that the completion of a cycle thus at no time has
current moving other than in a single direction."

We have a "cycle" because the current alternates or reverses direction
twice each cycle.

Hams likely agree with Terman that radio waves are produced to some
extent whenever a wire in open space carries a high-frequency current.
(Page 864, opus of 1955)

Kraus says on page 12 in the 3rd edition of "Antennas":
"Antennas convert electrons to photons, or vice versa."
Also: "Thus, time-changing current radiates and accelerated charge
radiates."
Also: The currents on the transmission line flow out on the antenna and
end there, but the fields associated with them keep on going.

Best regards, Richard Harrison, KB5WZI