Standing-Wave Current vs Traveling-Wave Current
 

Roy Lewallen wrote:
People who are having trouble with the concept of a -1 voltage
reflection coefficient for a perfect voltage source might benefit from
the following exercise:

Look at my first analysis, where the perfect source was connected
directly to the transmission line. Make no assumptions about the
impedance or reflection coefficient it presents. Then, when the
reflection of the initial forward wave returns, calculate the value the
re-reflected wave must have in order to make the sum of the waves
present, which is the line input voltage, equal to the perfect source
voltage. The voltage of the perfect source can't change, by definition.
The ratio of the re-reflected wave to the returning wave is the voltage
reflection coefficient (since we're dealing with voltage waves).

I'll do it for you:

The forward wave was vf(t, x) = sin(wt - x)
The returning wave was vr(t, x) = sin(wt + x)

The returning wave will strike the input end of the line and create a
new forward wave with value vf2(t, 0) = Gs * vr(t, 0) at the input,
where Gs is the source voltage reflection coefficient.

Before the first forward wave returns, we have only vf(t, 0) = sin(wt)
at the input end of the line. This is of course the source voltage.


After the wave arrives and re-reflects, we have at the input end

vtot(t, 0) = vf(t, 0) + vr(t, 0) + vf2(t, 0)
= vf(t, 0) + vr(t, 0) * (1 + Gs)

This must equal the source voltage, which is the line input voltage, and
cannot change. So plugging in values:

sin(wt) = sin(wt) + sin(wt) * (1 + Gs)

Solving for Gs = Gs = -1


I have made no statement about the "impedance" of the perfect source.
The only thing I've required is that the voltage remains constant, which
is the very definition of a perfect source. You can do a similar
exercise to show that the voltage reflection coefficient of a perfect
current source is +1.

Roy Lewallen, W7EL

I see your example as identical to Keith's example of two wave pulses
traveling in opposite directions.

At the point of interaction, Keith's example has a reflection factor of
1 or zero, depending upon whether the waves bounce or pass. Keith's
example is not a short circuit because two pulses of identical polarity
are interacting so a reflection factor of -1 could never exist.

In your example, the presence of voltage from the ideal source creates
conditions identical to Keith's example for the returning reflected
wave. Accepting this premise, then the reflection factor must be either
1 or zero, depending upon whether the waves bounce or pass.

By assuming that the waves reflect at the ideal source, you proved that
the reflection factor is -1, which is the factor for a short circuit.
This can not be the case, so waves must not reflect at the ideal voltage
source, they must pass.

vtot(t, 0) = vf(t, 0) + vr(t, 0)

= sin(wt)tot = sin(wt) + sin(wt)

= 2*sin(wt)

Someone will certainly say the the vtot(t,0) at the source location is
the source voltage, because it is defined that way. The conditions at
point (t,0) itself is actually unknown (because vf mysteriously appears,
and vr disappears by going off the transmission line), but point (t, 0)
is defined by assumptions. Therefore, at

vtot(t, 0) = vf(t, 0) + vr(t, 0)

= sin(wt)tot = sin(wt) + sin(wt)

= 2*sin(wt)


73, Roger, W7WKB

Standing-Wave Current vs Traveling-Wave Current
 

Keith Dysart wrote:
Can photons explain the state of a transmission
line driven with a step function after the line
has settled to a constant voltage?

Of course, photons can be used to explain all
EM wave action. A step function accelerates
electrons which then emit photons as EM waves.
Hint: electrons cannot move at the speed of
light. EM waves move at the speed of light.
--
73, Cecil http://www.w5dxp.com

Standing-Wave Current vs Traveling-Wave Current
 

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.

So that settles it, then.

No, it is not settled. Please reply to my posting
where I proved the reflection coefficient is plus
or minus one, the exact opposite of what you assert.

And the arguments that I have seen between Mr. Maxwell
and Dr. Bruene are on a completely different matter.

If you think that, it's prima facie evidence that you
are really confused.
--
73, Cecil http://www.w5dxp.com

Standing-Wave Current vs Traveling-Wave Current
 

Keith Dysart wrote:
My attempt to confuse!? We were discussing the determination
of reflection coefficients for Thevenin equivalent circuits.

No, we were discussing destructive and constructive
interference. Sorry about your confusion.

But in another post, you have agreed that there
is a complete lack of references supporting your
position, ...

No, I said I am not going to look for them. Sorry
about your confusion.

Enjoy the new ability to solve problems that
were previously outside your grasp.

I'm sorry, Keith, delusions of grandeur are a problem
outside of my field of expertise. Perhaps a professional
shrink could help you better than I.
--
73, Cecil http://www.w5dxp.com

Standing-Wave Current vs Traveling-Wave Current
 

On Thu, 3 Jan 2008 06:28:19 -0500, "David J Windisch"
wrote:

You were writing of Tesla over Edison, a-c over d-c power transmission, and
I was reminded that Prez Kennedy started some work years ago somewhere in
the West, involving hvdc transmission. I wasn't intentionally obscuring
things in that earlier post. 73 Dave N3HE

Hi Dave,

I am well practiced mixing it up with the masters of obscurity, and I
certainly don't confuse you with Prior Art, or Cecileo. However, the
HVDC project you allude to is new to me (even if it is/was decades
old).

As an aside, I have worked with the lawyer who was instrumental in
closing down WPPS (a nuclear power consortium we generally called
WHOOPS). That consortium agonized that Washington's credit rating
would go down the toilet if we defaulted - barely a blip in the
interest rate resulted when we mothballed several nuclear reactors.

Even further aside, I lived in Japan in the early 50s when they used
DC residential power (and we had to be careful to buy AC/DC
appliances). Woe to those living at the end of the block where the
street lights were dim.

Edison used to portray AC as being the killer current (eventually
selected for use on death row). Actually there were a mix of
characteristics that lent either the death potentiality. DC will
cause the muscles to clamp, and if you seized a hazardous wire, you
could never let go. AC, on the other hand (no pun), would cause
fibrillation, and you stood some chance of releasing the same
hazardous wire. AC, on the third hand (again, no pun), would also
cause the sweat glands to excrete (due to the same fibrillation) and
lower your path resistance (more lethal current).

73's
Richard Clark, KB7QHC

Standing-Wave Current vs Traveling-Wave Current
 

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.
In another thread, I proved your assertion 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
--
73, Cecil http://www.w5dxp.com

Standing-Wave Current vs Traveling-Wave Current
 

On 3 Jan, 09:51, Cecil Moore wrote:
Keith Dysart wrote:
If such analysis is appropriate, then it seems
to me that a pulse can be viewed as a chunk of
charge moving down the line.
Q2. Is this an appropriate view?

No.

Q3. If so, then what happens when two such chunks
of charge collide in the middle of the line?

They don't "collide". Clouds of photons collide
and their behavior is well known.

Q5. If no charge crosses the mid-point, then how
do the pulses, made up of chunks of charge.
pass the mid-point?

How can two water waves pass through each other
while the water molecules are only moving up and
down?
--
73, Cecil *http://www.w5dxp.com

Cecil
The current only changes direction at the behest
of the frequency. I know you think that the
current changes at the behest of the
length of radiator used i.e. at the top but
frankly that is lunacy. It is only when the capacitor charge
starts to move to the inductance and starts to store energy
which generates a diamagnetic field can propagation can occur
You must get a pencil and paper and think things out for
yourself instead of relying on books. With your IQ it should
be a cake walk because unlike some others you are able to get
back to first principles instead of learning by rote.
No personal offence intended. Being stubborn is not all that bad
but only if you are willing to think out alternatives given.
Otherwise it is a not invented in my back yard aproach which
amounts to an over inflation of one's ability
Best regards
Art

Standing-Wave Current vs Traveling-Wave Current
 

Keith Dysart wrote:
The example was carefully chosen to illustrate the
point, of course. But that is the value of particular
examples.

When the pulses are not identical, the energy that crosses
the point is exactly sufficient to turn one pulse
into the other.

The remainder of the energy must bounce
because it does not cross the mid-point.
...Keith

So it really is almost as though the pulses travel through one
another, rather than bounce off one another.

I have seen the concept that energy doesn't cross nodal points alluded
to in some texts. However there are so many exceptions to it found in
physical systems as to render it a dubious notion at best. Useful
perhaps for illustration purposes.

In the discussion of standing waves on a string, Halliday and Resnick
says "It is clear that energy is not transported along the string to
the right or to the left, for energy cannot flow past the nodal points
in the string, which are permanently at rest. Hence the energy
remains "standing" in the string, although it alternates between
vibrational kinetic energy and elastic potential energy."

So the idea is valid for a simple harmonic oscillator in which there
are no losses. In such a case, once the system begins oscillating, no
further input of energy is required in order to maintain oscillation.
Clearly there is no flow of energy into or out of such a system.
What is clear is that energy doesn't pass through the nodes. It is
less clear that there exists an inherent mechanism which prevents the
movement of energy.

And so it appears in cases where there is no transfer of energy that
one might claim that waves bounce off of one another. There are no
other examples, and no supporting mechanism for it of which I am
aware, and so one might be equally justified in claiming that waves
pass through each other in all cases.

73, ac6xg

Standing-Wave Current vs Traveling-Wave Current
 

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.

Please see Maxwell's equations for further
enlightenment.
--
73, Cecil http://www.w5dxp.com

Standing-Wave Current vs Traveling-Wave Current
 

On Jan 3, 1:48*pm, Cecil Moore wrote:
Keith Dysart wrote:
Can photons explain the state of a transmission
line driven with a step function after the line
has settled to a constant voltage?

Of course, photons can be used to explain all
EM wave action. A step function accelerates
electrons which then emit photons as EM waves.
Hint: electrons cannot move at the speed of
light. EM waves move at the speed of light.

Please describe the final state of the step
excited open circuited line using photons.

Thanks,

Keith

Standing-Wave Current vs Traveling-Wave Current
 

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

Standing-Wave Current vs Traveling-Wave Current
 

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

Standing-Wave Current vs Traveling-Wave Current
 

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

Standing-Wave Current vs Traveling-Wave Current
 

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

Standing-Wave Current vs Traveling-Wave Current
 

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

Standing-Wave Current vs Traveling-Wave Current
 

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

Standing-Wave Current vs Traveling-Wave Current
 

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

Standing-Wave Current vs Traveling-Wave Current
 

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

Standing-Wave Current vs Traveling-Wave Current
 

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

Standing-Wave Current vs Traveling-Wave Current
 

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

Standing-Wave Current vs Traveling-Wave Current
 

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

Standing-Wave Current vs Traveling-Wave Current
 

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

Standing-Wave Current vs Traveling-Wave Current
 

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

Standing-Wave Current vs Traveling-Wave Current
 

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

Standing-Wave Current vs Traveling-Wave Current
 

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

Standing-Wave Current vs Traveling-Wave Current
 

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

Standing-Wave Current vs Traveling-Wave Current
 

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.

Standing-Wave Current vs Traveling-Wave Current
 

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

Standing-Wave Current vs Traveling-Wave Current
 

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]

Standing-Wave Current vs Traveling-Wave Current
 

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

Standing-Wave Current vs Traveling-Wave Current
 

On Thu, 3 Jan 2008 12:19:32 -0800 (PST), art
wrote:

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

My dear Prior Art, sainted Guru of the RF, sir,

By your response, you certainly must admit that your celebrity status
abhors being ignored. After all, who else has responded to your
nonsense?

My idiocy serves yours by keeping your words alive! Who needs civily
when it draws only flies?

73's
Richard Clark, KB7QHC

Standing-Wave Current vs Traveling-Wave Current
 

Gene Fuller wrote:
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.

HP would be interested in knowing your theory that the
s-parameter equations cannot be used on a transmission
line. How can you possibly be that ignorant?

Actually, s-parameter equations are an ideal way to
analyze an impedance discontinuity in a transmission
line since the voltages are normalized to SQRT(Z0).
Squaring the normalized voltages yields power.

----50 ohm line--+--1/2WL 300 ohm line--50 ohm load
a1--|--b2
b1--|--a2

b1 = s11*a1 + s12*a2 b2 = s21*a1 + s22*a2

|a1|^2 is forward power, |b1|^2 is reflected power
|b2|^2 is forward power, |a2|^2 is reflected power

Squaring the s-parameter equations yields the power
density (irradiance) equation from the field of optics.
I'm sorry that technical fact upsets you so.
--
73, Cecil http://www.w5dxp.com

Standing-Wave Current vs Traveling-Wave Current
 

Richard Clark wrote:
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?

There, there, Richard, everything is going to be OK.
(Somebody get the net!)
--
73, Cecil http://www.w5dxp.com

Standing-Wave Current vs Traveling-Wave Current
 

Cecil Moore wrote:
Gene Fuller wrote:
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.

HP would be interested in knowing your theory that the
s-parameter equations cannot be used on a transmission
line. How can you possibly be that ignorant?

Actually, s-parameter equations are an ideal way to
analyze an impedance discontinuity in a transmission
line since the voltages are normalized to SQRT(Z0).
Squaring the normalized voltages yields power.

----50 ohm line--+--1/2WL 300 ohm line--50 ohm load
a1--|--b2
b1--|--a2

b1 = s11*a1 + s12*a2 b2 = s21*a1 + s22*a2

|a1|^2 is forward power, |b1|^2 is reflected power
|b2|^2 is forward power, |a2|^2 is reflected power

Squaring the s-parameter equations yields the power
density (irradiance) equation from the field of optics.
I'm sorry that technical fact upsets you so.

It's a swing and a miss. Strike three! You're out!

The entire point of s-parameter analysis is that the "network" can be
treated as a black box, characterized by the various parameters at the
ports. How does that work to analyze what is happening *inside* the box,
such as somewhere along the transmission line?

73,
Gene
W4SZ

Standing-Wave Current vs Traveling-Wave Current
 

Richard Clark wrote:
There is a vast gulf between seeming and proving.

Richard, you seem to exist. Please prove that you
indeed do exist.
--
73, Cecil http://www.w5dxp.com

Standing-Wave Current vs Traveling-Wave Current
 

Gene Fuller wrote:
The entire point of s-parameter analysis is that the "network" can be
treated as a black box, characterized by the various parameters at the
ports. How does that work to analyze what is happening *inside* the box,
such as somewhere along the transmission line?

That's pretty simple to answer, Gene. Simply make
the black box infinitesimally thin so that events
are transparent. Ignorance is immediately alleviated.
--
73, Cecil http://www.w5dxp.com

Standing-Wave Current vs Traveling-Wave Current
 

Cecil Moore wrote:
Roy Lewallen wrote:
2. I don't understand the mechanism which causes waves to bounce.

I don't understand the mechanism which causes waves to slosh.
Would you mind posting the sloshing equation?

Cecil,

I am not Roy, but I will answer your question.

It is really *very* straightforward. As I hope you will agree, a
standing wave is characterized by voltage and current that are in
quadrature, both in time and in space. In other words, the current loops
are spaced 1/4 wave from the voltage loops, and the current maximum
occurs 1/4 cycle ahead of (or behind) the voltage maximum.

Let's look at conservation of energy, one of your favorite topics.

At some arbitrary starting time, the voltage loop is at a maximum, and
the current is zero. At that time all of the energy in the standing wave
is contained in the capacitive or E^2 term. The physical location of
that energy is centered on the voltage loops.

As the cycle proceeds, the voltage decreases and the current increases.
After 1/4 cycle the current is at a maximum and the voltage is zero.
Now all of the energy in the standing wave is contained in the inductive
or H^2 term. The physical location of that energy is centered on the
current loops.

There are two important observations. First, the location of the
standing wave energy has physically moved by 1/4 wave during the 1/4
cycle. That is the "sloshing" that some of us use as a description. Call
it what you like.

The second observation is that the standing wave is not at all static.
Energy is moving back and forth at the speed of light in the medium.

Some people like to treat standing waves as poor distant cousins to
"real" waves, or perhaps only as "envelopes". It is their option to do
so, but it misses the actual behavior of standing waves and "sloshing"
energy.

The equations can be found in many places, such as Johnson's book on
Transmission Lines and Networks.

73,
Gene
W4SZ

Standing-Wave Current vs Traveling-Wave Current
 

Gene Fuller wrote:
The second observation is that the standing wave is not at all static.
Energy is moving back and forth at the speed of light in the medium.

Sorry Gene, The speed of water is "sloshing".
The speed of light is 10^10 faster than "sloshing".
--
73, Cecil http://www.w5dxp.com

Standing-Wave Current vs Traveling-Wave Current
 

Gene Fuller wrote:
The entire point of s-parameter analysis is that the "network" can be
treated as a black box, ...

Gene, I cannot find anywhere in the s-parameter information
where some of the network must be hidden inside a black box.
I always thought an s-parameter analysis could be done without
a black box. Could you help us out here and point out exactly
where it says a black box is a requirement. Seems to me the
purpose of an s-parameter analysis is to alleviate ignorance
which obviously doesn't match your agenda.
--
73, Cecil http://www.w5dxp.com

Standing-Wave Current vs Traveling-Wave Current
 

Cecil Moore wrote:
Gene Fuller wrote:
The second observation is that the standing wave is not at all static.
Energy is moving back and forth at the speed of light in the medium.

Sorry Gene, The speed of water is "sloshing".
The speed of light is 10^10 faster than "sloshing".

Cecil,

If your only concern is the definition of "sloshing", then about 100,000
messages have been wasted.

My definition of "sloshing" is as I stated. I believe that Roy would
have the same definition. If your definition involves the speed of
water, then I have no idea why that topic would be relevant here.

73,
Gene
W4SZ