Standing-Wave Current vs Traveling-Wave Current
 

Roy Lewallen wrote:
Roger wrote:

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

Not quite. The voltage following a pulse is zero; with my sine wave
example the source continues producing a sine wave at all times while
wave interaction is occurring.

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.

Because there are no mathematics and, as far as I can see, no mechanism
for wave interaction, I can't discuss Keith's example except from the
standpoint that the waves don't interact. (Excluding, of course,
superposition from the meaning of "interact".)

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.

Sorry, I don't accept the premise.

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.

Yes, that is correct.

This can not be the case, so waves must not reflect at the ideal
voltage source, they must pass.

Well, yes it can be the case, and is. But let's see where rejection of
this fact leads.

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.

I certainly say that.

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)

So by rejecting the fact that the waves reflect at the ideal source,
you're forced to conclude that the voltage at the output of a perfect
source is unknown, with waves "mysteriously" appearing and disappearing.
It also appears that you've violated Kirchoff's voltage law at the
input. My analysis requires no such contradictions, unknown voltages, or
mysteries. As an engineer, I like my analysis a whole lot better,
because it can give me answers which are testably correct.

Can you use your theory to show the voltage at all times at the input
or, if you choose, just inside the input of the line, as I've done using
conventional theory?
Yes. Here is the difference. I am using a short cut by looking at the
way waves travel and superimpose. My assumption is that the input
impedance matches the impedance of the line. Thus 1v generates a 1v
sine wave. For a simple example, such as any multiple of 1/2 wave, this
causes current proportional to applied voltage and impedance of the
line. For example, 50v applied to a 50 ohm line gives a 1 amp current.

We know a line 1/2 wavelength long will absorb energy for the length of
time it takes for a wave to travel the length of line and return, so
power must be applied during the entire time. Obviously, 2 half waves
will be applied over that time period so the amount of energy on the
line is equal to the amount of energy contained in the power applied
over time by two half waves, so total power applied is 2 * initial power.

Equally obvious, by leaving the source connected after the reflected
wave has returned, the question of how the source reacts to the external
application of power must be answered. This is a second question,
unrelated to traveling waves on a transmission line. My answer to this
question is to disconnect the source and replace the source with a
connection identical to the condition at the far end of the transmission
line. You can see that this leaves the transmission line isolated with
power circulating on the line (is is POWER, time is involved).

Your solution (as presented in Analysis 2) and the Power Analysis was to
keep the source supplied at all times in an attempt to create continuous
real world conditions, as contrasted to my method of a quick and
correctly timed disconnect. Both experiments are real world simulations
and can be performed.

Your analysis is identical to what happens at a real world transmitter.
A very good, automatically tuned transmitter! Your choice of an
ideal voltage source amounts to setting conditions that automatically
adjust for impedance changes as fast as the change happens. The
transmission line sees an outgoing impedance change going from 150 ohms
to infinity in your example. The source sees an outgoing impedance
change from 200 ohms to infinity during the same analysis period.

It is interesting to compare the final voltage ratios ( directly
proportional to energy levels and directly proportional to power levels)
on the transmission line to the SWR ratio. You used a SWR ratio of 3:1
and I used 1:1. You found that the total energy stored on the line was
4 times the initial energy applied, and I predicted that the total
energy stored was 2 times the initial. The link here is that the power
storage factor(SF) (I know you dislike the term) is found from

SF = SWR + 1.

Your storage factor was SF = 3 + 1 = 4. My storage factor was
SF = 1 + 1 = 2.

If we can accept that the storage factor is always SF = SWR + 1 for all
situations, then we have a very convient way to predict the voltage
increase that results form impedance mismatch at a discontinuity. I am
not yet ready to accept this generalization, but we should look at it
more carefully.

Ending comment. Both your and my analysis are correct. Congratulations
to both us!

73, Roger, W7WKB

Standing-Wave Current vs Traveling-Wave Current
 

Roger wrote:
If we can accept that the storage factor is always SF = SWR + 1 for all
situations, then we have a very convient way to predict the voltage
increase that results form impedance mismatch at a discontinuity.

The s-parameter equations do that automatically.
That's part of their usefulness.

Incidentally, the power reflection coefficients,
rho^2, and the power transmission coefficient,
(1-rho^2), also do the same thing. That's why
a power analysis is often easier than a voltage
analysis.
--
73, Cecil http://www.w5dxp.com

Standing-Wave Current vs Traveling-Wave Current
 

Roger wrote:
Roy Lewallen wrote:

Can you use your theory to show the voltage at all times at the input
or, if you choose, just inside the input of the line, as I've done
using conventional theory?
Yes. Here is the difference. I am using a short cut by looking at the
way waves travel and superimpose. My assumption is that the input
impedance matches the impedance of the line.

That's a different analysis than I did. I did two, one in which the
mismatch was infinite (perfect voltage source at the input) and one in
which the voltage reflection coefficient at the source was 0.5. The case
where the input is matched is simpler, because there is no re-reflection
from the source.

Can your method be used to analyze a circuit which is mismatched at the
source?

Thus 1v generates a 1v
sine wave.

Sorry, I don't understand that sentence.

For a simple example, such as any multiple of 1/2 wave, this
causes current proportional to applied voltage and impedance of the
line. For example, 50v applied to a 50 ohm line gives a 1 amp current.

It does for the time for one round trip. Then the current drops to zero
if the line is matched at the source.

We know a line 1/2 wavelength long will absorb energy for the length of
time it takes for a wave to travel the length of line and return, so
power must be applied during the entire time.

Yes.

Obviously, 2 half waves
will be applied over that time period so the amount of energy on the
line is equal to the amount of energy contained in the power applied
over time by two half waves, so total power applied is 2 * initial power.

You've lost me here. What's "total power"? Can you express it as an
equation?

Equally obvious, by leaving the source connected after the reflected
wave has returned, the question of how the source reacts to the external
application of power must be answered.

Well, the bit about "total power" wasn't obvious, so I'm not surprised
that what follows isn't either. You're apparently assuming that there
are waves of traveling power, which has led to demonstrably
self-contradictory results before, but let's see where it leads this time.

This is a second question,
unrelated to traveling waves on a transmission line. My answer to this
question is to disconnect the source and replace the source with a
connection identical to the condition at the far end of the transmission
line.

Which I presume is the open circuited example.

You can see that this leaves the transmission line isolated with
power circulating on the line (is is POWER, time is involved).

The power at any point can easily be calculated, and it shows that
energy is moving back and forth on the line. There are no waves of power
moving about.

Your solution (as presented in Analysis 2) and the Power Analysis was to
keep the source supplied at all times in an attempt to create continuous
real world conditions, as contrasted to my method of a quick and
correctly timed disconnect. Both experiments are real world simulations
and can be performed.

Ok. My method correctly predicts what will happen at any instant in
either case. Let's see how yours does.

Your analysis is identical to what happens at a real world transmitter.
A very good, automatically tuned transmitter!

I won't attempt to argue that, because of the great deal of controversy
about what the characteristics of a "real world transmitter" are. I make
no representation for my analysis other than what I stated: it uses an
ideal voltage source, which is rigorously defined and with
characteristics which are completely known, in conjunction with a
perfect resistance.

Your choice of an ideal
voltage source amounts to setting conditions that automatically adjust
for impedance changes as fast as the change happens.

No, there is no change in impedance. The perfect source has a zero
impedance at all times. The resistance has an impedance equal to its
resistance at all times. The voltage of the voltage source stays
constant (that is, it puts out a sine wave of constant amplitude and
phase) at all times.

The transmission
line sees an outgoing impedance change going from 150 ohms to infinity
in your example.

No, it sees a constant 150 ohms at all times.

The source sees an outgoing impedance change from 200
ohms to infinity during the same analysis period.

If you consider the analysis period to extend to steady state, then yes,
that's correct. You might recall that the power into the line increased
for one of the reflection periods because of the improved impedance
match between the source and line (impedances of 150 and 250 ohms
respectively) compared to the impedance match at other periods.

It is interesting to compare the final voltage ratios ( directly
proportional to energy levels and directly proportional to power levels)
on the transmission line to the SWR ratio.

The energy stored = 1/2 * C * v^2 + 1/2 * L * i^2, where v and i are the
voltage and current at a very short section of line, and C and L are the
capacitance and inductance, respectively, per that short section. As you
can see, energy is not directly proportional to V by any means. There
are places and times where energy is stored entirely in the magnetic
field, and v is zero.

You used a SWR ratio of 3:1
and I used 1:1.

No, the SWR on the line was infinite in my example. It's determined
entirely by the load impedance and the line's Z0. If you're assuming an
open ended line, the SWR is infinite in your example also.

You found that the total energy stored on the line was
4 times the initial energy applied,

No, the "initial energy" applied was zero. The analysis began with the
assumption that the line was initially completely discharged. It sounds
like you're confusing power, which had an initial value, and the energy,
which accumulated over time.

and I predicted that the total
energy stored was 2 times the initial.

Where and when did you predict that? What initial energy did you apply
to the line? Can you do an analysis with the line initially discharged
as I did?

The link here is that the power
storage factor(SF) (I know you dislike the term) is found from

SF = SWR + 1.

Your storage factor was SF = 3 + 1 = 4. My storage factor was
SF = 1 + 1 = 2.

First of all, the SWR in my example was infinite, so by your
calculation, the "power storage factor" was infinite.

If we can accept that the storage factor is always SF = SWR + 1 for all
situations, then we have a very convient way to predict the voltage
increase that results form impedance mismatch at a discontinuity. I am
not yet ready to accept this generalization, but we should look at it
more carefully.

I still don't know what you mean by "power storage factor" or why you
need to coin a new and vague term. So I guess I can accept that it's
whatever you want it to be. But it has no relationship to power or
energy that I can see.

Ending comment. Both your and my analysis are correct. Congratulations
to both us!

I've presented an analysis which gives, quantitatively, the exact
voltage at any point on the line, at any instant, from turn-on until
steady state. I've also shown the power being applied at any instant and
the total energy on the line at any time.

You've presented an analysis whose only result is a poorly defined
"power storage factor", derived from invalid premises.

Yes, congratulations to us both.

By the way, I didn't see from your "analysis" how many watts are stored
in the line. If power is stored as you say, you surely must be able to
determine how much.

Roy Lewallen, W7EL

Standing-Wave Current vs Traveling-Wave Current
 

Cecil Moore wrote:
Gene Fuller wrote:
* Did you perhaps notice that the surfer typically travels much faster
than the water?

Surfers even travel faster than the wave energy.
Sailboats travel faster than the wind.

Unfortunately, for your argument, nothing in the
universe (AFAWK) travels faster than an EM wave.
But maybe you can invent Warp Drive or Slip-Stream
Drive. :-)

Huh??? What argument of mine requires superluminal velocities? Did I
post something while I was sleeping? I hate it when I do that.

8-)

73,
Gene
W4SZ

Standing-Wave Current vs Traveling-Wave Current
 

On 4 Jan, 10:31, "Dave" wrote:
art spewed:

snip

A radiator in equilibrium is a full wave length

long ago, and far away... well maybe a couple months, and still in this
group, you said 1/2 wave was the equilibrium size? *so which is it, full
wave or half wave? *and where did gauss go?? *how do these funny cosmic dust
things fit into his equation??

David
you are quite correct in that I do fluctuate between a full wave
length and a half wave length antenna which is obviously wrong. My
problem in describing a wave length is to consider it in electric
terms thus a physical halfwave radiator is electrically a full wave
antenna since it handles one period. Ofcourse I should use standard
terminology and refer to physical length
even tho it is not exactly correct to my mind because it actually
varies in the real world where as electrical length doesn't. However,
I should stay with the accepted terminology to the best of my ability
even tho I believe that is why this arguement on colliding current is
realising postings and mass confusion in excess of 1000 each time it
comes up. I apologise for misleading terminolgy.
With reference to Gauss.
This is my starting point where with the addition of a time variable
to the static law became the same as Maxwells laws where Gauss formula
was firmly in theIsaac Newton camp
by revolving around gravitational fields.
funny cosmic dust come into the equation
Cosmic or galactic dust is everywhere on this earth .m. This dust is
what Gauss refers to
as static particles in a gravitational field in equilibrium. Thus I
was able to add elements to this field as long as the elements were in
equilibrium, together with a time vaying addition which brings it into
Maxwells court. This transition to a dynamic form then allows the use
of computor programs based only around Maxwells laws. Thus radiators
and arrays could be designed based on a equilibrium format instead of
a planar format as with a Yagi using existing standard computor
radiator design programs. From the results the following becomes
correct i.e A radiator can be any shape, size or elevation as long it
is in aquilibrium
and made from a diamagnetic material. Thus now helix radiators would
have to have counter windings to fall with in these parameters, where
the present helical antennas are absent the contra wound windings. All
this is what Prof Hattely was looking for in combining the EH field
which is unsuccesfull as is the latest RI antenna. None of these
inventors followed the same trail that I did that revealed the
equilibrium requirement for contra wound coil additions. Ask any other
questions that you have difficulty with. As an aside, this was what
Einstein was looking for during his last twenty years of life under
the title of a Grand Unification theory but without success
Regards
Art Unwin....KB9MZ

Standing-Wave Current vs Traveling-Wave Current
 

Keith Dysart wrote:

On Jan 3, 12:55pm, Mike Monett wrote:

[...]

Your explanation is easily proven false. Let's suppose it was
true.

Suppose it was possible to introduce a pulse of charge onto a
conductor.

Since like charges repel each other, what keeps the pulse
together? In other words, what prevents it from destroying
itself?

Then, when the first pulse meets the second, what mechanism
allows them to bounce off each other?

Then, after they have bounced off each other, what mechanism
keeps them together?

All good questions.

For which you have no answers.

But there's more bad news. Here's your original post of Dec 29:

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ ~~~~~~~~~~~~~~~~~~
Subject: Standing-Wave Current vs Traveling-Wave Current
Date: Sat, 29 Dec 2007 14:33:46 -0800 (PST)
From: Keith Dysart
Newsgroups: rec.radio.amateur.antenna

Keith Dysart wrote:

[...]

Consider a 50 ohm transmission line that is 4 seconds long with a
pulse generator at one end and a 50 ohm resistor at the other.

The pulse generator generates a single 1 second pulse of 50 volts
into the line. Before and after the pulse its output voltage is 0.

While generating the pulse, 1 amp (1 coulomb/s) is being put into
the line, so the generator is providing 50 watts to the line.

After one second the pulse is completely in the line.

The pulse is one second long, contains 1 coulomb of charge and 50
joules of energy. It is 50 volts with 1 amp: 50 watts.

Let's examine the midpoint (2 second) on the line.

At two seconds the leading edge of the pulse arrives at the
midpoint. The voltage rises to 50 volts and the current becomes 1
amp. One second later, the voltage drops back to 0, as does the
current. The charge and the energy have completely passed the
midpoint.

When the pulse reaches the end of the line, 50 joules are
dissipated in the terminating resistor.

Notice a key point about this description. It is completely in
terms of charge. There is not a single mention of EM waves,
travelling or otherwise.

Now we expand the experiment by placing a pulse generator at each
end of the line and triggering them to each generate a 50V one
second pulse at the same time. So after one second a pulse has
completely entered each end of the line and these pulse are racing
towards each other at the speed of light (in the line). In another
second these pulses will collide at the middle of the line.

What will happen? Recall one of the basics about charge: like
charge repel. So it is no surprise that these two pulses of charge
bounce off each and head back from where they came. At the center
of the line, for one second the voltage is 100 V (50 V from each
pulse), while the current is always zero. No charge crossed the
mid-point. No energy crossed the mid-point (how could it if the
current is always zero (i.e. no charge moves) at the mid-point.

[...]

So do the travelling waves "reflect" off each other? Save the term
"reflect" for those cases where there is an impedance
discontinuity and use "bounce" for those cases where no energy is
crossing a point and even Cecil may be happy. But bounce it does.

Keith

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ ~~~~~~~~~~~~~~~~~~
In order for the pulses to bounce off each other, your theory
requires that electrons move at nearly the speed of light.

Unfortunately, they do not. The classical formula for the drift
velocity of electrons in a conductor is:

v = I / (n * a * q)

where

v is the drift velocity of the electrons

I is the current in Amperes

n is the number of electrons per cubic metre, copper = 8.5e28 / m^3

A is the cross sectional area of the wire

Q is the charge of an electron, 1.6e-19C

References:

http://amasci.com/miscon/speed.html

http://physicsplus.blogspot.com/2007...electrons.html

http://resources.schoolscience.co.uk...elech2pg3.html

http://hyperphysics.phy-astr.gsu.edu...ic/ohmmic.html

Using the 1 Amp from your post, if the center conductor has an area
of 0.5 mm^2 (0.5e-6 m^2), the drift velocity is:

v = 1 / (8.5e28 * 0.5e-6 * 1.6e-19)

= 0.0001470 meters per second

So in one second, the electrons move 0.147 mm.

If the propagation constant of your line is 66%, the signal will
propagate at:

Vel = 3e8 * 0.66
= ~200e6 meters/sec

If the pulses meet in the center of the line, that point is about
400 million meters away from the electrons that originally carried
your pulse of charge.

This raises many questions. How can the pulses bounce off each other
if the electrons that carried the charge are 400 million meters
away? Obviously, they can't.

There's more bad news. If two collections of 1 Coulomb each were
concentrated one meter apart, the force between them could be
calculated from Coulomb's Law. In this example, the force is 1.01
million tons:

http://hyperphysics.phy-astr.gsu.edu...ic/elefor.html

This means the pulses could not even come close enough to bounce off
each other. This would certainly wreck any timing analysis you try
to do on the signals.

So your theory fails simple logic tests, it requires invalid
electron velocities, and it fails Coulomb's law.

It is clear the pulses cannot bounce off each other, as you claim
above when you state "But bounce it does."

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ ~~~~~~~~~~~~~~~~~~
But it appears that your underlying suggestion is that charge and
charge flow in the distributed capacitance and inductance can not
be used to analyze transmission lines.

That is not what you proposed. Your post states:

Notice a key point about this description. It is completely in
terms of charge. There is not a single mention of EM waves,
travelling or otherwise.

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ ~~~~~~~~~~~~~~~~~~
And yet I commonly see discussion of current in transmission
lines. Current is charge flow per unit time. Is this all invalid?

Must we abondon measurements of current? Voltage? These are all
based on the assumption of charge being a useful concept.

You are just trying to fog the issue. You cannot use charge by
itself as you claim above.

Keith

Regards,

Mike Monett

Standing-Wave Current vs Traveling-Wave Current
 

On Jan 5, 1:07*am, mike wrote:
* Keith Dysart wrote:

* On Jan 3, 12:55pm, Mike Monett wrote:

* [...]

* Your explanation *is *easily proven false. Let's *suppose *it was
* true.

* Suppose it *was *possible to introduce a pulse of *charge *onto a
* conductor.

* Since like *charges *repel *each * other, *what *keeps *the pulse
* together? In *other *words, * what *prevents *it *from destroying
* itself?

* Then, when *the *first *pulse meets *the *second, *what mechanism
* allows them to bounce off each other?

* Then, after *they *have bounced off *each *other, *what mechanism
* keeps them together?

* All good questions.

* For which you have no answers.

* But there's more bad news. Here's your original post of Dec 29:

* ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ ~~~~~~~~~~~~~~~~~~
* Subject: Standing-Wave Current vs Traveling-Wave Current
* Date: Sat, 29 Dec 2007 14:33:46 -0800 (PST)
* From: Keith Dysart
* Newsgroups: rec.radio.amateur.antenna

* Keith Dysart wrote:

* [...]

* Consider a 50 ohm transmission line that is 4 seconds long *with a
* pulse generator at one end and a 50 ohm resistor at the other.

* The pulse generator generates a single 1 second pulse of *50 volts
* into the line. Before and after the pulse its output voltage is 0.

* While generating the pulse, 1 amp (1 coulomb/s) is being *put into
* the line, so the generator is providing 50 watts to the line.

* After one second the pulse is completely in the line.

* The pulse is one second long, contains 1 coulomb of charge *and 50
* joules of energy. It is 50 volts with 1 amp: 50 watts.

* Let's examine the midpoint (2 second) on the line.

* At two *seconds *the *leading edge of *the *pulse *arrives *at the
* midpoint. The voltage rises to 50 volts and the current *becomes 1
* amp. One *second later, the voltage drops back to 0, *as *does the
* current. The *charge *and the energy *have *completely *passed the
* midpoint.

* When the *pulse *reaches *the *end *of *the *line, *50 *joules are
* dissipated in the terminating resistor.

* Notice a *key *point about this description. It *is *completely in
* terms of *charge. *There *is not a *single *mention *of *EM waves,
* travelling or otherwise.

* Now we expand the experiment by placing a pulse generator *at each
* end of *the *line and triggering them to each generate *a *50V one
* second pulse *at *the same time. So after one second *a *pulse has
* completely entered each end of the line and these pulse are racing
* towards each other at the speed of light (in the line). In another
* second these pulses will collide at the middle of the line.

* What will *happen? *Recall one of the *basics *about *charge: like
* charge repel. So it is no surprise that these two pulses of charge
* bounce off each and head back from where they came. At *the center
* of the *line, for one second the voltage is 100 V (50 V *from each
* pulse), while *the current is always zero. No *charge *crossed the
* mid-point. No *energy crossed the mid-point (how could *it *if the
* current is always zero (i.e. no charge moves) at the mid-point.

* [...]

* So do the travelling waves "reflect" off each other? Save the term
* "reflect" *for * those * cases * where * there * is * an impedance
* discontinuity and use "bounce" for those cases where no *energy is
* crossing a point and even Cecil may be happy. But bounce it does.

* Keith

* ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ ~~~~~~~~~~~~~~~~~~
* In order *for *the *pulses to bounce *off *each *other, *your theory
* requires that electrons move at nearly the speed of light.

It might best be called a hypothesis, but I don't think
it requires that electrons move at the speed of light,
rather charge move at the speed of light.

This would seem consistent with normal explanations where
charge starts to enter the line at some time T and starts
to exit the line at T + distance/speedOfLight later.

[snipped, an interesting computation of the speed of
electros]


* There's more *bad *news. If two collections of 1 *Coulomb *each were
* concentrated one *meter *apart, *the *force *between *them *could be
* calculated from *Coulomb's Law. In this example, the *force *is 1.01
* million tons:

* *http://hyperphysics.phy-astr.gsu.edu...ic/elefor.html

* This means the pulses could not even come close enough to bounce off
* each other. *This would certainly wreck any timing analysis *you try
* to do on the signals.

Another interesting analysis. It raises the question of how
a pulse containing one coulomb over a measurable length of
line actually maintains its shape and does not immediately
disperse.

This is a detail well beyond my knowledge, but I could
speculate that it is related to the inductance and
forces of the resulting magnetic field.

I'd further speculate that the force between the pulses
can not travel faster than the speed of light, and since
the pulse itself is travelling at the speed of light,
the two pulses reach each other at the same time that
force does.

* So your *theory *fails *simple * logic *tests, *it *requires invalid
* electron velocities, and it fails Coulomb's law.

* It is *clear the pulses cannot bounce off each other, *as *you claim
* above when you state "But bounce it does."

* ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ ~~~~~~~~~~~~~~~~~~
* But it appears that your underlying suggestion is that *charge and
* charge flow in the distributed capacitance and inductance *can not
* be used to analyze transmission lines.

* That is not what you proposed. Your post states:

Perhaps I was not clear in my post.

In any case, the question is fundamental...

Can charge and charge flow in the distributed capacitance and
inductance be used to analyze transmission lines?

* Notice a *key *point about this description. It *is *completely in
* terms of *charge. *There *is not a *single *mention *of *EM waves,
* travelling or otherwise.

* ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ ~~~~~~~~~~~~~~~~~~
* And yet *I *commonly *see discussion *of *current *in transmission
* lines. Current is charge flow per unit time. Is this all invalid?

* Must we *abondon measurements of current? Voltage? *These *are all
* based on the assumption of charge being a useful concept.

* You are *just *trying *to fog the issue. You *cannot *use *charge by
* itself as you claim above.

No. I am just trying to make clear the consequences of choosing
"no" as the answer to the question I directly posed above.

...Keith

Standing-Wave Current vs Traveling-Wave Current
 

Keith Dysart wrote:
It might best be called a hypothesis, but I don't think
it requires that electrons move at the speed of light,
rather charge move at the speed of light.

Anything moving at the speed of light in a transmission
line is photonic in nature. The nature of photons is
well known and they do not repel each other. They have
electric and magnetic fields but no charge. So there's
no charge moving at the speed of light in a transmission
line.
--
73, Cecil http://www.w5dxp.com

Standing-Wave Current vs Traveling-Wave Current
 

"Keith Dysart" wrote in message
...
On Jan 5, 1:07 am, mike wrote:
Keith Dysart wrote:


Can charge and charge flow in the distributed capacitance and
inductance be used to analyze transmission lines?

yes. that is the classical derivation shown in any decent text book.

You are just trying to fog the issue. You cannot use charge by
itself as you claim above.

No. I am just trying to make clear the consequences of choosing
"no" as the answer to the question I directly posed above.

moving charge is current. current traveling waves are well defined and
analyzed by existing methods. as stated before and well documented in
existing texts you need only analyze the current OR the voltage traveling
waves in a transmission line to get the full picture since one is always
linearly related to the other by Z0. waves traveling in opposite directions
do not interact, they don't 'bounce' off of each other, they just pass right
through each other as long as the conditions required for superposition
apply, which is almost all the time in amateur installations.

Standing-Wave Current vs Traveling-Wave Current
 

Cecil, W5DXP wrote:
"Unfortunately for your argument, nothing in the universe (AFAWK)
travels faster than an EM wave. But maybe you can invent Warp Drive or
Slip-Stream Drive. :-)"

Nothing to do with the argument, but I believe there is an exception
that comes to exceeding the speed of light, and that is when the wave is
slowed in a medium to a speed less than c.