Standing-Wave Current vs Traveling-Wave Current
 

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?
--
73, Cecil http://www.w5dxp.com

Standing-Wave Current vs Traveling-Wave Current
 

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
--
73, Cecil http://www.w5dxp.com

Standing-Wave Current vs Traveling-Wave Current
 

Gene Fuller wrote:
No one has ever said anything different. No one has ever denied
interference.

Denying that you ever argued about something is a first
step in the direction of understanding. Before long, you
will be arguing that you knew all of this stuff long ago.
--
73, Cecil http://www.w5dxp.com

Standing-Wave Current vs Traveling-Wave Current
 

On Jan 2, 2:32*pm, Cecil Moore wrote:
Keith Dysart wrote:
But this same information has been repeatedly provided
and ignored. Will this time be different?

And Cecil provides the answer: No!

The requested information is provided but then
completely ignored in the response.

I'm not the one who is ignoring that information.
Where are your calculations involving destructive
and constructive interference? Until you provide
that information, you are just blowing smoke.

...Keith

Standing-Wave Current vs Traveling-Wave Current
 

On Jan 2, 2:33*pm, Cecil Moore wrote:
Keith Dysart wrote:
Cecil Moore wrote:
There can be a large difference in the output
impedance of an amplifier designed to drive a 50 ohm
load and a 50 ohm Thevenin equivalent circuit.

Then your Thevenin circuit is not an equivalent
for the amplifier, is it?

No it isn't! So why are you trying to stuff it down
my throat?

I must have missed something.

You brought up the 50 ohm Thevenin equivalent circuit
which turned out not to be an equivalent circuit.

And now you think I am trying to stuff it down
your throat.

Amusing.

...Keith

Standing-Wave Current vs Traveling-Wave Current
 

On Jan 2, 2:36*pm, Cecil Moore wrote:
Keith Dysart wrote:
You should really spend some time looking for a
reference to support your assertion that "It will
not be the impedance needed to calculate the
reflection coefficient seen by the reflected
waves."

You will not find one.

You have got to be kidding, Keith. Even some of the
people on your side will admit that the effective
reflection coefficient for a source supplying zero
power is |1.0| nowhere near the value you calculated.
I believe that is what Roy said.

Bluster!

I am just waiting for a reference.

Still can't find one, can you?

...Keith

Standing-Wave Current vs Traveling-Wave Current
 

Keith Dysart wrote:
I am just waiting for a reference.
Still can't find one, can you?

Since the reflected wave is reflected, the reflection
coefficient cannot be zero.

If the reflected wave is not reflected, there
would exist current in the source, but there
is none.

There are no references for pathological thinking.
--
73, Cecil http://www.w5dxp.com

Standing-Wave Current vs Traveling-Wave Current
 

On Jan 2, 4:57*pm, Jim Kelley wrote:
Keith Dysart wrote:
On Dec 29, 2:31 pm, Cecil Moore wrote:

Roger wrote:

Are there reflections at point "+"? *Traveling waves going in opposite
directions must pass here, therefore they must either pass through one
another, or reflect off one another.

In the absence of a real physical impedance discontinuity,
they cannot "reflect off one another". In a constant Z0 transmission
line, reflections can only occur at the ends of the line and only
then at an impedance discontinuity.

Roger: an astute observation. And Cecil thinks he has
the ONLY answer. Allow me to provide an alternative.

Many years ago, when I first encountered this news group
and started really learning about transmission lines, I
found it useful to consider not only sinusoidallly
excited transmission lines, but also pulse excitation.
It sometimes helps remove some of the confusion and
clarify the thinking. So for this example, I will use
pulses.

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.

It is a minor extension to have this model deal
with sinusoidal excitation.

What happens when these pulses arrive back at the
generator? This depends on generator output
impedance. If it is 50 ohms (i.e. equal to Z0),
then there is no reflection and 1 joule is
dissipated in each generator. Other values
of impedance result in more complicated
behaviour.

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

It's fairly safe to make this argument when both pulses are identical.
* I challenge you to obtain this result when they are not. *:-)

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

Standing-Wave Current vs Traveling-Wave Current
 

On Jan 2, 11:58*pm, Cecil Moore wrote:
Keith Dysart wrote:
I am just waiting for a reference.
Still can't find one, can you?

Since the reflected wave is reflected, the reflection
coefficient cannot be zero.

If the reflected wave is not reflected, there
would exist current in the source, but there
is none.

There are no references for pathological thinking.

Still can't find one? ?

...Keith

Standing-Wave Current vs Traveling-Wave Current
 

On 2 Jan, 20:18, 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?
--
73, Cecil *http://www.w5dxp.com

Cecil,
All of the bickering come to a halt if you consider a full wave
radiator
instead of a half wave antenna. Yes, both can be resonant but only
one is in a state of equilibrium. The underpinnings of all laws
whether
mechanical or electrical is that all is in a state of equilibrium
otherwise the laws do not apply.
When there is a state of equilibrium one cannot have a collision of
waves.
The sequence for equilibrium is a magnetic field followed by a
electric field
which equals one period. Sloshing is a poor word. When current is
moving forward
you are charging up the `capacitor when the current changes direction
the capacitor
discharges as if it was shorted and creates a near instantaneous
electrostatic field
which dissapates while charging up the inductance which is also the
transition of
generating a magnetic field around the inductance.
During this time there is also a side ways force that affects or
depletes the current
by deflection which allows for propagation.This force is what is
termed "curl"
amongst many other names
At no time are there any counter waves because, not only are we
resonant but
we are also in a state of equilibrium. Move away from a fractional
wavelength
to a full wave length so both sides starts from a common point ie
resonant
and in a state of equilibrium and your difference will then be
resolved quickly.
Hope the above helps to stop posters from talking past each other.
My best regards
Art Unwin KB9MZ....XG(UK)

Standing-Wave Current vs Traveling-Wave Current
 

Keith Dysart wrote:

[...]

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.

[...]

Keith

Keith, your model is not realistic. As you know, any signal you
impose on a conductor will form an electromagnetic wave. This is the
combination of electrostatic and electromagnetic fields, and it
propagates at the normal velocity for that medium.

However, electromagnetic waves do not interact with each other, and
they cannot bounce off each other.

Recall that light from stars is electromagnetic. It travels many
light-years before it reaches your eyes. If electromagnetic waves
interacted, you would not be able to see individual stars - they
would merge into a blur.

Similarly, the signals reaching your antenna and traveling down the
coax to your receiver do not interact with each other. As long as
your receiver is not overloaded, the signals remain separate no
matter how many stations are on the air at the moment.

So the statement that like charges repel does not apply to
electromagnetic waves, and the pulses cannot bounce off each other.

Regards,

Mike Monett

Standing-Wave Current vs Traveling-Wave Current
 

Corrections:

Roy Lewallen wrote:
. . .

vtot(t, x)(steady state) = (sin(wt - x) + sin(wt + x)) / (1 - 0.5) = 2
* sin(wt - x) + sin(wt + x)

. . .

Should read

vtot(t, x)(steady state) = (sin(wt - x) + sin(wt + x)) / (1 - 0.5) = 2
* (sin(wt - x) + sin(wt + x))

This was a typo and has no effect on the steps which follow or the
conclusions.

Thanks very much to the kind person who brought this error to my attention.

Roy Lewallen, W7EL

Standing-Wave Current vs Traveling-Wave Current
 

Corrections:

Roy Lewallen wrote:
. . .
From the voltage analysis and the SPICE plot, the initial voltage at
the input of the line is sin(wt). So the voltage across the input
resistor is 3 * sin(wt) (+ toward the source), and the current flowing
into the line is (3 * sin(wt)) / 150 = 20 * sin(wt) mA. The average
power being delivered to the line is Vin(rms) * Iin(rms) (since the
voltage and current are in phase) = (0.7071 v. * 14.14 mA) = 10 mW.
Since the line initially presents an impedance of Z0, this should also
be Vin(rms)^2 / Z0 or Iin(rms)^2 / Z0. . . .

The last sentence should read:

Since the line initially presents an impedance of Z0, this should also
be Vin(rms)^2 / Z0 or Iin(rms)^2 * Z0. . . .

The calculations which follow were done using the correct formula, so
the mistake had no effect on the following steps or the conclusions.

I also carelessly and incorrectly used j as an abbreviation for joules
throughout the posting. The correct abbreviation is J. (And yes, the
name of the unit is correctly joule, not capitalized, as is the case for
most if not all SI units named after people.) This mistake is made
potentially worse because of the possible confusion with the imaginary
operator j (as used by electrical engineers). I apologize for any
confusion the mistake might have caused.

Thanks very much to the careful and thoughtful reader who brought these
errors to my attention.

Roy Lewallen, W7EL

Standing-Wave Current vs Traveling-Wave Current
 

On 2 Jan, 22:07, Mike Monett wrote:
* Keith Dysart wrote:

* [...]

* 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.

* [...]

* Keith

* Keith, your *model *is not realistic. As you *know, *any *signal you
* impose on a conductor will form an electromagnetic wave. This is the
* combination of *electrostatic *and *electromagnetic *fields, *and it
* propagates at the normal velocity for that medium.

* However, electromagnetic waves do not interact with each *other, and
* they cannot bounce off each other.

* Recall that *light *from stars is *electromagnetic. *It *travels many
* light-years before *it reaches your eyes. *If *electromagnetic waves
* interacted, you *would *not be able to see individual *stars *- they
* would merge into a blur.

* Similarly, the signals reaching your antenna and traveling *down the
* coax to *your receiver do not interact with each other. *As *long as
* your receiver *is *not overloaded, the *signals *remain *separate no
* matter how many stations are on the air at the moment.

* So the *statement *that *like * charges * repel *does *not *apply to
* electromagnetic waves, and the pulses cannot bounce off each other.

* Regards,

* Mike Monett

Mike
They will not listen to you because they are following books that are
incorrect.
They cannot get into their minds that for a given frequency the time
for
a cycle is always the same. And as I have pointed out many times in
different ways
that goes for 1/2 wave antennas too.
If one bends a half wave antenna such that the ends are close together
you
can feed at the wire ends.
So now you apply a DC current at one end and it will continue to go
forward
all the way to the other end of the wire.Wen it gets to the end of the
wire
the current reverses direction and goes back to the starting point
which
completes one wave length of travel which also represents one
wavelength of time.
Now the other side of the debate wants a half wave antenna.So let us
lay
that half wave antenna out in a similar way that we bent the half wave
length
antenna
Start a DC current at one end and it travels to the other end of the
wire
BUT TIME HAS NOT RUN OUT FOR FLOW IN THE SAME DIRECTION,and this is
the
crux of the debate since the current still has to move forward and
thus
can only procede down the center of the wire .
IT CANNOT GO BACK AT THIS TIME.
The time for forward travel runs out when it gets to the end of the
wire.
When time runs out it reverses direction and goes back up the inside
of
the wire and down the outside of the wire to the end or shall we say
return to the beginning to complete the cycle.
It can be seen that with a half wave antenna the current flows on the
surface
only half of the time such that it only encounters inductance and
capacitance
for half of a cycle such that it can radiate. The rest of the time
because
the current flow is not exposed to the surface it does not radiate.
At no time is the current going two different ways.
Unfortunately people ignore the requirements of time and for some
reason want to change current direction to half the time required
for that frequency thus creating collisions in the current flow.
THIS DOES NOT HAPPEN. So gentlemen layout a bent full wave
antenna and along side it lay out a half wave antenna in the same
fashion and start current flow in both at the same time where
length of travel is commensurate with time. You will now realise
that all participants in this debate have been talking past each
other.
Another way of looking at it is to draw a full wave length of
communicationline in ladder form with its component inductances
and capacitances. One side of the ladder drawn line has zero
energy storage components only resistance and it is this
line that represents travel down the center of the wire.
The bottom line is that a half wavelength antenna represents
a full cycle with respect to frequency.
This will be the last time that I will try to explain the radiation
characteristics of a antenna.
Best regards to all
Art Unwin KB9MZ.....XG

Standing-Wave Current vs Traveling-Wave Current
 

On Jan 2, 2:48*pm, Cecil Moore wrote:
Keith Dysart wrote:
I assume that you have not provided a reference to support
this assertion because you have not been able to find one.

I provided the reference a number of times and you
chose to ignore it. The reference is the chapter on
interference in "Optics", by Hecht.

I am suprised that a book on optics would discuss the
output impedance of Thevenin equivalent circuits.

Be that as it may, could you kindly provide the brief
extract from "Optics", by Hecht, that clearly states
that the impedance in a Thevenin equivalent circuit
can not be used to compute the reflection coefficient.

...Keith

Standing-Wave Current vs Traveling-Wave Current
 

On Jan 2, 3:44*pm, Cecil Moore wrote:
Keith Dysart wrote:
In a stub driven with a step function, where is the
energy stored?

Consider the state of the open circuited line
after settling.

Depends upon which valid model one is using.

1. Reflection Model - the energy is stored in the
forward and reflected traveling waves.

So it is stored only in the E field. Is it an
EM wave when only an E field is present?

2. The LCLCLC transmission line model - the energy
is alternately stored in the L's and C's.

Since there is no current, it is stored only in
the capacitance of the line.

3. The Sloshing Model - I'll let Roy handle that one.

The sloshing has stopped, so the answer is the same
as 2.

But the important question is: Do you consider it
to be an EM wave when only an E field is present?

...Keith

Standing-Wave Current vs Traveling-Wave Current
 

Keith Dysart wrote:
On Jan 1, 9:03 pm, Cecil Moore wrote:
Roger wrote:
The principles of superposition are mathematically usable, not too hard,
and I think very revealing. Yes, if we use part of the model, we must
use it all the way. To do otherwise would be error, or worse.
Roy and Keith don't seem to realize that the zero source
impedance for the ideal voltage source is only when the
source is turned off for purposes of superposition.

I am not sure you have the methodology quite correct.
The source is not turned off; its output is set to 0.

It does what every ideal voltage source will do when
set to a voltage; maintain that voltage. Through all
of this, the impedance of the ideal source remains 0.

Now it turns out that an ideal voltage source set
to zero volts can be replaced by a short which also
has an impedance of 0 and produces no volts. But this
does not alter that the ideal source always has an
impedance of 0.

Analogously, an ideal current source always has an
infinite impedance. When set to 0 amps, it behaves
exactly like an open circuit.

They
conveniently avoid turning the source voltage on to complete
the other half of the superposition process. When the
source signal and the reflected wave are superposed at
the series source resistor, where the energy goes becomes
obvious. Total destructive interference in the source
results in total constructive interference toward the load.
See below.

You have been a supporter of this theory for a long time.
Yes, I have. I am a supporter of the principles and laws of
physics. Others believe they can violate the principle of
conservation of energy anytime they choose because the
principle of conservation of energy cannot be violated -
go figure.

You should really stop repeating this to yourself. No
one is attempting to violate the principle of conservation
of energy.

By continually repeating this mantra, you convince
yourself that you do not need to examine the claims
of those who disagree with you. So you do not
examine and understand their claims. This seriously
limits your capability to learn.

If you truly wish to demolish the claims, you should
study them in great detail, then write an even better
and more persuasive description of the claim than did
the original author. Then identify and point out the
flaws.

As it stands, you do not examine the claims, but
immediately coat them with the tar of "violates
conservation of energy" or some other mantra and
walk away.

It does not lead to learning.

...Keith
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 have posted several times on the subject of impedance of an ideal
source, and I have learned from your words. You may find however, that
I have still not completely grasped an important component of the
concept. If that happens, please try again, using a different argument.

Learning is a meshing of words, ideas, concepts, experience, and more.
You can see that I am inexperienced. I can see that many of the posters
are very experienced. Experience is not necessary for presenting an
argument, but it certainly helps in presenting the argument wisely,
coherently, and convincingly. Correctness is always a judgment by the
reader.

73, Roger, W7WKB

Standing-Wave Current vs Traveling-Wave Current
 

Roy Lewallen wrote:
Roger wrote:

By storage factor, I simply mean the ratio of
forward power to total power on the transmission media under standing
wave conditions.


Power is neither stored nor conserved, so a power "storage factor" is
meaningless. Consider a very simple example. Let's charge a capacitor
with a constant current DC source. We'll apply 1 amp to a 1 farad
capacitor for 1 second. During that time, the power begins at zero,
since the capacitor voltage is zero, then it rises linearly to one watt
as the capacitor voltage rises to one volt at the end of the one second
period. So the average power over that period was 1/2 watt, and we put
1/2 joule of energy into the capacitor. (To confirm, the energy in a
capacitor is 1/2 * C * V^2 = 1/2 joule.) Was power "put into" or stored
in the capacitor?

Now we'll connect a 0.1 amp constant current load to the capacitor, in a
direction that discharges it. We can use an ideal current source for
this. The power measured at the capacitor or source terminals begins at
0.1 watt and drops linearly to zero as the capacitor discharges. The
average is 0.05 watt. Why are we getting less power out than we put in?

"Where did the power go?" is heard over and over, and let me assure you,
anyone taking care with his mathematics and logic is going to spend a
long time looking for it. So in this capacitor problem, where did the
power go?

It takes 10 seconds to discharge the capacitor, during which the load
receives the 1/2 joule of energy stored in the capacitor. Energy was
stored. Energy was conserved. Power was neither stored nor conserved.

Roy Lewallen, W7EL

By my using the words 'power' "storage factor", you got my point, hence
the reaction.

Before dismissing the concept of "storing power", consider that when
discussing a transmission line, it could be a useful description.

As you know, power is energy delivered over a time period. It always
carries a time dimension having beginning and end. Power(watt)
=v*i/(unit time) = 1 joule/second.

In the example you give of charging a capacitor, the time dimension is
lost, so you are correct that only energy is conserved. Power is lost.

With a transmission line, we have an entirely different case. Here
power is conserved because the time information is maintained. Power is
stored on the line during the period it resides on the line. For
example, we excite the line at one end and some time period later find
that power is delivered to some destination. During the time period
that the power was on the line, the information that defines the energy
distribution over time has been preserved.

If power is stored, we implicitly store energy. Energy is v*i measured
in joules without a time factor. Obviously we store energy on a
transmission line when we store power.

So if in the future, I use the term "power storage", please take it to
mean that energy distributed over time is under consideration. I hope
the term might be useful to you as well.

73, Roger, W7WKB

Standing-Wave Current vs Traveling-Wave Current
 

On Jan 3, 1:07*am, Mike Monett wrote:
* Keith, your *model *is not realistic. As you *know, *any *signal you
* impose on a conductor will form an electromagnetic wave. This is the
* combination of *electrostatic *and *electromagnetic *fields, *and it
* propagates at the normal velocity for that medium.

* However, electromagnetic waves do not interact with each *other, and
* they cannot bounce off each other.

That is the standard description, but it seems to have
some weaknesses.

* Recall that *light *from stars is *electromagnetic. *It *travels many
* light-years before *it reaches your eyes. *If *electromagnetic waves
* interacted, you *would *not be able to see individual *stars *- they
* would merge into a blur.

This would seem to me to depend on the nature of the
interaction. Clearly the interaction represented by
the term "bounce" (for lack of a better word) would
have to be such as to not violate any of these
observed behaviours.

* Similarly, the signals reaching your antenna and traveling *down the
* coax to *your receiver do not interact with each other. *As *long as
* your receiver *is *not overloaded, the *signals *remain *separate no
* matter how many stations are on the air at the moment.

* So the *statement *that *like * charges * repel *does *not *apply to
* electromagnetic waves,

Q1. Are you saying that it is inappropriate to view
a transmission line as distributed capacitance and
inductance and analyze its behaviour using charge
stored in the capacitance and moving in the
inducatance?

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?

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

The existing analysis techniques tell us that
no current ever flows at the mid-point of the
line, this means no charge crosses the mid-point.

Q4. Is this correct?

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

Q6. If they do not pass the mid-point, then what
happens to them?

I have offerred a somewhat intuitive explanation.
Other explanations are welcome.

Any explanation that does not involve charge will
immediately cause me to ask Q1 again.

...Keith

Standing-Wave Current vs Traveling-Wave Current
 

On Jan 3, 7:22*am, Roger wrote:
Keith Dysart wrote:
On Jan 1, 9:03 pm, Cecil Moore wrote:
Roger wrote:
The principles of superposition are mathematically usable, not too hard,
*and I think very revealing. *Yes, if we use part of the model, we must
use it all the way. *To do otherwise would be error, or worse.
Roy and Keith don't seem to realize that the zero source
impedance for the ideal voltage source is only when the
source is turned off for purposes of superposition.

I am not sure you have the methodology quite correct.
The source is not turned off; its output is set to 0.

It does what every ideal voltage source will do when
set to a voltage; maintain that voltage. Through all
of this, the impedance of the ideal source remains 0.

Now it turns out that an ideal voltage source set
to zero volts can be replaced by a short which also
has an impedance of 0 and produces no volts. But this
does not alter that the ideal source always has an
impedance of 0.

Analogously, an ideal current source always has an
infinite impedance. When set to 0 amps, it behaves
exactly like an open circuit.

They
conveniently avoid turning the source voltage on to complete
the other half of the superposition process. When the
source signal and the reflected wave are superposed at
the series source resistor, where the energy goes becomes
obvious. Total destructive interference in the source
results in total constructive interference toward the load.
See below.

You have been a supporter of this theory for a long time.
Yes, I have. I am a supporter of the principles and laws of
physics. Others believe they can violate the principle of
conservation of energy anytime they choose because the
principle of conservation of energy cannot be violated -
go figure.

You should really stop repeating this to yourself. No
one is attempting to violate the principle of conservation
of energy.

By continually repeating this mantra, you convince
yourself that you do not need to examine the claims
of those who disagree with you. So you do not
examine and understand their claims. This seriously
limits your capability to learn.

If you truly wish to demolish the claims, you should
study them in great detail, then write an even better
and more persuasive description of the claim than did
the original author. Then identify and point out the
flaws.

As it stands, you do not examine the claims, but
immediately coat them with the tar of "violates
conservation of energy" or some other mantra and
walk away.

It does not lead to learning.

...Keith

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

Standing-Wave Current vs Traveling-Wave Current
 

Keith Dysart wrote:

On Jan 3, 1:07am, Mike Monett wrote:

Keith, your model is not realistic. As you know, any signal you
impose on a conductor will form an electromagnetic wave. This is
the combination of electrostatic and electromagnetic fields, and
it propagates at the normal velocity for that medium.

However, electromagnetic waves do not interact with each other,
and they cannot bounce off each other.

That is the standard description, but it seems to have some
weaknesses.

No, there are no weaknesses. Maxwell's equations have stood the test
of time.

Recall that light from stars is electromagnetic. It tra vels many
light-years before it reaches your eyes. If electromagnetic waves
interacted, you would not be able to see individual stars they
would merge into a blur.

This would seem to me to depend on the nature of the interaction.
Clearly the interaction represented by the term "bounce" (for lack
of a better word) would have to be such as to not violate any of
these observed behaviours.

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

Similarly, the signals reaching your antenna and traveling down
the coax to your receiver do not interact with each other. As
long as your receiver is not overloaded, the signals remain sep
arate no matter how many stations are on the air at the moment.

So the statement that like charges repel does not apply to
electromagnetic waves,

Q1. Are you saying that it is inappropriate to view a transmission
line as distributed capacitance and inductance and analyze its
behaviour using charge stored in the capacitance and moving in the
inducatance?

That is not what you are saying. You are ignoring the magnetic
field.

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. You need to include the associated magnetic field.

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

Electromagnetic signals do not collide. They superimpose.

The existing analysis techniques tell us that no current ever
flows at the mid-point of the line, this means no charge crosses
the mid-point.

Q4. Is this correct?

That statement has no meaning.

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

The pulses are not chunks of charge. They are the combination of
electrostatic and electromagnetic fields. You cannot separate the
two.

Q6. If they do not pass the mid-point, then what happens to them?

That statement has no meaning.

I have offerred a somewhat intuitive explanation.

Your explanation does not work.

Other explanations are welcome.

Any explanation that does not involve charge will immediately
cause me to ask Q1 again.

Please study Maxwell's equations and how they are derived.

Keith

Regards,

Mike Monett

Standing-Wave Current vs Traveling-Wave Current
 

Keith Dysart wrote:
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.

All you have proved is that you cannot tell one photon
from another. Your whole charge repulsion argument
falls apart when dealing with photons (which constitute
EM waves). I suggest you study and discover what is
possible with photons and what is not possible. You
seem to be concentrating on the carriers of the waves
rather than the EM waves themselves. Photons do NOT
and cannot bounce off of each other under ordinary
circumstances. You are simply illustrating the limitations
of ignoring the basic physics of the situation and wasting
a lot of time and effort in the process.

I have sat on a cliff overlooking the Pacific Ocean
at Fitzgerald's Marine Reserve north of Santa Cruz, CA
and have seen waves rolling in, reflecting off the beach,
and rolling back out to sea. Those waves pass through
each other as if the other wasn't there. The wave energy
is moving in both directions. The H2O carriers move
hardly at all. You can argue that the energy in the
waves is equal and therefore no average energy is being
transferred, but I still see the waves with people
riding on those waves. I do not see waves bouncing off
of each other although one could, as you have, delude
oneself into creating a mental illusion of such.

When I look out into my back yard, I am seeing reflections.
If there were a thousand people here, they would all be
seeing different reflections all passing through each other.
Photonic waves pass through each other unimpeded. It would
be a weird looking world if they bounced off each other.

In a wire, photons do not bounce off each other. However,
superposition can cause a redistribution of photon energy
at an impedance discontinuity. We call that redistribution
of energy a "reflections".
--
73, Cecil http://www.w5dxp.com

Standing-Wave Current vs Traveling-Wave Current
 

On 3 Jan, 07:38, Keith Dysart wrote:
On Jan 3, 1:07*am, Mike Monett wrote:

* Keith, your *model *is not realistic. As you *know, *any *signal you
* impose on a conductor will form an electromagnetic wave. This is the
* combination of *electrostatic *and *electromagnetic *fields, *and it
* propagates at the normal velocity for that medium.

* However, electromagnetic waves do not interact with each *other, and
* they cannot bounce off each other.

That is the standard description, but it seems to have
some weaknesses.

* Recall that *light *from stars is *electromagnetic. *It *travels many
* light-years before *it reaches your eyes. *If *electromagnetic waves
* interacted, you *would *not be able to see individual *stars *- they
* would merge into a blur.

This would seem to me to depend on the nature of the
interaction. Clearly the interaction represented by
the term "bounce" (for lack of a better word) would
have to *be such as to not violate any of these
observed behaviours.

* Similarly, the signals reaching your antenna and traveling *down the
* coax to *your receiver do not interact with each other. *As *long as
* your receiver *is *not overloaded, the *signals *remain *separate no
* matter how many stations are on the air at the moment.

* So the *statement *that *like * charges * repel *does *not *apply to
* electromagnetic waves,

Q1. Are you saying that it is inappropriate to view
a transmission line as distributed capacitance and
inductance and analyze its behaviour using charge
stored in the capacitance and moving in the
inducatance?

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?

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

The existing analysis techniques tell us that
no current ever flows at the mid-point of the
line, this means no charge crosses the mid-point.

Q4. Is this correct?

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

Q6. If they do not pass the mid-point, then what
happens to them?

I have offerred a somewhat intuitive explanation.
Other explanations are welcome.

Any explanation that does not involve charge will
immediately cause me to ask Q1 again.

...Keith

The above is correct omly when the mid point you are refering to
is at the point where the time to complete half a peried is reached.
On a half period length that is at the end of a half wave radiator
the return path is the second half.The ant6enna only radiates
when the current travels on the surface towards the inductance
when the capacitors are discharging
Physical radiator length is half a cycle or period because half
the time it is travelling forward and the rest of the time backwards.
Hams must stop equating antenna length with a full wave length,
it equates to half a wave length. It is obvious then that the
completion of a cycle thus at no time has current moving
other than a single direction. Until hams get this distinction
into their heads they will never understand radiation.
Art

Standing-Wave Current vs Traveling-Wave Current
 

Keith Dysart wrote:
Still can't find one? ?

I haven't looked for one because I don't want to waste
my time. If there was a reference, Mr. Maxwell or Dr.
Bruene would have reported it by now but their argument
continues to rage, just like yours.
--
73, Cecil http://www.w5dxp.com

Standing-Wave Current vs Traveling-Wave Current
 

On Jan 3, 11:01*am, Mike Monett wrote:
* Keith Dysart wrote:

* On Jan 3, 1:07am, Mike Monett wrote:

* Keith, your *model is not realistic. As you know, any *signal you
* impose on a conductor will form an electromagnetic wave. *This is
* the combination of electrostatic and electromagnetic *fields, and
* it propagates at the normal velocity for that medium.

* However, electromagnetic *waves do not interact with *each other,
* and they cannot bounce off each other.

* That is *the *standard *description, but *it *seems *to *have some
* weaknesses.

* No, there are no weaknesses. Maxwell's equations have stood the test
* of time.

* Recall that light from stars is electromagnetic. It tra vels many
* light-years before it reaches your eyes. If electromagnetic waves
* interacted, you *would not be able to see *individual *stars they
* would merge into a blur.

* This would seem to me to depend on the nature of *the interaction.
* Clearly the interaction represented by the term "bounce" (for lack
* of a *better word) would have to be such as to not violate *any of
* these observed behaviours.

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

* Similarly, the *signals reaching your antenna and *traveling down
* the coax *to *your receiver do not interact with *each *other. As
* long as *your receiver is not overloaded, the signals *remain sep
* arate no matter how many stations are on the air at the moment.

* So the *statement *that *like charges *repel *does *not *apply to
* electromagnetic waves,

* Q1. Are you saying that it is inappropriate to view a transmission
* line as *distributed *capacitance and inductance *and *analyze its
* behaviour using charge stored in the capacitance and moving in the
* inducatance?

* That is *not *what *you are saying. You *are *ignoring *the magnetic
* field.

* 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. You need to include the associated magnetic field.

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

* Electromagnetic signals do not collide. They superimpose.

* The existing *analysis *techniques tell us *that *no *current ever
* flows at *the mid-point of the line, this means no *charge crosses
* the mid-point.

* Q4. Is this correct?

* That statement has no meaning.

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

* The pulses *are *not chunks of charge. They are *the *combination of
* electrostatic and *electromagnetic fields. You *cannot *separate the
* two.

* Q6. If they do not pass the mid-point, then what happens to them?

* That statement has no meaning.

* I have offerred a somewhat intuitive explanation.

* Your explanation does not work.

* Other explanations are welcome.

* Any explanation *that *does not *involve *charge *will immediately
* cause me to ask Q1 again.

* Please study Maxwell's equations and how they are derived.

* Keith

* Regards,

* Mike Monett

You did not directly answer Q1, but I take if from all
the other responses that you are saying the answer is
"no, it is not appropriate to view a transmission
line as distributed capacitance and inductance and
analyze its behaviour using charge stored in the
capacitance and moving in the inducatance?"

Taking this invalidates all the subsequent questions
since they are based on the premise that this kind
of analysis is appropriate.

Or have I misinterpreted and your only concern with
Q1 was that I did not mention that energy is stored
in the electric field created by charge in the
capacitance and the magnetic created by charge
flowing in the inductance?

...Keith

Standing-Wave Current vs Traveling-Wave Current
 

Mike Monett wrote:
However, electromagnetic waves do not interact with each other, and
they cannot bounce off each other.

Mike, can we add "do not interact with each other" in
*an unchanging medium*? Coherent waves can apparently
interact at a change in mediums (impedance discontinuity).

One of the s-parameter equations illustrates that apparent
fact.

b1 = s11*a1 + s12*a2 = 0

Assuming a1 and a2 are non-zero forward and non-zero reflected
voltages, the only way they could superpose to zero is to
interact at the impedance discontinuity.
--
73, Cecil http://www.w5dxp.com

Standing-Wave Current vs Traveling-Wave Current
 

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.
...
Regards,

Mike Monett

EM fields act that same as static magnetic fields.

Why not just get some iron filings, a paper and a couple of magnets?

Move the magnets about below the paper with the iron filings above and
actually get a visual on some magnetic fields and how they react to each
other?

I like things simple ... then the math can follow ...

Regards,
JS

Standing-Wave Current vs Traveling-Wave Current
 

Keith Dysart wrote:
On Jan 2, 2:48 pm, Cecil Moore wrote:
Keith Dysart wrote:
I assume that you have not provided a reference to support
this assertion because you have not been able to find one.
I provided the reference a number of times and you
chose to ignore it. The reference is the chapter on
interference in "Optics", by Hecht.

I am suprised that a book on optics would discuss the
output impedance of Thevenin equivalent circuits.

The "Optics", by Hecht reference is for destructive
and constructive interference, not Thevenin equivalent
circuits, but your attempt to confuse everyone is noted.
"Thevenin" is not even in the index of "Optics" so your
attempted diversion is ridiculous on the face of it.

I will repeat an earlier posting that you conveniently
chose to ignore. If we measure the forward power and
reflected power with a Bird wattmeter at the output of
your source during steady-state, it will tell you that:

forward power = reflected power

From that we can calculate the reflection coefficient.

rho = SQRT(Pref/Pfor) = plus or minus 1.0
--
73, Cecil http://www.w5dxp.com

Standing-Wave Current vs Traveling-Wave Current
 

On Jan 3, 11:10*am, Cecil Moore wrote:
Keith Dysart wrote:
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.

All you have proved is that you cannot tell one photon
from another. Your whole charge repulsion argument
falls apart when dealing with photons (which constitute
EM waves). I suggest you study and discover what is
possible with photons and what is not possible.

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

If not, there would seem to be some difficulty
with the applicability.

...Keith

Standing-Wave Current vs Traveling-Wave Current
 

On Jan 3, 11:13*am, Cecil Moore wrote:
Keith Dysart wrote:
Still can't find one? ?

I haven't looked for one because I don't want to waste
my time. If there was a reference, Mr. Maxwell or Dr.
Bruene would have reported it by now but their argument
continues to rage, just like yours.

Excellent. So there is NO reference that claims
that the output impedance can not be used to
compute the reflection coefficient.

There are many references that do.

So that settles it, then. Many references on one side,
none on the other. It is time for you to accept the
standard methodology for computing the reflection
coefficient at a generator. And no, I am not holding
my breath while I wait.

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

...Keith

Standing-Wave Current vs Traveling-Wave Current
 

Keith wrote:
"I am surprised that a book on optics would discuss the output impedance
of Thevenin`s equivalent circuits."

Hecht is a physicist.

On page 74 of Terman`s 1955 0pus, he writes:
"According to Thevenin`s theorem, any linear network containing one or
more sources of voltage and having two terminals behaves, in so far as a
load impedance connected across these terminals is concerned, as though
the network and its generators were equivalent to a simple generator
having an internal impedance Z and a generated voltage E, where E is the
voltage that appears across the terminals when no load is connected and
Z is the impedance that is measured between the terminals when all
sources of voltage are short-circuited."

On page 87 of his 1955 opus, Terman writes:
"The vector ratio of E2/E1 of the voltage of the reflected wave to the
voltage of the incident wave at the load is termed the "reflection
coefficient" of the load."

On page 97, Terman writes:
"The standing-wave ratio S is one means of expressing the magnitude of
the reflection coefficient;"

On page 214 of "Schaum`s Outline of College Physics", Bueche & Hecht
write:
"Standing waves---These might better not be called waves at all since
they do not transport energy and momentum.---"

Cecil is vindicated.

Best regards, Richard Harrison, KB5WZI

Standing-Wave Current vs Traveling-Wave Current
 

On Jan 3, 12:04*pm, Cecil Moore wrote:
Keith Dysart wrote:
On Jan 2, 2:48 pm, Cecil Moore wrote:
Keith Dysart wrote:
I assume that you have not provided a reference to support
this assertion because you have not been able to find one.
I provided the reference a number of times and you
chose to ignore it. The reference is the chapter on
interference in "Optics", by Hecht.

I am suprised that a book on optics would discuss the
output impedance of Thevenin equivalent circuits.

The "Optics", by Hecht reference is for destructive
and constructive interference, not Thevenin equivalent
circuits, but your attempt to confuse everyone is noted.

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

But in another post, you have agreed that there
is a complete lack of references supporting your
position, so the question is now settled and you
can use the standard methodology to compute
reflection coefficient at a generator where
the output impedance is well defined.

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

...Keith

Standing-Wave Current vs Traveling-Wave Current
 

Keith Dysart wrote:
Cecil Moore wrote:
1. Reflection Model - the energy is stored in the
forward and reflected traveling waves.

So it is stored only in the E field.

Wrong! Please reference a book on EM waves. An
EM wave CANNOT have a stationary E-field in an
ordinary transmission line or in free space.

But the important question is: Do you consider it
to be an EM wave when only an E field is present?

No, that violates the definition of an EM wave. If
only an E-field is present in a transmission line or
in free space, it is NOT an EM wave, by definition.

For a single EM wave to exhibit a zero H-field, the
Z0 would have to be infinite which is clearly impossible
for free space or ordinary transmission lines.
--
73, Cecil http://www.w5dxp.com

Standing-Wave Current vs Traveling-Wave Current
 

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

Standing-Wave Current vs Traveling-Wave Current
 

On Jan 3, 12:43*pm, Cecil Moore wrote:
Keith Dysart wrote:
Cecil Moore wrote:
1. Reflection Model - the energy is stored in the
forward and reflected traveling waves.

So it is stored only in the E field.

Wrong! Please reference a book on EM waves. An
EM wave CANNOT have a stationary E-field in an
ordinary transmission line or in free space.

But the important question is: Do you consider it
to be an EM wave when only an E field is present?

No, that violates the definition of an EM wave. If
only an E-field is present in a transmission line or
in free space, it is NOT an EM wave, by definition.

For a single EM wave to exhibit a zero H-field, the
Z0 would have to be infinite which is clearly impossible
for free space or ordinary transmission lines.

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).

...Keith

Standing-Wave Current vs Traveling-Wave Current
 

Keith Dysart wrote:
Q5."If no charge crosses the mid-point, then how do the pulses, made up
of charge, pass the mid-point?"

Radio waves propagate through empty space. No charges are needed for
propagation, but a wire contains free electrons which are urged to move
in-synch by a passing wave. That does not mean they migrate. In the
wire, electrons go nowhere fast. Their motion is mostly in-place. It the
energy which is traveling, not the electrons.

Best regards, Richard Harrison, KB5WZI

Standing-Wave Current vs Traveling-Wave Current
 

Keith Dysart wrote:

[...]

You did not directly answer Q1, but I take if from all the other
responses that you are saying the answer is "no, it is not
appropriate to view a transmission line as distributed capacitance
and inductance and analyze its behaviour using charge stored in
the capacitance and moving in the inducatance?"

That is not what you originally stated.

Taking this invalidates all the subsequent questions since they
are based on the premise that this kind of analysis is
appropriate.

Yes, it does.

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?

[...]

Keith

Regards,

Mike Monett

Standing-Wave Current vs Traveling-Wave Current
 

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

But a learning experience nonetheless. All one has
to do regarding false information is to produce
valid technical references to the contrary.

Ad hominem attacks are not technical references
and the mere assertion that the information is
misleading implies some level of omniscience in
the asserter that is not in evidence.

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.

A couple of technical web pages assert that wave
cancellation can be the cause of a redistribution
of photonic energy to areas that allow constructive
interference.

Since there is only one other direction available
in a transmission line, the constructive interference
must occur in the opposite direction from the
direction of wave cancellation.

That seems like a no-brainer to me.
--
73, Cecil http://www.w5dxp.com

Standing-Wave Current vs Traveling-Wave Current
 

On 22 Dec 2007, 12:57, John Smith wrote:
Cecil Moore wrote:
...
Why is the ignorance level about traveling waves so high
on this newsgroup? It's the result of those inadequate
lumped circuit models.

In Einsteins' spirit, let's have a real look at waves (basically the
KISS rule):

http://www.colorado.edu/physics/2000...ing_wave1.html

you must go to the bottom of each page and click to view the next of the
series.

The standing wave is "driven" by the forward & reverse traveling waves,
yet best thought of as being "separate in existence" (there are a total
of 3 waves!) ... and can only/really exist within strict confines of
design--or, resonance ...

But then, this is nothing new, or, you already knew that ... I just like
the way this is all presented--on those pages, or, even newbies are
introduced to the depth of the argument ...

Regards,
JS

John, There is no way you have a standing wave. I know the books
suggest
that but the books are wrong. 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 The turning around point is determined by frequency and
frequency alone regardles of the length of the radiator. I f the
forward time has not run out the current must continue going forward
and that means going down the center of the wire which is resistive
only and does not generate radiation. The current traverses the center
twice and it is when it starts to return on the outside is when
propagation comences. So at no time is the current changing direction
except at a time of half a period. It only has one pulse per period
and that pulse is dependent on the surface content that it traverses.
If it is a full wave where the length equates to half a period then
the capacitance and inductance is comensurate with length.
If it is a fractional wavelength it only radiates when returning on
the surface where the energy storage content is half that of a full
wave
This particular point has arrested the advances in antennas for 100
years just because man does not accept change.
Regards
Art

Standing-Wave Current vs Traveling-Wave Current
 

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. 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.
--
73, Cecil http://www.w5dxp.com