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