Standing-Wave Current vs Traveling-Wave Current
 

Roy Lewallen wrote:
I'm not seeing Cecil's comments in context, so this might be irrelevant,
but there is a convention involved with the direction of
reverse-traveling current waves. The common convention used in
transmission line analysis is that the positive direction of both
forward and reverse current is from the generator toward the load end of
the line. The consequences of this is that the current reverses sign --
really meaning only that it reverses direction -- upon reflection from
an open circuit (+1 voltage reflection coefficient), and it allows
calculation of the total current as the sum of the forward and reflected
currents. An equally valid convention is to define the positive
direction of both forward and reflected currents to be the direction of
travel. If this convention is used, then the current undergoes no change
in sign upon reflection. But the total current then equals the forward
current minus the reverse current. Either convention will produce
correct results, of course, as long as it's carefully and consistently
applied.

Thanks Roy, the former is the convention usually used for RF
wave analysis. The latter is the convention usually used for
light wave analysis. They are obviously different conventions.
--
73, Cecil http://www.w5dxp.com

Standing-Wave Current vs Traveling-Wave Current
 

On Dec 29, 3:16*pm, Cecil Moore wrote:
Roy Lewallen wrote:
In fact, current does quit flowing from the source.

In your example, the only way for the conservation of
energy principle to work, along with the laws of physics
governing EM waves, is for the reflected wave incident
upon the source to be reflected back toward the open
end of the stub. The forward and reflected energy simply
remains in the stub flowing end to end at the speed of
light in the medium. Anything else is impossible.

When the source impedance is the same as Z0 there
is no impedance discontinuity to produce a reflection.

Is the reflected wave reflected even without a
discontinuity?

...Keith

PS. And a circulator is not needed for the source
impedance to match Z0.

Standing-Wave Current vs Traveling-Wave Current
 

On Dec 29, 9:59*am, Cecil Moore wrote:
Keith Dysart wrote:
Still sidetracking away from your demons rather
than confronting them?!

No, you are the one who believes in the supernatural,
not I. I have confronted those supernatural demons
and decided they don't even exist in reality.

Well something causes you to latch up and bail with
"its not real world", though there was no protest
when the initial experiment is specified using
ideal elements.

But when the line is cut, it is clear that no
energy is moving between the separate sections.

Just as it is clear that before the cut, energy
was moving between the separate sections. It
requires belief in a supernatural to assert that
is not a change.

But you previously agreed that P(t) = 0 for all t,
and therefore no energy was moving between the sections.
Recall that P(t) = V(t) * I(t) and that at the
point on the line in question, I(t) is zero for all
t, therefore P(t) is zero for all t. So no energy
flow between the sections.

When the lines are joined, the voltage, current and
power distributions on the line remain the same.
Therefore, no energy is being transferred between
the now joined sections.
QED

Change the QED to BS and you will have it right.

This is the kind of comment that suggests stress,
rather than rational examination.

When the lines are joined, there is no longer a
physical impedance discontinuity so reflections
are impossible and energy starts flowing again
in both directions.

But as previously discussed, no energy flows. The
the voltage, current and power distributions are
the same, whether the line is cut or joined.

Believing that reflections can occur where there
exists no physical impedance discontinuity is a
religion, not a science.

Red herring. Straw man. I do not recall anyone
making the claim that reflections exist with no
physical impedance discontinuity. (Although you
did raise the possibility in another post).

At that point, I draw
the line - but you are free to have the religion
of your choice. Just please don't try to force
your religion on this technical newsgroup.

With or without reflections, no energy crosses
the points on the line with zero current.

Make that no *NET* energy and you will be so
technically correct that I will agree with you.

Sometimes when writers write NET, they mean time
averaged, but that is not your intent here, is it?

You do mean that P(t) is zero for all t at the
points on the open circuited line where the voltage
or current is always zero. Right?

...Keith

Standing-Wave Current vs Traveling-Wave Current
 

Cecil Moore wrote:
Gene Fuller wrote:
Cecil Moore wrote:
Incidentally, this is a different convention from the field
of optics.

That is a rather curious comment. Why do you say the convention is
different for optics? There are no commonly used "voltage" or
"current" descriptions in optics, so analysis is done using E-fields
and H-fields. Otherwise there is no difference in convention between
optical and RF.

In optics, to the best of my knowledge, there is no separate
reflection coefficient for the E-field and the H-field. If
true, that fact alone implies a different convention from
the field of RF.


There is no difference. The H-field is related to the E-field by the
properties of the medium. The current is related to the voltage by the
properties (Z0) of the medium. It is therefore possible to calculate a
single reflection coefficient. It is the same situation at RF as at
optical frequencies.

If you are referring to the sign change of the current reflection for
open vs. short circuit, then there is still no difference. The formula
for the reflection coefficient contain terms that change sign depending
on the relative properties of the media. It matters not whether one is
dealing with HF or visible light.

73,
Gene
W4SZ

Standing-Wave Current vs Traveling-Wave Current
 

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

Standing-Wave Current vs Traveling-Wave Current
 

Roy Lewallen wrote:
Roger wrote:
Roy Lewallen wrote:
. . .
at any point x, in degrees, along the line. This is also the total
voltage since there's no reflection yet.

"x" is referenced from the leading edge of the wave. At time 2pi, a
complete rotation has occurred and the wave front traveled to the open
circuit point. Understood.

No. x is referenced to the input end of the line. This is very
important. I'm sorry my statement that it is "any point x, in degrees,
along the line" didn't make this clear.
Yes, it is critical. I am sorry that I misunderstood this.

In our example then, "x" will always be positive. How am I to interpret
the meaning of vf(t, x) = sin(wt-x)?

As your argument is developed below, you begin using positive x.
What is the zero point it is referenced to? I will assume that it is
leading edge of original reference wave.

No, it's the input end of the line.

Sin(+x) represents a different polarity from the -x reference we were
using prior to this. I will remember this as I move through the
argument.

Sorry, but I don't understand this statement.

I understand that the convention for displaying a sin wave is that one
rotation is 2pi radians with positive rotation being counter clockwise
beginning with sin(x) = 0. Sin(90) = 1. Counter clockwise rotation
would indicate the the sin immediately becomes negative,
so sin(-90) = -1.

You begin the following argument using a reflection coefficient of -1,
which reverses the polarity of the wave. Am I to understand that your
model treats the input as a short circuit for the reflected wave?
Maybe I am missing an important point.

Yes, that is correct. The impedance of the source (a perfect voltage
source) is zero, so the reflection coefficient seen by the reverse
traveling wave is -1.

The logic of this assumption eludes me. In fact, it seems completely
illogical and counter to the concept of how voltage and current waves
are observed to move on a transmission line.

Maybe my concept of voltage being a concentration of positive (or
negative) charges is leading me astray. If two waves move in opposite
directions, but both of a positive character, at the time of crossing
paths, the voltages add. It happens at the open ends when the direction
reverses. It MUST happen identically when the reflected positive wave
returns to the source (at time 720 degrees in our one wavelength
example) and encounters the next positive wave just leaving the source.

In my model, the source voltage must change when the returning wave
hits the input end.

Then we've been using a different model. The one I've been using is the
one proposed by "Dave" -- a half wavelength open circuited line driven
by a voltage source -- except with your change in line length to one
wavelength. You cannot cause the voltage of a perfect voltage source to
change.

Are you assuming that vr is always propagating from the source as if the
source always supplied vf and vr simultaneously? As if vf was supplied
for time = 4pi, and then vr was applied?

Let's follow the returning wave as it hits the input end and
re-reflects. The reflection coefficient at the source is -1 due to the
zero-impedance ideal voltage source, so the re-reflected wave is

vf2(t, x) = -sin(wt - x)

Earlier, we defined the forward wave as vf(t, x) = sin(wt - x), so it
is logical to define vf2(t, x) = sin(wt-x). I do not see the logic in
reversing the voltage polarity with the minus sign.

The original forward wave is sin(wt - x). The reflected wave is sin(wt +
x), where the change in sign of x is a consequence of the reversal of
direction. Reflection from the source causes an inversion of the wave
polarity because of the -1 reflection coefficient, and another sign
change of x due to the reversal of direction, resulting in the equation
above.

and the total voltage anywhere along the transmission line just before
the re-reflection reaches the far end is vf(t, x) + vr(t, x) + vf2(t, x)
= sin(wt - x) + sin(wt + x) - sin(wt - x) = sin(wt + x). Now this is
interesting. When the second forward wave vf2 is added into the total,
the standing wave disappears and the total voltage is just a plain sine
wave with peak amplitude of 1. It's identical, in fact, to the original
forward voltage wave except reversed in phase.

Wow! This would happen if the re-reflection actually reverses. How
could this occur if we consider that voltage is a collection of
positive or negative particles?

Well, if it couldn't, then your concept of voltage is flawed. You might
re-think it.

We would have a positive reflection meeting with a positive outgoing
wave (or negative meeting negative). The situation would be the same
as at the open circuit end immediately following initial reversal, or
current would simply stop flowing from the source.

In fact, current does quit flowing from the source. The line is fully
charged and there is no load to dissipate any further energy from the
source. Any analysis showing continued current from the source is
obviously wrong for that reason.

I think you would agree that a steady state standing wave would form
immediately upon reflected wave reaching the initiating source if the
wave did not reverse.

I'm sorry, I don't understand that question.

I admire the time and effort spent on this analysis Roy. Very well
done no matter how history judges the merits of the argument. I think
I followed it all, and understood.

As I've mentioned, there are other valid ways of analyzing such a
circuit. At the end of the day, any analysis must produce the correct
result. Getting the correct result doesn't prove that the analysis is
valid, but failure to get the correct result proves that an analysis is
invalid. SPICE uses fundamental rules for analysis, so is a good
authority of what the answer should be, and it shows that my analysis
has produced the correct result.

The gif's certainly made it clear why you are skeptical of the power
of traveling wave analysis.

The SPICE results are simply a way of verifying that my analysis is
correct. The concept of traveling waves of average power has other,
serious problems.

Could we further discuss the merits of reversing the wave polarity
when the reflected wave returns to the source?

Sure. First please review the concept of reflection coefficient.

The behavior of the returning waves when they reach the source is often
not included in transmission line analysis because it plays no part in
determining the steady state SWR, impedance, or relationship between
voltages and currents at the ends or anywhere else along the line. The
only thing it impacts is the way steady state is reached during turn-on,
and not the final steady state condition itself, and this isn't
generally of interest. (An exception is the contrived and actually
impossible case of a completely lossless system such as the one I
analyzed, and it's not an exception either if viewed as a limiting
case.) As I mentioned in my posting, the steady state result is exactly
the same for any non-zero source resistance; the only effect of the
resistance is in determining how steady state is reached.

Roy Lewallen, W7EL

Your idea of a 5 wavelength long example was a good one Roy. It may
provide a way out of what seems to be a logical impasse (reversal at the
voltage source may be uncompromisable).

We could allow our future discussions (if any) to consider an extremely
long line, but consider only the 1/2 or 1 wavelength at the end for our
discussions. Thus, the source (and source for major disagreement) is
far removed from our discussion section. We could then consider the
input source as just another node for as long as we wanted.

Traveling waves easily explain standing waves on a 1/2 wavelength
section, as you demonstrated. Maybe they can explain or clarify more
things if we can get past "hang ups" such as the " -1 reflection at a
perfect voltage source".

73, Roger, W7WKB

Standing-Wave Current vs Traveling-Wave Current
 

Keith Dysart wrote:
When the source impedance is the same as Z0 there
is no impedance discontinuity to produce a reflection.

Here's a quote from "Fields and Waves ..." by
Ramo and Whinnery: "... significance cannot be
automatically attached to a calculation of power
loss in the internal impedance of the equivalent
circuit."

And from "T-Lines and Networks" by Walter C. Johnson:
"Although the Thevenin equivalent produces the correct
exterior results, its internal power relations may be
quite different from those of the network it replaces.
The power dissipated in the equivalent impedance of the
Thevenin circuit is not the same as the power dissipated
in the resistance of the actual network."

Translation: Do not use a Thevenin source impedance
to try to track power dissipation within the Thevenin
equivalent box - it won't work. Your argument that
there is zero power dissipation in the source resistor
inside the Thevenin equivalent box is bogus.

Is the reflected wave reflected even without a
discontinuity?

The incident reflected wave is either reflected by the
source or it isn't. You cannot have it both ways. Since
the internal conditions inside a Thevenin equivalent
circuit are a complete unknown, either choice is
a possibility.

What you will find is that, for a real-world source,
the destructive interference on one side of the source
equals the constructive interference on the other side
of the source. The conservation of energy principle
prohibits having it any other way.
--
73, Cecil http://www.w5dxp.com

Standing-Wave Current vs Traveling-Wave Current
 

Gene Fuller wrote:
Cecil Moore wrote:
In optics, to the best of my knowledge, there is no separate
reflection coefficient for the E-field and the H-field. If
true, that fact alone implies a different convention from
the field of RF.

There is no difference.

Please see Roy's posting. He explained the differences
in the two conventions. I don't need to repeat it.
--
73, Cecil http://www.w5dxp.com

Standing-Wave Current vs Traveling-Wave Current
 

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.

I completely concur with your analysis.

No doubt you have fine tuned the analysis to notice that the current
stops (meaning becomes unobservable) at the identical instant that the
voltage spike (to double) is observed. You would have noticed that the
zone of unmeasurable current spreads equally both ways from the
collision point at the velocity of the wave(s). The voltage spike
spreads in lock step with the loss of current detection. The maximum
width of the loss of current and voltage spike is the width of either of
the pulses.

Now did the two pulses reflect, or pass through one another? I have
considered the question and can not discern a difference in my analysis
either way. IT SEEMS TO MAKE NO DIFFERENCE!


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.

Roy and I are talking about this on other postings. I guess the purest
might point out that a 50 ohm generator only has a voltage to current
ratio of 50, but we don't know if it also has a resistor to absorb
energy. It is like a black box where the only thing we know about it is
that when we connect a 50 ohm resistor to it through a 50 ohm
transmission line, there are no standing waves.

In this case, a reflected wave could be used like a radar pulse to learn
what might be inside the box.

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


We certainly think similarly Keith. Thanks for the posting.

73, Roger, W7WKB

Standing-Wave Current vs Traveling-Wave Current
 

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.

Cecil, this sounds more like a pronouncement from God than like an
conclusion from observations.

Does this match your own concept of the traveling waves acting at the
{+} point Cecil? If not, where do we differ?

Where we differ is that you allow traveling waves to "reflect
off one another". There are no laws of physics which allow
that in the absence of a physical impedance discontinuity.
EM waves simply do not bounce off each other.

I am not aware of any laws of physics that prevent it either. I don't
see any evidence that it happens in open space, like light bouncing off
light. It might happen on transmission lines however. I just cannot
find any convincing evidence either way. What I have deduced so far
indicates that it makes no difference which happens.

Maybe both things happen (both reflect and pass). This because the EM
field travels very close to the speed of light. It is a little hard to
see how one wave could "see" the other coming. On the other hand, the
charges move slowly, far below the speed of light. It is easy to see
how they might "see or feel" each other coming.

73, Roger, W7WKB