Standing-Wave Current vs Traveling-Wave Current
 

Roger wrote:
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.

Not speaking for Roy, but x is usually 0 at the feedpoint
of a 1/2WL dipole, for instance. For any fixed time t=N,
the phase of the standing-wave signal is constant all up
and down the antenna. Please ask Roy how that fixed phase
can possibly be used to measure the phase shift through
a 75m bugcatcher loading coil.

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? 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.

RF engineering convention in phasor notation: If the wave
reflects from an open circuit, the current reverses phase
by 180 degrees such that If+Ir=0. The voltage does not
reverse phase so Vf+Vr=2Vf=2Vr.

If the wave reflects from a short circuit, the voltage reverses
phase by 180 degrees such that Vf+Vr=0. The current does not
reverse phase so If+Ir=2If=2Ir.

Incidentally, this is a different convention from the field
of optics.

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.

This is a tricky subject covered by many discussions among
experts. My take is that since the impedance "seen" by the
reflections is usually unknown (possibly even unknowable) the
discussion is a moot point. The convention is that if reflected
energy enters the source, it was, by definition, never generated
in the first place. All that is important at the output of a
source is the *NET* power output. Energy flow back into the
source is, by definition, completely ignored.

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

The Poynting vector yields the power density of any EM traveling
wave. The power density equations from the field of optics are
just as valid for RF waves as they are for light waves. If one
wants to understand the redistribution of energy, one will need
to understand constructive and destructive interference during
superposition. Roy is on record as not caring where the energy
goes.
--
73, Cecil http://www.w5dxp.com

Standing-Wave Current vs Traveling-Wave Current
 

Cecil Moore wrote:
Roger wrote:
Cecil Moore wrote:
Are there any reflections at point '+'?

If not, how is energy stored in the stub?

If so, what causes those reflections?

I am not sufficiently familiar with circulators to respond.

If the circulator is bothering you, forget it and assume the
following lossless conditions:

Ifor = 1 amp --
------------------------------+
-- Iref = 1 amp | 1/4
| WL
All Z0 = 50 ohms | shorted
| stub

Please think about it and answer the questions above.
The main point to remember is that there is no physical
impedance discontinuity at '+'.

OK. Let's begin by recognizing that this circuit is identical to a
straight transmission line. The purpose of identifying the stub is to
clearly locate the point 1/4 wavelength from the end of the line. The
line is shorted at the end.

We further assume that the peak current is 1 amp.

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.

Is it important to decide this issue? Yes, if it will affect the answer
to questions such as what is the voltage or current at this point.

Will it affect the answers? No. Under the conditions described, the
waves passing in opposite directions will have equal voltages and
opposite currents. If they pass through one another, the voltages will
add, but the currents will subtract. If they reflect, the voltage of
each component (Vf and Vr) will add on itself, and the individual
currents will reverse on themselves and therefore subtract. Either way,
the total voltage will double, and total measured current would be zero.
There is no reason to decide the issue.

How is energy stored in the stub? We have defined current as entering
an leaving the stub. Current is thought of as movement of charged
particles, but not as a concentration of particles. A concentration of
charged particles exhibits voltage. Energy is present when EITHER
current or voltage are shown to be present. Here, current is defined as
one amp so energy must be present some place on the line. The stub is
1/4 wavelength long physically, but it is 1/2 wavelength long
electrically, so that if we have energy present in the time-distance
shape of a sine wave, we would have an entire 1/2 wave's worth of energy
present on the stub at all times. The location of peak voltage (or peak
current) will depend upon the time-distance reference used to describe
the moving wave. (We would have equal voltage(but opposite polarity)
peaks located at the point {+} if we assumed the center of the forward
and reflected wave each to located 90 degrees from the shorted end.)

The circuit shows forward current Ifor and reflected current Iref as if
each were only one current. When we consider traveling waves, we need
to remember that Ifor and Iref can be measured on either of the two
wires composing a transmission line. The forward wave exists on both
wires, but the sides display opposite polatity and direction of current
despite both moving in the same direction. It is best to consider the
forward traveling wave as two waves, each carrying half the power, with
one wave per wire.

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

73, Roger, W7WKB

Is this the kind of answer you were looking for? The answer could be
given mathematically but that might be even more confusing.

Standing-Wave Current vs Traveling-Wave Current
 

Cecil Moore wrote:

If the wave reflects from a short circuit, the voltage reverses
phase by 180 degrees such that Vf+Vr=0. The current does not
reverse phase so If+Ir=2If=2Ir.

Incidentally, this is a different convention from the field
of optics.

Cecil,

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 any case this reversal or non-reversal is not a "convention". It is
the mathematical result that comes out of a proper solution to the
boundary value problem. The "convention" exists only if one considers
Maxwell equations to be a "convention".

73,
Gene
W4SZ

Standing-Wave Current vs Traveling-Wave Current
 

Cecil Moore wrote:
Roy Lewallen wrote:
v(t, x) = 2 * cos(x) * sin(wt)

At any particular time t=N, what is the variation
in phase for any x between 0 and 90 degrees? If
the variation is zero, how can such a signal be
used to measure delay?

The equation is a correct equation for describing a standing wave.
It is not describing a signal. It is a voltage-time-location(not
distance) relationship for a defined situation.

Having measured the voltage at two x points, from the equation I could
find the rotational angle of the wave at both points. The difference
between the two angles would be the distance of wave travel assuming the
standing wave was composed of traveling waves.

I don't think I would call that a measurement of delay, nor of phase. I
would call it a measurement of distance traveled on the transmission line.

73, Roger, W7WKB

Standing-Wave Current vs Traveling-Wave Current
 

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.

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.

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.

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.

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

Standing-Wave Current vs Traveling-Wave Current
 

Gene Fuller wrote:
Cecil Moore wrote:

If the wave reflects from a short circuit, the voltage reverses
phase by 180 degrees such that Vf+Vr=0. The current does not
reverse phase so If+Ir=2If=2Ir.

Incidentally, this is a different convention from the field
of optics.

Cecil,

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 any case this reversal or non-reversal is not a "convention". It is
the mathematical result that comes out of a proper solution to the
boundary value problem. The "convention" exists only if one considers
Maxwell equations to be a "convention".

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.

Roy Lewallen, W7EL

Standing-Wave Current vs Traveling-Wave Current
 

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.

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

Standing-Wave Current vs Traveling-Wave Current
 

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

Standing-Wave Current vs Traveling-Wave Current
 

Roger wrote:
Cecil Moore wrote:
Roy Lewallen wrote:
v(t, x) = 2 * cos(x) * sin(wt)

At any particular time t=N, what is the variation
in phase for any x between 0 and 90 degrees? If
the variation is zero, how can such a signal be
used to measure delay?

The equation is a correct equation for describing a standing wave.
It is not describing a signal.

Roy thinks it is describing a signal and so does W8JI.
It's the signal current they both used to "measure" the
delay through a 75m bugcatcher loading coil.

Having measured the voltage at two x points, from the equation I could
find the rotational angle of the wave at both points. The difference
between the two angles would be the distance of wave travel assuming the
standing wave was composed of traveling waves.

But that's not what Roy did. He measured the difference
in phase between two current measurements in a standing-
wave environment and declared that lack of phase difference
to be the "delay". He did NOT measure amplitude and calculate
backwards to get the delay. That's what I said he should
have done and he ploinked me.
--
73, Cecil http://www.w5dxp.com

Standing-Wave Current vs Traveling-Wave Current
 

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.

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.

My analysis produces the correct result but you have rejected
it out of hand. My analysis is the method by which EM light
waves are analyzed.
--
73, Cecil http://www.w5dxp.com