Standing-Wave Current vs Traveling-Wave Current
 

Cecil Moore wrote:
Roger wrote:
Cecil Moore wrote:
Roger wrote:
Stored in the 1/4 WL between the short and mouth. No more current
needed once stability is reached.

EM RF current is stored in the stub? In what form?

Come on Cecil! Let's not go around in circles! You know very well how
it happens.

Here's an example using a circulator and load in
a 50 ohm system. Please think about it.

SGCL---1---2------------------------------+
\ / | 1/4
3 | WL
| everything is 50 ohms | shorted
R | stub

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. My present
level of understanding is that they can only be built using ferrite
inductors which have an ansiotropic (non-linear) magnetic response. If
so, they could not be compared to transmission lines without adding that
non-linear factor. Apparently energy is stored in these inductors only
if the power is moving in one direction, so it never reaches one branch.
I don't understand how a ferrite could do that.

Is there such a thing as a "all transmission line" circulator? If so,
where could I find the circuit?

Thanks,

73, Roger, W7WKB

Standing-Wave Current vs Traveling-Wave Current
 

On Dec 28, 9:56*am, Cecil Moore wrote:
Keith Dysart wrote:
The cuts changed nothing about the conditions in the circuit.

This reminds me of the guru who asserted that he could
replace his 50 ohm antenna with a 50 ohm resistor
without changing the conditions.

And yet the claim is made that before the cuts there are no
reflections and after the cut there are a bunch. And yet the
conditions in the circuit are EXACTLY the same.

No, conditions are not exactly the same. Before the
cut, there were no reflections. After the cut, there
are reflections. Conditions have changed. I'll bet
the change in the natural noise pattern, which exists
in every system, could be detected at the time of
the cut.

It is good that you agree that the voltage, current
and power distributions are exactly the same, with
and without the cuts. These are the conditions to
which I refer.

It is no different than a connected capacitor and inductor
which will ring for ever given the appropriate initial
conditions.

If we are talking about things that can happen only in your
mind, why stop with irrelevant ringing assertions? Why not
assert that you can leap tall buildings at a single bound
(in your mind)?

I suppose this non-sequitor means that you agree that
an ideal section of transmission line, just like an
ideal inductor and capacitor, will store energy forever
if provided with the appropriate initial conditions,
but do not wish to admit it.

An intriguing thought experiment is to take several
of these sections with stored energy (with the proper
phase relationship) and connect them together. Do
the reflections present at the ends of the short
sections suddenly disappear when the sections
are connected to form a longer line? Are the
reflections now only occurring at the ends of
the longer section?

...Keith

Standing-Wave Current vs Traveling-Wave Current
 

Keith Dysart wrote:
Cancelling is a very iffy thing. Better to decide
before clicking send.

Unfortunately, after two double scotches, it's
extremely easy to click send. :-)
--
73, Cecil http://www.w5dxp.com

Standing-Wave Current vs Traveling-Wave Current
 

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

Standing-Wave Current vs Traveling-Wave Current
 

Keith Dysart wrote:
I suppose this non-sequitor means that you agree that
an ideal section of transmission line, just like an
ideal inductor and capacitor, will store energy forever
if provided with the appropriate initial conditions,
but do not wish to admit it.

I freely admit that it can happen in your mind (like
leaping tall buildings at a single bound). I do not
believe it can happen in a real-world situation. But
if you can demonstrate lossless transmission lines
and lossless inductors on the bench, be my guest and
probably win a Nobel Prize in the process.

An intriguing thought experiment is to take several
of these sections with stored energy (with the proper
phase relationship) and connect them together. Do
the reflections present at the ends of the short
sections suddenly disappear when the sections
are connected to form a longer line? Are the
reflections now only occurring at the ends of
the longer section?

Reflections are impossible except at physical impedance
discontinuities. There are zero reflections at a point
in a smooth fixed Z0 section of transmission line. Cut
the line and you get 100% reflection. That's why your
assertions of "no change" don't make sense. You would
have us believe that short circuit to open circuit is
"no change"??? If that is true, there's "no change"
between a shorted 1/4WL stub and an open 1/4WL stub.
--
73, Cecil http://www.w5dxp.com

Standing-Wave Current vs Traveling-Wave Current
 

Cecil Moore wrote:
Keith Dysart wrote:
Cancelling is a very iffy thing. Better to decide
before clicking send.

Unfortunately, after two double scotches, it's
extremely easy to click send. :-)

CECIL!

BTDT (Been There, Done That) ;-)

Regards,
JS

Standing-Wave Current vs Traveling-Wave Current
 

On Dec 28, 2:44*pm, Cecil Moore wrote:
Keith Dysart wrote:
I suppose this non-sequitor means that you agree that
an ideal section of transmission line, just like an
ideal inductor and capacitor, will store energy forever
if provided with the appropriate initial conditions,
but do not wish to admit it.

I freely admit that it can happen in your mind (like
leaping tall buildings at a single bound).

Thought experiments do usually occur within the mind.
This one is no different.

I do not
believe it can happen in a real-world situation.

That is good, for it is unlikely, though with superconductors
one might come close.

But
if you can demonstrate lossless transmission lines
and lossless inductors on the bench, be my guest and
probably win a Nobel Prize in the process.

More intrigue. This path has been trod before. Begin
with an experiment using ideal elements. Get
uncomfortably close to some truth. Declare the
experiment invalid because it could not happen
in the "real world".

It would be more valuable were you to confront
the demons rather than take the "real world"
escape.

An intriguing thought experiment is to take several
of these sections with stored energy (with the proper
phase relationship) and connect them together. Do
the reflections present at the ends of the short
sections suddenly disappear when the sections
are connected to form a longer line? Are the
reflections now only occurring at the ends of
the longer section?

Reflections are impossible except at physical impedance
discontinuities. There are zero reflections at a point
in a smooth fixed Z0 section of transmission line. Cut
the line and you get 100% reflection.

That is why it is so intriguing. The voltage and current
conditions have not changed, and yet, befo reflections,
after: none.

That's why your
assertions of "no change" don't make sense.

Are you claiming that the voltages or currents
have changed? Identify a measurable value that has
changed and the proof will be yours.

You would
have us believe that short circuit to open circuit is
"no change"???

If it does not change the circuit conditions, i.e.
voltages or currents.

If that is true, there's "no change"
between a shorted 1/4WL stub and an open 1/4WL stub.

Invalid generalization. We were discussing a specific
circuit, not any old 1/4WL stub.

...Keith

Standing-Wave Current vs Traveling-Wave Current
 

Roger wrote:
OK. I think I should tweak the example just a little to clarify that
our source voltage will change when the reflected wave arrives back
at the source end. To do this, I suggest that we increase our
transmission line to one wavelength long. This so we can see what
happens to the source if we pretended that we had not moved it all.

We pick our lead edge at wt-0 and define it to be positive voltage.
The next positive leading edge will occur at wt-360. Of course, a
half cycle of positive voltage will follow for 180 degrees following
points wt-0 and wt-360.

Initiate the wave and let it travel 540 degrees down the transmission
line. At this point, the leading edge wt-0 has reflected and has
reached a point 180 degrees from the full wave source. This is the
point that was originally our source point on the 1/2 wave line.
Mathematically, wt-0 is parallel/matched with wt-360, but because the
wt-0 has been reflected, the current has been reversed but the
voltage has not been changed.

After the initial wave has been propagating 540 degrees along the one
wavelength line, it will be back at the input end of the line, not 180
degrees from the source. (I assume that by "full wave source" you mean
the source connected to the input end of the line.)

Lets move to wave point wt-1 and wt-361 so that we will have non-zero
voltage and current.

vt = vr(wt-1) + vf(wt-361) = 2*.5*sin(1)

I'm sorry, you've lost me already. Where exactly are these "wave
points"? By "wave point wt-1" do you mean 1 physical degree down the
line from the source, or 1 degree from the leading edge of the intial
wave? As it turns out, those two points would be the same after 540
degrees of propagation. But "wt-361" would be one degree beyond the end
of the line by the first interpretation, or 1 degree short of the end of
the line by the other. What's the significance of the sum of the
voltages at these two different points?

I'm increasingly lost from here. . .

Here's my analysis of what happens after the initial step is applied,
using your 360 degree line and notation I'm more familiar with:

If the source is sin(wt) (I've normalized to a peak voltage of 1 volt
for simplicity) and we turn it on at t = 0, a sine wave propagates down
the line, described by the function

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

At time t = 2*pi/w (one period after t = 0), it arrives at the far end.
Just before the wave reaches the far end, we have:

vf(t, x) = sin(wt - x) evaluated at t = 2*pi/w, = sin(-x) = -sin(x)

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

Then the forward wave reaches the far end. The reflection coefficient of
an open circuit is +1, so the reflected voltage wave has the same
magnitude as the forward wave. It arrives at the source at t = 4*pi/w
(two periods after t = 0), where it's in phase with the forward wave.
The reverse wave (for the special case of a line an integral number of
half wavelengths long) is:

vr(t, x) = sin(wt + x)

So at any point x (in degrees) along the line,

v(t, x) = vf(t, x) + vr(t, x) = sin(wt - x) + sin(wt + x)

Using a trig identity,

v(t, x) = 2 * cos(x) * sin(wt)

Notice that a standing wave pattern has been formed -- the cos(x) term
describes the envelope of the voltage sine wave as a function of
position along the line. But notice that the peak amplitude of the total
voltage sine wave is 2 rather than 1 volt, except that it's now
modulated by the cos(x) position function. Also note that the time
function sin(wt) has no x term, which means that the voltage changes all
along the line at the same time.

At the moment the returning wave arrives at the source (t = 4*pi/w),
sin(wt) = 0, so

v(x) = 0

So the returning wave arrives at the source at the very moment that the
voltage is zero everywhere along the line. (For those interested in
energy, this means that the line's energy is stored entirely in the
magnetic field, or the equivalent line inductance, at this instant.)

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)

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.

And what happens when vf2 reaches the far end and reflects? Well,

vr2 = -sin(wt + x)

So just before it reaches the input end of the line, the total is now

vf + vr + vf2 + vr2 = sin(wt - x) + sin(wt + x) - sin(wt - x) - sin(wt
+ x) = 0

For the interval t = 6*pi/w to t = 8*pi/w, the total voltage on the line
is zero at all points along the line which the second reflected wave has
reached! Further analysis shows that the line continues to alternate among:

v(t, x) = sin(wt - x) [vf only] [Eq. 1]
v(t, x) = 2 * cos(x) * sin(wt) [vf + vr] [Eq. 2]
v(t, x) = sin(wt + x) [vf + vr + vf2] [Eq. 3]
v(t, x) = 0 [vf + vr + vf2 + vr2] [Eq. 4]

How about the value at the input end (x = 0) at various times?

When we had only the first forward wave, it was sin(wt). When we had the
original forward wave, the reflected wave, and the new forward wave, it
was also sin(wt). And as it turns out, it stays at sin(wt) at all times
as each wave returns and re-reflects. I was incorrect earlier in saying
that this example resulted in infinite currents. It doesn't, but a
shorted line, or open quarter wave line for example, would, when driven
by a perfect voltage source.

Just looking at the source makes it appear that we've reached
equilibrium. But we haven't. The total voltage along the line went from
a flat forward wave of peak amplitude of 1 to a standing wave
distribution with a peak amplitude of 2 when the reflection returned to
a flat distribution with peak amplitude of 1 when the first
re-reflection hit, then to zero when it returned. Maybe we'd better take
a look at what's happening at the far end of the line.

The voltage at the far end is of course zero from t = 0 to t = 2*pi/w,
when the initial wave reaches it. It will then become 2 * sin(wt), or
twice the source voltage where it will stay until the first re-reflected
wave (vf2) reaches it. The re-reflected forward wave vf2 will also
reflect off the end, making the total at the end:

vf + vr + vf2 + vr2 = sin(wt) + sin(wt) - sin(wt) - sin(wt) = 0

The voltage at the far end of the line will drop to zero and stay there
for the next round trip time of 4*pi/w! Then it will jump back to 2 *
sin(wt) for another period, then to zero, etc.

In real transmission line problems, some loss will always be present, so
reflections will become less and less over time, allowing the system to
reach an equilibrium state known as steady state. Our system doesn't
because it has no loss. The problem is analogous to exciting a resonant
circuit having infinite Q, with a lossless source. It turns out that
adding any non-zero series resistance at the source end, no matter how
small, will allow the system to converge to steady state. But not with
zero loss.

I set up a simple SPICE model to illustrate the line behavior I've just
described. I made the line five wavelengths long instead of one, to make
the display less confusing. The frequency is one Hz and the time to go
from one end to the other is five seconds. The perfect voltage source at
the input is sin(wt) (one volt peak).
http://eznec.com/images/TL_1_sec.gif is the total voltage at a point one
wavelength (one second) from the input end.
http://eznec.com/images/TL_5_sec.gif is the total voltage at the far end
of the line (five wavelengths, or five seconds from the input end).

First look at TL_1_sec.gif. During the time t = 0 to t = 1 sec., the
initial forward wave hasn't arrived at the one-wavelength sample point.
Then from t = 2 to t = 9, only the forward wave is present as in Eq. 1.
At t = 9 sec. the first reflected wave arrives, resulting in a total
voltage amplitude of 2 volts peak as predicted by Eq. 2. The
re-reflected wave arrives at t = 11 sec., at which time the amplitude
drops back to 1 volt peak as per Eq. 3. And at t = 19 seconds, vr2
arrives, dropping the voltage to 0 as predicted by Eq. 4. The pattern
then repeats, forever.

TL_5_sec.gif shows the total voltage at the open end of the line,
alternating between a 2 volt peak sine wave and zero as predicted.

A nearly identical analysis can be done for the line current.

Let me say once again that the introduction of any series source
resistance at all at the input will result in convergence rather than
the oscillating behavior shown here, so a steady state analysis of the
zero resistance case can be done as a limit as the source resistance
approaches zero. But any analysis of the start up conditions on the zero
source resistance line should produce the same results derived here
mathematically and confirmed by time domain modeling.

Roy Lewallen, W7EL

Standing-Wave Current vs Traveling-Wave Current
 

Roy Lewallen wrote:


Roy Lewallen, W7EL

Huh ...

Ever take a time out?

Look at some wave(s) on sting. Waves on a puddle? Acoustic waves?
Waves on a sand dune--left by the wind? etc?

Hmmmm...

Regards,
JS

Standing-Wave Current vs Traveling-Wave Current
 

Keith Dysart wrote:
On Dec 28, 2:44 pm, Cecil Moore wrote:
Keith Dysart wrote:
I suppose this non-sequitor means that you agree that
an ideal section of transmission line, just like an
ideal inductor and capacitor, will store energy forever
if provided with the appropriate initial conditions,
but do not wish to admit it.
I freely admit that it can happen in your mind (like
leaping tall buildings at a single bound).

Thought experiments do usually occur within the mind.
This one is no different.

I do not
believe it can happen in a real-world situation.

That is good, for it is unlikely, though with superconductors
one might come close.

But
if you can demonstrate lossless transmission lines
and lossless inductors on the bench, be my guest and
probably win a Nobel Prize in the process.

More intrigue. This path has been trod before. Begin
with an experiment using ideal elements. Get
uncomfortably close to some truth. Declare the
experiment invalid because it could not happen
in the "real world".
. . .

Oh, boy, the thought of Cecil doing his "proofs" without using lossless
lines, pure resistances, inductances, or capacitances, lossless antenna
conductors, or any other non-real-world components is enough to tempt me
to de-plonk him just to watch the show. But I'm afraid it'll just add
more DOO (Degrees Of Obfuscation) to his already formidable toolbox of
obscuring and misdirecting techniques. Oh well.

Roy Lewallen, W7EL

Standing-Wave Current vs Traveling-Wave Current
 

John Smith wrote:

ahhh, "sting" means "string" ...

Regards,
JS

Standing-Wave Current vs Traveling-Wave Current
 

Keith Dysart wrote:
More intrigue. This path has been trod before. Begin
with an experiment using ideal elements. Get
uncomfortably close to some truth. Declare the
experiment invalid because it could not happen
in the "real world".

When a thought experiment deviates far enough from
reality to become impossible, it is necessary to
recognize that one has crossed the line between
reality and mental masturbation. Would you like to
debate how many angels can dance on the head of
a pin?

It would be more valuable were you to confront
the demons rather than take the "real world"
escape.

I have confronted the supernatural and don't believe
in it. Your mileage may vary.

That is why it is so intriguing. The voltage and current
conditions have not changed, and yet, befo reflections,
after: none.

So changes have indeed occurred. A video signal is
a very good one to use to actually see the changes.
If you want to sweep the technical facts under the
rug, now is the time to remind me that a video signal
is not steady state.

Are you claiming that the voltages or currents
have changed?

Of course not. I am claiming that reflections have
*changed* and you seem to agree.

If ignorance is really your goal, why not drop the
experiment in the deepest part of the Pacific Ocean
where nobody can know anything about it?
--
73, Cecil http://www.w5dxp.com

Standing-Wave Current vs Traveling-Wave Current
 

A followup:

If we have a transmission line which is an integral number of half
wavelengths long, open circuited at the far end, and driven by a perfect
voltage source of Vs*sin(wt) in series with *any* non-zero resistance:

The amplitude of the wave reflected from the source will decrease each
time, resulting in convergence at the following steady state conditions:

vf(t, x) = (Vs/2) * sin(wt - x)
vr(t, x) = (Vs/2) * sin(wt + x)

Where x is the position from the source in electrical degrees or
radians, and vf and vr are the totals of all forward and reverse
traveling waves respectively.

The total voltage along the line at any time and position is:

v(t, x) = vf(t, x) + vr(t, x) = Vs * sin(wt) * cos(x)

This clearly shows that the total voltage at any point along the line is
sinusoidal and in phase at all points. The "standing wave" is the
description of the way the peak amplitude of the sine time function
differs with position x.

So the amplitude of the voltage at both ends of the line (where |cos(x)|
= 1) will equal the source voltage. The number of reflections (length of
time) it will take for the system to converge to within any specified
closeness of the steady state depends on the amount of source
resistance. If the source resistance is the same as the line Z0,
convergence is reached in a single round trip. The time increases as the
source resistance gets greater or less than this value. If the source
resistance in zero or infinite, convergence to this steady state will
never be reached, as shown in the earlier posting.

Roy Lewallen, W7EL

Standing-Wave Current vs Traveling-Wave Current
 

On Dec 28, 7:48*pm, Cecil Moore wrote:
Keith Dysart wrote:
More intrigue. This path has been trod before. Begin
with an experiment using ideal elements. Get
uncomfortably close to some truth. Declare the
experiment invalid because it could not happen
in the "real world".

When a thought experiment deviates far enough from
reality to become impossible, it is necessary to
recognize that one has crossed the line between
reality and mental masturbation. Would you like to
debate how many angels can dance on the head of
a pin?

Still sidetracking away from your demons rather
than confronting them?!

It would be more valuable were you to confront
the demons rather than take the "real world"
escape.

I have confronted the supernatural and don't believe
in it. Your mileage may vary.

That is why it is so intriguing. The voltage and current
conditions have not changed, and yet, befo reflections,
after: none.

So changes have indeed occurred. A video signal is
a very good one to use to actually see the changes.
If you want to sweep the technical facts under the
rug, now is the time to remind me that a video signal
is not steady state.

You've got that right. And nor is it the experiment
under discussion.

Are you claiming that the voltages or currents
have changed?

Of course not.

So whether the line is cut or not, the same voltage,
current and power distribution exist.

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

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

...Keith

Standing-Wave Current vs Traveling-Wave Current
 

Roy Lewallen wrote:
A followup:

If we have a transmission line which is an integral number of half
wavelengths long, open circuited at the far end, and driven by a perfect

But then again, if we have a coaxial circuit of infinite length, to
where any frequency can be expressed in a wavelength, or wavelengths,
blah, blah, blah ...

And, if those faeries could just spin straw into gold ... we'd all be
rich ...

Some arguments take me in the "general direction" I wish to go, some
don't ... you will excuse me ... I am sure ...

Regards,
JS

Standing-Wave Current vs Traveling-Wave Current
 

Roy Lewallen wrote:
Roger wrote:
OK. I think I should tweak the example just a little to clarify that
our source voltage will change when the reflected wave arrives back
at the source end. To do this, I suggest that we increase our
transmission line to one wavelength long. This so we can see what
happens to the source if we pretended that we had not moved it all.

We pick our lead edge at wt-0 and define it to be positive voltage.
The next positive leading edge will occur at wt-360. Of course, a
half cycle of positive voltage will follow for 180 degrees following
points wt-0 and wt-360.

Initiate the wave and let it travel 540 degrees down the transmission
line. At this point, the leading edge wt-0 has reflected and has
reached a point 180 degrees from the full wave source. This is the
point that was originally our source point on the 1/2 wave line.
Mathematically, wt-0 is parallel/matched with wt-360, but because the
wt-0 has been reflected, the current has been reversed but the
voltage has not been changed.

After the initial wave has been propagating 540 degrees along the one
wavelength line, it will be back at the input end of the line, not 180
degrees from the source. (I assume that by "full wave source" you mean
the source connected to the input end of the line.)

Sorry that I was not clear at the beginning. I am proposing that we
change the experiment to increase the transmission line length from 1/2
wavelength long (180 degrees), to one wavelength long (360 degrees). On
the longer line, the initial point on the wave (wt-0) will travel 720
degrees before it returns to the source. Making that assumption, I
think my description is correct.

Lets move to wave point wt-1 and wt-361 so that we will have non-zero
voltage and current.

vt = vr(wt-1) + vf(wt-361) = 2*.5*sin(1)

I'm sorry, you've lost me already. Where exactly are these "wave
points"? By "wave point wt-1" do you mean 1 physical degree down the
line from the source, or 1 degree from the leading edge of the intial
wave?

One degree from the leading edge.

As it turns out, those two points would be the same after 540
degrees of propagation. But "wt-361" would be one degree beyond the end
of the line by the first interpretation, or 1 degree short of the end of
the line by the other. What's the significance of the sum of the
voltages at these two different points?

Good point. I should have respecified 541 degrees of propagation time.
My goal was to look at the voltage at a single point because of the
presence of two parts of the wave at the same place and time.

I'm increasingly lost from here. . .

Here's my analysis of what happens after the initial step is applied,
using your 360 degree line and notation I'm more familiar with:

If the source is sin(wt) (I've normalized to a peak voltage of 1 volt
for simplicity) and we turn it on at t = 0, a sine wave propagates down
the line, described by the function

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

At time t = 2*pi/w (one period after t = 0), it arrives at the far end.
Just before the wave reaches the far end, we have:

vf(t, x) = sin(wt - x) evaluated at t = 2*pi/w, = sin(-x) = -sin(x)

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.

Then the forward wave reaches the far end. The reflection coefficient of
an open circuit is +1, so the reflected voltage wave has the same
magnitude as the forward wave. It arrives at the source at t = 4*pi/w
(two periods after t = 0), where it's in phase with the forward wave.


The reverse wave (for the special case of a line an integral number of
half wavelengths long) is:



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.

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.



vr(t, x) = sin(wt + x)

So at any point x (in degrees) along the line,

v(t, x) = vf(t, x) + vr(t, x) = sin(wt - x) + sin(wt + x)

Using a trig identity,

v(t, x) = 2 * cos(x) * sin(wt)

Notice that a standing wave pattern has been formed -- the cos(x) term
describes the envelope of the voltage sine wave as a function of
position along the line. But notice that the peak amplitude of the total
voltage sine wave is 2 rather than 1 volt, except that it's now
modulated by the cos(x) position function. Also note that the time
function sin(wt) has no x term, which means that the voltage changes all
along the line at the same time.

At the moment the returning wave arrives at the source (t = 4*pi/w),
sin(wt) = 0, so

v(x) = 0
No, v(t, x) = 0 We need to remember we are carrying two variables here,
like always keeping track of apples and oranges in the same equation.

So the returning wave arrives at the source at the very moment that the
voltage is zero everywhere along the line. (For those interested in
energy, this means that the line's energy is stored entirely in the
magnetic field, or the equivalent line inductance, at this instant.)


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.

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

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.

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? 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.

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.

And what happens when vf2 reaches the far end and reflects? Well,

vr2 = -sin(wt + x)

So just before it reaches the input end of the line, the total is now

vf + vr + vf2 + vr2 = sin(wt - x) + sin(wt + x) - sin(wt - x) - sin(wt
+ x) = 0

For the interval t = 6*pi/w to t = 8*pi/w, the total voltage on the line
is zero at all points along the line which the second reflected wave has
reached! Further analysis shows that the line continues to alternate among:

v(t, x) = sin(wt - x) [vf only] [Eq. 1]
v(t, x) = 2 * cos(x) * sin(wt) [vf + vr] [Eq. 2]
v(t, x) = sin(wt + x) [vf + vr + vf2] [Eq. 3]
v(t, x) = 0 [vf + vr + vf2 + vr2] [Eq. 4]

How about the value at the input end (x = 0) at various times?

When we had only the first forward wave, it was sin(wt). When we had the
original forward wave, the reflected wave, and the new forward wave, it
was also sin(wt). And as it turns out, it stays at sin(wt) at all times
as each wave returns and re-reflects. I was incorrect earlier in saying
that this example resulted in infinite currents. It doesn't, but a
shorted line, or open quarter wave line for example, would, when driven
by a perfect voltage source.

Just looking at the source makes it appear that we've reached
equilibrium. But we haven't. The total voltage along the line went from
a flat forward wave of peak amplitude of 1 to a standing wave
distribution with a peak amplitude of 2 when the reflection returned to
a flat distribution with peak amplitude of 1 when the first
re-reflection hit, then to zero when it returned. Maybe we'd better take
a look at what's happening at the far end of the line.

The voltage at the far end is of course zero from t = 0 to t = 2*pi/w,
when the initial wave reaches it. It will then become 2 * sin(wt), or
twice the source voltage where it will stay until the first re-reflected
wave (vf2) reaches it. The re-reflected forward wave vf2 will also
reflect off the end, making the total at the end:

vf + vr + vf2 + vr2 = sin(wt) + sin(wt) - sin(wt) - sin(wt) = 0

The voltage at the far end of the line will drop to zero and stay there
for the next round trip time of 4*pi/w! Then it will jump back to 2 *
sin(wt) for another period, then to zero, etc.

In real transmission line problems, some loss will always be present, so
reflections will become less and less over time, allowing the system to
reach an equilibrium state known as steady state. Our system doesn't
because it has no loss. The problem is analogous to exciting a resonant
circuit having infinite Q, with a lossless source. It turns out that
adding any non-zero series resistance at the source end, no matter how
small, will allow the system to converge to steady state. But not with
zero loss.

I set up a simple SPICE model to illustrate the line behavior I've just
described. I made the line five wavelengths long instead of one, to make
the display less confusing. The frequency is one Hz and the time to go
from one end to the other is five seconds. The perfect voltage source at
the input is sin(wt) (one volt peak).
http://eznec.com/images/TL_1_sec.gif is the total voltage at a point one
wavelen.com/images/TL_5_sec.gifgth (one second) from the input end.
eznehttp://c is the total voltage at the far end
of the line (five wavelengths, or five seconds from the input end).

First look at TL_1_sec.gif. During the time t = 0 to t = 1 sec., the
initial forward wave hasn't arrived at the one-wavelength sample point.
Then from t = 2 to t = 9, only the forward wave is present as in Eq. 1.
At t = 9 sec. the first reflected wave arrives, resulting in a total
voltage amplitude of 2 volts peak as predicted by Eq. 2. The
re-reflected wave arrives at t = 11 sec., at which time the amplitude
drops back to 1 volt peak as per Eq. 3. And at t = 19 seconds, vr2
arrives, dropping the voltage to 0 as predicted by Eq. 4. The pattern
then repeats, forever.

TL_5_sec.gif shows the total voltage at the open end of the line,
alternating between a 2 volt peak sine wave and zero as predicted.

A nearly identical analysis can be done for the line current.

Let me say once again that the introduction of any series source
resistance at all at the input will result in convergence rather than
the oscillating behavior shown here, so a steady state analysis of the
zero resistance case can be done as a limit as the source resistance
approaches zero. But any analysis of the start up conditions on the zero
source resistance line should produce the same results derived here
mathematically and confirmed by time domain modeling.

Roy Lewallen, W7EL

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.

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

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

73, Roger, W7WKB

Standing-Wave Current vs Traveling-Wave Current
 

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

Standing-Wave Current vs Traveling-Wave Current
 

Roy Lewallen wrote:
Oh, boy, the thought of Cecil doing his "proofs" without using lossless
lines, pure resistances, inductances, or capacitances, lossless antenna
conductors, or any other non-real-world components is enough to tempt me
to de-plonk him just to watch the show. But I'm afraid it'll just add
more DOO (Degrees Of Obfuscation) to his already formidable toolbox of
obscuring and misdirecting techniques. Oh well.

Roy, why must you resort to ad hominem attacks?
Does it mean that you are incapable of winning the
argument on technical merit?

You have even proved yourself and W8JI wrong about
using standing-wave current to "measure" the delay
through a 75m loading coil and don't even seem to
realize it. Here's the equation you posted:

v(t, x) = 2 * cos(x) * sin(wt)

The equation for I(t, x) would be similar with a
90 degree offset. Please come down from your ivory
tower and explain how that current can be used to
measure delay through a coil.

The point I was making is when imagination is allowed
to run wild in religion or in technical arguments,
anything is possible in the human mind. There simply
has to be a limit oriented to reality.

When a cable is cut at a point where it is known to
be transferring energy in both directions, it is no
longer transferring energy in both directions. That
is reality. No flights of fantasy will change that
technical fact.
--
73, Cecil http://www.w5dxp.com

Standing-Wave Current vs Traveling-Wave Current
 

Roy Lewallen wrote:
v(t, x) = vf(t, x) + vr(t, x) = Vs * sin(wt) * cos(x)

This clearly shows that the total voltage at any point along the line is
sinusoidal and in phase at all points.

It also clearly implies that statement is true for current
as well.

SOMEONE PLEASE PASS THIS CHALLENGE ON TO ROY. HE HAS
DUCKED AND DODGED AND REFUSED TO ANSWER FOR MUCH TOO LONG.
ONE OF HIS PLOYS IS TO SAY HE NEVER SAW THE CHALLENGE
BECAUSE HE HAS PLOINKED ME.

EXACTLY HOW CAN A CURRENT WHICH YOU ADMIT IS IN PHASE AT
ALL POINTS BE USED TO MEASURE THE DELAY THROUGH A 75M
BUGCATCHER LOADING COIL?

It's pretty obvious from experience that Roy would rather
mount an ad hominem attack against me rather than answer
the simple question above.
--
73, Cecil http://www.w5dxp.com

Standing-Wave Current vs Traveling-Wave Current
 

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.

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.

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.
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.

Believing that reflections can occur where there
exists no physical impedance discontinuity is a
religion, not a science. 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.
--
73, Cecil http://www.w5dxp.com

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

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