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