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