Analyzing Stub Matching with Reflection Coefficients
 

I'm not sure how many times it's worthwhile to keep repeating this, but
I guess I'll give it another couple of tries before giving up.

Richard Harrison wrote:
Roy Lewallen, W7EL wrote:
"If you`ll read what I`ve written, you`ll hopefully see that my only
point of contention is with your claim that waves reflect from a
"virtual short". They do not."

Seems to me they do.

If you are lucky enough to have a copy of Terman`s 1955 opus, we can
reason together.

Sorry, I'm not. All I have is a 1947 Third Edition of _Radio
Engineering_. I'm unfortunately stuck with having to think for myself.
But I trust you to quote him accurately, and Terman is to be trusted.

On page 91 is found Fig. 4-3 Vector (phasor) diagrams showing manner in
which incident and reflected waves combined to produce a voltage
distribution on the transmission line.

I'm sure they're correct, and similar diagrams can be found in many of
my other texts.

At an open circuit, the voltage phasors are in-phase.

Yes. And the current phasors are out of phase.

E2, the reflected phasor, rotates clockwise as it travels back toward
the source.

E1, the incident phasor, rotates counter-clockwise as we look back
toward the source.

Yes. These of course follows from the mathematical analysis of
transmission lines, found in many texts, and with which I'm very familiar.

Looking 1/4-wavelength back from the open-circuit, E2 and E1, each
having rotated 90-degrees, but in opposite directions, are now
180-degrees out-of-phase.

On page 92, Fig. 4-4 shows the current, which summed to zero at the open
circuit, has risen to its maximum value at 1/4-wavelength back from the
open-circuit while the voltage dropped to its minimum, nearly zero,
maybe close enough to declare a "virtual short-circuit", 1/4-wavelength
back from the open-circuit.

Yes, this is universally known.

What`s a short-circuit? Little voltage and much current.

Well, at a short circuit you'll find zero volts and any current. You'll
also find this at other places which aren't short circuits, such as
where multiple voltage waves add to zero and at the summing junction of
a perfect op amp. These aren't short circuits, but they are points of
zero voltage. Saying they are all the same is like saying that because
you find water in a creek, any place you find water must be a creek.
What sort of logic is that?

What`s the difference between a physical short and the virtual short?
Nothing except the shunting conductor.

Well, yes. For one thing, waves won't reflect from a virtual short. They
will, from a real short. Another difference is that a real short will
prevent any waves from proceeding beyond it; they pass right through a
virtual short. Good thing, too, or you wouldn't get any power to your
load. Another is behavior at other frequencies and with other
waveshapes. Walt has mentioned another, that a virtual short acts like a
real short only in one direction, even when all the other conditions for
similarity are met.

Is there current flowing at the open-circuit end of the 1/4-wave line
segment? No, the open-circuit won`t support current.

Correct, of course.

If a high-impedance generator of the same frequency were connected to
the virtual short point on the line, would it also be shorted? Yes.
Where? At the virtual short, not the open-circuit at the end of the
line.

Well, yes and no. When you first hook it to the virtual short, it won't
be shorted -- it'll see just the Z0 of the cable. Only when its output
reaches the end of the stub, reflects back, and adds to the forward wave
will it be short circuited. So it's the open end of the line which is
essential to creating the apparent short at the generator.

Now let us, as you say, reason together.

You're pointing out some similarities between a virtual short and a real
one, and giving that as evidence that waves reflect from a virtual
short. So consider a point on a 50 ohm line at which the forward and
reverse waves add to a V/I of, say, 10 ohms, purely resistive to keep it
simple. If you connect a generator (of the correct frequency) at that
point, it will see 10 ohms after things settle down to steady state,
just like your generator saw a short circuit at the "virtual short" in
steady state. So can we conclude that a traveling wave will partially
reflect when it encounters the 10 ohm point? The effective or "virtual"
reflection coefficient can be calculated as -2/3, from which the
reflected wave can be calculated. And, in fact, if we assume that such a
reflection takes place, we can calculate the magnitude and phase of the
resulting wave and, sure enough, there it will really be.

But if the wave does really reflect from this "virtual discontinuity",
we might have a problem. That point is a ways away from the "virtual
short" point (check your Terman diagram if you don't follow), so we have
a partial reflection occurring at this point as well as the full
reflection from the "virtual short". In fact, unless the line is
matched, we'll have reflections from every point along the line, or at
best everywhere except an infinitesimally short spot every half
wavelength! What a mess! Does Terman describe this problem in his book?
A diagram, perhaps, showing the infinite number of partial reflections
taking place all along the line?

No? Well, then, maybe it takes a perfect "virtual short" to get a
reflection, and even a tiny, tiny imperfection will prevent it. So that
would mean that you'd get no reflection at all from a "virtual
almost-short" on even a very slightly lossy line, right? The whole idea
goes to pot when you add even a tiny amount of loss? Or is a little loss
ok? Then we get a full reflection from a good or pretty good "virtual
short", but nothing if it gets too far from perfection. Do me a favor
and check your Terman for an equation or graph which shows just where
this abrupt transition point is (that is, at what "virtual resistance"
the reflection ceases), and why it exists.

Help me out here with my reasoning.

Roy Lewallen, W7EL

Analyzing Stub Matching with Reflection Coefficients
 

Roy Lewallen wrote:
Richard Harrison wrote:
If the virtual short were replaced with a real short, would anything
change? Not a thing except the line voltage distribution diagram would
lose its final 1/4-wavelength.

Don't you consider it a significant difference that no voltage, current,
or power would reach the load?

:-) :-) That's what I asked when you said the
virtual load on a transmitter could be replaced
with a lumped circuit "without changing anything".
--
73, Cecil http://www.w5dxp.com

Analyzing Stub Matching with Reflection Coefficients
 

Roy Lewallen wrote:
It's imperative to
constantly be aware of the range over which those models are valid, and
alert to any situation which might make the model invalid.

Roy, if your model prohibits interaction of coherent
waves, it is seriously flawed.
--
73, Cecil http://www.w5dxp.com

Analyzing Stub Matching with Reflection Coefficients
 

Richard Harrison wrote:
Roy Lewallen, W7EL wrote:
"If you`ll read what I`ve written, you`ll hopefully see that my only
point of contention is with your claim that waves reflect from a
"virtual short". They do not."

Seems to me they do.

I wonder if we can agree that if there is no physical
impedance discontinuity, there can be no reflections?

For instance, given a piece of 50 ohm open-ended coax
with a driving source:

source-----50 ohm coax-----+---1/4WL 50 ohm coax--open

There is a virtual short at point '+' and that virtual
short exists at a point where there is no physical
impedance discontinuity. Can we agree that the forward
wave is unaffected by that virtual short? Can we agree
that the reflected wave is unaffected by that virtual
short? After all, there is absolutely nothing there
that can physically disrupt any waves.

Or given one wavelength of coax being driven by a signal
generator equipped with a circulator load.

SGCL---1/4 WL---x---1/4WL---y---1/4WL---x---1/4WL---open

There are obviously reflections at the open. Are there
any reflections at the virtual open at 'y'? Are there
any reflections at the virtual shorts at 'x'? I would
submit that in the above example, the *only* reflections
in the entire system are happening at the open end of
of the coax and that the virtual shorts and opens are
themselves effects and not the cause of anything (except
maybe arguments) :-)
--
73, Cecil http://www.w5dxp.com

Analyzing Stub Matching with Reflection Coefficients
 

Roy Lewallen wrote:
"Don`t you consider it a significant difference that no voltage,
current, or power would reach the load?"

No, because the example`s load is a perfect open-circuit despite Richard
Clark`s disdain for good insulators. From the shorting point all the way
back to the generator, the voltage distribution is unchanged, real short
or virtual short. (I did not say unchanging.)

Best regards, Richard Harrison, KB5WZI

Analyzing Stub Matching with Reflection Coefficients
 

On Sun, 15 Apr 2007 19:23:56 +1000, Alan Peake wrote:



Walter Maxwell wrote:
.
.


We have thus proved that the virtual short circuit established at the stub point is actually performing as a
real short circuit.
.
.

Walt, W2DU

It is interesting to look at a single short pulse propagating along the
TL. At the stub point, the pulse must encounter a discontinuity in
impedance and therefore there will be a reflection. This can been seen
on a TDR. So there is a real reflection from a stub regardless of
whether or not it is a virtual short.
Alan
VK2ADB

I thank you for that, Alan, because, to continue, when the pulse is replaced with a sine wave, there is also a
reflection from the stub. And when going still further, since the stub presents a susceptance equal to the
line susceptance of the opposite sign at the stub point, some of the sine wave continues along the line and
reaches the mismatched termination, which also produces a real reflection. When the stub is placed at the
proper place on the line relative to the SWR (mismatch), the phase of the waves (voltage and current)
reflected from the load are opposite, respectively, to those of the waves reflected from the stub. The sum of
the voltage waves then yield a resultant reflection coefficient of 180° and the sum of the current waves yield
a resultant of 0°, establishing a short circuit to both sets of reflected waves, but an open circuit to the
source waves.

For people who understand that fields (voltage and current) radiating from two vertical radiators that are of
the same magnitude and of opposite sign result in a null in their radiation pattern in a particular direction
must also understand that for the null to be established there must also be interaction, or interference, or
summing between the fields to cause the travel to cease in the null direction. These people also understand
that energy that was traveling in the null direction has been re-directed in another direction, raising the
level of the energy in that direction from the original level.

IMHO, these people can't have it both ways. If the fields interact, or interfere in space, such as in those
radiated from two radiators, then coherent fields traveling in a transmission line must also interact,
interfere, or sum. This is the concept on which I'm basing my impedance matching analogy using the summation
of reflection coefficients.

Walt

Analyzing Stub Matching with Reflection Coefficients
 

Walter Maxwell wrote:
... then coherent fields traveling in a transmission line must also interact,
interfere, or sum.

There is no doubt that Roy is absolutely wrong when he
asserts that coherent EM waves do not interact. Every
time we tune our antenna tuners to zero reflected energy,
we are causing EM waves to interact following the rules
of *linear* interference. All those waves, inductors,
and capacitors within the antenna tuner are operating
within a linear environment. If they weren't, we would
generate lots of harmonics.

Seems to me, the only valid point of argument is whether
a purely virtual impedance is a cause or an effect.
--
73, Cecil http://www.w5dxp.com

Analyzing Stub Matching with Reflection Coefficients
 

Cecil, W5DXP wrote:
"I wonder if we can agree that if there is no physical impedance
discontinuity, there can be no reflection?"

Terman says on page 118 of his 1955 opus:
"When a wave traveling along a transmission line encounters an isolated
discontinuity, it is partially reflected; i.e., while a portion of the
wave continues to travel down the line, another portion of the wave is
reflected backwards."

On page 119 Terman says:
"A traveling wave passing through such a section (a tapered transmission
line to gradually and continuously change its impedance from one value
to another) will have its ratio of voltage to current transformed in
accordance with the ratio of the characteristic impedances involved."

Abrupt changes in impedance are discontinuities which produce
reflections. These are secondary energy sources.

Total reflection produced the wave distribution diagrammed in Fig. 4-3
on page 91, from an open-circuit on the line.

I see that a virtual short-circuit results 1/4-wave back from the open
circuit and the "short" is repeated 1/2-wave back from the first virtual
short. Through all the virtual shorts and opens, Terman shows the
incident and reflected waves progressing unimpeded.

I believe that if a generator is connected through 1/4-wave of
transmission line to a real short, 360-degrees of phase rotation
presents the generator with a voltage which is almost of the same phase
and magnitude as the generator`s output. Almost no current flows either
into the short or back into the generator. It is similar to connecting
nearly identical transformer windings in parallel.

Hybrid ring isolators are also constructed of 1/4-wave transmission line
sections, and whether a port accepts or rejects energy results from
voltage distributions at the ports, I think.

Best regards, Richard Harrison, KB5WZI

Analyzing Stub Matching with Reflection Coefficients
 

Richard Harrison wrote:
Through all the virtual shorts and opens, Terman shows the
incident and reflected waves progressing unimpeded.

Suggesting that a virtual short or virtual open, by itself,
doesn't cause reflections which is what I think others are
trying to say. Reflections associated with virtual shorts
and virtual opens always occur with some extra ingredients,
like physical impedance discontinuities, the very existence
of which makes one wonder if they might somehow be
associated with the reflections. Reflections always occur
at physical impedance discontinuities.
--
73, Cecil http://www.w5dxp.com

Analyzing Stub Matching with Reflection Coefficients
 

On Apr 14, 9:31 pm, Cecil Moore wrote:
Roy Lewallen wrote:
K7ITM wrote:
. . .
It's a useful visualization tool and design aid; it's a poor analysis
tool at best. At worst, it will lull you into building something that
just won't work, wasting time and resources.

In my opinion, the potential harm can be much worse. If it causes you to
buy into the notion that traveling waves interact in a linear medium,
that opens the door to a whole universe of invalid conclusions.

Here is how Hecht described interference in "Optics":
"... interference corresponds to the *interaction* of two or
more lightwaves yielding a resultant irradiance that deviates
from the sum of the component irradiances."

If traveling waves cannot interact in a linear medium, why
does Hecht say they do indeed interact?

It is exactly that kind of misleading terminology that has caused his
text to fall out of favor among many physics faculty.

To deny the body of laws of physics regarding EM waves from
the field of optics is an example of extreme ignorance.

You really aren't qualified to speak on behalf of the field of optics,
Cecil. You aren't quaified to speak on behalf of Eugene Hecht either,
for that matter. However, I think Dr. Hecht is still around so
perhaps you can persuade him to back you up. Be sure to ask him what
he thinks about the 4th mechanism of reflection.

ac6xg