Jim Kelley wrote:

"W5DXP" wrote in message
Yes, it does, which is not at all if we already know the
reflected irradiance which is a given.

Obviously, the load determines the boundary conditions and so it is not
irrelevant. You said that it was, and that's not correct. The load
impedance is what determines the reflectivity. Go ahead and disagree.

I probably should have used the word "redundant" instead of "irrelevant".
If the reflected power (irradiance) in a Z0-matched system is a given,
then the value of the load is redundant information and is NOT needed
for a solution.


:-) Yes, very technical. If a math question was posed, I must have
missed it.


What is the superposed sum of the
two above waves?


Zero.


What happens to the intrinsic energy pre-
existing in those waves before they cancel each other?


The answer the intrinsic energy in the waves where the waves exist is stored
in the transmission line, and nothing happens to energy where waves don't
exist. The waves in question don't convey energy from the source to the
load - obviously because they don't propagate from the source to the load.
It ain't rocket science - as you're so fond of saying.


Using the values above, calculate the rate of flow of energy equal to
V*I. That's how much energy is involved in your dilema here.


The rate of flow of energy has to be 100 joules/sec since the energy
in those two waves cannot stand still and cannot be destroyed.


The rate of flow of energy "in" those two waves is not 100 Joules per
second.


We
already know that the energy in those two waves joins the forward-
traveling power wave toward the load.


The energy does travel forward, but not by way of those two waves.

73, Jim AC6XG

Jim Kelley wrote:
What happens to the intrinsic energy pre-
existing in those waves before they cancel each other?

The answer the intrinsic energy in the waves where the waves exist is stored
in the transmission line, and nothing happens to energy where waves don't
exist. The waves in question don't convey energy from the source to the
load ...

Of course not because they are destroyed at the cancellation point. But
the energy in the canceled waves is indeed conveyed to the load.

The rate of flow of energy has to be 100 joules/sec since the energy
in those two waves cannot stand still and cannot be destroyed.

The rate of flow of energy "in" those two waves is not 100 Joules per
second.

Of course it is. We know that 50W of reflected power from a mismatched
load has not been re-reflected and tried to continue to flow toward the
source. We know it never gets past the impedance discontinuity. There's
only one thing that can stop a wave in its tracks without dissipation of
energy. That's another wave traveling in the same direction with equal
amplitude and opposite phase as explained on the Melles-Griot web page.
We are therefore forced to deduce that the other wave exists and indeed
it is predicted by Pfwd(|rho|^2) and the s-parameter term, s11*a1. It
doesn't last very long because it is instantaneously canceled as it
is reflected but we know it has to exist and indeed Pfwd1(|rho|^2)
equals 50W, the exact amount of energy we need to accomplish the
wave cancellation process.

The energy does travel forward, but not by way of those two waves.

If it doesn't come from the two canceled waves, where does it come from?
Besides the two canceled waves, only Pfwd1(1-|rho|^2) and Pref2(|rho|^2)
exist and there is not enough energy in those two other wave components
to account for the magnitude of Pfwd2. Hecht, in _Optics_ says the
constructive interference energy (flowing toward the load) comes from
the destructive interference event (toward the source). If the canceled
waves contain no energy then there is no destructive interference energy -
without which constructive interference is not possible.
--
73, Cecil, W5DXP

W5DXP wrote:

Jim Kelley wrote:
Why should someone believe what you say when there's
no evidence you've even worked the problem yourself?

Here's an interesting reply to one of my postings on
sci.physics.electromag:

Cecil said:
Seems to me that destructive interference event occurring in the direction
of the source feeds energy to the constructive interference event occurring
in the direction of the load.

The reply was:
Yes, just so. In fact, the phase change/non-change when the waves are
reflected/transmitted by low-to-high or high-to-low interfaces ensures
that this will always happen - if in phase on one side, they must be out
of phase on the other.

Which is exactly what I say in my article.
--
73, Cecil, W5DXP

Hi Cecil,

Obviously he's describing the phase relationships between reflected and
transmitted waves, and the fact that destructive interference occurs on
one side of the boundary and destructive interference occurs on the
other side of the boundary. I agree, and I've always told you I liked
that part of your paper.

73, Jim AC6XG

Cecil, W5DXP wrote:
"That was the definition of a traveling wave in a Zo enviroment,---."

All this discussion over the insipid subject of flat lines?

A virtual short appears in a line with 100% reflection. The virtual
short appears where the sum of incident and reflected waves produces a
concurrence of zero volts and maximum current, just as in a true short.
The line has zero loss to enable good repetitions of an actual short or
open.

Best regards, Richard Harrison, KB5WZI

Jim Kelley wrote:
Obviously he's describing the phase relationships between reflected and
transmitted waves, and the fact that destructive interference occurs on
one side of the boundary and destructive interference occurs on the
other side of the boundary. I agree, and I've always told you I liked
that part of your paper.

In the earlier example, the destructive interference occurs in the
direction of the source. The constructive interference occurs in
the direction of the load.

So what do you think generates the destructive interference energy?
What is the origin of the destructive interference energy that feeds
the constructive interference event? Exactly what wave components are
involved in the generation of the destructive interference energy?

In other words, how does the reflected power Poynting vector get
turned around at the Z0-match impedance discontinuity?
--
73, Cecil, W5DXP

Jim Kelley wrote:
You're making it sound as though the energy gets to the load by some
means other than waves.

No, I'm not. The energy in the canceled waves reverses direction and
joins the forward wave. That's what I said over on sci.physics.electromag
and those guys are amazed that anyone would disagree with that assertion.

In other words, the disappearance of two waves during a wave cancellation event
can result in reflected energy coherent with those two canceled waves. I don't
find that in the literature anywhere. Do you know of a reference?

The reply was:
I'm curious as to what kind of objections people had. It all makes perfect sense to me.

So the disappearance of two waves during a wave cancellation event and
a subsequent energy reflection makes perfect sense to real physicists.
Not one person on that newsgroup objected to my assertion.

There is no flow of energy from source to load via those waves, ...

Of course there is. All energy comes from the source. Therefore, the
energy in the reflected power Poynting vector originates from the source
and is reflected by the load. ERGO, that energy has flowed from the
source to the load. *ALL* energy incident upon the load comes from the
source, even the energy rejected by the load as reflected energy. The
above statement is simply a steady-state shortcut mantra that bears no
resemblance to reality.

That's your story, and you're sticking with it. I know. ;-)

You have not refuted it. Until you provide an iota of proof to
the contrary, I'll be sticking with it. Incidentally, your mantras
are not proof. Try uttering your mantras on sci.physics.electromag
and see what happens.

Like we haven't been through this already? The superpostion of V1 and
V2 accounts for all the energy that moves from source to load.

We've been through all this before. There is not enough energy in
|V1|^2/Z02 and |V2|^2/Z02 to support the constructive interference
between those two waves. Therefore, an equal magnitude of destructive
interference must be occurring somewhere else - according to Hecht.
Therefore, the destructive interference energy is generated by the
wave cancellation event between the two rearward-traveling reflected
waves just as described on the Melles-Griot web page and agreed
to by experts over on sci.physics.electromag.

How about you reply to my latest posting on sci.physics.electromag
so we can obtain opinions from some real experts? I predict our
differences would be settled in a matter of days on that newsgroup.
--
73, Cecil, W5DXP

W5DXP wrote:
So what do you think generates the destructive interference energy?
What is the origin of the destructive interference energy that feeds
the constructive interference event? Exactly what wave components are
involved in the generation of the destructive interference energy?

Not parsing 'generating destructive interference energy'. I know the
words individually, but I don't know what they're supposed to mean in
such a concatination. Not familiar with the expression. Sorry.

In other words, how does the reflected power Poynting vector get
turned around at the Z0-match impedance discontinuity?

In still other words, until you can answer that you're arguing an
unsupportable theory.

ac6xg

Richard Harrison wrote:
A virtual short appears in a line with 100% reflection. The virtual
short appears where the sum of incident and reflected waves produces a
concurrence of zero volts and maximum current, just as in a true short.

A physical short causes the voltage to go to zero. If a virtual
short causes the voltage to go to zero, what causes the virtual
short?

The virtual short cannot cause itself so ...

Either the virtual short causes the voltage to go to zero in
which case: What causes the virtual short?

Or the voltage going to zero causes the virtual short in which
case: What caused the voltage to go to zero?
--
73, Cecil, W5DXP

Jim Kelley wrote:


W5DXP wrote:

So what do you think generates the destructive interference energy?
What is the origin of the destructive interference energy that feeds
the constructive interference event? Exactly what wave components are
involved in the generation of the destructive interference energy?

Not parsing 'generating destructive interference energy'. I know the
words individually, but I don't know what they're supposed to mean in
such a concatination. Not familiar with the expression. Sorry.

:-) Non-response #127. Why am I not surprised?

In other words, how does the reflected power Poynting vector get
turned around at the Z0-match impedance discontinuity?

In still other words, until you can answer that you're arguing an
unsupportable theory.

:-) Non-response #128. Why am I not surprised?
--
73, Cecil, W5DXP

Cecil, W5DXP wrote:
"What caused the voltage to go to zero?"

Equal and opposite voltages. Reaction to connecting wires together
generates an opposite voltage which adds to zero with the incident
voltage. Current doubles at the short.

1/4-wave back from the short, a virtual open circuit appears. Cecil
claims this open circuit does not impede current.

1/4-wave short-circuit stubs are used as metallic insulators. They have
the characteristics of resonant circuits constructed of a
parallel-connected capacitor and coil, a very high impedance at
resonance.

From King, Mimno, and Wing, "Transmission Lines, Antennas, and Wave
Guides" page 29:

"A short-circuited line, one-quarter wavelength long at the desired
output frequency may be connected across the output terminals of a
transmitter or across the antenna feeder at any point without placing
much load on the transmitter at this fundamental or desired output
frequency, since at this frequency such a section has an impedance
ideally infinite, actually about 400,000 ohms."

Since I = E/Z, how much current do you think will flow into 400,000
ohms?

King, Mimno, and Wing`s impedance might scale down to only 33,333 ohms
on a 50-ohm line, still high, as they may have been considering a
600-ohm line.

All my radar texts say resonant transmission line sections have the same
characteristics as resonant lumped circuits and I trust them because the
radar circuits using tuned transmission lines to route the signal, work.

Best regards, Richard Harrison, KB5WZI