Conservation of Energy
 

Cecil, W5DXP wrote:
" ***Before they cancel***, they will burn a hole in the black paper
proving that they contain energy."

Anyone who can obtain a copy of Arnold B. Bailey`s "TV and Other
Receiving Antennas" can turn to page 158 to see a diagram of how waves
interfere. Interaction between components of the same signal from two
sources are shown in the figure to produce a combined signal which
varies regularly with the height above the earth. This has a
corresponding effect on the signal received by an antenna placed at a
particular altitude.

The average signal at all altitudes is the same as if there was no
interference. But, that is because the layers of strengthened signal are
exactly offset by the layers of reduced signal strength.

Best regards, Richard Harrison, KB5WZI

W5DXP wrote:

Jim Kelley wrote:

Well, I've already provided the quote from: "Fields and Waves in
Communication Electronics", by Ramo, Whinnery, and Van Duzer.


The topic of converstaion was: Poynting Vector Energy Anaylsis of An
Impedance Discontinuity.


Yes it was. So what happens to the reflected power Poynting Vector
(that the above reference says exists) when it encounters the
impedance discontinuity?

100W source--50 ohm feedline--+--1/2WL 150 ohm feedline--50 ohm load

The reflected power Poynting vector on the left side of the impedance
discontinuity is zero. The reflected power Poynting vector on the right
side of the impedance discontinuity is 33.33W pointing toward the left.
For about the tenth time, what happens to that reflected power Poynting
vector?

W5DXP wrote:

Jim Kelley wrote:

Well, I've already provided the quote from: "Fields and Waves in
Communication Electronics", by Ramo, Whinnery, and Van Duzer.


The topic of converstaion was: Poynting Vector Energy Anaylsis of An
Impedance Discontinuity.


Yes it was. So what happens to the reflected power Poynting Vector
(that the above reference says exists) when it encounters the
impedance discontinuity?

100W source--50 ohm feedline--+--1/2WL 150 ohm feedline--50 ohm load

The reflected power Poynting vector on the left side of the impedance
discontinuity is zero. The reflected power Poynting vector on the right
side of the impedance discontinuity is 33.33W pointing toward the left.
For about the tenth time, what happens to that reflected power Poynting
vector?

Like I said, you haven't done a Poynting vector energy analysis. If you
had, I wouldn't have to solve the problem for you. Why should someone
believe what you say when there's no evidence you've even worked the
problem yourself?

73, ac6xg

W5DXP wrote:

I have told you time after time, that the load impedance doesn't matter
and the feedline length doesn's matter.

It matters in exactly the same way that the index of refraction of the
glass substrate matters.

50V at zero degrees and 1A at 180 degrees = 50W
50V at 180 degrees and 1A at zero degrees = 50W


Non-sequitur. The topic of conversation had been: Examples where
"something is actually destroyed and gives up all it's energy."


Dodging another technical question, I see, and we all know why.

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

You know that the sum of the two above waves is zero volts and
zero amps, i.e. destruction of the two waves in their direction
of travel. Where does the energy go when the waves go to V=0 and
I=0 in their direction of travel.

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.

It seems that everything you don't know the answer to is a
non-sequitur. I fully understand your attitude.

Another one of your technical remarks?

ac6xg

Jim Kelley wrote:

W5DXP wrote:
I have told you time after time, that the load impedance doesn't matter
and the feedline length doesn's matter.

It matters in exactly the same way that the index of refraction of the
glass substrate matters.

Yes, it does, which is not at all if we already know the
reflected irradiance which is a given. If the reflected
irradiance is a given, you don't need to know the index of
refraction of the glass. In fact, if it is unknown, you can
calculate it from the given reflected irradiance.

50V at zero degrees and 1A at 180 degrees = 50W
50V at 180 degrees and 1A at zero degrees = 50W

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

It ain't rocket science. What is the superposed sum of the
two above waves? What happens to the intrinsic energy pre-
existing in those waves before they cancel each other?

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. We
already know that the energy in those two waves joins the forward-
traveling power wave toward the load.
--
73, Cecil, W5DXP

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

Peter Brackett wrote:
Cecil:
It certainly does resolve any problems. All waves contain energy. All
waves transfer that energy when they are dissipated or radiated or when
they are destroyed by wave cancellation.

No they don't, and...

Changing the energy in a TEM wave to heat is obviously a transfer of
energy. Radiating TEM energy from an antenna is obviously a transfer
of energy. When TEM wavefronts are destroyed by wave cancellation in
a transmission line, they cease to exist and must therefore transfer
their energy somewhere. There is simply no other possibility.

At an impedance discontinuity match point with reflections, the
destructive interference energy is transferred in phase to the
constructive interference event occurring in the other direction.
Please see the "Mental Exercise" thread cross posted from
sci.physics.electromag

What *exactly* is your definition of a "wave"?

"traveling wave - The resulting wave when the electric variation
in a circuit takes the form of translation of energy along a
conductor, such energy being always equally divided between
current and potential forms." i.e. P=|V*I| and Z0=V/I
--
73, Cecil http://www.qsl.net/w5dxp



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Cecil, W5DXP wrote:
"---energy being always equally divided between current and potential
forms---."

Except at opens and shorts, real and virtual, where amps or volts for an
instant however brief are brought down to zero, which is a transfer of
all energy to either the electric or magnetic field.

Best regards, Richard Harrison, KB5WZI

"W5DXP" wrote in message
...
Jim Kelley wrote:

W5DXP wrote:
I have told you time after time, that the load impedance doesn't matter
and the feedline length doesn's matter.

It matters in exactly the same way that the index of refraction of the
glass substrate matters.

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.

:-) 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

Richard Harrison wrote:

Cecil, W5DXP wrote:
"---energy being always equally divided between current and potential
forms---."

Except at opens and shorts, real and virtual, where amps or volts for an
instant however brief are brought down to zero, which is a transfer of
all energy to either the electric or magnetic field.

That was the definition of a traveling wave in a Z0 environment, Richard,
and opens and shorts are not a Z0 environment. Can you give an example of
a virtual short and explain how it develops?
--
73, Cecil, W5DXP

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

Richard Harrison wrote:

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

Equal and opposite voltages.

What caused the rearward-traveling current to go to zero at the
same time?

The problem is one of cause and effect. You cannot say the virtual
short causes the voltage and current wave conditions and then say
the voltage and current wave conditions causes the virtual short.

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

If the virtual short causes reflections, why doesn't the virtual open
cause reflections?

1/4-wave short-circuit stubs are used as metallic insulators.

That's nice, but we are not discussing physical shorts. We are
discussing virtual shorts.
--
73, Cecil http://www.qsl.net/w5dxp



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It's actually easy to predict the (resistive) impedance seen looking
into a shorted quarter wavelength line, or any odd multiple. The
impedance is simply the Z0 of the line divided by the loss in nepers.
One neper is about 8.7 dB, so the impedance is about 8.7 * Z0 / dB loss.

All else being equal, the impedance gets higher as frequency increases.
That's because the length of a quarter wave stub decreases in inverse
proportion to frequency, while loss (up to 1 - 10 GHz or so, where
conductor loss dominates) increases only as the square root of
frequency. So the impedance of a stub increases as the square root of
frequency. For example, a quarter wave stub, made from solid
polyethylene dielectric coax (VF = 0.66) at 3.5 MHz is about 46 ft. That
length of RG-58 has a loss of about 0.3 dB, so the impedance looking
into a quarter wave stub of RG-58 at 3.5 MHz is about 1450 ohms. Quite a
far cry from the textbook's example of 400 k ohms or Richard's
extrapolation to 33 k ohms! An RG-58 stub at 350 MHz, or 100 times the
frequency, would have an input impedance of about 14,500 ohms.

A more typical VHF example would be a quarter wave of RG-8 at two
meters. It would be about 13.4 inches long and a loss of about 0.03 dB,
for an input Z of about 14,500 ohms.

Incidentally, the formula I'm using is actually on the same page of King
et al's text as the 400 k ohm value Richard quotes. They say the 400 k
value is for "a reasonably low-loss line" -- to get 400 k ohms with a
600 ohm line, the loss would have to be about 0.013 dB.

The input impedance of an open circuited quarter wavelength line or
shorted half wavelength line is Z0 times the loss in nepers, or about
Z0 * dB loss / 8.7.

I actually ran into a case where the finite resistance of an open stub
became a problem, and it illustrates the hazard of blindly following a
"rule of thumb" without checking to see under what conditions it's
valid. The "Field Day Special" antenna, similar to a ZL special, can be
fed at the center of either element. I connected a one wavelength
transmission line to the center of each element, and fed one or the
other to switch directions, leaving the other line open circuited. When
RG-58 was used, the current diverted into the finite resistance of the
open stub disturbed the element current enough to very significantly
degrade the front/back ratio. The lines were one wavelength at 14 MHz,
or about 46 feet. Loss was a seemingly trivial 0.8 dB, but that means
that the input impedance was only about 540 ohms! 400,000 or even 33,000
would be an awfully poor estimate! Changing to 300 ohm twinlead solved
the problem. (Although 300 ohm twinlead can easily be as lossy as RG-58
when wet, the higher Z0 resulted in an adequately high stub impedance
even when it was wet.)

Roy Lewallen, W7EL

Richard Harrison wrote:

. . .
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.
. . .

Roy Lewallen wrote:
All else being equal, the impedance gets higher as frequency increases.

Double the frequency and you have a shorted 1/2WL stub. Isn't the
impedance of a shorted 1/2WL stub lower than the impedance of a
shorted 1/4WL stub?
--
73, Cecil http://www.qsl.net/w5dxp



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Richard Harrison wrote:

Cecil, W5DXP wrote:
"What caused the rearward traveling current to go to zero?"

The rearward-traveling current did not change phase on reflection from a
load of too-few ohms.

True, there is a re-reflection event exactly as you describe. But the
reflection coefficient for that re-reflection event is not 1.0 so not
all of the reflected power gets re-reflected. What happens to the rest
of it? How do you explain the discrepancy between the physical reflection
coefficient less than 1.0 and the apparent 100% re-reflection of reflected
energy at a Z0-match point?
--
73, Cecil http://www.qsl.net/w5dxp
"One thing I have learned in a long life: that all our science, measured
against reality, is primitive and childlike ..." Guess who said that.



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I'm not sure if that's meant to be humorous, or if you really did
misinterpret what I meant. In case it's the latter, I'll amplify.

If a quarter wavelength shorted stub at frequency f1 has impedance R1,
then a quarter wavelength shorted stub at frequency f2 will have an
impedance of R1 * sqrt(f2 / f1), if it's made of the same type of
transmission line, and the frequency is in the range where conductor
loss dominates (below about 1 - 10 GHz for typical coax).

Roy Lewallen, W7EL

W5DXP wrote:
Roy Lewallen wrote:

All else being equal, the impedance gets higher as frequency increases.


Double the frequency and you have a shorted 1/2WL stub. Isn't the
impedance of a shorted 1/2WL stub lower than the impedance of a
shorted 1/4WL stub?

Roy Lewallen wrote:
I'm not sure if that's meant to be humorous, or if you really did
misinterpret what I meant. In case it's the latter, I'll amplify.

Whoops, I really did misinterpret what you meant. When you said, "All
else being equal," I inferred that you were including the physical
length of the stub. Changing the length of the stub didn't seem to
match the condition, "All else being equal,". Sorry.
--
73, Cecil http://www.qsl.net/w5dxp



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I apologize for the lower-case "C" used to initiate the last "Cecil" in
my most recent posting. My finger must have been tired and weak when I
hit the shift-key. No offense was intended.

It`s fixed phase and high transmitter volts which thwart energy return
to the transmitter according to my profs over 50 years ago in
circumstances posed in the problem I was considering. It seems
reasonable when the stand-off happens at an impedance discontinuity.

Best regards, Richard Harrison, KB5WZI

Richard Harrison wrote:

Cecil, W5DXP wrote:
"But the reflection coefficient for that re-reflection event is not 1.0
so not all of the reflected power gets re-reflected."

100% re-reflection was a given. It is my assumption and I`m sticking
with it..

And my question is, in the absence of a physical reflection coefficient
magnitude of 1.0, what causes 100% re-reflection?

I`m tired of explaining something cecil refuses to accept.

I accept everything you have explained, Richard, but you have not answered
my question. What causes 100% re-reflection? Please be specific. A voltage
to current ratio is a result and not the cause of anything.
--
73, Cecil http://www.qsl.net/w5dxp
"One thing I have learned in a long life: that all our science, measured
against reality, is primitive and childlike ..." Albert Einstein



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Richard Harrison wrote:
It`s fixed phase and high transmitter volts which thwart energy return
to the transmitter according to my profs over 50 years ago in
circumstances posed in the problem I was considering. It seems
reasonable when the stand-off happens at an impedance discontinuity.

Why doesn't the same thing happen 1/2WL back up the transmission line
where the impedance is exactly the same?
--
73, Cecil http://www.qsl.net/w5dxp
"One thing I have learned in a long life: that all our science, measured
against reality, is primitive and childlike ..." Albert Einstein



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Cecil, W5DXP wroyte:
"Why doesn`t the same thing happen 1/2WL back up the line where the
impedance is exactly the same?"

Uniform distribution of inductance and capacitance 1/2WL back up the
line.

John E. Cunningham in "The Complete Broadcast Antenna Handbook" wrote as
I recall, an attempt to explain the reflection operation on a
transmission line as caused by impedance discontinuities.

If I recall, he said a short on a line vitiates capacitance at that
point. It shorts it out. Increased current in the final segment of the
line inductance sets off the reflection. The extraordinary volts
generated, reverse the wave direction and reverse the phase of the
volts. What happens to the current is a turnaround sooner than later,
but it still comes and goes in the same phase.

The discontinuity is essential to the turnaround. Lacking a reduction
point or growth point in inductance to capacitance ratio, you lack an
essential to generate the reversed signal. That`s why the virtual short
does not turn the signal around. It has no discontinuity.

Best regards, Richard Harrison, KB5WZI

Richard Harrison wrote:
If I recall, he said a short on a line vitiates capacitance at that
point. It shorts it out.

But there is NO actual short. The ratio of voltage to current is not
even zero. The ratio of voltage to current at a Z0-match point is Z0.

The discontinuity is essential to the turnaround.

Exactly! That's how b1 = s11*a1 + s12*a2 gets to equal zero. s11*a1 and s12*a2
are both reflections from the physical impedance discontinuity. No "virtual
short" required. s11*a1 and s12*a2 are reflections that are equal in magnitude
and opposite in phase.
--
73, Cecil http://www.qsl.net/w5dxp
"One thing I have learned in a long life: that all our science, measured against
reality, is primitive and childlike ..." Albert Einstein



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Cecil, W5DXP wrote:
"The ratio of voltage to current at a Zo-match point is Zo."

There is one voltage at a match point produced by the waves arriving
from opposite directions.

There is one power, the power delivered by the transmitter, feeding the
match point. The same power total flows from the match point to the
load. It`s the forward power minus the reflected power.

Though there is but one voltage at the match point, it is clear that the
currents in lines connected at the match are not equal because the Zo`s
of the connected lines are not equal.

We need the reflected power to reconcile our mismatch. The mismatch is a
catalyst that makes a reflection happen.

Say Zo is 150 ohms. 100 watts into Zo is 123.5 volts. 10 watts facing
the 100 watts presents 38.7 volts. Power direction is determined by the
123.5 volt source, not by the 38.7 volt reflection, at the match point.

Best regards, Richard Harrison, KB5WZI

Richard Harrison wrote:
Say Zo is 150 ohms. 100 watts into Zo is 123.5 volts.

I get 122.5 volts. :-)

10 watts facing
the 100 watts presents 38.7 volts. Power direction is determined by the
123.5 volt source, not by the 38.7 volt reflection, at the match point.

Essentially all you are saying is that net energy flows from the source
to the load. I agree but it has nothing to do with "facing off". Coherent
voltage waves flowing in opposite directions are unaffected by each other.
Coherent voltage waves flowing in the same direction merge into one wave.
The events that happens at an impedance discontinuity are reflections which
causes coherent waves to flow in the same direction and therefore merge. It
is explained in my part 1 article on my web page.
--
73, Cecil http://www.qsl.net/w5dxp
"One thing I have learned in a long life: that all our science, measured against
reality, is primitive and childlike ..." Albert Einstein



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Cecil, W5DXP wrote:
"I get 122.5 volts."

Dang cheap Chinese calculator operator!
Cecil is correct.

Cecil also wrote:
"Coherent voltage waves flowing in opposite directions are unaffected by
each other."

I agree, except at an impedance bump which shakes things up. Incoherent
voltage waves flowing in opposite directions are also unaffected by each
other. :-)

Best regards, Richard Harrison, KB5WZI

Richard Harrison wrote:

Cecil, W5DXP wrote:
"I get 122.5 volts."

Dang cheap Chinese calculator operator!
Cecil is correct.

Cecil also wrote:
"Coherent voltage waves flowing in opposite directions are unaffected by
each other."

I agree, except at an impedance bump which shakes things up.

Yes, and what we are discussing is exactly how an impedance bump
shakes things up.
--
73, Cecil http://www.qsl.net/w5dxp
"One thing I have learned in a long life: that all our science, measured against
reality, is primitive and childlike ..." Albert Einstein



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