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Old July 10th 03, 11:20 PM
Jim Kelley
 
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Default Cecil's Math a Blunder



W5DXP wrote:

Jim Kelley wrote:

Waves that propagate to a dissipative load transfer energy to that
dissipative load.



How does the wave about to be dissipated differ from the wave that is
about to be canceled? Hint: they are identical.



The waves themselves are indeed identical, i.e. you can't measure a
difference between them on a transmission line. But one might indeed be
transferring energy while the other is not. For example, take a length
of 50 ohm transmission line with a short at one end. Think about this
circuit with and without a circulator at the source. With the
circulator in circit, energy is transferred from the source to the
circulator load. Without the circulator, the source transfers no energy
to a load. If there are no re-reflections from the source, is there a
measureable difference?

How does the wave know ahead of time whether it is going to encounter a
dissipative load or not? It obviously cannot know ahead of time so all
waves with the same V and I in phase carry the same amount of energy.


This is a misconception on your part. I don't claim anything has to
know anything. That is your claim, and I don't agree. I think it's silly.

A wave about to encounter a dissipative load or the same wave about to be
canceled carry the same amount of energy.


They have the same potential to transfer energy. The do not necessarily
transfer that energy. That depends on boundary conditions.

The wave gives up that energy in
both cases. In the first case it gives up its energy as heat. In the second
case, it gives its energy to the constructive interference.


I understand that is your theory. I even like the sound of your theory.
But you need to understand that it is only your theory. It is an
unproven, and unsupported theory. I've tried to come up with support
for it, but find none. Yes, the energy which would have been reflected
does appear in the transmitted direction. We know this from
conservation of energy - energy incident equals energy transmitted plus
energy reflected. What does not appear to happen is all the bouncing
around you describe.

It can be shown that at the first boundary, two reflected waves
destructively interfere, producing zero reflected energy. It can also
be seen at that boundary that the two waves traveling in the forward
direction yield all of the forward energy due to their constructive
interference.

I understand that if you take the interference term and change its sign
you get the same number, and its not just a coincidence. Yes, you can
say the amount cancelled in the reflected direction equals the amount of
enhancement in the forward direction. But that doesn't mean energy had
to turn around in order to accomplish that. That would only be true if
energy were indeed traveling in the reverse direction to begin with.
When we realize it's not, there's nothing left to account for Cecil.
There's no conservation of energy problem to solve. It's all there,
just as it says in the physics books, and enegineering books as well.
If you want to believe that energy bounces all over the place like, so
be it. Just don't expect to be able to sell paraphysical phenomena as
science.

But as I've tried to tell you many times before, when you can show that
Poynting vector solution changing direction solely as a result of
destructive interference, you'll have a case - and my vote. Until
then, it looks like a non-starter.

73, ac6xg

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Old July 11th 03, 01:01 AM
W5DXP
 
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Jim Kelley wrote:
The waves themselves are indeed identical, i.e. you can't measure a
difference between them on a transmission line. But one might indeed be
transferring energy while the other is not.


How can two waves that are identical possibly be doing different things.
It doesn't make any sense at all. Waves cannot exist without intrinsic
energy!!! If any wave is destroyed, it gives up its energy. There is
simply no other place for the energy to go.

How does the wave know ahead of time whether it is going to encounter a
dissipative load or not? It obviously cannot know ahead of time so all
waves with the same V and I in phase carry the same amount of energy.


This is a misconception on your part. I don't claim anything has to
know anything. That is your claim, and I don't agree. I think it's silly.


That the wave has to be smart falls out from your assertions. Your waves
appear to exist without intrinsic energy so they have to know ahead of time
whether to exist without energy or not. Only a wave existing without energy
could be destroyed and not give up any energy in the process.

But you need to understand that it is only your theory. It is an
unproven, and unsupported theory. I've tried to come up with support
for it, but find none.


Please reference the cross-posting from sci.physics.electromag. Somebody
predicted that those guys would tear me apart. On the contrary, they
tend to agree with me.

It can be shown that at the first boundary, two reflected waves
destructively interfere, producing zero reflected energy. It can also
be seen at that boundary that the two waves traveling in the forward
direction yield all of the forward energy due to their constructive
interference.


Those reflected waves not only destructively interfere, they are
destroyed so they MUST give up any intrinsic energy in the waves.
Nothing else is possible.

Yes, you can
say the amount cancelled in the reflected direction equals the amount of
enhancement in the forward direction. But that doesn't mean energy had
to turn around in order to accomplish that.


Huh??? "Reflected" to "forward" isn't a turn around? Can I have a hit
off whatever you are smokin'? :-)

That would only be true if
energy were indeed traveling in the reverse direction to begin with.
When we realize it's not, there's nothing left to account for Cecil.


If reflected energy from a mismatched load is not traveling in the
reverse direction, pray tell what direction is it traveling in? I
can see you haven't referenced the reflected power flow vector.

There's no conservation of energy problem to solve.


There certainly is in your scenario. The reflected energy doesn't ever
change directions but it somehow joins the forward wave anyway. Can
you spell B-A-F-F-L-E-G-A-B? We know reflected energy is traveling
rearward from a mismatched load. Your mantras cannot change that
fact of physics.
--
73, Cecil http://www.qsl.net/w5dxp



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Old July 11th 03, 07:43 PM
Jim Kelley
 
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W5DXP wrote:

Jim Kelley wrote:
The waves themselves are indeed identical, i.e. you can't measure a
difference between them on a transmission line. But one might indeed be
transferring energy while the other is not.


How can two waves that are identical possibly be doing different things.
It doesn't make any sense at all. Waves cannot exist without intrinsic
energy!!! If any wave is destroyed, it gives up its energy. There is
simply no other place for the energy to go.


Here's a quote from Born and Wolfe, pg. 48, talking about waves
encountering the interface between dense and less dense media:

"Although there is a field in the second medium, it is easy to see that
no energy flows across the boundary. More precisely it will be shown
that, although the components of the Poynting vector in the direction
normal to the boundary is in general finite, its time average vanishes;
this implies that the energy flows to and fro, but that there is no
lasting flow into the second medium."


Please reference the cross-posting from sci.physics.electromag. Somebody
predicted that those guys would tear me apart. On the contrary, they
tend to agree with me.


I tended to agree with what you said there too. You can be very
agreeable when you want to be. :-)

I notice you didn't mention anything about photons dematerializing and
rematerializing over there. Give that one a try. :-)

Those reflected waves not only destructively interfere, they are
destroyed so they MUST give up any intrinsic energy in the waves.
Nothing else is possible.


If you mean they give up their energy to the two waves traveling in the
other direction, then I agree.

Can I have a hit
off whatever you are smokin'? :-)


I've got a pack of Camel lights out in the car. You can have 'em.

That would only be true if
energy were indeed traveling in the reverse direction to begin with.
When we realize it's not, there's nothing left to account for Cecil.


If reflected energy from a mismatched load is not traveling in the
reverse direction, pray tell what direction is it traveling in?


P2 is energy that had been traveling in the reflected direction and was
re-reflected. That's it. Good thing too, because any more than P2 and
P1 and you'd have a perpetual motion machine! :-)

I
can see you haven't referenced the reflected power flow vector.


I've seen you reference it, but I have seen you define one - or write it
out mathematically.

There's no conservation of energy problem to solve.


There certainly is in your scenario. The reflected energy doesn't ever
change directions but it somehow joins the forward wave anyway. Can
you spell B-A-F-F-L-E-G-A-B?


You wrote the bafflegab - whatever that is. I certainly never said
reflected energy joins the forward energy. That's your invention.

73, ac6xg
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Old July 12th 03, 11:20 AM
Richard Harrison
 
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Cecil, W5DXP wrote:
"They", above, are reflected waves. The two waves traveling in the other
direction are forward waves."

Forward waves? Being, I suppose, a voltage wave and a current wave, as
it is impossible to have two voltage waves or two current waves of
identical frequency traveling in the same direction on the same line.
Two identical frequency waves, voltage or current, form a resultant and
merge. They can`t maintain a separate existence and become inseparable
in a single waveform.

Best regards, Richard Harrison, KB5WZI

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Old July 12th 03, 04:09 PM
W5DXP
 
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Richard Harrison wrote:
Cecil, W5DXP wrote:
"They", above, are reflected waves. The two waves traveling in the other
direction are forward waves."

Forward waves? Being, I suppose, a voltage wave and a current wave, as
it is impossible to have two voltage waves or two current waves of
identical frequency traveling in the same direction on the same line.
Two identical frequency waves, voltage or current, form a resultant and
merge. They can`t maintain a separate existence and become inseparable
in a single waveform.


Of course they superpose. In the s-parameter equation, b1=s11*a1+s12*a2,
those two reflected waves superpose to b1. In the s-parameter equation,
b2=s21*a1+s22*a2, those two forward waves superpose to b2. They superpose
at the point where they both appear together for the first time.

But the point is that s11*a1 and s12*a2 are flowing in the direction
of the source when they cancel (interfere destructively). s21*a1 and
s22*a2 are flowing in the direction of the load when they interfere
constructively.
--
73, Cecil http://www.qsl.net/w5dxp



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