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

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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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

W5DXP wrote:

Jim Kelley wrote:
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.

This is just another one of your *NET* energy statements.

Actually it's just a quote from an optics book describing, among other
things, how a field can cross a boundary without energy necessarily
flowing across it. It happens to be the same thing I've been trying to
explain to you.

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

I did, and no one objected. But we don't need quantum physics to
explain HF RF waves.

I searched the entire thread. Not one mention of photons
dematerializing or rematerializing.

Upon what
technical point do we still disagree? If the rearward-traveling
waves give up their energy to the two waves traveling in the
forward direction, the energy has changed direction, by definition.

Actually you have it backwards. If energy had been reflected, the
forward waves would have given up their energy to the reflected waves.
The reflected waves only give up their energy in a sense by not taking
it in the first place.

There is not enough energy in P1 and P2 to support Pfwd2. Constructive
interference energy must be added into Pfwd2.

P1 and P2 are the only terms in the equation, Cecil. Obviously they
have enough energy to produce the correct answer. But I know what you
are getting at, Cecil, and it is interesting. But the example Joseph
Legris gave you (the same one I described previously) should illustrate
for you the problem with what you are saying. It's apparent from that
example that energy from the lasers goes directly to the constructive
interference side without having to come from somewhere else (other than
the source). And it's also true that that only occurs when destructive
interference is also occurring elsewhere.

According to Hecht, that
constructive interference energy can only originate from destructive
interference.

What he's saying is that you can't have one without the other.

73, Jim AC6XG

Jim Kelley wrote:

Actually it's just a quote from an optics book describing, among other
things, how a field can cross a boundary without energy necessarily
flowing across it. It happens to be the same thing I've been trying to
explain to you.

But nobody cares about your *NET* energy quotations. This thread is
about the energy components underlying the steady-state solution.
What is it about that statement that you don't understand? If you
want to discuss net energy, please start another thread.

I searched the entire thread. Not one mention of photons
dematerializing or rematerializing.

It was in another thread, Jim.

Actually you have it backwards. If energy had been reflected, the
forward waves would have given up their energy to the reflected waves.

The forward wave *DOES* give up (Pfwd*|rho|^2) energy to the reflected wave
at a mismatched load. I'm sorry to pull an argumentum ad populum on you
but everybody on this newsgroup knows that except you.

The reflected waves only give up their energy in a sense by not taking
it in the first place.

More Bafflegab. Please research the reflected power Poynting vector from
the previously provided reference of Ramo and Whinnery and get back to
us.

P1 and P2 are the only terms in the equation, Cecil.

Which leads you to believe that P1+P2 P1+P2? What is it about destructive
interference energy feeding energy to the constructive interference event
that you do not understand?

Assume Wave(1) is associated with P1 and Wave(2) is associated with P2.

P1 + P2 + 2*Sqrt(P1*P2) is the proper equation when Wave(1) and Wave(2)
are in zero phase. On the other side of the impedance discontinuity we
have a similar equation: P3 + P4 - 2*Sqrt(P3*P4). Agree so far?

Pref1 = P3 + P4 - 2*Sqrt(P3*P4) = zero reflected power toward the source.

So P3 + P4 = 2*Sqrt(P3*P4) = 100% destructive interference during wave
cancellation, i.e. Wave(3) and Wave(4) are destroyed. The destructive
interference equals the total of P3+P4 *because* those waves were
destroyed. Wave(3) and Wave(4) give up their combined energy components
as destructive interference energy.

Hecht tells us that the magnitude of the destructive interference must
equal the magnitude of the constructive interference. Guess what? He is
right! |2*Sqrt(P3*P4)| does indeed equal |2*Sqrt(P1*P2)| and we can simply
say that according to Hecht, |2*Sqrt(P1*P2)| comes from |2*Sqrt(P3*P4)|.
The destructive interference event supplies the power for the constructive
interference event. I'm getting tired of typing the following quote from
_Optics_, by Hecht:

"The principle of Conservation of Energy makes it clear (to everyone except
Jim) that if there is constructive interference at one point, the 'extra'
energy at that location must have come from ... destructive interference
somewhere else." I don't know how Hecht could have said it any clearer.
The constructive interference energy is supplied from the destructive
interference event.

So the valid equation becomes P1+P2+2*Sqrt(P3*P4) = P1+P2+P3+P4, i.e.
all the power in the entire Z0-matched system winds up flowing toward
the load but we already knew that. How you can argue with that fact
is simply unbelievable.

What he's saying is that you can't have one without the other.

Yes, and the destructive interference event *supplies* energy to the
constructive interference event. It certainly cannot be vice versa
as you are attempting to imply. Hecht was very careful to say that
constructive interference requires a source of energy from a
destructive interference event. See above quote.
--
73, Cecil http://www.qsl.net/w5dxp



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Roy Lewallen wrote:
So where do you draw the required surface over which to integrate the
Poynting vector and thus obtain the "reflected power"?

As you know, a one-dimensional transmission line is a lot easier to
work with than the 3D space involved with radiation from antennas.
In fact, all we need for a transmission line Poynting vector is
a plane. Since the field area of the coax is constant and known, we
can treat it as a constant normalized value. It allows us to come
up with a simplified version of what you posted. Quoting from,
"Fields and Waves in Communications Electronics", by Ramo, Whinnery,
and Van Duzer: "... we are often most interested in the ratio of power
in the reflected wave to that in the incident wave, and this ratio is
given by ...

Pz-/Pz+ = |rho|^2"

where Pz- is the reflected power Poynting vector, Pz+ is the forward
power Poynting vector, and rho is the reflection coefficient. Pretty
simple stuff that I learned 45 years ago. Again from the same book:

Pz+ = (E+)x(H+) and Pz- = -(E-)x(H-)
--
73, Cecil http://www.qsl.net/w5dxp



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Jim Kelley wrote:

W5DXP wrote:
But nobody cares about your *NET* energy quotations.

Nobody said anything about *NET* energy in the quotation.

But it is obvious that they are referring to *NET* energy. Please
stop trying to sneak *NET* energy stuff into the argument.

The forward wave *DOES* give up (Pfwd*|rho|^2) energy to the reflected wave
at a mismatched load. I'm sorry to pull an argumentum ad populum on you
but everybody on this newsgroup knows that except you.

At a single boundary, yes.

If it exists at a single boundary, then it exists.

Which leads you to believe that P1+P2 P1+P2?

I don't believe that.

If you believe that all the power comes from P1 and P2 in the following
equation, then yes, you do believe that.

P1 + P2 + 2*Sqrt(P1*P2) = Pfwd2

Hint: the 2*Sqrt(P1*P2) power does NOT come from P1 and P2. If it did
you would be violating the conservation of energy principle and creating
free energy. That term is the *INTERFERENCE TERM* which is supplied from
the destructive interference going on on the other side of the impedance
discontinuity. Why do you have such a problem with the concept of
conservation of energy?

The destructive
interference equals the total of P3+P4 *because* those waves were
destroyed.
Wave(3) and Wave(4) give up their combined energy components
as destructive interference energy.

There exists no mechanism for them to "give up" their energy in the way
you describe.

The mechanism is conservation of energy involving destructive interference
feeding energy to the constructive interference event. The guys over on
sci.physics.electromag can explain it to you.
--
73, Cecil http://www.qsl.net/w5dxp



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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

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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Jim Kelley wrote:
I've never disputed the solutions produce the same number. I disputed
the validity of the approach you took to arrive at the equation.

I guess I owe the group an apology. The guys over on sci.physics.electromag
are inferring things that I never meant to imply. I thought the meaning
of my words was clear but I guess that's not the case. I apologize to
anyone I accused of deliberately misinterpreting what I said.
--
73, Cecil http://www.qsl.net/w5dxp



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