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