Jim Kelley wrote:
Just cut the BS, Cecil. In order to prove your assertion you must first
be able to describe how two co-linear, coherent waves that are 180
degrees out of phase at every point along their path and traveling in
the same direction can under those circumstances at any time produce
measureable energy. In addition, you must be able to measure it. Let
me know when you do.

It is an indirect measurement, Jim. Given the s-parameter
equation, b1 = s11*a1 + s12*a2 = 0, s11 is not zero, a1
is not zero, s12 is not zero, and a2 is not zero. Although
HP cannot measure those quantities either, they tell us
that |s11*a1|^2 is in watts, e.g. 100 watts. They tell us
that |s12*a2|^2 is in watts, e.g. 100 watts. When all energy
is accounted for, it is obvious that those 200 watts are no
longer in the direction of the source but have changed
direction toward the load. This ain't rocket science.

If reflections are eliminated toward the source by wave
cancellation, the reflected energy is redistributed back
toward the load just as explained on the Melles-Groit and
FSU web pages. If it weren't headed for the source in the
first place, they wouldn't say it was "REDISTRIBUTED". If
200 joules/sec disappear toward the source and there are
only two directions in a transmission line, do you really
want to tell us that you can't figure out in which
direction those joules go? Do you need help from my
10 year old grandson?

You clearly fail to understand the process defined by the
wave reflection distributed network model. Until you are
in a position to discredit that model, you are just blowing
smoke.
--
73, Cecil http://www.w5dxp.com

Cecil Moore wrote:

Jim Kelley wrote:

Just cut the BS, Cecil. In order to prove your assertion you must
first be able to describe how two co-linear, coherent waves that are
180 degrees out of phase at every point along their path and traveling
in the same direction can under those circumstances at any time
produce measureable energy. In addition, you must be able to measure
it. Let me know when you do.


It is an indirect measurement, Jim.

:-) Sure thing. Like I said, let me know when you measure energy in
the canceled waves.

ac6xg

Jim Kelley wrote:

Cecil Moore wrote:
It is an indirect measurement, Jim.

:-) Sure thing. Like I said, let me know when you measure energy in
the canceled waves.

Let me know when you figure out an explanation for the
reversal of momentum in those reflected waves. So far,
you have absolutely refused to provide any explanation.
--
73, Cecil http://www.w5dxp.com

Cecil Moore wrote:

Jim Kelley wrote:


Cecil Moore wrote:

It is an indirect measurement, Jim.


:-) Sure thing. Like I said, let me know when you measure energy in
the canceled waves.


Let me know when you figure out an explanation for the
reversal of momentum in those reflected waves. So far,
you have absolutely refused to provide any explanation.

The momentum in reflected waves changes direction upon reflection.
What part of that do you need to have explained?

So, back to you. Let's hear more about your measurement of canceled
waves.

ac6xg

Jim Kelley wrote:
The momentum in reflected waves changes direction upon reflection. What
part of that do you need to have explained?

What causes 100% reflection when the power reflection
coefficient (reflectance) is only 0.5?
--
73, Cecil http://www.w5dxp.com

Cecil Moore wrote:

Jim Kelley wrote:

The momentum in reflected waves changes direction upon reflection.
What part of that do you need to have explained?


What causes 100% reflection when the power reflection
coefficient (reflectance) is only 0.5?

If you would just work the problem the hard way, you would see where
you're misconception lies. Any given wave front will never reflect
100% from a surface which is only 50% reflective, no matter how
vicious your insults become, how may URLs you cut and paste, or how
furiously you wave your hands. But when you work the problem as has
been suggested you will see how energy gets from source to load. It
does not rely on macroscopic layman's explanations or mathematical
shortcuts in order to get there.

The only energy "lost" by partial reflection in the process is that
which is reflected back toward the source or stored in the system
during the transient period. The sum of all the partial reflections
equals the energy stored in the system (less the portion of energy
admitted to the load or reflected back to the source). After the
transient period, no energy is reflected back to the source, and the
energy entering the system from the source equals the energy existed
the system through the load.

I know that you understand the difference between potential and
kinetic energy in mechanics. Please try to consider that the concepts
are no less valid in electromagnetism. (Note that we even use the
word 'potential' to describe voltage.)

73, ac6xg

Jim Kelley wrote:
Any given wave front will never reflect 100%
from a surface which is only 50% reflective,

That's all you have to say, Jim, to defeat your argument.
If you would stop refusing to perform a simple calculation
involving my example at:

http://www.w5dxp.com/thinfilm.GIF

you would understand. When the internal (0.009801w)
wave reflection arrives at t3 and interferes with the
(0.01w) external reflection wave, what is the resulting
reflected power back toward the source. When you calculate
the results and realize that it is not 0.01 - 0.009801 watts,
you will begin to understand the nature of interference.
--
73, Cecil http://www.w5dxp.com

Jim Kelley wrote:


Cecil Moore wrote:

Jim Kelley wrote:

Just cut the BS, Cecil. In order to prove your assertion you must
first be able to describe how two co-linear, coherent waves that are
180 degrees out of phase at every point along their path and
traveling in the same direction can under those circumstances at any
time produce measureable energy. In addition, you must be able to
measure it. Let me know when you do.


It is an indirect measurement, Jim.

:-) Sure thing. Like I said, let me know when you measure energy in
the canceled waves.


At the risk of being both a dullard and messing up all the fun, does
not every destructive interference have to be balanced by a constructive
interference, which in turn leads to a condition of "Okey dokey?"

A canceled wave needs a reinforced wave, and then nothing is lost,
nothing is gained.

- 73 de Mike N3LI -

Michael Coslo wrote:
At the risk of being both a dullard and messing up all the fun, does
not every destructive interference have to be balanced by a constructive
interference, which in turn leads to a condition of "Okey dokey?"

Yep, any destructive interference toward the source is exactly
offset by constructive interference toward the antenna. If one
takes time to calculate the component phasor voltages on both
sides of a Z0-match located away from the source, the constructive
and destructive interference is obvious.

A canceled wave needs a reinforced wave, and then nothing is lost,
nothing is gained.

Exactly. The reflected energy that appears to be lost as
destructive interference in the direction of the source
when a Z0-match is achieved, is recovered in the forward
wave as constructive interference energy traveling toward
the antenna.
--
73, Cecil http://www.w5dxp.com

Michael Coslo wrote:
At the risk of being both a dullard and messing up all the fun, does
not every destructive interference have to be balanced by a constructive
interference, which in turn leads to a condition of "Okey dokey?"

A canceled wave needs a reinforced wave, and then nothing is lost,
nothing is gained.

- 73 de Mike N3LI -

Hi Mike -

One could suggest a number of different possible scenarios in which
nothing is lost or gained. But an impossible scenario is one which
violates thermodynamic principles. Another might describe phenomena
which is not in accord with Maxwell's equations. One should therefore
feel comfortable discarding any description which is inconsistent with
both thermodynamics and Maxwell's equations.

73, ac6xg