Richard Clark wrote:
And true for ALL accumulated reflections there after. Reflections do
not add any energy to the cup when the first interface is draining it
more quickly.

If the first interface is a Z0-match, it is not "draining it"
at all. That's what you are missing. The examples have been
for matched systems. If you change your medium 3 to an index
of refraction of 4.0 instead of 4.04, your example is *perfectly
matched* and the first interface is NOT draining the system
because all reflections are canceled. 100% of the source power
is being delivered through the first interface because of that
wave cancellation. At that point it is being joined by the
reflected power so the forward power in the second medium is
greater than the source power. In the second medium, forward
power equals source power plus reflected power. In your example,
source power is one watt, forward power in the second medium is
1.125 watt, and reflected power in the second medium is 0.125 watt.

My analysis allowed ALL of the energy in the reflection from the
second interface ( 0.098X) to combine with the first reflection
(0.11X). This total superposition ...

Power cannot be superposed. If you are going to deal with power,
you need to use the interference equations for power calculations.
Or alternately, convert the powers to voltages, superpose the
voltages, and then calculate the power from the superposed voltage.
That's what most RF engineers do and it works well.

As an optical engineer, I've dealt with the harsh reality of this myth
of total reflection cancellation.

Since you don't understand the physics, it is no wonder that you
failed to accomplish total reflection cancellation. And total
reflection cancellation is easier to accomplish in a transmission
line than it is with light in air.
--
73, Cecil http://www.qsl.net/w5dxp

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Richard Harrison wrote:

Jim Kelley wrote:
"That`s because power doesn`t propagate, and hasten to add - neither do
Poynting vectors."

Words are not objects. They merely represent objects. We use
abstractions for brevity and clarity. Even the best say: "--the Poynting
vector or power density (watts per square meter)." See Kraus` 3rd
edition of "Antennas" page 73, under "Power Patterns".

So I guess that must be where he talks about power and Poynting vectors
that pretend to be electromagnetic fields and propagate along waveguides?

73, ac6xg

Richard Clark wrote:

On Fri, 29 Jul 2005 16:58:29 -0700, Jim Kelley
wrote:

It is quite evident that through the actions of the first interface,
that there is less energy incident upon the second interface.
Further, given that both interfaces operate with identical reflective
and transmissive properties, it follows the second interface could not
reflect enough to totally negate the reflections of the first.

True for any one reflection.


Hi Jim,

And true for ALL accumulated reflections there after. Reflections do
not add any energy to the cup when the first interface is draining it
more quickly.


Yes, that was my point. You seemed to have momentarily overlooked that
detail.

My analysis allowed ALL of the energy in the reflection from the
second interface ( 0.098X) to combine with the first reflection
(0.11X).

Single (first) reflection only.

This total superposition was both more than generous, and at
the same time very unlikely; and yet with this generous allowance
there is still excess reflection from the first interface.

Your generosity is hardly the issue. Previously reflections are more to
the point.

Hence for
something less than total superposition of ALL energies, it hardly
bodes a better yield in total cancellation - the energy just isn't
there in the first place. 0.098X 0.11X is the simple economics of
the balance.

Sounds good. It's wrong, but the sound of it is good nevertheless.

As an optical engineer, I've dealt with the harsh reality of this myth
of total reflection cancellation.

Failure can tend to make some people bitter. ;-)

Basically these claims are for first year students where demanding too
much inquiry would push them into switching majors to Business school.
Simple optics with simple, ordinary glasses exhibit quite useful
results, but they do not embody a proof. To anyone following the math
of my presentation, it is quite obvious what WOULD tend towards a more
complete cancellation - and such a subtle shift in the formula
diverges only slightly from the choir book hymn. It's not that hard
when the interface ratios drive the answer.

Another good sounding thing.

73, ac6xg

Cecil Moore wrote:

Roy Lewallen wrote:

Since you say that power superposes, we should expect power waves to
add and cancel just like voltage waves.


Strawman Alert!!! Richard H. did NOT say power superposes.
It is obvious that what he was disagreeing with was the
statement: "You CANNOT ... even talk about the "power"
of various reflections in the same media."

I think you should be allowed to talk about whatever you want, Cecil.
That sort of thing can be highly therapeutic. ;-)

73, ac6xg

W5DXP wrote:

Why don't you visit your local power company, call a meeting of their
engineers, and inform them that there is no power flowing in their
transmission lines? :-)

Are you talking about the guys with the clipboards, or the guys with the
hard hats?

73, ac6xg

On Sat, 30 Jul 2005 13:08:40 -0700, Jim Kelley
wrote:

Hence for
something less than total superposition of ALL energies, it hardly
bodes a better yield in total cancellation - the energy just isn't
there in the first place. 0.098X 0.11X is the simple economics of
the balance.

Sounds good. It's wrong, but the sound of it is good nevertheless.

Hi Jim,

As the total energy returning from the second interface is 0.098X, not
0.11X, and not even all of that passes through the back of the first
interface, where do you come up with the remaining difference to
accomplish a complete cancellation?

Even if an infinite summation of ALL reflections kept any energy from
transiting the second interface (a generous allowance not likely to be
observed anywhere), then it cannot exceed what energy was initially
introduced into the system. With that generous allowance, the
infinite summation can only equal 0.098X which remains less than the
reflection from the first interface of 0.11X. Without that generous
allowance, it must be something less, which can only increase the
total, uncancelled reflection product. Either way, "total
cancellation" is not total.

73's
Richard Clark, KB7QHC

Jim Kelley wrote:

W5DXP wrote:
Why don't you visit your local power company, call a meeting of their
engineers, and inform them that there is no power flowing in their
transmission lines? :-)

Are you talking about the guys with the clipboards, or the guys with the
hard hats?

I've seen power engineers with clipboards and hardhats. Their
title is "Power Engineer". What will they think when you tell
them that power generators don't generate power, power transmission
lines don't transfer power, and power meters don't meter power?
They may all end up in a white coat because of you.

The rest of the world simply does not share your definition of
"power". Did you see my question? If the power associated with
an EM pulse is not in the pulse, where is it? A power meter
registers power during the pulse but nothing before and after
the pulse. The power was there during the pulse. Where was it
before the pulse arrived? Where did it go after the pulse left?
--
73, Cecil http://www.qsl.net/w5dxp

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As the total energy returning from the second interface is 0.098X, not
0.11X, ...

The total energy returning from the second interface is 0.125 watts.
Your solution is incorrect. Why don't you take a poll to see who
agrees with you?

Even if an infinite summation of ALL reflections kept any energy from
transiting the second interface (a generous allowance not likely to be
observed anywhere), then it cannot exceed what energy was initially
introduced into the system.

It cannot exceed the energy. It can certainly exceed the steady-state
source power (joules/sec) and it does. Some energy is sourced during
the transient stage that does not reach the load. It is stored in the
second medium as standing waves. The forward power in the first medium
is one watt with no reflections. The forward power in the second medium
is 1.125 watts. The reflected power in the second medium is 0.125 watts.
The forward power in the third medium is one watt with no reflections.

Have you never seen the forward power in a transmission line exceed
the source power? If the system is Z0-matched and the SWR on a
1/4WL transformer is 5.83:1, the steady-state forward power on the
1/4WL transformer is *double* the steady-state source power.

Either way, "total cancellation" is not total.

In your n=1,2,4 example, reflected wave cancellation is total in
medium 1 (assuming a lossless system). If n2=sqrt(n1*n3), the
reflected wave cancellation is total.
--
73, Cecil http://www.qsl.net/w5dxp

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Jim Kelley wrote:
"That`s because power doesn`t propagate, I hasten to add - neither do
Poynting vectors.:

Some world-class experts disagree with Jim. Here is a sample quotation
using the words "power flow".

From E.M. Purcell writing about "Antenna Gain and Receiving Cross
Section" on page 19 of "Radar System Engineering" edited by Louis M.
Ridenour:
"If the transmitting antenna were to radiate energy isotropically-that
is, uniformly in all directions-the power flow through unit area at a
distance R, from the antenna could be found by dividing P, the total
radiated power, by 4piRsquared."

Best regards, Richard Harrison, KB5WZI

On Sun, 31 Jul 2005 15:52:19 -0500, (Richard
Harrison) wrote:

Jim Kelley wrote:
"That`s because power doesn`t propagate, I hasten to add - neither do
Poynting vectors.:

Some world-class experts disagree with Jim. Here is a sample quotation
using the words "power flow".

From E.M. Purcell writing about "Antenna Gain and Receiving Cross
Section" on page 19 of "Radar System Engineering" edited by Louis M.
Ridenour:
"If the transmitting antenna were to radiate energy isotropically-that
is, uniformly in all directions-the power flow through unit area at a
distance R, from the antenna could be found by dividing P, the total
radiated power, by 4piRsquared."

Best regards, Richard Harrison, KB5WZI

Richard, in addition to Ridenour, I quote from Reflections 1 and 2,
Chapter 8, "A semantic problem with the term "power flow" also fuels
the erroneous belief that reflected power is fictitious. This brings
us to the question, "Does power flow?" To help us understand the
answer, let's examine an analogy that involves current.
When we talk about "current flow," we take the meaning for
granted. However, does current really flow? The basic electricity
sections of engineering textbooks (also The ARRL Handbook) say that
current does not flow--charge flows. Current is defined as the
quantity of charge flowing past a point per unit time. However, once
we leave basic electricity and move on to circuit analysis, the term
"current flow" is used almost exclusively--and
yet we know exactly what is meant.
The same problem exists with the term "power flow." Engineering
textbooks define power as the "quantity of energy passing a point per
unit time." Thus, power does not flow--energy flows. However, except
when reciting the definition of power, textbooks and journals on wave
propagation use the term "power flow" almost exclusively, with only an
occasional use of "energy flow." As with "current flow," we know what
is meant because of the common usage which generally overshadows the
strict definition."
Perhaps this explanation will satisfy Jim, but perhaps not. We'll see.

Walt, W2DU