On 26 Jul 2005 14:54:49 -0500, wrote:

Not sure which indices you are now referring to: n_3=4.0 or 4.04? With
4.0, there is zero reflected light. With 4.04 there is a small amount.

Hi Tor,

It doesn't matter one iota. You can attempt to achieve a perfect
1:2:4 correlation, or any similar relationship that satisfies the
square laws so demanded.

The point of the matter is with such design characteristics fulfilled,
then each interface reflects/transmits in the same proportion. With
each reflecting/transmitting in the same proportion, the second
interface, by the actions of the first, must have less incident upon
it. It then follows that the same proportion of less incident is not
enough to "Totally Cancel" the first interface reflection.

Observing the conservation of energy at the first interface:
X = 0.89X + 0.11X
Observing the conservation of energy at the second interface:
0.89X = 0.792X + 0.098X

0.098X 0.11X

What is even more obvious is that following the second interface, you
have lost roughly 20% of that incident upon the first interface.

There is not enough energy at the second interface, reflected back, to
"Totally Cancel" the energy in the reflection of the first interface.
There can be no other outcome.

That difference yields first interface reflection products that have:
1200 TIMES MORE POWER THAN THE SUN!

73's
Richard Clark, KB7QHC

A B
1000mW----| |
|----888.9mW----|
111.1mW----| |----790.1mW
|----98.76mW----|
87.79mW----| |
|----10.97mW
^^^^^

I would not even dream of superposing powers. Perhaps you
don't understand the role of the interference term in the

Then don't draw diagrams like above labeled with powers of
multiple reflections bouncing around in the same media

Eugene Hecht would be surprised to hear you say such since he
talked about the powers (irradiance) of various reflections in
the same media.

Only when dealing with the interference of TWO waves.
He works with electric field when discussing anti-reflection
layers. See 9.7.1

I'm not saying the power equation you give is wrong, but it
is correct for just two waves. With more than two waves, it
obviously becomes more complicated


Tor
N4OGW

Richard Clark wrote:

On 26 Jul 2005 14:54:49 -0500, wrote:
Not sure which indices you are now referring to: n_3=4.0 or 4.04? With
4.0, there is zero reflected light. With 4.04 there is a small amount.

It doesn't matter one iota.

Of course it matters. The 1,2,4 combination meets the necessary and
sufficient conditions for the elimination of reflections in medium 1
assuming a steady-state lossless system, i.e. perfectly matched.

The point of the matter is with such design characteristics fulfilled,
then each interface reflects/transmits in the same proportion. With
each reflecting/transmitting in the same proportion, the second
interface, by the actions of the first, must have less incident upon
it.

That is simply not true, Richard. You obviously have not read my
Melles-Griot web posting and/or simply don't understand it. The
forward power in medium 2 is greater than the forward power in
either medium 1 or medium 3. You keep making that same mistake
over and over and over. Why don't you simply take time to
understand the truth?

What is even more obvious is that following the second interface, you
have lost roughly 20% of that incident upon the first interface.

But you have gained the energy involved in the wave cancellation
in medium 1. The forward power in medium 2 is higher than it is
in medium 1. You don't actually believe the BS by some gurus on
this newsgroup that standing waves don't contain any energy, do you?

There is not enough energy at the second interface, reflected back, to
"Totally Cancel" the energy in the reflection of the first interface.
There can be no other outcome.

Only in your mind, Richard, only in your mind. The outcome in reality
is complete cancellation of the reflections in medium 1 assuming
1,2,4 indices of refraction.

That difference yields first interface reflection products that have:
1200 TIMES MORE POWER THAN THE SUN!

Somebody get the net!
--
73, Cecil http://www.qsl.net/w5dxp

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Tor, by misunderstanding my posting, you have just fed the monster. :-)

I see :O Well, I can't get everyone to understand basic physics.

Tor
N4OGW

Ah but you must avoid upsetting GOD (oops I mean Richard.). A simple answer
about grounding given to a guy a while back put him into apoplectic fits,
necessitating the doubling of his meds. You see, "ground" ISN'T "ground".
That isn't "clever" enough for Richard. He promptly claimed that rf currents
flowing in a ground system were a "bad thing' and proceeded to "explain" it
all to us. I wish my elmer 30 years ago had been as "clever" as Richard. All
these books on the shelves simply are irrelevant, because nothing can be
simple, you see... It must be complicated, even if it IS wrong........

"W5DXP" wrote in message
oups.com...
Richard Clark wrote:
Needless to say, that same first interface is going to conserve energy
by the total of refraction and reflection being equal to the energy
incident upon it. I will skip that to allow ALL of this second
reflection to "try" to totally cancel the first reflection:
110mW - 92mW
or
18mW

Richard, if you follow the leads of other gurus on this newsgroup,
you will 'ploink' me, never admit a mistake, and dupe yourself
into believing that you were right all along. Your "Extreme Failures
of Poor Concepts" could have been avoided if you guys had just read and

understood the following two quotes that I have posted multiple times.
What is it about, "... all 'lost' reflected intensity will appear as
enhanced intensity in the transmitted beam.", that you guys don't
understand?

www.mellesgriot.com/products/optics/oc_2_1.htm

"Clearly, if the wavelength of the incident light and the thickness
of the film are such that a phase difference exists between reflections

of p, then reflected wavefronts interfere destructively, and overall
reflected intensity is a minimum. If the two reflections are of equal
amplitude, then this amplitude (and hence intensity) minimum will be
zero."

"In the absence of absorption or scatter, the principle of conservation

of energy indicates all 'lost' reflected intensity will appear as
enhanced intensity in the transmitted beam. The sum of the reflected
and
transmitted beam intensities is always equal to the incident intensity.

This important fact has been confirmed experimentally."

http://micro.magnet.fsu.edu/primer/j...ons/index.html

"... when two waves of equal amplitude and wavelength that are 180-
degrees out of phase with each other meet, they are not actually
annihilated. All of the photon energy present in these waves must
somehow be recovered or redistributed in a new direction, according to
the law of energy conservation ... Instead, upon meeting, the photons
are redistributed to regions that permit constructive interference, so
the effect should be considered as a redistribution of light waves and
photon energy rather than the spontaneous construction or destruction
of light."
--
73, Cecil http://www.qsl.net/w5dxp

Cecil Moore wrote:

Powers obey the equation P1 + P2 + 2*(P1*P2)cos(theta)

I thought you said nature doesn't obey math models. :-)

P1 + P2 would be superposing powers. I didn't do that.

P1 + P2 + 2*sqrt(P1*P2)cos(theta) is NOT superposing powers.

You do understand that interference is the result of the superposition
of waves traveling in the same direction, right?

Eugene Hecht would be surprised to hear you say such since he
talked about the powers (irradiance) of various reflections in
the same media.

Again Cecil, you're really in no position to speak on his behalf. He
might not think that highly of your doing so in fact. Perhaps you
should obtain his permission first.

I did not superpose powers. I used Hecht's equations for powers
(irradiance).

Actually, the fact that those equations appear in his book (as well as
just about any other optics book) does not convey any rights of
ownership or invention to the author. They were around a long time
before Dr. Hecht.

The only difference in irradiance and power is
the unit-area which, for our example, is arbitrary.

You should show him your equation with all the power terms (P1-P4) in
it! You had some of the powers going one way, and the others going the
other way. You needed them all to add up to some particular number, so
you just added them. - or maybe subtracted some of them. I don't recall
exactly.

wrote:

W5DXP wrote:
Eugene Hecht would be surprised to hear you say such since he
talked about the powers (irradiance) of various reflections in
the same media.

Only when dealing with the interference of TWO waves.
He works with electric field when discussing anti-reflection
layers. See 9.7.1

I'm not saying the power equation you give is wrong, but it
is correct for just two waves. With more than two waves, it
obviously becomes more complicated

medium 1 medium 2 medium 3
A B
1000mW----| |
|----888.9mW----|
111.1mW----| |----790.1mW
|----98.76mW----|
87.79mW----| |
|----10.97mW

I appologize if the diagram wasn't clear. It is extremely
difficult to draw graphics in ASCII.

But I only ever deal with TWO waves. Above in medium 1,
the two waves are the 111.1 wave and the 87.79 wave.
There are only TWO rearward-traveling waves there. Hecht's
irradiance equation works for those TWO waves.

The TWO waves in medium 2 are the 888.9 wave and the
10.97 wave. There are only TWO forward-traveling waves
there. Hecht's irradiance equation works for those
TWO waves.

Since I am not trying to calculate the power in the standing
waves, there are only ever TWO values of power to deal with.

Sorry for the confusion.
--
73, Cecil http://www.qsl.net/w5dxp

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Jim Kelley wrote:
You do understand that interference is the result of the superposition
of waves traveling in the same direction, right?

Of course, there are two superposition terms in each direction
represented by the two S-parameter equations.

b1 = s11*a1 + s12*a2 toward the source

b2 = s21*a1 + s22*a2 toward the load

The two terms in the first equation interfere with each other
toward the source. The two terms in the second equation interfere
with each other toward the load. When b1=0, the system is matched,
and total destructive interference exists toward the source. In
that case, total constructive interference exists toward the load.

Actually, the fact that those equations appear in his book (as well as
just about any other optics book) does not convey any rights of
ownership or invention to the author. They were around a long time
before Dr. Hecht.

Didn't mean to imply that he owns them. "Hecht's equations" simply
means they appear in his book. They were around a long time before
anyone on this newsgroup called them "gobblegook". The interference
equations also appear in Chipman's 1968 "Transmission Lines".

You should show him your equation with all the power terms (P1-P4) in
it! You had some of the powers going one way, and the others going the
other way.

Nope, you are confused. All powers are going the same direction for any
individual equation in my article. Destructive interference energy in one
direction equals constructive interference energy in the opposite direction.
If you will wade through the transient buildup to steady-state of the recent
examples, you won't be confused anymore by direction. During the transient
buildup phase, when the first destructive interference occurs toward the source,
an equal amount of constructive interference energy flows toward the load -
same energy, just flowing in the opposite direction. If destructive
interference energy in one direction equals P3+P4, constructive interference
energy in the opposite direction equals P3+P4 to satisfy the conservation
of energy principle.

Perhaps you are remembering the equation (Pload = Pfor - Pref) where Pfor
and Pref are flowing in different directions. But that equation doesn't
appear in my article.
--
73, Cecil http://www.qsl.net/w5dxp

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wrote:
You CANNOT superimpose POWERS, ...

Hi Tor, I appologize that the earlier energy flow chart
was not drawn as well as it might have been. I've expanded
that chart in the following and provided the power management
equations associated with the superposition of EM fields. The
'X' event is the initial reflection event at the first
discontinuity, point A. The 'Y' event is the initial reflection
event at the second discontinuity, point B. The 'Z' event is the
initial re-reflection event back at point A.

100W 1/4WL
XMTR---50 ohm---+--------100 ohm---------+---200 ohm---200 load
feedline A feedline B feedline

Just before the first reflection arrives back at A

Pfor(50) | |
100W----X |
X----+ |
11.11W----X \ Pfor(100) |
Pref(50) | +----88.89W----Y
| Y----79.01W
| +----9.876W----Y Pfor(200)
| / Pref(100) |
| --+ |
| |

Just after the first reflection arrives back at A

Pfor(50) | Pfor(100) |
100W----X 109.67W |
X----+ |
11.11W----X \ P1 |
P3 | +----88.89W----Y
| Y----79.01W
Pref(50) | +----9.876W----Y Pfor(200)
0.139W | / Pref(100) |
Z----+ |
8.779W----Z |
P4 Z----1.097W |
| P2 |

P1 = Pfor(50)*(1-rho^2), P2 = Pref(100)*rho^2

P3 = Pref(100)*(1-rho^2), P4 = Pfor(50)*rho^2

How much destructive interference between wave(P3) and wave(P4)?

Pref(50) = P3 + P4 - 2*sqrt(P3*P4) = 11.11W + 8.779W - 19.75W = 0.139W

19.75W of destructive interference, negative since cos(180)=-1

How much constructive interference between wave(P1) and wave(P2)?

Pfor(100) = P1 + P2 + 2*sqrt(P1*P2) = 88.89W + 1.097W + 19.75W = 109.67W

19.75W of constructive interference, positive since cos(0)=+1

Destructive interference energy = constructive interference energy
(toward the source) (toward the load)

It follows that |P1*P2| = |P3*P4| in order to satisfy the conservation
of energy principle.
--
73, Cecil, http://www.qsl.net/w5dxp


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Tor, N4OGW wrote:
"You CANNOT superimpose POWERS, or even talk about the "power" of
various reflections in the same media."

It is done all the time. P=Esquared / R = Isquared x R.

On page 99 of the 1955 edition of "Electronic and Radio Engineering"
Terman writes:

"The standing-wave ratio S is one means of expressing the magnitude of
the reflectiom coefficient; the exact relation between the two is:

S=1+absolute value of the reflection coefficient / 1-absolute value of
the reflection coefficient

or

absolute value of the reflection coefficient= S-1/S+1

Standing-wave ratio=S=Emax/Emin

This definition of standing-wave ratio is sometimes called voltage
standing-wave ratio (VSWR) to distinguish it from the standing-wave
ratio expressed as a power ratio, which is (Emax/Emin)squared."

Best regards, Richard Harrison, KB5WZI