Roy Lewallen wrote:
If you put the perfect voltage source and the source resistor into a box
and label it "Source", you have a Source whose impedance perfectly
matches the transmission line. It's a Z0-matched system.

This is exactly like the ham who puts up a random length dipole,
feeds it through a 9:1 balun using 450 ohm ladder-line, and
declares that he has a Z0-match.

50 ohm
100v
source--+--1/2WL 50 ohm feedline--+--200 ohm load

Measuring the 50 ohm SWR at your source output yields 4:1.
Measuring the 200 ohm SWR at your source output yields 1:1.
It is *NOT* Z0-matched to the 50 ohms inside your source.

In a lossless system, a Z0-match to the source impedance
would imply a conjugate match and maximum power transfer.
Your example obviously does not transfer maximum power.
--
73, Cecil http://www.qsl.net/w5dxp

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Cecil Moore wrote:
Ian White GM3SEK wrote:
In other words, Cecil's circulator is a device that takes a perfectly
straight-forward argument... and makes it circular.

Cecil's circulator is a device that gets hot when reflected
waves are present, proving that reflected waves possess
energy and power.

Cecil's TDR allows everything about a reflected pulse to
be known and measured.

Cecil's TV ghosting allows one to see the reflections with
one's own eyes.


None of those observations proves the existence of "waves of energy and
power". They can all be explained more simply and easily in terms of
waves of voltage and/or current (or in circulators and waveguide, waves
of E-field and/or H-field).

The world is still waiting for an explanation of the detailed internal
workings of *any* of those devices, done entirely in terms of "waves of
energy and power". A newsgroup limited to ASCII text is not the best
medium to attempt this... so will the forthcoming QEX article oblige?



I'm a professional teacher

Then this is far more serious than I thought! :-)



--
73 from Ian GM3SEK 'In Practice' columnist for RadCom (RSGB)
http://www.ifwtech.co.uk/g3sek

Cecil, W5DXP wrote:
"The reflection from surface "A" is canceled by an equal magnitude and
opposite phase reflection from surface "B:."

Is this not analogous to what happens on a short-circuited 1/4-wave
stub? The hard short reverses the phase. That, combined with travel to
and from the short, produces a total phase rotation of 360-drgrees.

The result is that the open end of the short-circuited stub, the
incident voltage is in-phase and of the same magnitude (no stub loss) so
that no current flows between the incident and reflected sources.

It is as if one connects identical battery cells ib parallel. The
impedance is, in effect, infinite between sources of identical voltage.

Optical experts must have siezed upon the opposite of this somehow.
Their quarter-wave must have ben terminated in the equivalent of an
open-circuit. This 1/4 wave would accept 100% of light presented at its
surface, or would it need to present 377 ohms at its surface?

I am ignorant of optics and find the analogy difficult to understand.

Best regards, Richard Harrison, KB5WZI

Ian White GM3SEK wrote:
None of those observations proves the existence of "waves of energy and
power". They can all be explained more simply and easily in terms of
waves of voltage and/or current (or in circulators and waveguide, waves
of E-field and/or H-field).

Ian, I haven't seen you, or anyone else, offer proof that
"waves of voltage and/or current" can exist without energy.
I always thought ExH equals the power in the Poynting Vector
for any EM wave whether it be an RF wave or a light wave.

The world is still waiting for an explanation of the detailed internal
workings of *any* of those devices, done entirely in terms of "waves of
energy and power". A newsgroup limited to ASCII text is not the best
medium to attempt this... so will the forthcoming QEX article oblige?

The QEX article's present title is: "An Energy Analysis at
an Impedance Discontinuity in an RF Transmission Line".
It is mostly a review of the principles of the wave
reflection model which was invented before anyone hatched
the stupid idea that waves can exist without energy. :-)
It will be mostly old hat stuff for people who have ever
done an analysis using power reflection coefficients but
that seems to be a lost art lately.

Incidentally, a complete analysis of a Z0-matched impedance
discontinuity can be done using only an energy analysis.
If voltages and currents are actually required, their
magnitudes and phases can be calculated from the energy
components.
--
73, Cecil, http://www.qsl.net/w5dxp

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On Fri, 10 Jun 2005 12:08:48 -0500, (Richard
Harrison) wrote:

I am ignorant of optics and find the analogy difficult to understand.

Hi Richard,

You stand second only to the source that drew this difficulty.

In fact, the "technical" quality of all presentations in this thread
have plunged so far south that even the penguins avoid them.

73's
Richard Clark, KB7QHC

Richard Harrison wrote:
Cecil, W5DXP wrote:
"The reflection from surface "A" is canceled by an equal magnitude and
opposite phase reflection from surface "B:."

Is this not analogous to what happens on a short-circuited 1/4-wave
stub? The hard short reverses the phase. That, combined with travel to
and from the short, produces a total phase rotation of 360-drgrees.
The result is that the open end of the short-circuited stub, the
incident voltage is in-phase and of the same magnitude (no stub loss) so
that no current flows between the incident and reflected sources.

No net current flows but the forward current and reflected current
are constant RMS values as are the forward and reflected voltage
values. The virtual impedance at the mouth of a lossless 1/4WL
shorted stub is (Vfor+Vref)/(Ifor-Iref) where Vfor/Ifor = Z0
and Vref/Iref = Z0. Since Ifor=Iref for a lossless stub, the
impedance is zero. But the current is quite high at the shorted
end of the stub where it is Ifor+Iref.

You can estimate that current if you measure the voltage at
the mouth of the stub. 0.5*V/Z0 will yield the estimated forward
or reflected current.

It is as if one connects identical battery cells ib parallel. The
impedance is, in effect, infinite between sources of identical voltage.

Optical experts must have siezed upon the opposite of this somehow.
Their quarter-wave must have ben terminated in the equivalent of an
open-circuit. This 1/4 wave would accept 100% of light presented at its
surface, or would it need to present 377 ohms at its surface?

The phase of reflections follows a different convention in optics.
And the index of refraction is inversely proportional to Z0. But
a 1/4WL of thin film is akin to 1/4WL of transmission line used
as a series matching section - not parallel but series. This is
essentially how non-glare glass works.

source---50 ohm coax---+---1/4WL 61 ohm coax---+---75 ohm load

50/50 = 1.00, the index of refraction for air
61/50 = 1.22, a good index of refraction for a thin-film
75/50 = 1.50, a good index of refraction for glass

I am ignorant of optics and find the analogy difficult to understand.

I've learned more about reflections and superposition from "Optics",
by Hecht, than from any other single source. I highly recommend it.
--
73, Cecil http://www.qsl.net/w5dxp

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Cecil Moore wrote:

Jim Kelley wrote:

It was a tongue-in-cheek reply, Roy. These problems always seem to
end up with a circulator in them at some point - clearly illustrating
what happens under entirely different circumstances. :-)


The circulator, lossless feedlines of unreasonable length,
Time Domain Reflectometers, TV ghosting, etc. are all tools
for illustration purposes. However, a Z0-matched system is
an ordinary configuration in ham radio and is easy to analyze
since no reflected energy is allowed to reach the source. The
conservation of energy and momentum rules dictate where the
energy must go in such a case. We can debate why the reflected
energy is 100% re-reflected but there is no question that it
*is* 100% re-reflected because none reaches the source and
there are only two directions in a transmission line.

I wish we were able to discuss this without it becoming so
confrontational. As I've said to you many times, you've got 98% of this
thing nailed to a tee. But that 2% is a major error from a physical
standpoint. Energy is not flowing in the way you describe it. Power
doesn't flow at all, but that's a different discussion. When the fields
cancel, as in the anti-reflective/impedance matching scenario, energy is
not conveyed in the reflected direction. There is no conservation of
energy problem until you claim that that energy from cancelled waves IS
moving in the reflected direction. Once you make that claim, you're
forced to imagine a way for it to reverse its course, and that's where
the problem lies. We've been over this a hundred times and you just
refuse to accept it. It violates physics. It violates Maxwells
equations. It's wrong, and I hate arguing with you about it, but as a
fellow enthusiast I advise you not to print that part of your article in
QEX. It's an absurdity in the midst of brilliance, not unlike myself. ;-)

73, Jim AC6XG

Jim Kelley wrote:
I wish we were able to discuss this without it becoming so
confrontational.

Me too. Please treat me the way you would like to be treated
and we will get along just fine. Oh, and thanks for the doorknob
caps. I use them now instead of stubs.

As I've said to you many times, you've got 98% of this
thing nailed to a tee. But that 2% is a major error from a physical
standpoint. Energy is not flowing in the way you describe it. Power
doesn't flow at all, but that's a different discussion.

I've been very careful to consider power as the measured energy flowing
past a point even though many others, including the IEEE, present the
Power Flow Vector right along with the current phasor. But as far as
EM wave energy and power are concerned, they are virtually
interchangeable. EM wave energy always travels at the speed of light -
imagine that. :-) The energy in an EM wave can only be stored as an
EM wave traveling at the speed of light as in a delay line. It cannot
be put into an RF battery and used as needed for an old wives' tale.
Walter Johnson recognized the close relationship between EM energy and
EM power and actually presented the "conservation of power" principle in
his book. The only difference between EM energy and EM power is time.
The joules in the joules/sec must be conserved.

When the fields
cancel, as in the anti-reflective/impedance matching scenario, energy is
not conveyed in the reflected direction.

Of course, and I never said it was. That's your straw man left over from
our last argument. If the fields cancel, there is no energy available
for a rearward flow. That's why you cannot measure reflected power in
the coax on a Z0-matched system. All the reflected energy has changed
direction. But we know the internal reflected energy has traveled in
the rearward direction. What altered its momentum in that rearward
direction?

Energy must be conveyed from the far surface or the fields would not
cancel. That's called the internal reflection and its rearward flow
can be detected. Since the internal reflection wave possesses momentum
in the rearward direction, something has to reverse that momentum. We
can disagree on the mechanism that reverses the momentum of the
internal reflected wave, but the wave is there and can be detected.

The entire reason I cannot accept what you say is that you have never
given a reasonable explanation of what happens to the momentum of
the rearward-traveling internal reflection wave. Want to try again?
Please concentrate on that one topic. Resolving the momentum in the
internal reflection wave is the only way to convince me that I am wrong.
I think that is the concept that convinced the QEX editors that I was
right. If one accepts the conservation of momentum principle, then
one has to accept a cause for the reversal of that momentum.

The internal reflection wave is illustrated in section 9.4.1 of
"Optics". It has obviously traveled in the rearward direction across
the width of the thin-film. What happens to its momentum and energy
if it gets canceled?

There is no conservation of
energy problem until you claim that that energy from cancelled waves IS
moving in the reflected direction.

Something has to reverse the momentum in the wave reflected from the
far surface, i.e. the "internal reflection". If you don't know what
that is, reference section 4.3 in "Optics". You have never given a
reasonable explanation of what reverses the momentum of the internal
reflection which, I'm sure you would agree, doesn't appear in the
glare so it had to be reversed. Exactly how was it reversed?

Once you make that claim, you're
forced to imagine a way for it to reverse its course, and that's where
the problem lies.

No, since it is obvious that the momentum of that internal reflection
wave reverses, a reversing mechanism is necessary. I've presented a
mechanism and you haven't.

We've been over this a hundred times and you just
refuse to accept it.

Yes, and you refuse to accept it. Momentum doesn't change by magic.
There has to be a physical reason. I have presented my take on that
physical reason. You have presented no explanation. Between having
an explanation and having none, guess what my choice will be?

Please just explain how the momentum in the internal reflection wave
gets reversed within the boundary conditions of the classical wave
reflection model. No quantum physics, please.

So let's concentrate on that narrow topic of what happens to the
momentum in the internal reflection wave, shall we? Here's a
diagram where 'n' is the index of refraction and the internal
reflection from surface 'B' is shown:

| |
laser--------air----|----thin-film-------|----glass---------
n-1.0 | n=1.2222 | n=1.4938
A --reflection----B

The internal reflection that I have been talking about occurs when
the forward wave in the thin-film encounters surface 'B'. The
reflectance is 0.01 at that surface so one percent of the forward
irradiance will be reflected. It will have momentum in the rearward
direction. Please tell us what alters the momentum of that reflection
such that it doesn't appear as glare at surface 'A' and instead
reverses direction and joins the forward wave.
--
73, Cecil http://www.qsl.net/w5dxp

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Ian White GM3SEK wrote:

In other words, Cecil's circulator is a device that takes a perfectly
straight-forward argument... and makes it circular.


Always thrusting staight into that very round bullseye, Ian.

73
Tom
K0TAR