Roy Lewallen wrote:
Cecil Moore wrote:
We are not going to get anywhere until you admit there is 68
joules/sec in the feedline that haven't yet made it to the load. Once
you admit that fact, everything else will be moot.

I never said a word about how much energy is stored in the feedline;
it's irrelevant.

Do you think it is random coincidence that the amount of energy
stored in the feedline is *EXACTLY* the amount of energy required
by the forward and reflected waves???? The fact that you think
it is irrelevant is simply a flight into a wet dream fantasy.

It's stored there during the initial charging to the
steady state condition, and the same amount remains there until the
steady state condition no longer exists. It's exactly like the DC charge
on a capacitor ...

What a coincidence! It's *EXACTLY* the amount of energy required
by the forward and reflected waves that you say don't exist - and
it's RF photons, not DC, so it must travel at the speed of light!
You cannot store photons in a capacitor.

Surely even you can do the basic circuit analysis which shows that the
source is continuously generating 40 watts, ...

Of course, but during the power-on transient period, 68 watts is NOT
delivered to the load. Please don't make me waste my time calculating
the forward and reflected power during the power-on transient period.
When you actually do those calculations, you will agree with me that,
during the power-on transient phase, 68 watts of power has been stored
in the feedline and remains there until the power-off transient phase.
It is *EXACTLY* the amount of power required by the forward and
reflected waves.

The line's stored energy will be dissipated in either the source or load
resistor or both when the source power is turned off. But we're doing a
steady state analysis here.

Which must necessarily include the energy stored in the feedline during
the power-on transient condition because it is *still there during
steady-state*. Ignoring the energy stored in the feedline during the
power-on transient phase is both irrational and illogical. It could
even be bad for your mental health.

What exactly is my premise, please?

from my earlier posting:

You are apparently assuming there is not enough energy in the system
during steady-state to support the forward and reflected power waves.

I'm making no such assumption.

I'm sorry, Roy, but that is just BS! Either you admit there is enough
energy to support the steady-state forward and reflected waves or you
don't.

My analysis doesn't need to consider the stored energy at all.

And that is exactly why your analysis is wrong. You have, once again,
been seduced by the steady-state model and are spreading old wives'
tales as a result. Hopefully, you don't really want to do that.

The source has supplied 68 joules/sec that has not reached the load.

I think you mean 68 joules.

68 joules/sec in your original example. 68 joules in my one-second-
long feedline example.

Where will you find to put those all-important 68 joules in an
s-parameter analysis? That's a steady state analysis.

The 68 joules were stored in the feedline during the power-on
transient phase. They are still there during steady-state. An
S-parameter analysis yields the correct results because it
includes the reflected power, |a2|^2, as an energy source,
something you deny. Wonder what the S-parameter analysis folk
know that you don't choose to admit?

Finally, an actual answer. So of the 18 watts of "reverse power", 11.52
watts is making it to the source to "engage in constructive
interference". Does any of it get dissipated in the source resistor, or
does it just slide through unscathed?

It never encounters the source resistor as it is re-reflected by
wave cancellation, not by an impedance discontinuity. I have a QEX
article coming soon that will explain the details. Stand by. With
an unprejudiced open mind, you might actually learn something.

What's (s22*a2)^2?

Same as (1-rho^2)*Pref2 in ham terms. Reflected power from the
load that is re-reflected back toward the load by an impedance
discontinuity.

What's s11*a1?

Same as Vfor1*rho in ham terms. Forward voltage that is reflected
back toward the source by an impedance discontinuity.

I'll leave you to the folks who regard this kind of gobbledegook as
convincing evidence.

Do you realize what intellectual shape you would be in if you had
adopted that attitude when you were one year old? :-)
--
73, Cecil http://www.qsl.net/w5dxp


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Cecil Moore wrote:

Roy Lewallen wrote:

Finally, an actual answer. So of the 18 watts of "reverse power",
11.52 watts is making it to the source to "engage in constructive
interference". Does any of it get dissipated in the source resistor,
or does it just slide through unscathed?


It never encounters the source resistor as it is re-reflected by
wave cancellation, not by an impedance discontinuity. I have a QEX
article coming soon that will explain the details.

The Journal of Irreproducible Results could also be persuaded to publish
that claim, Cecil. Any implication that almost half of Maxwell's
equations are superfluous should easily qualify as an irreproducible
result. :-)

ac6xg

Jim Kelley wrote:

Cecil Moore wrote:
It never encounters the source resistor as it is re-reflected by
wave cancellation, not by an impedance discontinuity. I have a QEX
article coming soon that will explain the details.

The Journal of Irreproducible Results could also be persuaded to publish
that claim, Cecil. Any implication that almost half of Maxwell's
equations are superfluous should easily qualify as an irreproducible
result. :-)

Don't know exactly what you are inferring but the editors of QEX
have seen the light, :-) even if at RF frequencies. Quite a few
sources from the field of optics indicate that the phenomenon is well
known in that field even if not well understood in the field of RF.

It is very simple physics, Jim. When two coherent EM waves of equal
amplitudes and opposite phases attempt to travel in the same direction
in the same path, they cancel each other in their original direction
of travel. This is explained under "total destructive interference"
in "Optics" by Hecht. Since the energy in the two waves cannot be
canceled, that energy goes somewhere else. In a transmission line,
there are only two directions. If two waves cancel in one direction,
their combined energy components head back in the only other direction.
Everyone has seen that light interference pattern with his/her own eyes
and some just never realized what was happening.

Here's an example. If you wade through it, you will be forced to admit
that the destructive interference/wave cancellation at the non-glare
surface 'A' causes a reversal in the direction of the reflected
irradiance. That, my friend, is a 100% re-reflection, just as Walter
Maxwell has been saying for decades. 'n' is the index of refraction:

n=1.0 | n=1.2222 | n=1.4938
Laser-------air-------|---1/4WL thin film---|---infinite glass----...
| |
A B

The reflection from surface 'A' is canceled by an equal magnitude
and opposite phase reflection from surface 'B'. The energy components
in those two canceled waves join the forward wave because energy
cannot be destroyed (even though that concept seems to have serious
consequences to your mental health :-). Here's a quote from the
following web page.

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.”

In an RF transmission line, since there are only two possible
directions, the only “regions that permit constructive interference”
and "redistribution in a new direction" at an impedance discontinuity
is the opposite direction from the direction of destructive
interference/wave cancellation.

The above laser example is virtually identical to the following:

RF XMTR--50 ohm coax--+--1/4WL 61 ohm coax--+--infinite 75 ohm coax

If we use a coherent laser beam, no coax is required, so the
behavior of the actual EM waves is relatively easy to analyze.
--
73, Cecil http://www.qsl.net/w5dxp

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Cecil Moore wrote:

Jim Kelley wrote:
The Journal of Irreproducible Results could also be persuaded to
publish that claim, Cecil. Any implication that almost half of
Maxwell's equations are superfluous should easily qualify as an
irreproducible result. :-)


Don't know exactly what you are inferring but the editors of QEX
have seen the light, :-) even if at RF frequencies. Quite a few
sources from the field of optics indicate that the phenomenon is well
known in that field even if not well understood in the field of RF.

Cecil,

The Journal of Irreproducable Results is a hilarious journal that has
had a number of interesting articles in it. One I remember had to do
with Peanut Butter and the Three Stooges and the Precession of the
Earth's Axis (someone correct me if I misremembered). I believe it has
been referred to as Mad Magazine for Stephen Hawking.

Another had to do with a nice compression algorithm that eventually
reduced the input to 1 bit, no matter the input, and since that bit was
predictable as just a 1, we could eliminate that also.

tom
K0TAR

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

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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