Walter Maxwell wrote:

Cecil Moore wrote:
IMO, you and Steve were much closer in principles than either one of you
realized.

Sorry Cecil, I don't think so. Steve has missed the most vital aspect of the
phenomenon--what happens to the energy, or power in the reflected waves on
return to the match point.

Well, once Steve admitted that the two reflected waves completely cancel
each other in a matched system, what happens to the pre-existing energy in
those two waves before they cancel is obvious. Energy cannot be destroyed
and if it doesn't flow toward the source, it must flow toward the load
as explained at the bottom of the Melles Griot Web Page:

http://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." (emphasis mine)

"In the absence of absorption or scatter, THE PRINCIPLE OF CONSERVATION OF ENERGY
indicates ALL (rearward-traveling) "LOST" REFLECTED INTENSITY will appear as ENHANCED
INTENSITY IN THE (forward-traveling) 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." (emphasis mine)

Steve at first said the energy in the canceled waves continues to flow toward the
source without a voltage and current and that interference was not involved. He later
changed his mind. All that should be archived on r.r.a.a on Google for the summer
of 2001. Here's an excerpt. Steve said: "The total forward power increases as a direct
result of the vector superposition of forward voltage and current. This DOES NOT
require a corresponding destructive interference process ..." thus contradicting Hecht
in _Optics_ who says any constructive interference process must be accompanied by
an equal magnitude of destructive interference.

Now to continue what Steve said is: "The result of this
wave cancellation is that the total steady-state rearward-traveling wave has a
net voltage of 0 V nd 0 A, respectively, and an impedance match occurs." No No
No.

As we've discussed earlier, voltage and current cannot both go to zero
simultaneously, except in the rearward direction.

But that's what he said above. The rearward-traveling wave indeed does have a
net voltage of 0 V and 0 A and the reflections toward the source disappear at
the match point. I think you and Steve really agree on about 98% of this match
point stuff but you two obviously disagree on the definition of "re-reflection".

I know you don't agree with me that a one-way virtual short is what causes the
re-reflection, but in a short time I'll be able to prove it to you in a manner
you'll not be able to rebut. Stay tuned.

You might want to check your work against an s-parameter analysis. The s-parameter
equations a b1 = s11(a1) + s12(a2) and b2 = s21(a1) + s22(a2)

Given that a match point in a transmission line can be considered to be a two-
port network, |s22|^2 is the Power (re)reflected from the network output divided
by the (reflected) Power incident on the network output. s22 is the reflection
coefficient looking into port 2 and is *not equal to 1.0 or zero*. In a matched
system with nothing but resistances, it is often the negative of the reflection
coefficient at the load.

This is covered in HP's AN 95-1 available on the web.

Those who don't believe will get quite a surprise when I reveal what the output
source resistance of the xmtr really is under this condition. Waddya think?

I think I am too ignorant of the subject to venture an opinion. Which of the
following systems do you think I would prefer, the conjugately-matched one at
50% efficiency or the non-conjucately one at 98% efficiency?

1 ohm XMTR----100V out---tuner---1/2WL 450 ohm line--50 ohm load

50 ohm XMTR---100V out---tuner---1/2WL 450 ohm line--50 ohm load

Arguing that Tesla/Westinghouse would have to conjugately match their 60 Hz
AC generators is what shot down Edison's dream of an all DC power distribution
system for the USA.

I would love to have a transmitter with a zero ohm internal impedance,
completely conjugately unmatched. :-)
--
73, Cecil http://www.qsl.net



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