On Fri, 12 Mar 2004 21:12:23 GMT, Walter Maxwell wrote:


My first publication of this issue appeared in QST, October 1973, entitled, "A
View Into the Conjugate Mirror." This article appears in both Eds 1 and 2 of
Reflections as Chapter 4, Steve also copied from another of my articles, this
one in QEX,, Mar/Apr 1998, entitled "Examining the Mechanics of Wave
Interference in Impedance Matching," which also appears in Reflections 2 as
Chapter 23.

Steve and I have been in contentious controversy on this subject for several
years. He continued this controversy by publishing this totally erroneous
material in QEX,, erroneous except for the portion in the single column where he
presented my material correctly. The remaining portion of his article is simply
an unsuccessful attempt to show that my position is incorrect, and therefore
calls it a 'fallacy'.

In fact, however, the entire portion following the correct portion he copied
from me is where the REAL fallacy lies--it proves that he knows very little
about the subject of his title, "Wave Mechanics of Transmission Lnes." It also
shows he doesn't have a clue concerning the superposition of two rearward
traveling waves that are conjugately related at the matching point. In fact,
the two waves cancel each other, and establish either a one-way open circuit or
a one-way short circuit that totally re-reflects the reflected power, with its
voltage and current components traveling in the same phase as t;hose from the
source, and therefore adding to the source power.

I know that many on this thread believe that no open or short circuit can be
established by the superposition of waves. It is true that forward and reflected
waves, traveling in OPPOSITE directions establish only the standing wave--no
open or short circuits. But it's a different ball game when two waves traveling
in the SAME direction are conjugately related. The waves are conjugately related
because the canceling wave generated by the matching device is tailored to have
the same magnitude but opposite phase as the wave reflected from the mismatched
load on the transmission line.

Here's why a short or open circuit is established when conjugately related waves
join at a matching point. From an analytic viewpoint the voltage appearing at
any point on the line can be replaced with a generator delivering the same
voltage at the same phase that appears at that point. This generator is called a
'point' generator that delivers an impedance-less EMF. Now consider one
generator delivering the voltage appearing in the wave reflected at the
mismatched load and a second generator delivering the voltage from the canceling
wave reflected by a matching stub, or whatever the matching device, at the same
point on the line as the first. The voltage from this second generator has the
same magnitude, but opposite phase from that of the first generator. When the
voltages delivered by the t wo generators are 180 degrees out of phase we have a
short circuit--if they're in phase we have an open circuit. As the result, in
either of these two conditions no reflected wave can pass rearward of the
matching point.

From the simple fact that the impedance at the input of an antenna tuner is 50+
j0 we know that no reflected power is traveling rearward further than the tuner
input. Where did the power in the reflected wave go? That energy cannot
disappear as if by some sort of magic--it is totally re-reflected by the open or
short circuit, and adds to the source power to establish a forward power equal
to the sum of the source and reflected power.

I hope this helps to end the confusion, and also gives Cecil what he deserves
for his attempt to give you guys the straight dope.

Walt, W2DU

This is a post script to go along with the above discussion.

Sorry, Guys, I forgot to mention that for those of you who don't have access to
my QEX article, or Chapter 23 in Reflections, Chapter 23 appears in PDF form for
downloading from my web page at http://home.iag.net/~w2du.

I would also like to add that from the first QST article I published on this
subject in 1973, and the QEX article in 1998, I have received more than one
hundred responses from RF engineers saying that my explanation of impedance
matching via wave mechanics gave them the first real understanding of how the
impedance matching process works. No one, other than Steve Best, has disputed my
explantion.

If the more than one hundred responses seems exaggerated please take a look at
the hit count on my web page. At this time the hit count reads 19,927, and I've
had NO responses disputing any of the material presented there.

So who do you want to believe? Steve Best or me?

Walt, W2DU

Richard Harrison wrote:
Cecil has argued that the high impedance at the open end of a shorted
1/4-wave stub does not inhibit current into the stub. The reflection
point is at the short, not at the high impedance point back 1/4-wave
from the short. That sounds reasonable to me, but I wonder if it makes
any difference.

Only to those who think the stub is a physical open-circuit at the
mouth of the stub and therefore, no current flows in the stub. :-)
--
73, Cecil http://www.qsl.net/w5dxp



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On Sat, 13 Mar 2004 15:53:34 -0600, Cecil Moore
wrote:

Richard Harrison wrote:
Cecil has argued that the high impedance at the open end of a shorted
1/4-wave stub does not inhibit current into the stub. The reflection
point is at the short, not at the high impedance point back 1/4-wave
from the short. That sounds reasonable to me, but I wonder if it makes
any difference.
Cecil wrote:
Only to those who think the stub is a physical open-circuit at the
mouth of the stub and therefore, no current flows in the stub. :-)

In a stub constructed of lossless material the only current that flows in the
stub is that required to bring it up to the steady statecondition.

In a practical stub with attenuation current flows into the stub continually,
but only that sufficient to compensate for the loss due to attenuation to retain
it's steady state condition.

Walt, W2DU

Richard Harrison wrote:
I was wrong when I wrote waves would be exhibitable on demand were they
annhilated. Cecil was right. It makes no difference whether two equal
and opposite signals are received or no signal is received. Same result.

Which means that if two coherent waves flowing in the same direction are
completely cancelled, the E-fields and the H-fields in that direction are
also cancelled and therefore, zero energy propagates in that direction.

So the question remains: What happens to the energy in those two rearward-
traveling cancelled waves? Melles-Griot says the energy is not lost and
instead appears as enhanced intensity in the forward-traveling wave. From
rearward-traveling energy to forward-traveling energy certainly sounds
like a direction reversal to me.

"Enhanced Intensity" is another name for constructive interference and
in a matched system, it is "total constructive interference". "Complete
Wave Cancellation" is another name for "total destructive interference".

Bottom line: In a Z0-matched system, the reflected energy flows rearward
to the Z0-match point, changes direction at the Z0-match point (because of
interference), and joins the forward energy wave flowing toward the load.

Anyone got a better word than "re-reflected" to describe that event?
From looking at a diagram of laser light encountering a perfect 1/4WL
thin-film coating on glass, it is apparent that all the rearward-traveling
reflected irradiance (power) encounters a conjugate "mirror" and joins the
forward-traveling energy wave.
--
73, Cecil http://www.qsl.net/w5dxp



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Walter Maxwell wrote:
In a stub constructed of lossless material the only current that flows in the
stub is that required to bring it up to the steady statecondition.

In a practical stub with attenuation current flows into the stub continually,
but only that sufficient to compensate for the loss due to attenuation to retain
it's steady state condition.

All that is true for *NET* current, Walt. But a full magnitude of forward and
reflected current is flowing in and out of the mouth of the stub. All of the
reflections in a stub occur at the shorted end. What happens at the mouth of
the stub is superposition and interference, not reflection. The forward and
reflected voltages superpose (in phase) to a high net value and the forward
and reflected currents superpose (out of phase) to a low net value.

(|Vfwd|+|Vref|)/(|Ifwd|-|Iref|) = very high V/I ratio = very high impedance,
but that high impedance is an effect and not the cause of anything (except
arguments on r.r.a.a :-)

Inside the stub, Vfwd/Ifwd = Z0 and Vref/Iref = Z0

Anyone can prove this to himself. Install a wattmeter 1/8WL down into the 1/4WL
shorted stub. One will read high forward power and high reflected power. From
those powers, one can actually calculate the forward and reflected voltages
and currents at that halfway point within the stub given that halfway point
is equal to 45 degrees.
--
73, Cecil http://www.qsl.net/w5dxp



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