On Sun, 30 May 2004 16:45:03 -0500, Cecil Moore wrote:

Walter Maxwell wrote:
In two Communications Quarterly issues he said as much, referring to my writings
in Reflections and in QEX.. During our last email communication he reiterated
his comment concerning incompetence, and said he was going to write a definitive
article concerning matching that would prove it, but that this time he wasn't
going to mention me by name.

Walt, as you know, I tangled with Steve over the subject of interference which
he claimed didn't exist at a match point. He later changed his mind and told
me in an email that he needed to rewrite part 3 of his article. Maybe he got
it right on his CDs. _Optics_, by Hecht proves Steve's equations in part 3 to
be the actual interference equations from optics with RF Power substituted for
light Irradiance.

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.

He said in Part 3: "... the two rearward traveling waves at the match point
(rearward waves 1 and 2) are 180 degrees out of phase with respect to each
other and a complete cancellation of both waves occurs."

Yes, but Cecil, the cancellation of the waves is only in the rearward direction,
because at the match point the waves and the energy they carry (volts x amps)
are totally reversed. 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. When voltage goes to zero at
the match point because the two returning voltages are equal magnitude and of
opposite phase, the current is doubled and the V x I energy in the rearward
traveling waves is totally re-reflected in the forward direction. Steve totally
ignores the energy in the reflected waves, except to say, "A total
re-reflection of the reflected voltage, current and power does not occur at the
match point and it (re-reflection) is not necessary for the impedance match to
occur." This statement is totally untrue, because as I said above, all of the
power in the reflected waves of voltage AND current is totally re-reflected in
the forward direction, the same as if the E and H fields had encountered a
physical short. We all know what happens in this case. The only difference is
that the virtual short established by the wave interference is one way only--to
the rearward traveling waves.

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 said in Reflections II: "With equal magnitudes and opposite phase at the
same point (Point A, the matching point), the sum of the two (rearward-
traveling) waves is zero."

Which means zero impedance, the boundary condition causing the total
re-reflection. This exactly what Slater is implying.

This agrees with J. C. Slater, from _Microwave_Transmissions_, "The
fundamental principle behind the elimination of reflections is then to have
each reflected wave canceled by another wave of equal amplitude and opposite
phase."

Cecil, the Slater reference is where I originally obtained this concept for my
QST article that appeared in Oct 1973, nearly 30 years ago. Check the ref number
in Reflections--No. 35.

All three above appear to me to be in agreement so the disagreements are really
about the down-in-the-noise details. Some gurus on this newsgroup disagree with
you, Dr. Best, and J. C. Slater.

Those who disagree with Slater need to refresh their memories with a review of
transmission lines 101. They ain't gonna win.

Others on this newsgroup have been asking about your opinion of conjugately-
matched transmitters. I have no interest in that particular discussion but you
might point out some references.

Cecil, I don't have a particular reference handy, but I can quote some of my own
measurements the others you mention might find of interest.. IMO they'll have a
hard time disgreeing with the data if they don't already believe a transmitter
is conjugately matched to its load when it's delivering all of its available
power at an arbitrarily selected drive level within the normal operating range.
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?

Walt

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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Cecil:

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

I would love it as well!

:-)

I fear that Walter is far too hung up on that conjugate match stuff...
Walt, you need to move
on to more important things. Don't continue to fuss with others over such a
trivial and ultimately
unimportant point. Have you left for the North yet, or still in Deland?
Sorry I've been far too
busy with consulting to get over to see you this past winter.

BTW, Cecil...

Faced with the possibility of owning such an ideal transmitter and given the
choice...

I wonder, considering the condition of most amateur antennas... which you
would love better in practice...

A completely mismatched zero Ohm impedance voltage transmitter

-OR-

A completely mismatched infinite Ohm impedance current transmitter.

--
Peter K1PO
Indialantic By-the-Sea, FL.

On Mon, 31 May 2004 13:26:58 GMT, "Peter O. Brackett"
wrote:

Cecil:

[snip]

I fear that Walter is far too hung up on that conjugate match stuff...
Walt, you need to move
on to more important things. Don't continue to fuss with others over such a
trivial and ultimately
unimportant point.

Hi Peter, I agree, but if you had read the last portion of Cecil's post you'd
have seen that apparently others have again asked my opinion of the subject.

Have you left for the North yet, or still in Deland?

Yep, been back in Michigan since April 12--coooold up here--wish we were still
in DeLand. Lot's a thunderstorms too.

Sorry I've been far too
busy with consulting to get over to see you this past winter.

We'll be back in DeLand in Nov. so please continue planning to come over.

Walt

On Mon, 31 May 2004 08:06:43 -0500, Cecil Moore wrote:

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 that case, of course all of the incident energy is transmitted.

"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)

Cecil, you're preaching to the choir here--see my above statement.

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.

Superposition of forward voltage and current? I didn't realize that voltage and
current superpose. But if Steve says so. I agree on the interference, but
doesn't the destructive have to occur before there can be constructive
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 don't recall Steve ever mentioning current. He simply says the voltages
cancel, resulting in 0 V. What Steve apparently doesn't understand is how the
energy direction is reversed when the rearward voltages and currents go to zero.
Since the energy cannot go to zero what happens at the match point is that the E
fields go to zero and the disappearing E field energy merges into the H field
energy, raising both the H field energy and its associated current to double
their values relative to those prior to re-reflection. The energy now propagates
forward and the E and H fields resume their normal relationship, half the total
energy in each. This is precisely what happens to the EM field when it
encounters a physical short circuit. But in our case the reflected EM fields
encounter a virtual short circuit, with the same result as with a physical short
except that the virtual short to the reflected waves is transparent to the
source wave.

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.

As I said above I'll prove to you conclusively that the virtual short circuit is
established by the wave interference, contrary to what you and many on this rraa
thread believe.

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.

I have no disagreement with the S parameter analysis, Cecil.

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

Cecil, your comparison above lacks logic. Your are assuming that the XMTR is a
classical generator, where the maximum efficiency cannot exceed 50% because the
internal resistance is dissipative. OTH, the souce resistance at the output of
the XMTR is non-dissipative, in which case it wouldn't matter which of the two
XMTRs you choose. The only dissipative resistance in the amp is that which heats
the plate. That dissipation is the only dissipation in the source--the other
dissipation is only in the load. If you don't agree with this concept please
review Chapter 19 in Reflections 2. I also have other measurements that prove
the concept is true.

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.

Doncha just love Tesla?

Walt

Walter Maxwell wrote:

wrote:
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.

Superposition of forward voltage and current?

I'm sure he meant "superposition of forward voltages and superposition of
forward currents."

I don't recall Steve ever mentioning current.

I think you are right re his article. The above quote is from an
r.r.a.a. posting circa Summer 2001.

What Steve apparently doesn't understand is how the
energy direction is reversed when the rearward voltages and currents go to zero.

"How" is not explained in any of the physics references. The closest
physics reference that explains it is _Optics_, by Hecht where he says
something like, at a point some distance from a source, constructive
interference must be balanced by an equal magnitude of destructive interference.
In a matched system, there is "complete destructive interference" toward the
source side of the match point and "complete constructive interference" toward
the load side of the match point. Energy is always displaced from the "complete
destructive interference" event to the "complete constructive interference"
event. (That's what you call a "virtual short" or "virtual open" capable of
re-reflecting the reflected energy.)

In s-parameter terms, b1 is the reflected voltage from port 1 toward the source.
Port 1 is the input to a matched tuner (transmatch). The equation is:

rearward-traveling voltage reflected toward the source b1 = s11(a1) + s12(a2)

For b1 to be zero, i.e. zero reflections toward the source, s11(a1) must be equal
in magnitude and opposite in phase to s12(a2). That is "complete destructive
interference". Since there are only two directions, "complete constructive interference"
must occur in the direction of b2 = s21(a1) + s22(a2) toward the load which is the
opposite direction from b1.

s11 is the port 1 reflection coefficient. a1 is the port 1 incident voltage.
s21 is the port 2 to port 1 transmission coefficient. a2 is the voltage
reflected from the load that is incident upon port 2.

Match-Point
Port1 Port2
Source------Z01--------x------------Z02------------load
a1-- --a2
--b1 b2--

The only dissipative resistance in the amp is that which heats
the plate. That dissipation is the only dissipation in the source--the other
dissipation is only in the load.

Why isn't the source impedance a negative resistance, i.e. a source
of power Vs a positive resistance, a sink of power?
--
73, Cecil http://www.qsl.net/w5dxp



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On Wed, 02 Jun 2004 08:37:30 -0500, Cecil Moore wrote:

Walter Maxwell wrote:

wrote:
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.

Superposition of forward voltage and current?

I'm sure he meant "superposition of forward voltages and superposition of
forward currents."

I don't recall Steve ever mentioning current.

I think you are right re his article. The above quote is from an
r.r.a.a. posting circa Summer 2001.

What Steve apparently doesn't understand is how the
energy direction is reversed when the rearward voltages and currents go to zero.

"How" is not explained in any of the physics references. The closest
physics reference that explains it is _Optics_, by Hecht where he says
something like, at a point some distance from a source, constructive
interference must be balanced by an equal magnitude of destructive interference.
In a matched system, there is "complete destructive interference" toward the
source side of the match point and "complete constructive interference" toward
the load side of the match point. Energy is always displaced from the "complete
destructive interference" event to the "complete constructive interference"
event. (That's what you call a "virtual short" or "virtual open" capable of
re-reflecting the reflected energy.)

Cecil, I explained the 'how', both in Reflections and in QEX. My explantion of
'how' is what Steve is continually stating is incorrect, especially in his last
3-part QEX article. Statements in that article prove he doesn't understand the
wave mechanism that reverses the direction of the reflected energy. Evidence of
this is that by simply saying the voltages cancel is insufficient description of
how the energies reverse direction. In fact, in his Oct 99 ComQuart article he
specifically states that both voltages and power cancel. This tell me that he
doesn't understand the wave action he's attempting to teach.

MIT's Slater and Harvard's Alford both explain it brilliantly, but Steve rejects
those references as 'irrelevant', and says I mistakenly used them as references
in Reflections.

What is really perplexing to me is that several posters on this subject said
that Steve's 3-parter is the best and most illuminating article they ever read
on the subject. How can they have missed some of the most egregious errors
appearing in that paper is unbelievable!

In s-parameter terms, b1 is the reflected voltage from port 1 toward the source.
Port 1 is the input to a matched tuner (transmatch). The equation is:

rearward-traveling voltage reflected toward the source b1 = s11(a1) + s12(a2)

For b1 to be zero, i.e. zero reflections toward the source, s11(a1) must be equal
in magnitude and opposite in phase to s12(a2). That is "complete destructive
interference". Since there are only two directions, "complete constructive interference"
must occur in the direction of b2 = s21(a1) + s22(a2) toward the load which is the
opposite direction from b1.

Cecil, if s11(a1) is equal in magnitude but in opposite phase with s12(a2) this
constitutes a short circuit. Assume two generators delivering harmonically
related output voltages equal to the two 's' voltages. When the generators are
connected with their output terminals reversed, causing their voltages to be 180
degrees out of phase--this configuration is a SHORT CIRCUIT. What I've been
trying to say is that this is the same condition as when the reflected waves of
voltage and current from a mismatched termination are of equal magnitude and
opposite phase with the voltage and current waves reflected by a matching device
such as a stub, the opposing voltages in those two sets of waves constitute a
short circuit the same as the voltages delivered by the two opposing
generators.

s11 is the port 1 reflection coefficient. a1 is the port 1 incident voltage.
s21 is the port 2 to port 1 transmission coefficient. a2 is the voltage
reflected from the load that is incident upon port 2.

Match-Point
Port1 Port2
Source------Z01--------x------------Z02------------load
a1-- --a2
--b1 b2--

The only dissipative resistance in the amp is that which heats
the plate. That dissipation is the only dissipation in the source--the other
dissipation is only in the load.

Why isn't the source impedance a negative resistance, i.e. a source
of power Vs a positive resistance, a sink of power?

Cecil, the source impedance is often correctly referred to as a negative
resistance. But it must be remembered that the source resistance of Class B and
C amps is non-dissipative, and thus totally re-reflect incident reflected power.
By this I mean that the dissipative resistance that heats the plate is entirely
separate from the output resistance represented by the load line. Remember, the
DC power goes to only two places: that which is dissipated as heat, and that
which is delivered to the load. The reflected power incident on the output
terminals of the tank has no effect on the power dissipated as heat.

Here's an example. First, adjust an amp to deliver 100 watts into a 50-ohm
resistive load. Second, change the load to a reactive 50 + j50 load and readjust
the pi-network to again deliver 100 watts into the new load. The plate current
will be exactly as in the first case, and the heat dissipated will be the same.
The difference is that in the first case the output impedance of the amp was 50
+ j0, while in the second case the output impedance is 50 - j50, due to
readjusting the reactive components in the pi-network to match the 50 + j50-ohm
load. Whether one likes it or not, this constitutes a conjugate match.

As for the plate temperature remaining the same in both cases, first, the
readjustment of the pi-network returned the input resistance of the network to
the same value as in the first case. Thus the plates saw no different condition
between the two cases. And second, Eric Nichols, KL7AJ, has measured
calorimetrically the temperature of the water cooling the tubes of megawatt
transmitters with greatly differing values of reflected power incident on the
xmtr. He has shown that the water temperature remains constant whatever the
value of the reflected power.

Since you mentioned earlier that some posters would like my opinion on the
nature of the source resistance in rf amps I'll put a paragraph or two together
with measurement data to support my opinion.

Walt

On Wed, 02 Jun 2004 16:49:09 GMT, Walter Maxwell wrote:

On Wed, 02 Jun 2004 08:37:30 -0500, Cecil Moore wrote:

Walter Maxwell wrote:

wrote:
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.

Superposition of forward voltage and current?

I'm sure he meant "superposition of forward voltages and superposition of
forward currents."

I don't recall Steve ever mentioning current.

I think you are right re his article. The above quote is from an
r.r.a.a. posting circa Summer 2001.

What Steve apparently doesn't understand is how the
energy direction is reversed when the rearward voltages and currents go to zero.

"How" is not explained in any of the physics references. The closest
physics reference that explains it is _Optics_, by Hecht where he says
something like, at a point some distance from a source, constructive
interference must be balanced by an equal magnitude of destructive interference.
In a matched system, there is "complete destructive interference" toward the
source side of the match point and "complete constructive interference" toward
the load side of the match point. Energy is always displaced from the "complete
destructive interference" event to the "complete constructive interference"
event. (That's what you call a "virtual short" or "virtual open" capable of
re-reflecting the reflected energy.)

Cecil, I explained the 'how', both in Reflections and in QEX. My explantion of
'how' is what Steve is continually stating is incorrect, especially in his last
3-part QEX article. Statements in that article prove he doesn't understand the
wave mechanism that reverses the direction of the reflected energy. Evidence of
this is that by simply saying the voltages cancel is insufficient description of
how the energies reverse direction. In fact, in his Oct 99 ComQuart article he
specifically states that both voltages and power cancel. This tell me that he
doesn't understand the wave action he's attempting to teach.

MIT's Slater and Harvard's Alford both explain it brilliantly, but Steve rejects
those references as 'irrelevant', and says I mistakenly used them as references
in Reflections.

What is really perplexing to me is that several posters on this subject said
that Steve's 3-parter is the best and most illuminating article they ever read
on the subject. How can they have missed some of the most egregious errors
appearing in that paper is unbelievable!

snip

Cecil, the following is a direct quote from Steve's Comm Quart Article, Oct
1999:
"For the impedance matching network to 'work', this analysis must demonstrate
that the steady-state traveling backward power developed at the matching network
input is equal in magnitude but 180 degrees out of phase with the initial power
reflected at the matching network input. For this to occur Vback must be the
negative of VR. In this case ALL POWER TRAVELING BACKWARD TOWARDS THE
TRANSMISTTER WILL BE CANCELED, resulting in the steady-state matched condition."

Emphasis mine.

Walt

Walter Maxwell wrote:
Cecil, the following is a direct quote from Steve's Comm Quart Article, Oct
1999:
"For the impedance matching network to 'work', this analysis must demonstrate
that the steady-state traveling backward power developed at the matching network
input is equal in magnitude but 180 degrees out of phase with the initial power
reflected at the matching network input. For this to occur Vback must be the
negative of VR. In this case ALL POWER TRAVELING BACKWARD TOWARDS THE
TRANSMISTTER WILL BE CANCELED, resulting in the steady-state matched condition."

Joules/sec possesses phase? Joules/sec can be canceled?
--
73, Cecil http://www.qsl.net/w5dxp

Walter Maxwell wrote:

Cecil Moore wrote:
"How" is not explained in any of the physics references.

Cecil, I explained the 'how', both in Reflections and in QEX.

Yes, I know you did, Walt. By "physics references" above, I meant
books like college physics textbooks, e.g. _Optics_, by Hecht.

What is really perplexing to me is that several posters on this subject said
that Steve's 3-parter is the best and most illuminating article they ever read
on the subject. How can they have missed some of the most egregious errors
appearing in that paper is unbelievable!

Not recognizing his power equations as classical EM physics interference
terms was a pretty huge mistake in Part 3. But alleged gurus on this
newsgroup have done the same thing. Apparently, power is simply ignored
in present-day transmission line theory.

Cecil, if s11(a1) is equal in magnitude but in opposite phase with s12(a2) this
constitutes a short circuit.

I agree it constitutes a "short circuit" for superposed rearward-
traveling voltages. But exactly the same thing happens to the current
as happens to the voltage. And an "open circuit" is what causes the
rearward-traveling currents to superpose to zero.

The two rearward-traveling superposing voltages might be:
(100v at zero degrees) superposed with (100v at 180 degrees)
The superposed sum of the two rearward-traveling voltages is zero.
This indeed acts like a short where voltages go to zero.

The two corresponding rearward-traveling superposing currents might be:
(2a at 180 degrees) superposed with (2a at zero degrees)
The superposed sum of the two rearward-traveling currents is zero.
This acts like an open where currents go to zero.

Or if you prefer, both the E-fields and the H-fields cancel to
zero when complete destructive interference occurs. In a transmission
line, it causes a surge of constructive interference energy in the
opposite direction, something you have called "re-reflection from a
virtual short".
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73, Cecil http://www.qsl.net/w5dxp