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".
--
73, Cecil http://www.qsl.net/w5dxp

On Thu, 03 Jun 2004 14:03:26 -0500, Cecil Moore
wrote:

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.

Cecil, this is exactly what I've been trying to persuade you of, but always said
no, there is no short developed. But you must also agree that under this
condition the current doubles.

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.

Of course, but the voltage doubles.

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".

Well, Cecil, here's where we part company to a degree. Unlike voltage and
current that can go to zero simultaneously only in the rearward direction, E and
H fields can never go to zero simultaneously. At a short circuit the E field
collaples to zero, but its energy temporarily merges with the H field, making
the H field double it normal value. But the changing H field immediately
reestablishes the E field, both now traveling in the forward direction. And yes,
this is called re-reflection.

If Steve understands the action of the fields in the EM wave it's hard to
understand why he finds it so erroneous to associate voltage and current with
the their respective fields in impedance matching. Apparently he can't conceive
that the voltages and currents in reflected waves can be considered to have been
delivered by separate generators connected with opposing polarities.

Walt

Walter Maxwell wrote:
Cecil, this is exactly what I've been trying to persuade you of, but always said
no, there is no short developed. But you must also agree that under this
condition the current doubles.

Nope, for complete destructive interference, both the E-field and
H-field associated with the two interfering waves collapse to zero.
For the resulting complete constructive interference, the ratio of
the E-field to the H-field equals the characteristic impedance of
the medium.

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.

Of course, but the voltage doubles.

Nope, again here are the two sets of reflected waves.

#1 100v at zero degrees and 2a at 180 degrees = 200W

#2 100v at 180 degrees and 2a at zero degrees = 200W

Superposing those two reflected waves yields zero volts and
zero amps.

Well, Cecil, here's where we part company to a degree. Unlike voltage and
current that can go to zero simultaneously only in the rearward direction, E and
H fields can never go to zero simultaneously.

For "complete destructive interference" as explained in _Optics_, by
Hecht, the E-field and B-field (H-field) indeed do go to zero
simultaneously. That is what causes a completely dark ring in a
set of interference rings. Of course, a resulting corresponding
complete constructive interference causes the brightest of rings.

If Steve understands the action of the fields in the EM wave it's hard to
understand why he finds it so erroneous to associate voltage and current with
the their respective fields in impedance matching. Apparently he can't conceive
that the voltages and currents in reflected waves can be considered to have been
delivered by separate generators connected with opposing polarities.

He pretty much ignored current. His power equations are exact copies of
the light irradiance interference equations from optics, but he apparently
didn't realize it until I pointed it out to him.
--
73, Cecil http://www.qsl.net/w5dxp

On Thu, 03 Jun 2004 16:18:36 -0500, Cecil Moore
wrote:

Walter Maxwell wrote:
Cecil, this is exactly what I've been trying to persuade you of, but always said
no, there is no short developed. But you must also agree that under this
condition the current doubles.

Nope, for complete destructive interference, both the E-field and
H-field associated with the two interfering waves collapse to zero.
For the resulting complete constructive interference, the ratio of
the E-field to the H-field equals the characteristic impedance of
the medium.

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.

Of course, but the voltage doubles.

Nope, again here are the two sets of reflected waves.

#1 100v at zero degrees and 2a at 180 degrees = 200W

#2 100v at 180 degrees and 2a at zero degrees = 200W

Superposing those two reflected waves yields zero volts and
zero amps.

Well, Cecil, here's where we part company to a degree. Unlike voltage and
current that can go to zero simultaneously only in the rearward direction, E and
H fields can never go to zero simultaneously.

For "complete destructive interference" as explained in _Optics_, by
Hecht, the E-field and B-field (H-field) indeed do go to zero
simultaneously. That is what causes a completely dark ring in a
set of interference rings. Of course, a resulting corresponding
complete constructive interference causes the brightest of rings.

If Steve understands the action of the fields in the EM wave it's hard to
understand why he finds it so erroneous to associate voltage and current with
the their respective fields in impedance matching. Apparently he can't conceive
that the voltages and currents in reflected waves can be considered to have been
delivered by separate generators connected with opposing polarities.

He pretty much ignored current. His power equations are exact copies of
the light irradiance interference equations from optics, but he apparently
didn't realize it until I pointed it out to him.

Well, Cecil, perhaps I don't understand Melles-Griot. However, when an EM wave
encounters a short circuit the E field goes to zero, but in its change from
normal level to zero, that changing field develops a corresponding H field that
adds to its original value, causing it to double--the H field does NOT go to
zero. But as it returns to its normal value that change in turn develops a new E
field propagating in the opposite direction. From there on the EM field
reconstitutes itself with normal E and H fields of equal value, each supporting
one half of the propagating energy.

This sequence must also prevail in optics--are you sure you are interpreting
your double zero at the correct point in the circles?

Walt

Walter Maxwell wrote:
Well, Cecil, perhaps I don't understand Melles-Griot. However, when an EM wave
encounters a short circuit the E field goes to zero, but in its change from
normal level to zero, that changing field develops a corresponding H field that
adds to its original value, causing it to double--the H field does NOT go to
zero. But as it returns to its normal value that change in turn develops a new E
field propagating in the opposite direction. From there on the EM field
reconstitutes itself with normal E and H fields of equal value, each supporting
one half of the propagating energy.

I absolutely agree, Walt, "when an EM wave encounters a short circuit ...".
But when "complete destructive interference" occurs, something else happens.
The E-field goes to zero AND the H-field (B-field) goes to zero at the same
time. If you concentrate on the voltage, it looks like a short. If you
concentrate on the current, it looks like an open. It is both or neither.

Complete destructive interference requires that both fields go to zero
simultaneously and emerge as constructive interference in the opposite
direction obeying the rule that E/H=V/I=Z0. It is an energy reflection
that is also an impedance transformation at an impedance discontinuity.

Let's assume that 100v at zero degrees with a current of 2a at 180 degrees
encounters another wave traveling in the same rearward direction of 100v
at 180 degrees with a current of 2a at zero degrees. These two waves cancel.
The voltage goes to zero AND the current goes to zero. Each wave was associated
with 200 watts. So a total of 400 watts reverses directions. Assuming the
destructive interference occurred in a 50 ohm environment and the resulting
constructive interference occurred in a 300 ohm environment, the reflected
wave would be 346 volts at 1.16 amps. It's pretty simple math.

346*1.16 = 400 watts 346/1.16 = 300 ohms

The above quantities represent the destructive/constructive interference.
These quantities must be added to the other voltages and currents that are
present to obtain the net voltage and net current.
--
73, Cecil http://www.qsl.net/w5dxp



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On Thu, 03 Jun 2004 21:40:02 -0500, Cecil Moore wrote:

Walter Maxwell wrote:
Well, Cecil, perhaps I don't understand Melles-Griot. However, when an EM wave
encounters a short circuit the E field goes to zero, but in its change from
normal level to zero, that changing field develops a corresponding H field that
adds to its original value, causing it to double--the H field does NOT go to
zero. But as it returns to its normal value that change in turn develops a new E
field propagating in the opposite direction. From there on the EM field
reconstitutes itself with normal E and H fields of equal value, each supporting
one half of the propagating energy.

I absolutely agree, Walt, "when an EM wave encounters a short circuit ...".
But when "complete destructive interference" occurs, something else happens.
The E-field goes to zero AND the H-field (B-field) goes to zero at the same
time. If you concentrate on the voltage, it looks like a short. If you
concentrate on the current, it looks like an open. It is both or neither.
Complete destructive interference requires that both fields go to zero
simultaneously and emerge as constructive interference in the opposite
direction obeying the rule that E/H=V/I=Z0. It is an energy reflection
that is also an impedance transformation at an impedance discontinuity.

Let's assume that 100v at zero degrees with a current of 2a at 180 degrees
encounters another wave traveling in the same rearward direction of 100v
at 180 degrees with a current of 2a at zero degrees. These two waves cancel.
The voltage goes to zero AND the current goes to zero. Each wave was associated
with 200 watts. So a total of 400 watts reverses directions. Assuming the
destructive interference occurred in a 50 ohm environment and the resulting
constructive interference occurred in a 300 ohm environment, the reflected
wave would be 346 volts at 1.16 amps. It's pretty simple math.

346*1.16 = 400 watts 346/1.16 = 300 ohms

The above quantities represent the destructive/constructive interference.
These quantities must be added to the other voltages and currents that are
present to obtain the net voltage and net current.

Cecil, at this point I'm not clear what's happening in your above example. Where
and what is the phase reference for these two waves? It appears to me that the
reference phase must be that of the source wave, because the voltage and current
in both rearward traveling waves are 180 degrees out of phase. Educate me on how
the cancellation takes place and how the energy reverses direction.

It seems to me the out of phase voltage yields a short circuit, while the out of
phase current yields an open circuit. How can both exist simultaneously? And
further, what circuit can produce these two waves simultaneously?

In addition, I believe your example has changed the subject. My discourse
concerns what occurs to an EM wave when it encounters a short circuit. In this
case you're going to have to prove to me that both E and H fields go to zero.
IMO it can't happen. The E field collapses to zero while the H field doubles,
because the energy in the changing E field merges into the H field. Before the
EM wave encountered the short both fields contained the same energy, thus the
E-field energy adding to the original H field energy, doubles it, and while that
H field was changing it was developing a new E field that launches a new wave in
the opposite direction, the reflected wave.

So my argument with you, Cecil, is that I maintain the H field doubles on
encountering a short circuit, while you maintain that both E and H fields go to
zero. What's your answer to this dilemma?

Walt
Walt

Walter Maxwell wrote:
Cecil, at this point I'm not clear what's happening in your above example. Where
and what is the phase reference for these two waves? It appears to me that the
reference phase must be that of the source wave, because the voltage and current
in both rearward traveling waves are 180 degrees out of phase. Educate me on how
the cancellation takes place and how the energy reverses direction.

J. C. Slater explains how the cancellation takes place.
"The method of eliminating reflections is based on the interference
between waves. Two waves half a wavelength apart are in opposite
phases, and the sum of them, if their amplitudes are numerically
equal, is zero. 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."

Note that the above applies to both voltage waves and current waves.
Both voltage and current go to zero during complete destructive interference,
i.e. both E-field and H-field go to zero during complete destructive
interference.

It seems to me the out of phase voltage yields a short circuit, while the out of
phase current yields an open circuit. How can both exist simultaneously?

They can't, and that's my argument. It's easy to understand. If the two rearward-
traveling voltages are 180 degrees out of phase then the two rearward-traveling
currents MUST also be 180 degrees out of phase, since the reflected current is
ALWAYS 180 degrees out of phase with the reflected voltage. If one looks only
at the voltages, one will say it's a short-circuit. If one looks only at the
currents, one will say it's an open-circuit.

And further, what circuit can produce these two waves simultaneously?

According to J. C. Slater (and Reflections II, page 23-9) a match point
produces these two waves simultaneously, two reflected voltages 180 degrees
out of phase and two reflected currents 180 degrees out of phase.

In addition, I believe your example has changed the subject. My discourse
concerns what occurs to an EM wave when it encounters a short circuit.

There is no argument about what happens at a short circuit. What I am saying
is that a match point is NOT a short circuit.

In this
case you're going to have to prove to me that both E and H fields go to zero.
IMO it can't happen.

J. C. Slater says that's what happens in the above quote. Voltages 1/2WL apart
in time cancel to zero. Currents 1/2WL apart in time cancel to zero.

So my argument with you, Cecil, is that I maintain the H field doubles on
encountering a short circuit, while you maintain that both E and H fields go to
zero. What's your answer to this dilemma?

My argument is that it is NOT a short circuit. It is "complete destructive
interference" as explained in _Optics_, by Hecht where both the E-field and
B-field go to zero. J. C. Slater says that the two rearward-traveling voltages
are 180 degrees out of phase and the two rearward-traveling currents are 180
degrees out of phase. So whatever happens to the voltage also happens to the
current, i.e. destructive interference takes both voltage and current to zero.
--
73, Cecil http://www.qsl.net/w5dxp



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Cecil, H. Adam Stevens sent you an email earlier today, with a cc to me. I
replied to all, but the copy to you came back host unknown. But your address
appeared as . What's the ONEDOT? I now see it in your return
address on the rraa.

What's going on?

Walt

Walter Maxwell wrote:
Cecil, H. Adam Stevens sent you an email earlier today, with a cc to me. I
replied to all, but the copy to you came back host unknown. But your address
appeared as . What's the ONEDOT? I now see it in your return
address on the rraa.

What's going on?

It's a spam preventative, Walt. Change '.ONEDOT.' to '.'
to get

It didn't bounce.
hmmmmm
I use spam filters.
Oh well, ........... takes all kinds.
73

BTW Walt, I've had one of your baluns for almost 20 years.
Damn thing just won't die.
H.

"Walter Maxwell" wrote in message
...
Cecil, H. Adam Stevens sent you an email earlier today, with a cc to me. I
replied to all, but the copy to you came back host unknown. But your
address
appeared as . What's the ONEDOT? I now see it in your
return
address on the rraa.

What's going on?

Walt