On Fri, 04 Jun 2004 16:37:33 -0500, Cecil Moore
wrote:
My dictionary says that a credit card is money.
Try stop payment on a $20 bill you spent yesterday.

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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On Fri, 04 Jun 2004 23:06:16 -0500, Cecil Moore
wrote:
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.
You know, Walt, just like a wave breaking against the reef. The coral
reflects a wave of water, and a wave of salt, and perhaps a wave of
brine shrimp... If you correspond along these lines long enough, you
may rummage up a chowder from Cecil, but I wouldn't expect
Chateâubriand.

73's
Richard Clark, KB7QHC

On Sat, 05 Jun 2004 05:14:49 GMT, Richard Clark wrote:

On Fri, 04 Jun 2004 23:06:16 -0500, Cecil Moore
wrote:
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.
You know, Walt, just like a wave breaking against the reef. The coral
reflects a wave of water, and a wave of salt, and perhaps a wave of
brine shrimp... If you correspond along these lines long enough, you
may rummage up a chowder from Cecil, but I wouldn't expect
Chateâubriand.

73's
Richard Clark, KB7QHC

Well, Richard, if I ain't gonna git Chateaubriand I ain't goin'.

Walt

On Fri, 04 Jun 2004 23:06:16 -0500, Cecil Moore wrote:

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.

Cecil, your reference below to Chapter 23 in Reflections has re-oriented me and
I'm now on your page. This is the case of a 1/4wl matching transformer. I didn't
recognize my own writing. Sorry.

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."

Yes, Cecil, this quote is my Ref 35, which I used to support this concept in my
QST article in Oct 1974, so I am well familiar with it. And yes, you are
correct in that the waves reflected from points A and B in Fig 6 are in the
phase you state.

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.

But Cecil, take another look at Fig 6 on page 23-5 to note that those two waves
arrive 180 out of phase at point A, which means only that the E and H fields
cancel in the rearward direction only, resulting in a Zo match to the source.
But what's important (and I inadvertantly omitted this fact in the book, which
will be corrected in III) is that when the waves reflected from B reach A they
find a discontinuity at A of an open-circuit type condition, because the line Zo
now goes from low to high (in the rearward direction). Thus when the waves
reflected at B arrive at A the voltage which is already at 360 (0) degrees does
not change phase, but the current which is at 180 degrees on arrival changes by
180 due to the open circuit at A to rearward traveling waves. Thus the current
is now also at 0 degrees. With both voltage and current at 0 degrees relative to
the source, all the power in the waves reflected at B are re-reflected at A and
add to the source power.

Consequently, Cecil, both the E and H fields go to zero only in rearward
direction at the match point, which is necessary for no reflections to travel
rearward of the match point, but the H field goes to zero only temporarily while
the E field is doubled temporarily, as they should when encountering an open
circuit as they do on arriving at A.

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?

I've answered my own dumb question here.

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.

I think I've addressed this paragraph above.

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

As I embarrassingy discovered it's the 1/4wl transformer--dumb me.

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.

I was confused here, Cecil, because you did change the subject from discussing
what happens to the E and H fields when encountering a short to the 1/4wl
transformer example without my catching on to the change. As I said, dumb me.

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 the match point at A is an open 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.

Yep, but only in the rearward direction.

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?

We've answered this.

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.

I'll not argue with the way Hecht describes the phenomena.

G'nite, Cecil, it's 3 :20 AM and I've got to go back to bed.

Walter Maxwell wrote:
But Cecil, take another look at Fig 6 on page 23-5 to note that those two waves
arrive 180 out of phase at point A, which means only that the E and H fields
cancel in the rearward direction only, resulting in a Zo match to the source.

Yes, and that is exactly my point. EXACTLY the same thing happens to the E-fields
and H-fields. That means exactly the same thing that happens to the rearward-
traveling voltages also happens to the rearward-traveling currents. Two equal-
magnitude/opposite-phase voltages cancel. Two equal-magnitude/opposite-phase
currents cancel. That doesn't happen at either an open or a short. If one
looks at just the voltages, it looks like a short. If one looks at just the
currents, it looks like an open.

With both voltage and current at 0 degrees relative to
the source, all the power in the waves reflected at B are re-reflected at A and
add to the source power.

Reflected waves ALWAYS have the voltage and current 180 degrees out of phase.
"With both voltage and current at 0 degrees" implies a forward wave which is
the *EFFECT* of re-reflection, i.e. after re-reflection. The rearward-traveling
voltage and current CANNOT be in phase before the re-reflection. After the re-
reflection, they are no longer rearward-traveling so they are in-phase and forward-
traveling. It's a cause and effect thing.

1. Proper adjustment of the tuner/stub/quarter-wave-transformer CAUSES all
voltages and currents at the perfect match point to be in-phase or 180 degrees
out of phase.

2. Destructive interference between two rearward-traveling waves of equal magnitudes
and opposite phases causes the E-field voltage and H-field current to go to zero
simultaneously, i.e. the E-fields superpose to zero in the rearward direction and
the H-fields superpose to zero in the rearward direction. EXACTLY the same thing
happens to the H-fields as happens to the E-fields in the rearward direction. Neither
a short nor an open acts like that.

"... 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." From the Melles-Groit web page.

3. "... the principle of conservation of energy indicates all "lost" reflected
intensity will appear as enhanced intensity in the transmitted beam." Quoted
from the Melles Groit web page. They don't say how but they do say why. In a
perfectly matched system, all rearward-traveling energy changes direction at
the match point and joins the forward wave as constructive interference. That
constructive interference energy is equal in magnitude to the energy involved
in the destructive interference event. It's a simple conservation of energy
process.

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 the match point at A is an open circuit.

Destructive interference causes the rearward-traveling voltage to go to zero.
Voltage doesn't go to zero at an open circuit. Exactly the same thing happens
to the rearward-traveling voltages and rearward-traveling currents. Voltage
goes to zero at a short-circuit. Current goes to zero at an open circuit. A
match point causes both to go to zero in the rearward direction. That means
the voltage sees a short-circuit looking rearward and the current sees an
open-circuit looking rearward.

"Consequently, all corresponding voltage and current phasors are 180 degrees
out of phase at the matching point. ... With equal magnitudes and opposite
phase at the (matching) point, the sum of the two waves is zero." _Reflections_II_,
page 23-9. This is true for both step-up or step-down impedance discontinuities.
At a perfectly matched point, the two rearward-traveling voltages are ALWAYS of
equal magnitude and opposite phase. The two rearward-traveling currents are
ALWAYS of equal magnitude and opposite phase. That's the only way complete
destructive interference of rearward-traveling waves can occur.

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.

Yep, but only in the rearward direction.

The rearward direction is what we are talking about. The point is that EXACTLY
the same thing happens to the two rearward-traveling current waves as happens
to the two rearward-traveling voltage waves. A short-circuit doesn't affect
voltages and currents in the same way. An open-circuit doesn't affect
voltages and currents in the same way. A match point affects the rearward-
traveling voltages and rearward-traveling currents in EXACTLY the same way.
The re-reflection at a match point is a conservation of energy reflection where
the rearward destructive interference energy supplies energy to constructive
interference in the opposite direction. For light, the equation a

Destructive Interference Irradiance = I1 + I2 - 2{SQRT[(I1)(I2)]} (9.16)

Constructive Interference Irradiance = I1 + I2 + 2{SQRT[(I1)(I2)]} (9.15)

_Optics_, by Hecht, fourth edition, page 388

Note the similarities to equations 13 and 15 in Dr. Best's QEX article,
Part 3.

PFtotal = P1 + P2 - 2{SQRT[(P1)(P2)]} (Eq 15)

PFtotal = P1 + P2 + 2{SQRT[(P1)(P2)]} (Eq 13)

Too bad he didn't label them as Hecht did, as "total destructive interference"
and "total constructive interference" equations.
--
73, Cecil http://www.qsl.net/w5dxp



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Cecil Moore wrote:

Walter Maxwell wrote:
But Cecil, take another look at Fig 6 on page 23-5 to note that those
two waves arrive 180 out of phase at point A, which means only that
the E and H fields cancel in the rearward direction only, resulting
in a Zo match to the source.

Yes, and that is exactly my point. EXACTLY the same thing happens to the
E-fields and H-fields. That means exactly the same thing that happens to
the rearward-traveling voltages also happens to the rearward-traveling
currents.

In my class in secondary school counseling, I learned a technique that
might be helpful here. It's called, "Be the thing." Whatever it is that
you are trying to understand, mentally become that thing. In other words,
assume that you are the reflected current to find out what you
would experience. Obviously, it is just a mental exercise, but one
that I have found quite useful throughout the years.

First, assume that you are the reflected voltage from a mismatched load.
What do you encounter back at the match point? You encounter another
reflected voltage with equal magnitude and opposite phase traveling in
the same rearward direction. What happens to you? Your momentum in the
rearward direction is reversed and your energy starts flowing toward the
load. As a reflected voltage, based on your necessarily limited knowledge,
you assume that you must have encountered a virtual short circuit.

Second, assume that you are the reflected current from a mismatched load.
What do you encounter back at the match point? You encounter another
reflected current with equal magnitude and opposite phase traveling in
the same rearward direction. What happens to you? Your momentum in the
rearward direction is reversed and your energy starts flowing toward the
load. As a reflected current, based on your necessarily limited knowledge,
you assume that you must have encountered a virtual open circuit.

There exists an apparent contradiction. A match point cannot simultaneously
be a virtual short and a virtual open. How is the apparent contradiction
resolved? Is there anything else in physics that can cause a total reflection
of energy besides a short, open, or pure reactance? The answer is, "yes", and
it happens all the time in the field of optics. In a system with only two
directions of energy travel available, total destructive interference in one
direction has to result in total constructive interference in the other
direction. That's the way perfect non-glare thin-film coated glass works in
the presence of a coherent single-frequency laser beam.
--
73, Cecil http://www.qsl.net/w5dxp



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On Sat, 05 Jun 2004 13:05:18 -0500, Cecil Moore wrote:

Cecil Moore wrote:

Walter Maxwell wrote:
But Cecil, take another look at Fig 6 on page 23-5 to note that those
two waves arrive 180 out of phase at point A, which means only that
the E and H fields cancel in the rearward direction only, resulting
in a Zo match to the source.

Yes, and that is exactly my point. EXACTLY the same thing happens to the
E-fields and H-fields. That means exactly the same thing that happens to
the rearward-traveling voltages also happens to the rearward-traveling
currents.

In my class in secondary school counseling, I learned a technique that
might be helpful here. It's called, "Be the thing." Whatever it is that
you are trying to understand, mentally become that thing. In other words,
assume that you are the reflected current to find out what you
would experience. Obviously, it is just a mental exercise, but one
that I have found quite useful throughout the years.

First, assume that you are the reflected voltage from a mismatched load.
What do you encounter back at the match point? You encounter another
reflected voltage with equal magnitude and opposite phase traveling in
the same rearward direction. What happens to you? Your momentum in the
rearward direction is reversed and your energy starts flowing toward the
load. As a reflected voltage, based on your necessarily limited knowledge,
you assume that you must have encountered a virtual short circuit.

Second, assume that you are the reflected current from a mismatched load.
What do you encounter back at the match point? You encounter another
reflected current with equal magnitude and opposite phase traveling in
the same rearward direction. What happens to you? Your momentum in the
rearward direction is reversed and your energy starts flowing toward the
load. As a reflected current, based on your necessarily limited knowledge,
you assume that you must have encountered a virtual open circuit.

There exists an apparent contradiction. A match point cannot simultaneously
be a virtual short and a virtual open. How is the apparent contradiction
resolved? Is there anything else in physics that can cause a total reflection
of energy besides a short, open, or pure reactance? The answer is, "yes", and
it happens all the time in the field of optics. In a system with only two
directions of energy travel available, total destructive interference in one
direction has to result in total constructive interference in the other
direction. That's the way perfect non-glare thin-film coated glass works in
the presence of a coherent single-frequency laser beam.

Yes, Cecil, I understand. However I don't particularly like the notion of saying
both fields go to zero, or both fields go to zero in the rearward direction.
Confusing. Remember, weeks ago I swore that both fields could never go to zero
simultaneously? The reason I disagreed with you is that you didn't mention the
'direction'. The reason I dislike hearing that both fields go to zero is that
it's really not true. Like I've said many times, on encountering a short,circuit
voltage and the E field go to zero and the current and H field doubles AND
REVERSES DIRECTION. To me, Reversing direction is more meaningful and less
confusing than both going to zero, and it still says there is no energy
propagating rearward of the match point.

Going now to the cancellation process when the voltages and currents of both
waves are mutually out of phase. You say that voltages 180 out yields a short
(agreed) and that currents 180 out yields an open. Sounds good, and I mistakenly
agreed a coupla days ago. But I don' think so. I believe voltage 180 out defines
a short--period. Look at it this way. Take a zip cord and put male plugs on both
ends. Plug one end into an outlet, say the top one, and then plug the other end
into the bottom outlet with the polarities reversed. With respect to voltage we
have a 'circuit breaker' short circuit, because the voltages entering the zip
cord at each end were 180 out. But so were the currents initially. Then why the
short circuit current flow? Certainly not because the circuit is open to
current.

Another scenario with the same initial conditions and results: Take two
identical generators delivering the same level of harmonically related output
voltages. Connect their terminals in phase.Voltages in phase--currents in phase.
Result? No current flow. Why? Zero voltage differential. Open circuit to
voltage--open circuit to current.

Now reconnect their terminals in the opposite manner. Voltages 180 out--currents
180 out. Do we have current flow? You bet--dead short! Because current results
from voltage, if voltages are 180 out of phase we have a short to both voltage
and curent. No open circuit to current.

Cecil, I hope we're both still on the same page on this one;

Walt

Walter Maxwell wrote:
Yes, Cecil, I understand. However I don't particularly like the notion of saying
both fields go to zero, or both fields go to zero in the rearward direction.

But Walt, that's exactly what happens when total destructive interference
occurs as explained by J. C. Slater in _Microwave_Transmission_.

I believe voltage 180 out defines a short--period.

That same belief is what got Dr. Best into trouble. He never considered
what happens to the reflected current waves. In a sense, your and his
disagreements are because you both made the same conceptual mistake and
arrived at different conclusions because of that common mistake. If you
and he had not made that shared mistake, you both would have arrived at
the same conclusions.

Another scenario with the same initial conditions and results: Take two
identical generators delivering the same level of harmonically related output
voltages. Connect their terminals in phase.Voltages in phase--currents in phase.
Result? No current flow. Why? Zero voltage differential. Open circuit to
voltage--open circuit to current.

Now reconnect their terminals in the opposite manner. Voltages 180 out--currents
180 out. Do we have current flow? You bet--dead short! Because current results
from voltage, if voltages are 180 out of phase we have a short to both voltage
and curent. No open circuit to current.

This is the problem with trying to use circuit analysis to replace network analysis.
Put the two sources at the two ends of a transmission line and please reconsider
the outcome. Equip the two sources with circulators and dummy loads so the outcome
cannot be in doubt.
--
73, Cecil http://www.qsl.net/w5dxp



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