Richard Harrison wrote:
"Create a reflected wave near the load that is equal in magnitude but
opposite in phase from the wave reflected by the load; in this way the
two reflected waves cancel each other."

Unfortunately, he didn't say what happens to the intrinsic energy
necessary for those two canceled waves to have existed in the first
place. The question is: At the moment that two waves are canceled,
do they give up their intrinsic energy components? Or, as in the sour
grapes tale, were they never associated with any intrinsic energy to
begin with?
--
73, Cecil http://www.qsl.net/w5dxp



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they're called "evanescent waves"
get a book

"Richard Harrison" wrote in message
news:4061-3F2059E0-171@storefull-yada

It is in there. The waves go away.

Terman said:
blah blah

Richard Harrison wrote:
Terman says the reflected energies cancel.

Does he define, "cancel"? Energy components cannot cease to exist.
I suspect he means they engage in destructive interference which
requires a constructive interference elsewhere according to Hecht.

Equal and opposite reflected energies at the junction are a short.

A short takes the voltage to zero and doubles the current. Are you
sure that is what happens during wave cancellation?

The
energy can`t disappear. It does not go back toward the load. Half has
already been there. The other half has alredy been reflected by Zstub.
Only one route is left to the reflected energy, but Terman says there`s
no reflection on the generator side of the junction. Your Bird wattmeter
won`t see any reflected power. What this says is that the stub produces
an impedance match and the incident energy becomes the same as the
generator output and load energies.

Would you please publish an example?
--
73, Cecil http://www.qsl.net/w5dxp



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H. Adam Stevens wrote:
they're called "evanescent waves" get a book

In one-dimensional transmission lines?
--
73, Cecil http://www.qsl.net/w5dxp



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Cecil, W5DXP wrote:
"Would you publish an example?"

I am error prone as anyone reading many of my postings may attest. The
1955 edition of Terman`s "Electronic and Radio Engineering" is the
culmination of about 25 or more years of Terman`s work which was
checked, rechecked, then checked again. It`s stood the test of time ever
since its publication too. In Sect. 4-11 are answers to questions of
impedance matching to transmission lines.

I think Terman should be consulted straight away for a simple logical
cause of no reflections. It is called impedance matching. The match does
away with reflections.

Best regards, Richard Harrison, KB5WZI

Richard Harrison wrote:

Cecil, W5DXP wrote:
"Would you publish an example?"

I am error prone as anyone reading many of my postings may attest. The
1955 edition of Terman`s "Electronic and Radio Engineering" is the
culmination of about 25 or more years of Terman`s work which was
checked, rechecked, then checked again. It`s stood the test of time ever
since its publication too. In Sect. 4-11 are answers to questions of
impedance matching to transmission lines.

I'll look up the book next time I am at Texas A&M.

I think Terman should be consulted straight away for a simple logical
cause of no reflections. It is called impedance matching. The match does
away with reflections.

Not directed at Richard, but I've seen the following logic used
often with EM waves.

What causes the match? The cancellation of reflections.

What causes the cancellation of reflections? The match.

Does anyone else detect the circular logic problem in the above?

Another one:

'rho' is zero because there are no reflections.

There are no reflections because 'rho' is zero.
--
73, Cecil http://www.qsl.net/w5dxp



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Cecil, W5DXP wrote:
"What causes the match? The cancellation of reflections."

This is like Abbot and Costello. You got it! Who`s on first!

You have a source which only delivers energy at a certain volts to amps
ratio, period!

You have a load which only accepts energy at a certain volts to amps
ratio, period!

When the above ratios are in fact identical there is no problem. When
the above ratios are different something has to give and it isn`t either
ratio, Zo or ZL. Instead, the load takes what the source can deliver and
rejects the surplus volts or amps whichever is the case, as are created
by ZL, and the limited deliverability created by the Zo. The surplus
generates a reflected wave as Cunningham recounted with his description
of missing inductance and capacitance in the missinng continuation of
the transmission line (A short on the line vitiates the capacitance
etc.). The explanation is logical and I recommend his broadcast antenna
book.

What`s not to understand about matching? When the load is not matched,
you add the incremental receptivity required to take the surplus current
or voltage at the load, so that the load, and the stub in Terman`s
example, are a perfect match and there is no longer any surplus to be
reflected. The impedance of the load has been adjusted, by the addition
of a stub in Terman`s example, to Zo. Terman says the wave from the stub
cancels the wave from the mismatched load. No doubt a shorted stub makes
a reflection. The reflected volts from the stub are exactly equal and
opposite to the reflected surplus volts from the load at the junction,
and Terman says there is no reflection toward the generator.

I never won an argument with Terman, though I tried. I believe there is
no reflection past the stub back toward the generator. If so, and it
certainly is so, I`ve put those stubs out at the antenna myself and they
create a match when adjusted properly, the energy on a lossless line
will be the same at the generator and anywhere on lhe line. It`s flat.

Best regards, Richard Harrison, KB5WZI

Richard Harrison wrote:
What`s not to understand about matching?

The energy and momentum in EM waves, particularly, the energy and
momentum in the wave reflected from a mismatched load. What happens
to that energy and momentum? We know it joins the forward wave. How
does that energy and momentum get reversed? A virtual impedance is
a result and not the cause of anything.

The reflected volts from the stub are exactly equal and
opposite to the reflected surplus volts from the load at the junction,
and Terman says there is no reflection toward the generator.

I'm not arguing that point. What happens to the energy in the two waves
that have their voltage and current amplitudes equal in amplitude and
opposite in phase? There is no doubt those waves cancel and there is
no reflection toward the generator. But what happens to the intrinsic
energy that pre-existed the cancellation event?

I never won an argument with Terman, though I tried. I believe there is
no reflection past the stub back toward the generator.

No argument about that. The question is where did the energy and momentum
go when those two waves were canceled thus eliminating reflections back
toward the generator. Waves can be canceled. The energy in those waves
cannot be destroyed.
--
73, Cecil http://www.qsl.net/w5dxp



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On Sat, 26 Jul 2003 11:06:46 -0500 (CDT),
(Richard Harrison) wrote:


These two waves continue to circulate. One in the short length of
transmission line between the mismatched load and the stub placed on the
line to correct the mismatch. The other continues to circulate in the
stub. This circulation is established during the transient period and
continues until it is disestablished. The reflection which would have
prevailed on the line between the generator and the stub junction with
the line, were the stub not there, is simply absent. The line sees a
combined load plus stub impedance at their junction of Zo.

This was confirmed at the bench by my example of a mismatched source
feeding a mismatched load (creating this characteristic of circulating
energy in a purely resistive network) illustrating the Mismatch Error
in the thread:
"The Cecilian Gambit, a variation on the Galilean Defense revisited"
This characteristic is also borne out by the necessity of having low R
components such that each successive pass of the circulating energy is
not power lost to that Ohmage.

There is no reflected wave in the line in the length between the
generator and the stub junction with the line. So, there`s no extra
energy to be accounted for.

Best regards, Richard Harrison, KB5WZI

Hi Richard,

This last point, too, has been a verity for Primary Standards Labs for
generations. Experience at the bench has been heavily discounted here
recently.

73's
Richard Clark, KB7QHC

Cecil wrote:

But that is not what happens when two waves cancel. For an idea of what
happens, please reference Reflections II, page 23-8,9. If the two voltages
are 180 degrees out of phase and the two currents are 180 degrees out of
phase, neither one of them can double. Calling that a virtual short or
virtual open is incorrect, IMO.

Cecil, this is the error on Reflections that I told you about last week when
you cited the above reference in Reflections. I inadvertantly called the
effect a 'short circuit', while it is actually an open circuit. If a 3rd
edition of Reflections is ever printed this error will be corrected. Now
when two waves interfere with their voltages 180 degrees out of phase, but
with their currents in phase, we have a short circuit. Either of these
conditions occur at the matching stub point, depending on the resistance
component of the load impedance and the distance from the load to the stub
point.

Cecil also wrote:
"What happens to the energy in the two waves that have their voltage and
current amplitudes equal in amplitude and opposite in phase?"

These two waves continue to circulate.

I forgot to say they are traveling in the same direction and that they
are coherent. The two waves disappear from existence. What happens to
the energy they contained?

I thought it was common knowledge that the reflected energy is totally
re-reflected in the forward direction when encountering the open or short
circuit at the matching point.

Walt, W2DU