Ian White, G3SEK wrote:
So now we have two cases, one where reflections happen and one where
they can't. The transmitter suffers exactly the same problem in both
cases, entirely because it sees the same wrong value of wrong load
impedance.

The transmitter possesses an IQ of zero. Hopefully, yours, mine, and
others exceeds that zero value. The goal is understanding. The goal
is (hopefully) NOT a reductio ad absurdum.

"Reflected power" simply doesn't enter into it.

If you don't care about the facts, reflected power doesn't enter into it.
But if you are trying to understand the physics, certainly the reflected
power enters into it. If you don't care about understanding physics, by
all means, go with your steady-state shortcuts. But please don't try to
talk all the people who are trying to understand the physics into just
accepting your steady-state religion on faith.
--
73, Cecil http://www.qsl.net/w5dxp



-----= Posted via Newsfeeds.Com, Uncensored Usenet News =-----
http://www.newsfeeds.com - The #1 Newsgroup Service in the World!
-----== Over 80,000 Newsgroups - 16 Different Servers! =-----

Ian White, G3SEK wrote:

W5DXP wrote:
Even if the 1625's can't tell the difference, W5DXP can. :-)

That's the whole point - the *only* difference is a conceptual one that
exists inside your mind.

BS, Ian. My pet cockroach can tell the difference between a V/I
ratio resistance and a resistor. Why can't you?
--
73, Cecil http://www.qsl.net/w5dxp



-----= Posted via Newsfeeds.Com, Uncensored Usenet News =-----
http://www.newsfeeds.com - The #1 Newsgroup Service in the World!
-----== Over 80,000 Newsgroups - 16 Different Servers! =-----

Ian White, G3SEK wrote:
Yes! That principle of impedance substitution is so simple, so
fundamental, some people never notice it's there at all.

And you would apparently like to pull the wool over the eyes of everyone
who notices that the definition of impedance has changed in the process.
Shame on you for that attempt at obfuscation!
--
73, Cecil http://www.qsl.net/w5dxp



-----= Posted via Newsfeeds.Com, Uncensored Usenet News =-----
http://www.newsfeeds.com - The #1 Newsgroup Service in the World!
-----== Over 80,000 Newsgroups - 16 Different Servers! =-----

Dave Shrader wrote:
Now the question becomes: "Am I thinking like Cecil or is Cecil thinking
like me?"

Maybe we are both thinking like Ramo and Whinnery, two pretty smart individuals,
who taught me most of what I know about fields and waves.
--
73, Cecil http://www.qsl.net/w5dxp



-----= Posted via Newsfeeds.Com, Uncensored Usenet News =-----
http://www.newsfeeds.com - The #1 Newsgroup Service in the World!
-----== Over 80,000 Newsgroups - 16 Different Servers! =-----

It's important not to confuse the sort of pulses or steps used in TDR
with transient sine wave conditions.

It's perfectly valid to derive the sinusoidal steady state conditions on
a transmission line by looking at the transient conditions that occur
from the time the source is first turned on. And because of the
transient nature of the signal, the most practical way to approach this
analysis is in the time domain.

TDR also (obviously) involves time domain analysis. But it's quite
different. Sinusoidal transient analysis assumes a sinusoidal source
that stays on once it's turned on. But TDR involves either a pulse type
source that's off when the pulse reflection returns, or a step type
source that provides a DC step to the transmission line. In this case,
the source voltage is a stable DC value from the time of the initial
step. In the case of the sinusoidal source, the source voltage continues
changing while the transients are propagating.

In both cases, the sum of all forward and reflected voltages or currents
have to sum to the correct values at all points, and this knowledge can
be used to derive various wave components. But the results and some of
the methods can be very different for the two cases. For example, when a
reactive load or impedance bump is present, a simple reflection
coefficient can be calculated for the sine wave, based on the reactance
at the sine wave's frequency. The reflected wave will be a simple
replica of the incident wave, altered only in phase and amplitude. You
can't do this with a pulse or step; a reactive load changes its shape,
defying a simply defined reflection coefficient. (Some confusion arises
because of the use in TDR of a reflection coefficient, usually denoted
rho. It's the same as the magnitude of the sine wave reflection
coefficient -- but only if the anomaly or load causing the reflection is
purely resistive and a constant value from DC or a low frequency up to
the equivalent maximum frequency contained in the TDR pulse and viewable
with the TDR system. With some TDR systems having equivalent bandwidths
of over 50 GHz, this can be an onerous requirement.) Another important
difference is what happens to a returning wave when it reaches the
source -- reaction to a source that's off, at a stable DC value, or at
some point in the cycle of a sinusoidal waveform is different.

TDR is a very valuable technique, providing important information and
illuminating insights about transmission line phenomena. But great care
has to be taken in extrapolating TDR observations to what happens in a
sinusoidal transient or steady state environment. As readers have seen,
I'm very wary of explanations of sinusoidal phenomena, either steady
state or transient, that depend on drawing parallels to TDR results. You
should be, too.

Roy Lewallen, W7EL

W5DXP wrote:
Tdonaly wrote:

I would like to know why Cecil, for instance, uses pulses, as in a
TDR, in
order to argue a steady state point.


Do steady-state signals obey one set of laws of physics and pulses
obey a different set of laws of physics? You seem to feel so but
I just don't have that much faith!

The useful steady-state shortcuts have developed into a religion that
has no place in science. I am not opposed to steady-state shortcuts.
I am opposed to the steady-state religion that has evolved based on
faith. "Have faith, there is no such thing as reflected waves."
"Have faith, photons can be exchanged between equivalent inductors
and capacitors in a transmission line so they move sideways at less
than the speed of light instead of lengthways at the speed of light."

Particle physicists would really be interested in any proof of that.

"Have faith, a V/I ratio is identical to a physical impedance because
a source, with an IQ of zero, cannot tell the difference." "Reflections
completely disappear the instant that steady-state conditions are reached."
There are many more faith-based characteristics of the steady-state model.
These are just the ones that come to mind.

Roy Lewallen wrote:
It's important not to confuse the sort of pulses or steps used in TDR
with transient sine wave conditions.

Why? Do they obey different laws of physics?

TDR also (obviously) involves time domain analysis. But it's quite
different.

Why? Does a TDR obey a different set of physics laws?

The reflected wave will be a simple
replica of the incident wave, altered only in phase and amplitude.

Not if it contains random noise and all waves contain random noise.

TDR is a very valuable technique, providing important information and
illuminating insights about transmission line phenomena. But great care
has to be taken in extrapolating TDR observations to what happens in a
sinusoidal transient or steady state environment.

Why? Does a TDR obey a different set of physics laws than sine waves?
--
73, Cecil http://www.qsl.net/w5dxp
"One thing I have learned in a long life: that all our science, measured
against reality, is primitive and childlike ..." Albert Einstein



-----= Posted via Newsfeeds.Com, Uncensored Usenet News =-----
http://www.newsfeeds.com - The #1 Newsgroup Service in the World!
-----== Over 80,000 Newsgroups - 16 Different Servers! =-----

It might be of interest to note that old time video amplifiers with peaking
coils sometimes caused ghosts on the screen that were indistinguishable
from ghosts caused by reflections.

Tam/WB2TT
"Ian White, G3SEK" wrote in message
...
W5DXP wrote:
Think what would have happened if you had measured the impedance at
the TX end of your o/c transmission line (very high or very low,
depending on the length) and replaced it with a resistor and
inductor/capacitor giving the same value of R +/- jX.
There's no transmission line, so no traveling waves of anything, and
no reflections - just a transmitter with a very wrong value of load
impedance. The 1625s would have burned up just the same.

Yes they would, but in that case reflections are not the cause of the
impedance.

Good... hold that thought.

In the first case, reflections are the *CAUSE* of the
impedance that burned up the transmitter.

Correct; and the values of R and +/-jX that the transmitter sees are
calculated by considering the reflected voltage and current waves.

So now we have two cases, one where reflections happen and one where
they can't. The transmitter suffers exactly the same problem in both
cases, entirely because it sees the same wrong value of wrong load
impedance.

"Reflected power" simply doesn't enter into it.


--
73 from Ian G3SEK 'In Practice' columnist for RadCom (RSGB)
Editor, 'The VHF/UHF DX Book'
http://www.ifwtech.co.uk/g3sek

I think the steady state solution would be a lot more palatable if the
analysis started at T=0, when you turn the source on. The extra step of
seeing how you reach steady state makes the latter more "real". One could
almost be convinced that there are no reflections in the steady state;
problem is , there would be no other explanation for standing waves, and I
can measure them with unambiguous instruments.

Tam/WB2TT

Tarmo Tammaru wrote:
I think the steady state solution would be a lot more palatable if the
analysis started at T=0, when you turn the source on. The extra step of
seeing how you reach steady state makes the latter more "real". One could
almost be convinced that there are no reflections in the steady state;
problem is , there would be no other explanation for standing waves, and I
can measure them with unambiguous instruments.

I keep wondering what laws of physics get repealed just as the system
transitions to steady-state. Do photons really start moving from side
to side instead of end to end? Do the standing waves magically sustain
themselves without any reflected waves?
--
73, Cecil http://www.qsl.net/w5dxp



-----= Posted via Newsfeeds.Com, Uncensored Usenet News =-----
http://www.newsfeeds.com - The #1 Newsgroup Service in the World!
-----== Over 80,000 Newsgroups - 16 Different Servers! =-----

Did you really read what I wrote?

Roy Lewallen, W7EL

W5DXP wrote:
Roy Lewallen wrote:

It's important not to confuse the sort of pulses or steps used in TDR
with transient sine wave conditions.


Why? Do they obey different laws of physics?

TDR also (obviously) involves time domain analysis. But it's quite
different.


Why? Does a TDR obey a different set of physics laws?

The reflected wave will be a simple replica of the incident wave,
altered only in phase and amplitude.


Not if it contains random noise and all waves contain random noise.

TDR is a very valuable technique, providing important information and
illuminating insights about transmission line phenomena. But great
care has to be taken in extrapolating TDR observations to what happens
in a sinusoidal transient or steady state environment.


Why? Does a TDR obey a different set of physics laws than sine waves?

Tarmo Tammaru wrote:
there would be no other explanation for standing waves,

May I suggest that you consider charge.
Consider how static charge can store energy in the capacitance of the
line.
Consider how moving charge can store energy in the inductance of the
line.
Consider how the charge moves to change the energy distribution within
the line.
Do it for a pulse of charge; then for multiple pulses.
Consider what happens when the pulses collide; consider pulses of the
same polarity and different.
Do it for a step of charge; then make the step so long it looks like DC.
Do it for sinusoids.
Do it for opens, shorts and terminated lines.
Do it for matched, unmatched and disconnected sources (disconnect just
after injecting the pulse, step or sinusoid).
Do it for sources at both ends of the lines.

And soon you will have an explanation which does not require waves
travelling up and down the line to explain the observed voltages and
currents of the standing 'wave'.

....Keith

W5DXP wrote:
Ian White, G3SEK wrote:
Yes! That principle of impedance substitution is so simple, so
fundamental, some people never notice it's there at all.

And you would apparently like to pull the wool over the eyes of everyone
who notices that the definition of impedance has changed in the process.
Shame on you for that attempt at obfuscation!

You are using that principle of impedance substitution whenever you
calibrate your antenna impedance bridge using known values of resistORS,
capacitORS and inductORS.

Of course *you* are aware of the difference in what's connected to the
instrument - you have more information than it has. The only claim Bill
and I have been making is that you cannot tell the difference from any
*electrical* measurement made at a single frequency in the steady
state... and those were exactly the conditions that burned up your
transmitter, so the substitution principle is valid for this branch of
the discussion.

That whole principle relies on the fact that, at the same frequency and
in the steady state, the "definition of impedance" in terms of its
electrical properties does *not* change. That's the whole point.


--
73 from Ian G3SEK 'In Practice' columnist for RadCom (RSGB)
Editor, 'The VHF/UHF DX Book'
http://www.ifwtech.co.uk/g3sek

wrote:
And soon you will have an explanation which does not require waves
travelling up and down the line to explain the observed voltages and
currents of the standing 'wave'.

Yes, soon you will have some mathematical shortcuts. But do mathematical
shortcuts really cause photons to flow sideways between an equivalent
capacitance and an equivalent inductance? Can you describe a bench
experiment where photons are transferred from a capacitor to an inductor
and back?
--
73, Cecil http://www.qsl.net/w5dxp



-----= Posted via Newsfeeds.Com, Uncensored Usenet News =-----
http://www.newsfeeds.com - The #1 Newsgroup Service in the World!
-----== Over 80,000 Newsgroups - 16 Different Servers! =-----

Roy Lewallen wrote:

W5DXP wrote:
In a nutshell, what laws of physics get repealed
just as steady-state is achieved?

Sorry, that's not worthy of a reply.

OK, I'll answer it. No laws of physics get repealed just as
steady-state is achieved. All the steady-state model does is to
allow some mathematical shortcuts. (But it does not allow photons
to oscillate locally between equivalent lumped circuit constants.)

Some people apparently believe that those steady-state mathematical
shortcuts turn around and effect reality. Causing something happen in
reality by just thinking about a mental model is the domain of religion,
not science. The steady-state model does NOT effect reality. Conditions
on a transmission line are exactly the same whether the steady-state
model is used or not. Nothing magic, like disappearing standing waves,
happens at the instant steady-state is achieved.

Too many engineers believe in the primacy of consciousness rather
than in the primacy of existence.
--
73, Cecil http://www.qsl.net/w5dxp



-----= Posted via Newsfeeds.Com, Uncensored Usenet News =-----
http://www.newsfeeds.com - The #1 Newsgroup Service in the World!
-----== Over 80,000 Newsgroups - 16 Different Servers! =-----

Ian White, G3SEK wrote:
You are using that principle of impedance substitution whenever you
calibrate your antenna impedance bridge using known values of resistORS,
capacitORS and inductORS.

Yes, but I comprehend what I am doing. For you to imply the "electrical
properties don't change" between a 50 ohm dummy load and a 50 ohm dipole
antenna is simply ridiculous.

That whole principle relies on the fact that, at the same frequency and
in the steady state, the "definition of impedance" in terms of its
electrical properties does *not* change. That's the whole point.

The electrical properties *can* change and that's the whole point. The
electrical properties of a 50+j0 dummy load and a 50+j0 antenna are
almost completely different.

A transmission line can transfer photons. Can a lumped constant L-C
model transfer photons?
--
73, Cecil http://www.qsl.net/w5dxp



-----= Posted via Newsfeeds.Com, Uncensored Usenet News =-----
http://www.newsfeeds.com - The #1 Newsgroup Service in the World!
-----== Over 80,000 Newsgroups - 16 Different Servers! =-----

Ian, G3SEK wrote:
"A related topic would be the effect of tank circuit Q on bandwidth of
HF amplifiers;"

Class A amplifiers are little used as HF finals, so in practical
amplifiers current is only part-time.

Impedance of a parallel resonant circuit is high. Circuit impedance
rises with inductance. Q rises with capacitance.

A Class C plate tank introduces a load on tube or transistor. It should
waste only a small percentage of the power generated. It should have
enough Q to linearize the output of the amplifier.

Terman says it is easy to show that the Class C tank circuit efficiency
is: 1 - Qloaded/Qunloaded.

Loaded Q is the ratio of the circulating volt-amperes to the transmitted
watts. If Q is too high, bandwidth is too narrow. If Q is too low,
harmonics are high.

As Q is ordinarily high, the tank circuit impedance is higher than the
load on the amplifier.

Impedance on the Class C amplifier has little effect on the tube or
transistor loading. Output impedance presented by the transmitter to the
load is determined in many cases by the percentage of the time the
amplifier is switched-off.

Best regards, Richard Harrison, KB5WZI

Cecil,

Try this out. For a line driven by a sin wave, there is a lumped parameter
equivalent circuit with impedance R + JX for any line length and any
termination. For a line exited by a pulse, this equivalent circuit is an
infinite series with nuls at n/PW; there is no length of line where the line
is resonant.

Tam/WB2TT
"W5DXP" wrote in message
...
Do steady-state signals obey one set of laws of physics and pulses
obey a different set of laws of physics?

W5DXP wrote:
Ian White, G3SEK wrote:
You are using that principle of impedance substitution whenever you
calibrate your antenna impedance bridge using known values of
resistORS, capacitORS and inductORS.

Yes, but I comprehend what I am doing. For you to imply the "electrical
properties don't change" between a 50 ohm dummy load and a 50 ohm dipole
antenna is simply ridiculous.

I didn't either say that or imply it. What's truly "ridiculous" is for
you to *infer* that I did.

I think I've already made my points well enough for other readers to
judge, so I really am done this time.

No doubt you'll have the last word, Cecil. Use it well.

--
73 from Ian G3SEK 'In Practice' columnist for RadCom (RSGB)
Editor, 'The VHF/UHF DX Book'
http://www.ifwtech.co.uk/g3sek

Was Descartes an engineer??

DD, W1MCE

W5DXP wrote:

Roy Lewallen wrote:

[SNIP]

Too many engineers believe in the primacy of consciousness rather
than in the primacy of existence.

I left out the word "tank" in the sentence: Tank impedance on the Class
C amplifier has little effect on tube or transistor loading." Sorry.
Sometimes I delete too much when I shuffle things on the screen, I
should write it first.

Best regards, Richard Harrison, KB5WZI

Tarmo Tammaru wrote:
Try this out. For a line driven by a sin wave, there is a lumped parameter
equivalent circuit with impedance R + JX for any line length and any
termination.

Yes, but that's ALL it is - just an equivalent circuit. "Equivalent"
doesn't mean equivalent radiation. "Equivalent" doesn't mean equivalent
forward and reflected waves. It ONLY means equivalent NET voltage and
equivalent NET current at a point. It is a shortcut. It does NOT effect
reality. Past the net voltage and net current point absolutely nothing
is equivalent.
--
73, Cecil http://www.qsl.net/w5dxp



-----= Posted via Newsfeeds.Com, Uncensored Usenet News =-----
http://www.newsfeeds.com - The #1 Newsgroup Service in the World!
-----== Over 80,000 Newsgroups - 16 Different Servers! =-----

Ian White, G3SEK wrote:

W5DXP wrote:
Yes, but I comprehend what I am doing. For you to imply the "electrical
properties don't change" between a 50 ohm dummy load and a 50 ohm dipole
antenna is simply ridiculous.

I didn't either say that or imply it. What's truly "ridiculous" is for
you to *infer* that I did.

I didn't have to infer anything, Ian, those words in quotes are *your
quoted words*. Here they are again:

That whole principle relies on the fact that, at the same frequency
and in the steady state, the "definition of impedance" in terms of
its electrical properties does *not* change.

That certainly implies that there is no difference between the electrical
properties of the impedance of a 50 ohm dummy load and a 50 ohm antenna.
--
73, Cecil http://www.qsl.net/w5dxp



-----= Posted via Newsfeeds.Com, Uncensored Usenet News =-----
http://www.newsfeeds.com - The #1 Newsgroup Service in the World!
-----== Over 80,000 Newsgroups - 16 Different Servers! =-----

Jim Kelley wrote:
Your
problem proves that your analysis and the notion of power flow are
flawed.

You haven't answered my question so you have not earned the right to
set up a new straw man. Why does Pref1 suddenly go to zero just as
Pref2 arrives at the impedance discontinuity?

As Timo and I have both said to you, it can be shown that that 50
joules/sec does not travel rearward.

You have not presented an iota of evidence that Pref2 does not travel
rearward. Half of that 50 joules/sec comes from rearward-traveling Pref2.
Sorry, until you address that contradiction in your concepts, you don't
have any creditability on this subject.

You have repeatedly refused to answer this simple question: How does
the energy in the reflected wave from the mismatched load get turned
around? We know it possesses momentum so it does turn around. What is
your physics mechanism for explaining the change in direction of
momentum of Pref2?

You have already admitted that wave cancellation is
responsible for Pref1 being zero.

"Admitted" is a funny word for it.

After months of denying it, you finally admitted it. Admitted seems
entirely appropriate.

When waves cease to exist, they are forced to give up their intrinsic energy.

Waves don't cease to exist.

That's where you are wrong. Waves cease to exist when they encounter a
matched dummy load, for instance. Waves also cease to exist when they
are destroyed by wave cancellation. Light waves cease to exist when they
encounter a perfectly flat black surface. Your assertion is false.
--
73, Cecil http://www.qsl.net/w5dxp



-----= Posted via Newsfeeds.Com, Uncensored Usenet News =-----
http://www.newsfeeds.com - The #1 Newsgroup Service in the World!
-----== Over 80,000 Newsgroups - 16 Different Servers! =-----

W5DXP wrote:

Jim Kelley wrote:
Your
problem proves that your analysis and the notion of power flow are
flawed.

You haven't answered my question so you have not earned the right to
set up a new straw man. Why does Pref1 suddenly go to zero just as
Pref2 arrives at the impedance discontinuity?

Has wave cancellation suddenly become a point of contention?

Your latest example obviously shows the weakness in your argument.

As Timo and I have both said to you, it can be shown that that 50
joules/sec does not travel rearward.

You have not presented an iota of evidence that Pref2 does not travel
rearward.

The absence of any evidence that it does presents a more formidable
problem, in my opinion.

You have repeatedly refused to answer this simple question: How does
the energy in the reflected wave from the mismatched load get turned
around?

That's a lie, and everyone here knows it. I've answered the question
every time it was posed. Timo answer it the same way I have. Energy
does not get turned around - it never flows to the left of the
discontinuity.

We know it possesses momentum so it does turn around.

Must be the Royal 'we'.

What is
your physics mechanism for explaining the change in direction of
momentum of Pref2?

What reason can you provide for continuing to beat your neice?

You have already admitted that wave cancellation is
responsible for Pref1 being zero.

"Admitted" is a funny word for it.

After months of denying it, you finally admitted it. Admitted seems
entirely appropriate.

That's an absolute fabrication. Revisionist history. I was the one who
INSISTED that no power ever flowed back from the discontinuity. In fact
you argued with me about it. You've got a major mental glich happening
there, Cecil.

Waves cease to exist when they encounter a
matched dummy load, for instance.

The wave is transformed at a load. It doesn't simply "cease to exist".
But I can see this is leading to another of your symantics arguments.
You could avoid them in the future by using convention terminology. But
I doubt you really want to avoid them. Seems they're the 'secret
weapon' of the newsgroup warrior.

73, ac6xg

Cecil begs to differ:

The electrical properties *can* change and that's the whole point. The
electrical properties of a 50+j0 dummy load and a 50+j0 antenna are
almost completely different.


Like what, as far as the source and feedline are concerned?
Ain't that basically "AC" RF power being produced by the TX, flowing through
the transmission line, to 50+j0? (Resistance/impedance)
Who gives the hoot if that 50+j0 is power eating resistor or power barfing
antenna. Call it virtual(y) non-radiating antenna or virtual(y) radiating
resistor. Simply transducers looking pretty to the feedline and doing their
assigned thing. F the SWR, matchit !
RIP

BUm

W5DXP wrote:

There are many more faith-based characteristics of the steady-state model.

Actually, the steady state "model" describes the observable aspects of
the phenomenon. According to a least one theorist, physics is a science
of measurement. Science is of course a study of observable phenomena.

73, AC6XG

W5DXP wrote:

Jim Kelley wrote:

W5DXP wrote:
Has wave cancellation suddenly become a point of contention?

It was our point of contention for months. Finally, you relented.

It's been my impression that wave cancellation and interference has
always been the one thing we've agreed upon.

Your latest example obviously shows the weakness in your argument.

It obviously shows the strength of my argument since you avoid
discussing it at all costs. :-)

If I wasn't discussing it, you obviously wouldn't have had anything to
post reply comments about. That your numbers are the same in two
scenarios, each involving a different input power level perfectly
illustrates your misinterpretion of the meaning of the numbers.

I've answered the question every time it was posed.

But your "answer" is always a bogus non-answer, an empty mantra.

So now you agree that I've answered the question. The fact you disagree
with the answer is irrelevant to that point.

Energy does not get turned around - it never
flows to the left of the discontinuity.

According to Ramo & Whinnery, reflected energy does indeed flow
rearward from a mismatched load. That reflected energy possesses
momentum in the rearward direction and changes direction at the
impedance discontinuity. You still have not offered an acceptable
explanation for that energy and momentum turn around.

Apparently, neither have Ramo & Whinnery. But that's understandable,
given that the idea is entirely your invention.

What reason can you provide for continuing to beat your neice?

Your only response is a diversion. That speaks volumes.

It speaks Cecilian, actually. You didn't notice the similarity in
technique? It was a response in kind to the exact form of the question
you asked.

I was the one who
INSISTED that no power ever flowed back from the discontinuity. In fact
you argued with me about it.

I NEVER argued with you about that.

I cannot recall a single instance of you ever agreeing with anything
I've written on the subject - including (V3 + V4) * (I3 + I4) = 0,
and P3+P4-(2*SQRT(P3*P4)=0. That's wave cancellation my friend, and I
posted these things at the beginning of this discussion. You've always
argued with the validity of these equations.

It was a mistake to try again to be civil to you.

When did that happen?

The wave is transformed at a load. It doesn't simply "cease to exist".

The energy in the wave is transformed from RF to heat by the dummy
load. The RF wave certainly does cease to exist.

As I said, we're now in an argument over semantics caused by your
creative use of terminology. "Cease to exist" in any case implies
something which isn't true.

But back to the point, for what amount of time do the cancelled waves
"exist" in order that they might then be able to "cease to exist"? If
you say during the transient period between T0 and steady-state, then
we're in agreement.

73, Jim AC6XG

Yuri Blanarovich wrote:
Who gives the hoot if that 50+j0 is power eating resistor or power barfing
antenna.
1
You don't care whether all your power is going into a dummy
load or into your antenna?
2
We are trying to figure out something about transmission lines
so we take away the transmission line and replace it with an
"equivalent circuit" that doesn't act like a transmission at all?
How in the world does that tell us anything about transmission
lines?
--
73, Cecil, W5DXP

1
Am I getting sucked into the bottomless black hole of transmission lines?
Yea, I care! If I want to test or tune my transmitter or amplifier and need a
dummy, I use dummy load. When I need to radiate killer signals I use my Razor
beams, but they are designed to have 50 ohm impedance and broad bandwidth
(possible, been there, done it) no matching crap, no reflections (ok very
small) and I don't give a Freak about nurturing reflections on the line. I hate
reflected waves and I suppress their generation/reflection/travel and endless
discussions about problem that nobody wants.

2
Huh? You haven't figured that SWR is bad for transmission lines?
All I want to know about transmission line that it doesn't have impedance
bumps, keeps its impedance, that it has lowest possible loss and doesn't
radiate or let the water in. (That goes for open wire feeders too :-)
Oh, color doesn't matter and it should be repulsive to chipmunks and other SWR
ignorant critters.
Why is this endless argument about reflections going on? We know they are bad,
we know how to eliminate them, so what's the problem? You love them so much
that you want to keep them in your coax?
There are no reflections around this shack, only in the mirror of my ugly face
after 48 hr contest.
OK hit me now that it is impossible to have reflectionless antenna-coax-tx.
My answer, yea, there are some (very little) and they are insignificant to me
to worry about and lose sleep and hours at the keyboard. Ins't it like eunuch
dreaming about sex? :-)

SK
73 Yuri da noSWR BUm

Jim Kelley wrote:

W5DXP wrote:
Has wave cancellation suddenly become a point of contention?

It was our point of contention for months. Finally, you relented.

Your latest example obviously shows the weakness in your argument.

It obviously shows the strength of my argument since you avoid
discussing it at all costs. :-)

You have not presented an iota of evidence that Pref2 does not travel
rearward.

The absence of any evidence that it does presents a more formidable
problem, in my opinion.

I quoted the evidence from Ramo & Whinnery. You just ignored it.

I've answered the question every time it was posed.

But your "answer" is always a bogus non-answer, an empty mantra.

Energy does not get turned around - it never
flows to the left of the discontinuity.

According to Ramo & Whinnery, reflected energy does indeed flow
rearward from a mismatched load. That reflected energy possesses
momentum in the rearward direction and changes direction at the
impedance discontinuity. You still have not offered an acceptable
explanation for that energy and momentum turn around.

What reason can you provide for continuing to beat your neice?

Your only response is a diversion. That speaks volumes.

I was the one who
INSISTED that no power ever flowed back from the discontinuity. In fact
you argued with me about it.

I NEVER argued with you about that. You are back to your lying ways.
I would NEVER argue with a Bird wattmeter? You obviously have me
confused with Dr. Best. In fact, I published part of my article
saying that everything takes place at exactly the impedance discontinuity
and nowhere else. Having to lie about what I have said is just proof of
the weakness of your argument.

I'm sorry, Jim. I refuse to engage in a discussion at your chosen ethical
and moral level. It was a mistake to try again to be civil to you.

The wave is transformed at a load. It doesn't simply "cease to exist".

The energy in the wave is transformed from RF to heat by the dummy
load. The RF wave certainly does cease to exist. You're not back to
your phantom waves that last forever without a source of energy, are
you?
--
73, Cecil, W5DXP

Yuri Blanarovich wrote:
Who gives the hoot if that 50+j0 is power eating resistor or power barfing
antenna.

You don't care whether all your power is going into a dummy
load or into your antenna?

We are trying to figure out something about transmission lines
so we take away the transmission line and replace it with an
"equivalent circuit" that doesn't act like a transmission at all?
How in the world does that tell us anything about transmission
lines?
--
73, Cecil, W5DXP

Yuri Blanarovich wrote:
Huh? You haven't figured that SWR is bad for transmission lines?

An SWR between 4:1 and 16:1 is what allows me to use a 130 ft.
$20 dipole fed with 400 ohm window line on all HF bands without
needing an antenna tuner. What's wrong with live and let live?
I find your approach to antennas boring as heck but I am not
going to rag on you about it. Different strokes for different
folks. One has to have a transmission line anyway - might as
well let it perform the matching function.
--
73, Cecil http://www.qsl.net/w5dxp



-----= Posted via Newsfeeds.Com, Uncensored Usenet News =-----
http://www.newsfeeds.com - The #1 Newsgroup Service in the World!
-----== Over 80,000 Newsgroups - 16 Different Servers! =-----

Jim Kelley wrote:
I cannot recall a single instance of you ever agreeing with anything
I've written on the subject - including (V3 + V4) * (I3 + I4) = 0,
and P3+P4-(2*SQRT(P3*P4)=0.

You really need to get your head checked. I am the person who first
posted P3+P4-[2*Sqrt(P3*P4)]=0. It is proof of wave cancellation, the
event you vehemently denied for about six weeks. And I agree that
(V3*I3) + (V4*I4) + interference = (V3+V4)*(I3+I4) = 0. In fact,
since you admitted that wave cancellation exists at an impedance
discontinuity in a Z0-matched line, we have very little disagreement
left. The only thing we disagree on now is how long it takes the two
rearward-traveling wavefronts to cancel. I say it happens in a dt of
time as dt approaches zero. You say it happens in zero time. Just how
far apart are those two concepts?

You've always argued with the validity of these equations.

BS! You argue loud and long, eventually change your mind, and then
come back in a few days with The Big Lie - that is what you believed
all the while. Anyone who has been following this discussion has
witnessed you using that underhanded technique any number of times.

It was a mistake to try again to be civil to you.

When did that happen?

I tried to be nice to you, Jim, and you spit in my face all over again.
Please find someone else to abuse.

But back to the point, for what amount of time do the cancelled waves
"exist" in order that they might then be able to "cease to exist"? If
you say during the transient period between T0 and steady-state, then
we're in agreement.

Exactly what laws of physics completely change during the transient period
between TO and steady-state? Photons start moving sideways instead of carrying
energy up and down the transmission line???? More bafflegab!

Pref2(1-|rho|^2) obviously exists all the way from the mismatched load back
to the impedance discontinuity. You have avoided explaining that momentum
reversing mechanism like the plague. How do you get those photons turned
around?
--
73, Cecil http://www.qsl.net/w5dxp



-----= Posted via Newsfeeds.Com, Uncensored Usenet News =-----
http://www.newsfeeds.com - The #1 Newsgroup Service in the World!
-----== Over 80,000 Newsgroups - 16 Different Servers! =-----

Cecil, W5DXP wrote:
"That`s where you are wrong."

This argument has evoked plain statements, i.e., "When waves cease to
exist, they are forced to give up their intrinsic energy." And, "Waves
don`t cease to exist."

The statements need qualifications. Perhaps waves "cancel" without
ceasing to exist.

My speculation is that two radiated fields which cancel don`t eliminate
each other at all. They simply coincide out-of-phase, and their
resultant is zero along an azimuth where cancellation of their effect
continues. If we had a way to identify the vectors composing the zero
resultant, we could prove them there. Separate modulation might be
contrived to perform identification. The modulation idea comes from what
happens as a null azimuth in a MW BC radiation pattern is approached.
Carrier and sideband frequencies don`t cancel exactly together and it
sounds weird.

On wires, it`s different. Connect same-frequency energy exactly
out-of-phase, and you have a short circuit. In space, you don`t have an
electric current. You may have zero electrons. You have only fields
until you encounter a conductor.

Best regards, Richard Harrison, KB5WZI

Yuri Blanarovich wrote:
Huh? You haven't figured that SWR is bad for transmission lines?

An SWR between 4:1 and 16:1 is what allows me to use a 130 ft.
$20 dipole fed with 400 ohm window line on all HF bands without
needing an antenna tuner. What's wrong with live and let live?
I find your approach to antennas boring as heck but I am not
going to rag on you about it. Different strokes for different
folks. One has to have a transmission line anyway - might as
well let it perform the matching function.
--
73, Cecil http://www.qsl.net/w5dxp



Did I deny to let you live with your SWRed piece of wire? Be happy.

If my 7 el Razor, Quad Yagi log driven design is boring, and your piece of wire
is exciting, may the SWR be with you!

Gee, what a revolutionary concept, let your transmission line do the matching
for your doublet with open wire feeders. Yo everybody, tear down your coax fed
beams, arrays, towers, just put up doublet and work the WAS in no time. Can you
picture all those stupid contesters, serious DXers with decent antennas wasting
their efforts to eliminate SWR from their feedlines? Yep, I can picture trying
to use coax feedlines for matching kilowats to non-resonant antennas. Good for
melting the ice in the gutters.

Be happy with your doublet, I am looking for "boring" killer antennas that can
hear and work stuff that others can't. To me much more fun and challenge than
your SWR endeavors.

da (Quagi) Razor BUm

Art Unwin wrote:
"Way back in this thread you alludedd to antennas as being
transformers."

Art seems ready to bring up #*+@%$! ("the thing" as Phil Harris would
say) yet another time.

I would like to see Art produce some hard numbers indicating improvement
by his "thing" as compared with other antennas. Art has indicated he
wants others to produce his numbers, even refine his design. I`d like
Art to even differentiate his "thing" from a "T" or "Delta" match,
regardless of where the feedline wires go.

Fresh participants might be persuaded to do Art`s work or salute his
design.

Best regards, Richard Harrison, KB5WZI

Richard Harrison wrote:
The statements need qualifications. Perhaps waves "cancel" without
ceasing to exist.

Some waves cancel without ceasing to exist. But if the cancellation
is permanent, the waves simply cease to exist.

My speculation is that two radiated fields which cancel don`t eliminate
each other at all.

That is true, but that is not what we are discussing. We are discussing
permanent wave cancellation within the confines of a transmission line.

On wires, it`s different. Connect same-frequency energy exactly
out-of-phase, and you have a short circuit.

No you don't, Richard. Maximum current occurs at a short circuit.
The net current from two canceled waves is zero. The net voltage
from two canceled waves is zero. It is neither a short circuit
nor an open circuit to the canceled waves. It is simply wave
cancellation. To the canceled waves, it looks like both a short
circuit to the two voltages and an open circuit to the two currents.

It is the same thing that happens at the air to thin-film interface
in perfect non-glare glass when the incident beam is coherent.
--
73, Cecil http://www.qsl.net/w5dxp



-----= Posted via Newsfeeds.Com, Uncensored Usenet News =-----
http://www.newsfeeds.com - The #1 Newsgroup Service in the World!
-----== Over 80,000 Newsgroups - 16 Different Servers! =-----

Thought I'd change the thread name to more accurately reflect its
content. This seems to be the fate of nearly all threads in this newsgroup.

Roy Lewallen, W7EL

On Tue, 22 Jul 2003 10:38:13 -0700, Roy Lewallen
wrote:

Thought I'd change the thread name to more accurately reflect its
content. This seems to be the fate of nearly all threads in this newsgroup.

Roy Lewallen, W7EL

Boy, ain't that the truth.

Danny, K6MHE

Roy, you've been getting blamed for everything lately.

Now we can blame you for the new thread ... you started it! grin

DD, W1MCE

Roy Lewallen wrote:

Thought I'd change the thread name to more accurately reflect its
content. This seems to be the fate of nearly all threads in this newsgroup.

Roy Lewallen, W7EL

Dave Shrader wrote:
Roy, you've been getting blamed for everything lately.
Now we can blame you for the new thread ... you started it! grin

Roy doesn't seem to appreciate me making hamburger out of ham radio's
sacred cows. :-) I actually enjoy the T-Bones best of all.
--
73, Cecil http://www.qsl.net/w5dxp



-----= Posted via Newsfeeds.Com, Uncensored Usenet News =-----
http://www.newsfeeds.com - The #1 Newsgroup Service in the World!
-----== Over 80,000 Newsgroups - 16 Different Servers! =-----