OK, if we shine a flashlight at a mirror the light bounces back and what
ever is caught by the reflector will be reflected. Take away the reflector
and the reflection just keeps going. If someone can tell me what the hot
filament does perhaps I can understand what happens in the finals, or
whatever.
tnx
Hank WD5JFR
"Cecil Moore" wrote in message
...
Henry Kolesnik wrote:
I just want to know the reflection physics in the Tx, no antenna tuner,
just
a mismatched antenna. I recall no pysics book that tell me how the
reflection sees the transmitter.

The reflected waves obey the laws of physics. The kicker is that we
don't know (and apparently cannot directly measure) the source impedance.
What the reflections can do is modify the designed-for load line through
superposition of the forward and reflected waves. Modification of the
designed-for load line is not desirable and, if unprotected, can cause
over-voltage, over-current, or phase problems.
--
73, Cecil http://www.qsl.net/w5dxp



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Henry Kolesnik wrote:
OK, if we shine a flashlight at a mirror the light bounces back and what
ever is caught by the reflector will be reflected. Take away the reflector
and the reflection just keeps going. If someone can tell me what the hot
filament does perhaps I can understand what happens in the finals, or
whatever.

The hot filament provides a source of electrons. How many
electrons are emitted depends on the instantaneous voltages
on the other elements, the plate and grids. Reflected waves
have an effect on those instantaneous voltages but IMO,
there's not much sense in pursuing the "filament" line of
reasoning.
--
73, Cecil http://www.qsl.net/w5dxp



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On Sun, 23 May 2004 17:18:52 GMT, "Henry Kolesnik"
wrote:

OK, if we shine a flashlight at a mirror the light bounces back and what
ever is caught by the reflector will be reflected. Take away the reflector
and the reflection just keeps going. If someone can tell me what the hot
filament does perhaps I can understand what happens in the finals, or
whatever.
tnx
Hank WD5JFR

Hi Hank,

Well I see in correspondence following this as "replies," that they
are far from satisfactory. All part of the arcana that precedes the
convolutions of math you would have had to endure.

The abandonment of this lead is simply a matter of Cecil's lack of
experience in the metaphor of light.

To answer your question above. Removing the reflector is unnecessary
as it is part of the initial condition and has nothing to do with it
serving as the correlative to a tuner that you want to remove from the
argument (which is a perfectly acceptable imposition of conditions).

Your question also goes to the heart of the matter. We begin with a
hot filament which emits radiation (and yes, no reflector is required
so we will skip that as one of your conditions). The amount of
radiation is directly correlated to the amount of heat. This will
simplify matters, but in the end it will yield a failure of metaphors
(which always occur if you cannot bridge the logic).

The radiation strikes a reflection (immaterial whether complete or
partial) and that portion which returns, impinges upon the filament,
the source. The filament absorbs the power, which in turn raises its
temperature (everyday experience proves the heat of such radiation).
This will, in turn, cause a higher radiation (given the quid-pro-quo
of heat and radiation). In a sense, this means the reflected power is
re-radiated. The confirmation of this is that if you achieved full
reflection, you then define total insulation of the radiation (no heat
escapes) and temperature rises accordingly, and this may lead to
catastrophic failure of the filament (a very bright illumination if
you could see it, and consequent fusing current - electric kilns use
this principle but tolerate the current by under rating the source).
You can imagine the correlative to transmitter failure for the same
conditions.

The failure of the metaphor? RF is not heat (common light is) and the
return of power to be rendered into heat does not result in a higher
RF output.

I will anticipate the sophomore's comments that RF reflections do not
become heat, in and of itself:

The returned power (either through wave mechanics or lumped circuitry)
must result in either a higher potential across the source, or a
higher current through it. Elevated potentials yield the everyday
experience of an arc (heat). Elevated currents yield the everyday
experience of current density through the same element (much like the
filament of our metaphor - heat). One failure mode comes with the
peak power snap, followed by the muttering of "Oh ****!" Or it comes
through the more progressive thermal runaway, followed by the
muttering of "what's that funny smell?"

73's
Richard Clark, KB7QHC

It's true in the case of transmission lines that there are standing
waves and reflections but, unfortunately, this concept has somehow
come to dominate and confuse the concept of matching a tranmitter to
an antenna -- a generator to a load. Like in many areas of science,
mathematicians and scientists often find convenient ways to
mathematically describe and predict physical phenomenon that hinders,
even misleads, the understanding of how it actually works.

If you leave out the complex part of impedences for the moment and
think of 100 volt generator that has a 50 ohm internal impedance
driving a 50 ohm load, current is 1 amp and the power dissipated by
the load is 50 watts. There is also 50 watts dissipated by the
generator's internal impedance, for a total of 100 watts dissipated by
the entire system. Therefore, the "available" power for this generator
is 50 watts.

Maximum available I^2*R power only occurs when the load impedance is
equal to the generator's characteristic impedance, 50 ohms (do the
math). Any load impedance higher or lower, ALWAYS produces less
"available" power.

Herein lies one of the big problems with the "reflection" definition,
conceptually. The generator (transmitter) is not a constant-power
device. When a manufacturer says that it's XYZ transmitter produces
100 watts, it only produces (has available) 100 watts (after internal
dissipation) into a 50 ohm load. Any other load *always* produces less
available power, due to simple I^2*R laws. It has nothing to do with
reflections or standing waves, although, mathmatically, reflection
formulas accurately describe it.

A couple of examples using a 100 volt constant voltage generator and
an internal impedance (RG) of 50 ohms:

1) Load (RL) = 50 ohms. Current = 100 / (50 + 50) = 1 amp
Power dissipated in RL (PL) = (1)^2*50 = 50 watts
Power dissipated in RG (PG) = (1)^2*50 = 50 watts
SWR = RL/RG = PL/PG = 1:1

2) Load (RL) = 100 ohms. Current = 100 / (50 + 100) = .667 amp
Power dissipated in RL (PL) = (.667)^2*100 = 44.5 watts
Power dissipated in RG (PG) = (.667)^2*50 = 22.25 watts
SWR = RL/RG = PL/PG = 2:1

3) Load (RL) = 25 ohms. Current = 100 / (50 + 25) = 1.34 amp
Power dissipated in RL (PL) = (1.34)^2*25 = 44.9 watts
Power dissipated in RG (PG) = (1.34)^2*50 = 89.8 watts
SWR = RG/RL = PG/PL = 2:1

Notice that the total power dissipated in all three examples is
different. The transmitter is NOT a constant-power source, but it's
also not a unlimited power source and has operational limits.

Therefore, what is commonly called "reflected power" is power that
never leaves the transmitter and is dissipated as heat by the
transmitter's internal 50 ohm impedance (if the transmitter's design
doesn't prematurely shut down first).

Al


"Henry Kolesnik" wrote in message ...
I know that any power not dissipated by an antenna is reflected back to the
transmitter. Then the transmitter "reflects" this reflection back to
antenna, ad nauseum until its all gone. I also know that a short or an open
is required to reflect power and I'm searching for which it is, an open or a
short. I'm inclined to think it's a virtual open but I'm at a loss to
understand that and I wonder if someone has a good explanation or analogy
and some math wouldn't hurt.
tnx
Hank WD5JFR

alhearn wrote:
Herein lies one of the big problems with the "reflection" definition,
conceptually.

That's why I often resort to a signal generator with a circulator/load
to illustrate my point. That signal generator *is* a constant power
source.

Therefore, what is commonly called "reflected power" is power that
never leaves the transmitter and is dissipated as heat by the
transmitter's internal 50 ohm impedance (if the transmitter's design
doesn't prematurely shut down first).

You can mount an argument that if the source doesn't see its
source impedance, then there is a reflection at that internal
mismatch. But that's not what is commonly called reflected power.

When we talk about reflected power on this newsgroup, we are usually
referring to the forward power rejected by a mismatch between the
transmission line Z0 and the antenna impedance (associated with mismatch
loss). In a typical ham radio antenna system, the "lost" reflected power
is forced to engage in destructive interference at the tuner and thus
joins the forward power wave.
--
73, Cecil http://www.qsl.net/w5dxp



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Cec et al, what have circulators, S-parameters, etc. to do with HF ?

Not much but neither does much of this thread regarding my original query.
I hope someone that understands the question can answer in a way that my
feeble brain can comprehend. At least we're eliminating what's not the
answer.
73
Hank WD5JFR

"Reg Edwards" wrote in message
...
Cec et al, what have circulators, S-parameters, etc. to do with HF ?

Henry Kolesnik wrote:
Not much but neither does much of this thread regarding my original query.

They have a lot to do with the subject of the thread. :-)

I hope someone that understands the question can answer in a way that my
feeble brain can comprehend. At least we're eliminating what's not the
answer.

Back a dozen years ago, or so, Mr. Bruene tried to 'ping' a final
amp with a slightly off-frequency signal to ascertain the output
impedance of the amp and he published his results in QST. His apparent
error was that he didn't do it at the frequency of operation of the
amp and he didn't know what the 'Q' of a final amp really is. There
has been a running argument ever since, probably best documented in
QEX magazine. There are basically two sides to the argument.

1. If load-pulling causes a falloff of power on each side of the
operating point, then the system is conjugately matched. That
conjugate match includes such things as non-dissipative resistances.

2. Conjugate matches do not exist in a typical amateur system.

Brilliant minds have been trying to prove one or the other and
both sides (IMO) have failed in that proof. There is no final,
definitive proof of either position. If there was such, there
would be no argument. That's why we cannot answer
your question. The picture is further clouded by a definition.
If reflected energy makes its way back into the final amp, it
was never generated in the first place, by definition.
--
73, Cecil http://www.qsl.net/w5dxp


"Reg Edwards" wrote:
Cec et al, what have circulators, S-parameters, etc. to do with HF ?



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Henry, the trouble is nobody can understand your questions. This is because
a "conjugate match" is not relevant. And that's because the internal
impedance of the transmitter is an unknown quantity.

The sole purpose of the so-called SWR meter is to indicate whether or not
transmitter is loaded with 50 ohms.

Can anyone suggest what else it USEFULLY indicates? What else does anyone
need to know?
----
Reg, G4FGQ


"Henry Kolesnik" wrote
Not much but neither does much of this thread regarding my original query.
I hope someone that understands the question can answer in a way that my
feeble brain can comprehend. At least we're eliminating what's not the
answer.
73
Hank WD5JFR

"Reg Edwards" wrote in message
...
Cec et al, what have circulators, S-parameters, etc. to do with HF ?

Reg Edwards wrote:
Cec et al, what have circulators, S-parameters, etc. to do with HF ?

What does scattering have to do with antenna systems with
reflections???? Maybe you should download the HP AN 95-1?
It's available at:

http://www.sss-mag.com/pdf/hpan95-1.pdf

From _Fields_and_Waves_in_Communication_Electronics_ by
Ramo, Whinnery, and Van Duzer. "11.09 Scattering and
Transmission Coefficients ... This article introduces two
of the *most* useful (two-port analysis) forms based
on wave quantities." This is a (c)1965 book, a later version
of a classic fields and waves (c)1944 book by Ramo & Whinnery.
--
73, Cecil http://www.qsl.net/w5dxp



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