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.
73
Hank WD5JFR
There are no stupid questions, just stupid people asking!
"Tdonaly" wrote in message
...
Henry wrote,

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



This post is guaranteed to get Cecil revivified. Here's a hint: quit
thinking
solely in
terms of power, that's for fellows who want to explain how it all works
without
going into
any of the complicated details. Get an undergraduate physics text that
discusses
waves, and read it, or, better yet, take a class. Take all explanations
you
read in
amateur publications with a grain of salt. After you've done this, you
still
won't be
able to argue with Cecil, because that requires an extensive knowledge of
the
moronic,
unfair, and downright pathalogical debating techniques of which Cecil is a
master.
But, it will be harder for you to fall for some of the crackpot ideas
you're
liable to
read on this newsgroup, and it will give you something to think about when
sipping
your after-dinner port.

73,
Tom Donaly, KA6RUH

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



-----= Posted via Newsfeeds.Com, Uncensored Usenet News =-----
http://www.newsfeeds.com - The #1 Newsgroup Service in the World!
-----== Over 100,000 Newsgroups - 19 Different Servers! =-----

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



-----= Posted via Newsfeeds.Com, Uncensored Usenet News =-----
http://www.newsfeeds.com - The #1 Newsgroup Service in the World!
-----== Over 100,000 Newsgroups - 19 Different Servers! =-----

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



-----= Posted via Newsfeeds.Com, Uncensored Usenet News =-----
http://www.newsfeeds.com - The #1 Newsgroup Service in the World!
-----== Over 100,000 Newsgroups - 19 Different Servers! =-----

I thought filaments produced photons/light waves as well. You took me to
light now you want to leave!
Come on I just want a good basic understand on what it is the the Tx
reflects the power , how it does it and a little simple math. My dad used
to say if you can't explain something you think you know to someone else it
might because you don't understand it yourself or lack command of the
language. In this case for me it's both.
73
Hank WD5JFR
"Cecil Moore" wrote in message
...
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



-----= Posted via Newsfeeds.Com, Uncensored Usenet News =-----
http://www.newsfeeds.com - The #1 Newsgroup Service in the World!
-----== Over 100,000 Newsgroups - 19 Different Servers! =-----

Henry Kolesnik wrote:
I thought filaments produced photons/light waves as well. You took me to
light now you want to leave!

Well, the light emitted by tube filaments is irrelevant to the RF
function. I use light examples because light and RF are both EM
waves and a lot more is known about light than about RF.

Come on I just want a good basic understand on what it is the the Tx
reflects the power , how it does it and a little simple math. My dad used
to say if you can't explain something you think you know to someone else it
might because you don't understand it yourself or lack command of the
language. In this case for me it's both.

What I am trying to say is that I don't know what happens inside
a transmitter. Under Z0-matched conditions, as with a tuner, I
believe that what happens inside a transmitter is pretty much
irrelevant. The transmitter sees its designed-for impedance and
that is essentially all that matters.

What I am willing to discuss in detail is what happens at a Z0-match
point (x) in an antenna system with reflections - something like the
following:

XMTR---51.5 ohm line---x---1/4WL 300 ohm line---1749 ohm load
100W -- 200W --
-- 0W -- 100W

For these typical conditions, all voltages and currents are either
in-phase or 180 degrees out of phase at the match point (x), which
makes a power analysis the most simple analysis 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 100,000 Newsgroups - 19 Different Servers! =-----

"Cecil Moore" wrote in message
...

What I am willing to discuss in detail is what happens at a Z0-match
point (x) in an antenna system with reflections - something like the
following:

what you are willing to discuss is irrelevent as it has nothing to do with
the original topic which was about what happens in the transmitter.


For these typical conditions, all voltages and currents are either
in-phase or 180 degrees out of phase at the match point (x), which
makes a power analysis the most simple analysis of all.

that should read "For these specific conditions", those conditions are
hardly 'typical', they are a very exactly contrived example which makes it
easy to compare powers. thus making the proper analysis of currents or
voltages seem unnecessary. taking simple cases like this and improperly
generalizing them is what leads to the worst mis-conceptions and circular
arguments in this group.

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