On Sun, 23 May 2004 10:06:15 -0500, Cecil Moore
wrote:

Richard Clark wrote:
As for the math, you will find it by
the reams, once you've been overwhelmed with the arcana of hyperbolic
descriptions of a novel physics that have to proceed its proof.

A scattering parameter analysis,...
arcana deleted as an obviously fulfilled prophecy.

On Sun, 23 May 2004 09:31:10 +0100, "Ian White, G3SEK"
wrote:

This is a basic cross-check that should always be applied... but
regrettably isn't.

Hi Ian,

Perhaps in this immediate thread. However, I have demonstrated both
sides coming to the same conclusions several many times, and one
example as recently as within this last week.

This issue is not about being right, it is about ego foremost else why
all the debate? Hank has asked a fairly straightforward question
with rather simple terms he could accept as a compelling case. To
this point (some 22 entries) that has been largely abandoned with each
scribbler answering their own imagining of how to discover the
philosopher's stone.

73's
Richard Clark, KB7QHC

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! =-----

Richard Clark wrote:

wrote:
A scattering parameter analysis,...

arcana deleted as an obviously fulfilled prophecy.

Richard, you are the only technical person I know of who
ever considered s-paramater analysis to be a secret or
mystery. It is one of the more technically popular methods
of analysis, ideally suited to transmission line analysis.
--
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! =-----

On Sun, 23 May 2004 02:07:13 GMT, "Henry Kolesnik"
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


Hi Hank,

Round Two.

A short or an open certainly reflects. However, as you have observed
through your question, so does a poorly matched antenna; thus you must
agree that it presents neither a short nor an open. As such,
reflection is not confined to these two conditions.

We can display a condition where we have a 2:1 mismatch. This is
fairly commonplace as a consideration, if not as a reality. Here, the
reflected power is less than the total applied (some 12% if dead
reckoning is correct). That is, if the antenna presents a 2:1
mismatch to the power applied to it, nearly 90% of that power will
proceed to be radiated with a trivial portion returned to the source
(or the tuner). If we boost that mismatch to 10:1, that increases the
reflection substantially (let's call it 90% in this spirit of dead
reckoning) and naturally less is radiated. The math is quite as
simple as a balance ledger.

10:1 for a 50 Ohm source would mean either the load presents a Z of
500 Ohms, or 5 Ohms. From this you can see that we are approaching
either an open (hi-Z) or a short (lo-Z) and either perform the same
job of reflection - short of total reflection. As such, there is no
distinction to the power whether it encounters either, the reflection
ensues by virtue of the simple mismatch, not by the literal condition.
The computation of mismatch defines how reflective the interface is.

73's
Richard Clark, KB7QHC

"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.

Richard Clark wrote:
On Sun, 23 May 2004 05:57:28 +0000 (UTC), "Reg Edwards"
wrote:

Dear Richard, you are confusing the matter even further, if that were
possible.


Old Son,

Have I stolen your thunder?

Like the pendulum with a dull and meagre swing, your predictable
observations on this subject return to an unremarkable SWR, a topic
never raised without the old wife discovering it under her bed....

Does the following strike a bell?

A QUOTATION:

"When you can measure what you are speaking about and express it in numbers
you know something about it. But when you cannot measure it, when you cannot
express it in numbers, your knowledge is of a meagre and unsatisfactory
kind. It may be the beginning of knowledge but you have scarcely in your
thoughts advanced to the state of science."


You have never offered a scintilla of quantized discussion to the
matter.

We had a pet scintilla when I was a kid. Cutest sarn thing you ever
saw, like a cross between a rabbit and a squirrel! 8^)


hehe, - Mike KB3EIA -

Cecil Moore wrote:

Richard Clark wrote:

As for the math, you will find it by
the reams, once you've been overwhelmed with the arcana of hyperbolic
descriptions of a novel physics that have to proceed its proof.


A scattering parameter analysis, described in HP Application Note
95-1 (available on the web) is ideal for analyzing what happens
at a match point in a typical ham radio antenna system.

b1 = s11(a1) + s12(a2)

b2 = s21(a1) + s22(a2)

b1 is the net forward voltage, b2 is the net reflected voltage
a1 is the incident forward voltage, a2 is the incident reflected voltage

Quoting from HP AN 95-1: Another advantage of s-parameters springs
from the simple relationship between the variables a1, a2, b1, and
b2, and various power waves:

|a1|^2 = Power incident on the input of the network.
(forward power incident on the match point)

|a2|^2 = Power reflected from the load.

|b1|^2 = Power reflected from the input port of the network.
(power reflected from the match point back toward the source)

|b2|^2 = Power incident on the load.

The previous four equations show that s-parameters are simply
related to power gain and mismatch loss, quantities which are
often of more interest than the corresponding voltage functions.

|s11|^2 = Power reflected from the network input divided by
power incident on the network input

|s22|^2 = Power reflected from the network output divided by
power incident on the network output

|s21|^2 = Power delivered to a Z0 load divided by power available
from a Z0 source

|s12|^2 = Reverse transducer power gain with Z0 load and source

End quote.

b2 is the voltage reflected back toward the source and

b2 = s21(a1) + s22(a2)

It should be obvious that b2 cannot be zero unless there exists
total destructive interference between s21(a1) and s22(a2), i.e.
s21(a1) is equal in magnitude and opposite in phase to s22(a2).
--
73, Cecil http://www.qsl.net/w5dxp

Richard is right, There is the first ream!

Sorry, I'm a bit pippish today..........

- Mike KB3EIA -

Dave wrote:

"Cecil Moore" wrote:
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.

Uhhhh Dave, the original topic is the Subject: line. If anything,
what happens inside a transmitter is the irrelevant subject since
appreciable reflections hardly ever reach the typical ham 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.

No, those are typical conditions, where the ham radio antenna system
is tuned to a Z0-match by a tuner, either external or internal. It is
not a "very exactly contrived example" at all. It is absolutely typical
of any ham radio installation where the final amp sees close to a 1:1 SWR
and that is the great majority. At the Z0-match point at the input of
every properly tuned transmatch, the voltages and currents are either in
phase or 180 degrees out of phase. If you don't know that, it is no
wonder that you label my power analysis stuff as "contrived".
--
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! =-----

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