Mike Coslo wrote:
Richard is right, There is the first ream!
Sorry, I'm a bit pippish today..........

Ignorance of s-parameter analysis, like ignorance of the
Smith Chart, is not a mortal sin.
--
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:

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,

At last count there are 130 responses to this post, this is #131, and
the question still hasn't been answered.

When energy or power is transmitted in any medium where the wavelength
and the length of the transmission medium are significant percentages of
one another some energy/power is reflected at any discontinuity in the
transmission medium. The reflected energy/power may be re-reflected if a
discontinuity exists in the backward path.

The simplest example that we can all understand is the common case of
the echo!! H E L L O ! .... HELLO ! .... hello ! .... etc. The
energy/power is re-reflected many times until we can't hear it. But is
is still re-reflecting at sub-audible levels until 100% dissipation
occurs. As long as the discontinuities exist the echoes exist!

DD, W1MCE


What you describe as reflection and re-reflection occurs between the
mismatched antenna and the tuner that has been adjusted to minimize
power returned to the transmitter. The sole function of the tuner is
to keep this power from being dissipated by the transmitter (common
experience of arcing, denoting a voltage reflection, or thermal
runaway, denoting a current reflection). The "virtual" reflection
(offered by the tuner) is generally know as the complex conjugate of
the remote load, seen at the near end of the line through which it is
returning. This means that the line transforms the phase and
amplitude of the reflection, and the tuner's job is to invert that
relationship to counteract it, and return it to the antenna.

There are both wave descriptions of this process, and lumped circuit
equivalents. Both work, and both describe the same process from
different points of view. One does not negate the other's validity
(unless, of course, you attempt to mix the points of view and demand
consistency in terms - a frequent rhetorical trap here).

There will no doubt be a flurry of denials to this simple example with
contortions of logic to match. 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.

Keep your eye on how your literal points in your question go abandoned
with these arcane theories.

73's
Richard Clark, KB7QHC

On Tue, 25 May 2004 01:45:19 GMT, Dave Shrader
wrote:
At last count there are 130 responses to this post, this is #131, and
the question still hasn't been answered.

Hi Dave,

Well, actually, it has been answered sufficiently, apparently it
hasn't been understood - quite a gulf between those two positions. As
such, your response doesn't necessarily constitute an answer either,
that is, until the gulf is spanned.

73's
Richard Clark, KB7QHC

Richard Clark wrote:

On Tue, 25 May 2004 01:45:19 GMT, Dave Shrader
wrote:

At last count there are 130 responses to this post, this is #131, and
the question still hasn't been answered.


Hi Dave,

Well, actually, it has been answered sufficiently, apparently it
hasn't been understood - quite a gulf between those two positions. As
such, your response doesn't necessarily constitute an answer either,
that is, until the gulf is spanned.

I've gotten some education from it, and had some fun too!

- Mike KB3EIA -

Richard

You say my question has been answered but I haven't seen the answer, too
old, too stupid, whatever but I still would like to understand. So, would
you be kind enough to give me a better understanding of the "mechanism" in
the transmitter final that can dissipate and transform yet it can't
dissipate a reflection because there's some kind of one way device that acts
like a checkvalve or diode that reflects.
tnx


--
73
Hank WD5JFR
"Richard Clark" wrote in message
...
On Tue, 25 May 2004 01:45:19 GMT, Dave Shrader
wrote:
At last count there are 130 responses to this post, this is #131, and
the question still hasn't been answered.

Hi Dave,

Well, actually, it has been answered sufficiently, apparently it
hasn't been understood - quite a gulf between those two positions. As
such, your response doesn't necessarily constitute an answer either,
that is, until the gulf is spanned.

73's
Richard Clark, KB7QHC

On Tue, 25 May 2004 02:59:52 GMT, "Henry Kolesnik"
wrote:

Richard

You say my question has been answered but I haven't seen the answer, too
old, too stupid, whatever but I still would like to understand. So, would
you be kind enough to give me a better understanding of the "mechanism" in
the transmitter final that can dissipate and transform yet it can't
dissipate a reflection because there's some kind of one way device that acts
like a checkvalve or diode that reflects.
tnx

Hi Hank,

There is no answer as you phrase the question because the transmitter
does dissipate power reflected to it. I can easily imagine you may
find no understanding as I go hyperbolic in the following lines. :-)

It can also reflect (some or all of) what it does not absorb, but only
if it does not match the line connected to it. The degree to which it
absorbs/reflects is determined by the ratio of its Z to the line/load
at the antenna terminal. In this case (bear with me), the source Z is
NOW found at the load's mismatch, transformed through the line. This
means matching works both ways (often the singlemost ignored aspect of
obvious reciprocity in any of these arguments). Such circuit analysis
is called superposition. This is merely an academic way of saying you
look at the problem from both points of view where you reverse roles
(source becomes load, and load becomes source). This is a common
practice learned in first quarter circuit analysis when Kirchhoff's
laws are developed (Thevenin/Norton equivalent circuits also emerge).

For example (a strained one at that), if you had a 500 Ohm Source
characteristic Z, you will never launch much power into a 50 Ohm
system because it would immediately hit a reflective interface at the
antenna connector. This means that a 500 Ohm source, when confronted
by power going towards it from a 50 Ohm system will reflect most of
that power (but how did we get this power into the system in the first
place - Karma?) This, by design, will never happen in any transistor
ham rig built in the last quarter century.

So, this is why you see such a vacuum of response when you (the
general readership "you") ask:
"What is the source Z if it is not 50 Ohms?"
Silence guarantees they either don't know (a fatal admission for egos)
or if they offered a value, we could all balance the checkbook and
that would end the game being played. Hence we are treated to all
these suppositions of shorts and opens or magic reflectors hidden
beneath the hood. The frequent toss-off comment of "no one knows" is
called projection, a psychological salve for the ego meaning if I
don't know, then certainly no one else does either - or they are
wrong.

The answer is the transmitter source Z is 50 Ohms at rated power. If
a watt of power is chooglin' down the line toward it, that 50 Ohms is
going to dissipate into a watt worth of calories. This can be argued
with wave mechanics, or lumped circuit equivalents - doesn't matter
because it's all the same calories. Modern rigs can tolerate this
watt through limited feedback and level controlling circuitry - if you
paid more than a kilobuck for your rig that is. For the rest of us,
it's a spin of the wheel and you take your chance. 1 watt hardly
amounts to much, but are you that lucky that it is ONLY 1 watt? Are
we to suppose those unfortunate souls who blasted their rig
transmitting into a mismatch lost it only because they lacked enough
magic pixie dust? I will bet no rig was lost to a cold snap with
frosty finals.

I've already answered about where the heat can be found, so we will
conserve bandwidth.

=== WARNING to the logic impaired, the following is a supposition ===

Now, lets simply accept those answers that require the magic pixie
dust of total reflection from the transmitter. Fine, the mismatched
antenna reflects some power, this power returns to the transmitter,
the transmitter simple routes it ALL back to the antenna,
round-and-round until the antenna finally radiates it. If this were
true, what do we need tuners for? Take a survey of everyone who
chants this mantra of the rig reflecting, and ask if they have a tuner
in the line. Is it a paper weight holding down their license? Is it
a line stretcher because they needed another foot extension between
the antenna lead and their rig? Dare I point out the utter failure of
their faith?

I will pause to allow those answers to emerge and enjoy the thrashing
dawn of a new era of metaphysics. Ooooohhhhhhmmmmms Laaaaaaaaaaw!

73's
Richard Clark, KB7QHC

Henry Kolesnik wrote:
You say my question has been answered but I haven't seen the answer, too
old, too stupid, whatever but I still would like to understand. So, would
you be kind enough to give me a better understanding of the "mechanism" in
the transmitter final that can dissipate and transform yet it can't
dissipate a reflection because there's some kind of one way device that acts
like a checkvalve or diode that reflects.

By definition, reflected energy dissipated in the source was never
generated in the first place.
--
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 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.

Hank, EM reflections are covered on this web page.

http://www.mellesgriot.com/products/optics/oc_2_1.htm

In particular: "Clearly, if the wavelength of the incident light and the
thickness of the film are such that a phase difference exists between
reflections of p, then reflected wavefronts interfere destructively, and
overall reflected intensity is a minimum. If the two reflections are of
equal amplitude, then this amplitude (and hence intensity) minimum will
be zero."

"In the absence of absorption or scatter, the principle of conservation of
energy indicates all "lost" reflected intensity will appear as enhanced
intensity in the transmitted beam. The sum of the reflected and transmitted
beam intensities is always equal to the incident intensity. This important
fact has been confirmed experimentally."

In order for (rearward-traveling) "reflected intensity" to "appear as
(forward-traveling) enhanced intensity in the transmitted beam", the
momentum of that (rearward-traveling) intensity must change directions.
Thus, it appears that total destructive interference between two rearward-
traveling reflected waves in a transmission line will reverse the direction
of momentum of the energy in those canceled reflected waves.

We need to change a few of your statements:

Any power not dissipated or radiated by an antenna is reflected back.
"Dissipation" means EM energy transformed into heat, according to
the IEEE Dictionary.

The transmitter/tuner end will not re-reflect 100% of the reflected energy
unless there exists a short, open, pure reactance, or "total destructive
interference" as explained in _Optics_, by Hecht.

Besides a short or an open, a purely reactive impedance will cause
100% energy reflection. Apparently, so will "total destructive
interference". Quoting from _Microwave_Transmission_, by J. C. Slater:

"The method of eliminating reflections is based on the interference
between waves. ... The fundamental principle behind the elimination
of reflections is then to have each reflected wave canceled by another
wave of equal amplitude and opposite phase."

That's pretty clear. We get one set of rearward-traveling reflections
at the match point. We get another set of rearward-traveling reflections
at the antenna. If these two sets of reflections are equal in magnitude
and opposite in phase at the match point, they cancel each other and the
rearward-traveling momentum energy in those two waves is conserved by
changing direction to become part of a forward-traveling wave.
--
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
...
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.

Hank, EM reflections are covered on this web page.

http://www.mellesgriot.com/products/optics/oc_2_1.htm

In particular: "Clearly, if the wavelength of the incident light and the
thickness of the film are such that a phase difference exists between
reflections of p, then reflected wavefronts interfere destructively, and
overall reflected intensity is a minimum. If the two reflections are of
equal amplitude, then this amplitude (and hence intensity) minimum will
be zero."

"In the absence of absorption or scatter, the principle of conservation of
energy indicates all "lost" reflected intensity will appear as enhanced
intensity in the transmitted beam. The sum of the reflected and
transmitted
beam intensities is always equal to the incident intensity. This important
fact has been confirmed experimentally."

In order for (rearward-traveling) "reflected intensity" to "appear as
(forward-traveling) enhanced intensity in the transmitted beam", the
momentum of that (rearward-traveling) intensity must change directions.
Thus, it appears that total destructive interference between two rearward-
traveling reflected waves in a transmission line will reverse the
direction
of momentum of the energy in those canceled reflected waves.

We need to change a few of your statements:

Any power not dissipated or radiated by an antenna is reflected back.
"Dissipation" means EM energy transformed into heat, according to
the IEEE Dictionary.

The transmitter/tuner end will not re-reflect 100% of the reflected energy
unless there exists a short, open, pure reactance, or "total destructive
interference" as explained in _Optics_, by Hecht.

Besides a short or an open, a purely reactive impedance will cause
100% energy reflection. Apparently, so will "total destructive
interference". Quoting from _Microwave_Transmission_, by J. C. Slater:

"The method of eliminating reflections is based on the interference
between waves. ... The fundamental principle behind the elimination
of reflections is then to have each reflected wave canceled by another
wave of equal amplitude and opposite phase."

That's pretty clear. We get one set of rearward-traveling reflections
at the match point. We get another set of rearward-traveling reflections
at the antenna. If these two sets of reflections are equal in magnitude
and opposite in phase at the match point, they cancel each other and the
rearward-traveling momentum energy in those two waves is conserved by
changing direction to become part of a forward-traveling wave.
--
73, Cecil http://www.qsl.net/w5dxp
Cecil,

I am not quite sure what you are saying. But, I ran a SPICE simulation of
the following:
1V 1MHz source with resistor R0 feeding a 50 Ohm 250 ns transmission line
shorted at the far end. Independent of R0, in steady state the voltage at
the input end of the transmission line will be 1V. The effect of R0 is to
limit how long it takes to reach steady state. For R0 = 50 Ohms, it is one
cycle; for R0 = 500 Ohms, it is about 8 cycles, as eyeballed off the
waveform display.

Tam/WB2TT wrote:
I am not quite sure what you are saying. But, I ran a SPICE simulation of
the following:
1V 1MHz source with resistor R0 feeding a 50 Ohm 250 ns transmission line
shorted at the far end. Independent of R0, in steady state the voltage at
the input end of the transmission line will be 1V. The effect of R0 is to
limit how long it takes to reach steady state. For R0 = 50 Ohms, it is one
cycle; for R0 = 500 Ohms, it is about 8 cycles, as eyeballed off the
waveform display.

Does SPICE report the steady-state forward and reflected waves
or just the superposition of those two waves? We all know what
they look like when superposed. The question is whether the
identity of the forward and reflected waves disappear after
they are superposed. To the best of my knowledge, the very
existence of standing waves requires the existence of a forward-
traveling wave and a rearward-traveling wave.

I have asked for examples of standing waves void of rearward-
traveling waves and none has been forthcoming.
--
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! =-----