On Fri, 3 Oct 2003 07:39:15 +0100, "Ian White, G3SEK"
wrote:

Sorry! Just to continue and further confuse the haggling, the forward
voltages are unknown because one does not know, in the case of amateur
systems, what is the internal voltage and internal impedance of the
transmitter.

It is this unknown voltage and internal impedance which the so-called SWR
(Rho) meter merely ASSUMES.


Reg, that can't possibly be you. Someone has hijacked your e-mail.

Where is either of those assumptions required? Those are transmitter
properties, and they only affect the overall level of
power/voltage/current on the line. Reflection coefficient (rho) and SWR
are properties exclusively of the line and its load, not the
transmitter.


Hi Ian,

There are one of two possible explanations for your posting:
1. You have not obtained that copy of Chipman that you ordered.
2. You have not read it.
Of course, you can add a third, fourth or fifth... in complete absence
of Chipman's discussion if his material does not agree with your
interpretations.

This is not an unexplored topic, and in fact dates back to earlier
discussions whose citations to Chipman were offered by me to no refute
- merely denial and the general wholesale abandonment of learned
posters who preferred to chase after specious claims (simpler game).

73's
Richard Clark, KB7QHC

Richard Clark wrote:
There are one of two possible explanations for your posting:
1. You have not obtained that copy of Chipman that you ordered.
2. You have not read it.
Of course, you can add a third, fourth or fifth... in complete absence
of Chipman's discussion if his material does not agree with your
interpretations.

Richard, you might be interested to know that HP's s-parameter ap note,
AN 95-1, page 22 under Transducer Power Gain, lists the power available
from the source as the (square of the magnitude of the source voltage)
divided by [one minus the (square of the magnitude of the source's complex
reflection coefficient)], i.e. |Vs|^2/(1-|rho|^2)=power available from
the source where presumably source-rho = (Zs-Z0)/(Zs+Z0)
--
73, Cecil, W5DXP

On Fri, 03 Oct 2003 12:01:08 -0500, Cecil Moore
wrote:

Richard Clark wrote:
There are one of two possible explanations for your posting:
1. You have not obtained that copy of Chipman that you ordered.
2. You have not read it.
Of course, you can add a third, fourth or fifth... in complete absence
of Chipman's discussion if his material does not agree with your
interpretations.

Richard, you might be interested to know that HP's s-parameter ap note,
AN 95-1, page 22 under Transducer Power Gain, lists the power available
from the source as the (square of the magnitude of the source voltage)
divided by [one minus the (square of the magnitude of the source's complex
reflection coefficient)], i.e. |Vs|^2/(1-|rho|^2)=power available from
the source where presumably source-rho = (Zs-Z0)/(Zs+Z0)

Hi Cecil,

-sigh- even when you offer confirmatory recitations you still miss the
details. There are only 11 pages in Application Note 95-1 and the
material you describe appears on page 4 not 22.

The voltage from the generator is also portrayed in Fig. 3 entitled
"Flow graph of network of Fig. 2." Figure 2, of course, shows the
generator complete with Zs which most here deny exists, or dismiss as
immaterial to any discussion. This is due entirely to their speed
reading past their own sources' discussion that ALL DISCUSSION OF SWR
assumes the source matches the line it feeds. Such an explicit or
implicit relationship is fundamentally required, or the entire text
that they cite is rendered useless gibberish. The most garbled of
those proclamations is that the source Z has no bearing on line SWR.

This same flowgraph is present in many similar works (AN 95-1 is
hardly unique) and being presented early in the work (like Chipman's
similar observation of requiring source-line matching) is skipped so
that the reader (sic) can scrounge their favorite snippet of math and
remove it from its required context. Chipman also presents much the
same treatment in non S-Parameter discussion, but that is quite
obviously from the part unread by the great mass of so called
adherents to his discussion.

However, to give some flexibility to the discussion; such shortfalls
of understanding how SWR works is simply through lack of experience in
the matter. It is understandable when the usual approach to this
topic is taken by employing a transmitter that both specifies its
output at a Z of 50 Ohms and exhibits a Z of 50 Ohms. Given such a
source, the casual debater is lulled into the comfortable illusion of
having been born on third base thinking they hit a triple in the
debate against source Z (no, the count is three strikes).

Simply because they encounter no ill consequence of source mismatch is
NOT evidence of the source Z being immaterial to the process of
measuring SWR. Luck counts for nothing in debate - unless it is
admitted to. None here count themselves lucky - it would diminish
their sense of erudition.

I don't expect there will be any substantive discussion following this
that will change physics to conform to those illusions (my comments
here will not "change their minds").

73's
Richard Clark, KB7QHC

Richard Clark wrote:

wrote:
-sigh- even when you offer confirmatory recitations you still miss the
details. There are only 11 pages in Application Note 95-1 and the
material you describe appears on page 4 not 22.

-sigh- The PDF version of HP ap note AN 95-1 contains 79 pages.

Simply because they encounter no ill consequence of source mismatch is
NOT evidence of the source Z being immaterial to the process of
measuring SWR.

A source mismatch affects the power available from the source. The
SWR does not depend upon the power available from the source. The
SWR is the same whether the source is 1% efficient or 99% efficient.
Efficiency depends upon Zs. SWR does not.
--
73, Cecil, W5DXP

On Fri, 03 Oct 2003 13:49:15 -0500, Cecil Moore
wrote:

A source mismatch affects the power available from the source. The
SWR does not depend upon the power available from the source. The
SWR is the same whether the source is 1% efficient or 99% efficient.
Efficiency depends upon Zs. SWR does not.

Hi Cecil,

So, a source exhibiting 200 Ohms Resistive feeding a 50 Ohm line that
is then terminated in a load of 200 Ohms Resistive exhibits what SWR?

The absence of a numeric answer is par for the course here. The
answer, of course, can be found in Chipman's text but that requires
the act of reading, not snipping (which would still be available to
the literate). Many here stumble when it comes to measuring SWR
employing (in this case) 2 resistors and a hank of line - how they
could imagine they respond faithfully to more elaborate enquiries is
quite amusing, especially when they argue the Source Z has nothing to
do with it.

My mental image of that assemblage of pundits is that of them crowded
on a small desert isle, each proclaiming it to be a vast, lush
continent. Another SWR Don added to that bunch will teeter someone
into the brine. ;-)

Do any of you know how to tread water? Seems to be the perfect
Darwinian thinning mechanism; but in fact most already tread water at
high tide.

73's
Richard Clark, KB7QHC

Richard Clark wrote:
So, a source exhibiting 200 Ohms Resistive feeding a 50 Ohm line that
is then terminated in a load of 200 Ohms Resistive exhibits what SWR?

Assuming a lossless 50 ohm line, the steady-state SWR is 4:1 no
matter what the source impedance (assuming there exists a source
voltage not equal to zero volts). Steady-state SWR in a lossless
feedline is a constant fixed voltage ratio from end to end.
--
73, Cecil, W5DXP

On Fri, 03 Oct 2003 14:58:15 -0500, Cecil Moore
wrote:

Richard Clark wrote:
So, a source exhibiting 200 Ohms Resistive feeding a 50 Ohm line that
is then terminated in a load of 200 Ohms Resistive exhibits what SWR?

Assuming a lossless 50 ohm line, the steady-state SWR is 4:1 no
matter what the source impedance (assuming there exists a source
voltage not equal to zero volts). Steady-state SWR in a lossless
feedline is a constant fixed voltage ratio from end to end.

Hi Cecil,

WRONG!

The method of computation you employ violates Chipman and any other of
a host of authoritative sources on the topic. If you actually
attempted to verify this at the bench you would "perhaps" find it at
only one point, or at harmonic distances (wavelength specific) along
the line. That means you would stand only a couple of percent chance
of that with a random choice. Guesses are not responsive to the
intent and no points are awarded.

I will leave you to discover Chipman's means to find SWR any where
along any line for yourself. You might enjoy the celebrity of being
the second to do so. ;-)

73's
Richard Clark, KB7QHC

On Fri, 03 Oct 2003 18:11:14 GMT, Richard Clark wrote:

On Fri, 03 Oct 2003 12:01:08 -0500, Cecil Moore
wrote:

Richard Clark wrote:
There are one of two possible explanations for your posting:
1. You have not obtained that copy of Chipman that you ordered.
2. You have not read it.
Of course, you can add a third, fourth or fifth... in complete absence
of Chipman's discussion if his material does not agree with your
interpretations.

Richard, you might be interested to know that HP's s-parameter ap note,
AN 95-1, page 22 under Transducer Power Gain, lists the power available
from the source as the (square of the magnitude of the source voltage)
divided by [one minus the (square of the magnitude of the source's complex
reflection coefficient)], i.e. |Vs|^2/(1-|rho|^2)=power available from
the source where presumably source-rho = (Zs-Z0)/(Zs+Z0)

Hi Cecil,

-sigh- even when you offer confirmatory recitations you still miss the
details. There are only 11 pages in Application Note 95-1 and the
material you describe appears on page 4 not 22.

The voltage from the generator is also portrayed in Fig. 3 entitled
"Flow graph of network of Fig. 2." Figure 2, of course, shows the
generator complete with Zs which most here deny exists, or dismiss as
immaterial to any discussion. This is due entirely to their speed
reading past their own sources' discussion that ALL DISCUSSION OF SWR
assumes the source matches the line it feeds. Such an explicit or
implicit relationship is fundamentally required, or the entire text
that they cite is rendered useless gibberish. The most garbled of
those proclamations is that the source Z has no bearing on line SWR.

This same flowgraph is present in many similar works (AN 95-1 is
hardly unique) and being presented early in the work (like Chipman's
similar observation of requiring source-line matching) is skipped so
that the reader (sic) can scrounge their favorite snippet of math and
remove it from its required context. Chipman also presents much the
same treatment in non S-Parameter discussion, but that is quite
obviously from the part unread by the great mass of so called
adherents to his discussion.

However, to give some flexibility to the discussion; such shortfalls
of understanding how SWR works is simply through lack of experience in
the matter. It is understandable when the usual approach to this
topic is taken by employing a transmitter that both specifies its
output at a Z of 50 Ohms and exhibits a Z of 50 Ohms. Given such a
source, the casual debater is lulled into the comfortable illusion of
having been born on third base thinking they hit a triple in the
debate against source Z (no, the count is three strikes).

Simply because they encounter no ill consequence of source mismatch is
NOT evidence of the source Z being immaterial to the process of
measuring SWR. Luck counts for nothing in debate - unless it is
admitted to. None here count themselves lucky - it would diminish
their sense of erudition.

I don't expect there will be any substantive discussion following this
that will change physics to conform to those illusions (my comments
here will not "change their minds").

73's
Richard Clark, KB7QHC

Richard, I'm dismayed with your statements above. Are you really serious? Or are
you just giving Cecil a bad time?

I've been grappling with your last email to me concerning the nature of the
source resistance of RF amps, and as with your statements above, I'm at a loss
as to how to respond, because we are 180 degrees apart on the source resistance
issue. I'm still going to respond to it, but right now I want to address the SWR
issue.

Richard, how can you possibly believe that the output impedance of the source
has any effect on the SWR on a transmission line? The only conditions
responsible for SWR are the Zo of the line and the ZL of the load--nothing else.
I've been bench measuring SWR for more than 50 years, beginning with using the
slotted line before more sophisticated machinery was available. It didn't matter
what the source impedance was, the SWR remained the same, whatever the source.
Ian told it like it is, and so does Walter C. Johnson in his "Transmission Lines
and Networks, Page 100, where he says:

"The steady state ratio Eplus/Eminus was determined in Eq 11 as the reflection
coefficient k...This ratio is determined only by the load and the line, not by
the generator. It is completely unaffected by the quantity kg = (Zg - Zo)/(Zg +
Zo), which is the reflection coefficient seen by an individual
backward-traveling wave as it reaches the generator terminals. ...the latter
affects the steady state solution only on its influence on the sending-end
voltage, i.e., through its influence on the magnitude of the entire solution."

Your reply comments, please.

Walt, W2DU

On Fri, 03 Oct 2003 21:41:15 GMT, Walter Maxwell wrote:


Richard, I'm dismayed with your statements above. Are you really serious? Or are
you just giving Cecil a bad time?

That wouldn't take much to push Cecil off dead center.


I've been grappling with your last email to me concerning the nature of the
source resistance of RF amps, and as with your statements above, I'm at a loss
as to how to respond, because we are 180 degrees apart on the source resistance
issue. I'm still going to respond to it, but right now I want to address the SWR
issue.

Richard, how can you possibly believe that the output impedance of the source
has any effect on the SWR on a transmission line?

Stephen Adam of HP using Beatty describes it quite well. The data you
have by email and has been posted here demonstrates it equally well.
It takes no more than two resistors and a length of line to confirm or
deny. My data confirms it, absolutely no one has offered negative
evidence, simply denials.

The only conditions
responsible for SWR are the Zo of the line and the ZL of the load--nothing else.
I've been bench measuring SWR for more than 50 years, beginning with using the
slotted line before more sophisticated machinery was available. It didn't matter
what the source impedance was, the SWR remained the same, whatever the source.
Ian told it like it is, and so does Walter C. Johnson in his "Transmission Lines
and Networks, Page 100, where he says:

If Walter Johnson was not explicit about it, he was certainly implicit
about the requirement that the source match the line it is driving for
any discussion of SWR. This is so commonplace that no one ever
examines the situation where the source is a mismatch.

Too many here simply flip to the section in their favorite book about
SWR and wholly neglect the fundamentals that present this simple
requirement. I have presented quotes, chapter and verse from Chipman
where he explicitly says as much, and those who hold Chipman have
abandoned discussion rather than refute those quotes or accept their
error. As one scribbler put it I was not going to "change his mind."
I have no doubt of that, such a statement paints one into an extremely
embarrassing corner once having uttered it. One thing I learned as a
Metrologist is that I am always wrong, the significance is in the
degree of error, not the philosophy of sin and the rejection in
ignorance.

Any number of correspondents here "might" have the capacity to simply
repeat my methods and report their data; but absolutely none
demonstrate it. I might be so far in error the meter is pegged, but
the quality of "sneer review" absolves me of sin. ;-)

Hi Walt,

I await your response by email for our last round of discussion. What
is presented above is old material already discussed. There is
nothing new presented by me in it that has not found its way to your
mailbox.

73's
Richard Clark, KB7QHC

Richard Clark wrote:
I have presented quotes, chapter and verse from Chipman
where he explicitly says as much, ...

And I have presented quotes, chapter and verse from Chipman,
that disprove your interpretations of what he said.
--
73, Cecil http://www.qsl.net/w5dxp



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