Jim Kelley wrote:
Here's an idea, Cecil. Instead of simply trying to discredit your
correspondent, why don't actually find something wrong with the
equations?

It's obvious you didn't comprehend the difference between
s11=(Z1-Z0)/(Z1+Z0) and rho=s'11=Vref/Vfwd, at least for
awhile. Anyone can copy equations out of a book while maintaining
a misunderstanding of a definition. There is no problem with the
equations. The only problem is with the correspondent's definitions
and it is a minor one that is easy to fix just by getting the
definitions correct.
--
73, Cecil, W5DXP

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

Jim Kelley wrote:
As I said, the reflection coefficient at '+' can be calculated
accurately using just the characteristic impedances, as shown by Born
and Wolf.

Why is this so hard for you to understand? What is the rho
of the following?

source---50 ohm feedline---+---150 ohm feedline---load150

"Just the characteristic impedances" are given. You say you
can "calculate rho accurately" from just that. So prove your
statement. (load150 means the load is not equal to 150 ohms)

If you say rho = (150-50)/(150+50) then you are mistaken. If
Born and Wolf say rho = (150-50)/(150+50) then Born and Wolf
are mistaken. rho = Vref/Vfwd and, contrary to what you say,
there is *NOT* enough information given to calculate rho.
--
73, Cecil, W5DXP

Cecil Moore wrote:

Jim Kelley wrote:
As I said, the reflection coefficient at '+' can be calculated
accurately using just the characteristic impedances, as shown by Born
and Wolf.

Why is this so hard for you to understand?

Answer: it isn't hard for me to understand.

Question: Why must you inevitably get personal in these technical
discussions?

What is the rho
of the following?

source---50 ohm feedline---+---150 ohm feedline---load150

"Just the characteristic impedances" are given. You say you
can "calculate rho accurately" from just that. So prove your
statement. (load150 means the load is not equal to 150 ohms)

That's a different problem, isn't it.

If you say rho = (150-50)/(150+50) then you are mistaken.

Let the record show that I didn't say it. ;-)

If
Born and Wolf say rho = (150-50)/(150+50) then Born and Wolf
are mistaken.

You mean to tell me you don't even know what you're arguing about here?
If you understood and had paid any attention at all you'd have known
what Born and Wolf would say, and you wouldn't be speculating so wildly
about it.

rho = Vref/Vfwd and, contrary to what you say,
there is *NOT* enough information given to calculate rho.

That's _exactly_ what I would say about it.

It wouldn't be possible to evaluate r23 without knowing the load
impedance. And as I pointed out before, the 150 ohm feedline must be
some known number of quarter wavelengths in order to know which form of
the equation to use.

73, Jim AC6XG

On Fri, 03 Oct 2003 14:48:38 -0500, Cecil Moore
wrote:

Jim Kelley wrote:
As I said, the reflection coefficient at '+' can be calculated
accurately using just the characteristic impedances, as shown by Born
and Wolf.

Why is this so hard for you to understand? What is the rho
of the following?

source---50 ohm feedline---+---150 ohm feedline---load150

"Just the characteristic impedances" are given. You say you
can "calculate rho accurately" from just that. So prove your
statement. (load150 means the load is not equal to 150 ohms)


Howzabout (as a variation of another posting):
source50---50 ohm feedline---+---150 ohm feedline---load150

If it is so easy for you, and difficult for Jim, this should be a
slam-dunk.... But I won't hold my breath for either of my posts to
find a literal, numeric answer.

I also promise no more follow-up responses to either thread where no
solution (a numeric one, not a philosophical treatise) is presented.
;-)

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

Richard Clark wrote:
Howzabout (as a variation of another posting):
source50---50 ohm feedline---+---150 ohm feedline---load150

If it is so easy for you, and difficult for Jim, this should be a
slam-dunk....

It is not easy for me or anyone else. It is impossible to accurately
calculate rho just from the above information. Jim is the one who
says it's easy. I say it's impossible. Did you read Roy's thoughts
on the subject?
--
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

Richard Clark wrote:
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.

-sigh- One cannot verify anything about lossless lines on the bench.
The only verifications about lossless lines that are possible have to
be done in one's head because that's the only place lossless lines
exist. One can come close with open-wire line and extrapolate the
results to lossless lines. You see the effects more on coax than on
open-wire line simply because coax is lossier than open-wire line.

I will leave you to discover Chipman's means to find SWR any where
along any line for yourself.

You told me to reference page 139 which I did. All that page talks
about is lossy feedline with a complex Z0. The purely resistive feedline,
given by you in your example, cannot have a complex Z0. So what Chipman
has to say is irrelevant to the problem you posed, i.e. purely resistive
Z0, purely resistive source impedance, and purely resistive load. You
apparently should have posed a complex Z0.

Chipman explains perfectly why the measured SWR may vary with a
lossy line, i.e. with a complex Z0. There are points of conjugate
matching up and down the line where an oscillation takes place. The
oscillation causes extra reflections and re-reflections at the
conjugate match point, an exchange of a third energy between the
capacitive reactance and the inductive reactance at that point,
that affects the SWR readings. But such is not possible with the
purely resistive Z0 that you posed.
--
73, Cecil http://www.qsl.net/w5dxp



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