Richard Clark wrote:

Cecil Moore wrote:
Ever notice how many SWR meters have 3.0 at half scale?

Amateur grade - 1 in 4;

Even my double-needle meters have 3:1 as a vertical line
at the center. All of my single-needle SWR meters have 3.0
at half scale since they are actually reading |rho|.

Check out this humoungous one. Click on the small meter
picture for a full-size picture.

http://www.mfjenterprises.com/produc...prodid=MFJ-868
--
73, Cecil http://www.qsl.net/w5dxp

But how does the attenuation length of non-TEM modes relate to the Zo of
a transmission line?

ac6xg

Cecil Moore wrote:

Actually, it is called an argumentum ad verecundiam, an appeal
to authority - a technical authority in this case. I don't know
Kevin G. Rhodes at Dartmouth. He merely answered my question on
s.p.e. that didn't find an answer on this newsgroup. Exactly what
did you find technically wrong with the following evidence?

****Quote****
Newsgroups: sci.physics.electromag
From: "Kevin G. Rhoads"
Date: Tue, 07 Oct 2003 12:49:14 -0400
Subject: Transmission Line Question

For 10 MHz I would expect that all other modes
would be non-propagating (i.e., evanescent) even though RG-213
is a large coax (improved RG-8 apparently). The speed of propagation
is listed as 66%, so the nominal wavelength is 3/2 times the free
space wavelength for the TEM mode. 3/2 x 30m = 45m, which implies
the decay rate in space for non-TEM modes is going to be large
as the cable diameter is .405" (jacket) which implies the
spacing from inner to outer conductors will be less than .203".
For order of magnitude estimate, assuming the lowest non-TEM mode
can be approximated using a characteristic equation that really
is only applicable in Cartesian geometries:
(1/45m)**2 = (1/.203")**2 + kz**2
Clearly, kz must be imaginary to make this work. thus an
evanescant, non-propagating wave:
kz**2 = (1/45m)**2 - (1/.203")**2
To the accuracy used to date, the first term on the right
is negligible, so the decay rate, alpha, can be estimated:
alpha**2 = - (kz)**2 = (1/2.03")**2
Or, the lowest order undesired mode should reduce intensity
by a factor of 1/e (0.37) in about 2.03"; power will reduce
by that factor squared in the same distance (.135). In
about four inches, undesired mode power is down to about
0.018, in six inches, .00248, and after a foot, 6.14x10-6

You should double check my algebra, but I think the estimate
is reasonable. To put it into other terms, since the wavelength
in the coax dielectric is 45m and the conductor to conductor
spacing is about 2", any non-TEM mode will suffer attenuation
in E-field intensity with a space-rate constant rounghly
equal to the conductor to conductor spacing. INtensity
drops by 1/e = 1/2.71828 every 2 inches. Power availalbe
drops faster, being square of intensity.

So unless almost all the power diverts into an undesireable
mode (by a factor of more than a million to one), one foot
of cable should see pure TEM at the end.
***End Quote***

On Tue, 11 Oct 2005 16:57:51 GMT, Cecil Moore wrote:
I dug up some calculations from sci.physics.electromag
which you recite here; then in sci.physics.electromag you can quote
their use by authorities (sic both times) in
rec.radio.amateur.antenna....
This appeal is called a circle of friendship - not evidence.
Actually, it is called an argumentum ad verecundiam, an appeal
to authority - a technical authority in this case.

Actually - you have said nothing that removes this from a circle of
old wives sitting around the stove in the kitchen.

Name dropping is not an appeal to authority, nor are third hand
quotes.

On Tue, 11 Oct 2005 17:11:15 GMT, Cecil Moore wrote:
Even my double-needle meters have 3:1 as a vertical line
You have a rather limited exposure to the field of instrumentation.

Jim Kelley wrote:

But how does the attenuation length of non-TEM modes relate to the Zo of
a transmission line?

And to whether it's parallel or coaxial? Good question.
Kevin Rhodes' email address can be had from Google.
--
73, Cecil http://www.qsl.net/w5dxp

Richard Clark wrote:
Name dropping is not an appeal to authority, nor are third hand
quotes.

Exactly what did you find technically wrong with that
third hand quote?
--
73, Cecil http://www.qsl.net/w5dxp

I would then assume you disregard anything written in books as it falls in
the same category.

"Richard Clark" wrote in message
...
On Tue, 11 Oct 2005 16:57:51 GMT, Cecil Moore

Name dropping is not an appeal to authority, nor are third hand
quotes.

"Cecil Moore" wrote
Reg, I dug up some calculations from sci.physics.electromag
from about a year ago that indicate one foot of 50 ohm coax
on each side of the Bird is enough to make the line real,
i.e. not imaginary, and that's a conservative estimate.
===========================================

Cec, I still have a 30-watt, 160m, portable transceiver which I made
about 20 years back. It's in an aluminium attache case and still
works. In my travelling days I used to toss a wire out of the hotel
bedroom window.

Lift the lid and on the front panel is a 1.5"-square moving-coil
meter. It is used as a TLI on transmit and as an S-meter on receive.

The meter scale is calibrated 0-500 microamps.

But my imagination doesn't fool ME.

On receive, when the meter indicates 50% of full-scale deflection I
know that the meter is actually measuring 250 microamps.

And on transmit, when the meter indicates 90% of full scale deflection
I know that the meter is actually measuring 450 microamps.

Let this little anecdote be a friendly warning to they who use meters
with a 0 to infinity SWR scale, or scaled in terms of forward and
reverse power.
----
Reg, G4FGQ.

Richard Clark wrote:
On Tue, 11 Oct 2005 09:11:19 +0100, Ian White G/GM3SEK
wrote:

The subsequent conversion to VSWR is a mathematical relationship only.

Hi Ian,

This seems to be a particularly notable difference - to which
absolutely NO ONE has ever deviated from in ANY determination of SWR!

That is to say, this "mathematical" distinction that some rely on to
differentiate their arguments has not got one scintilla of difference
over any other method.

The only way to claim you "directly" measure SWR is to find some way
to place two probes of a meter along the line such that one probe goes
into the trough and the other into the peak and the meter reads SWR
directly. Unfortunately for rhetoric's sake, this STILL renders the
determination in terms of a mathematical relationship. It cannot be
escaped.

Thank you, you're right. The key difference between direct and indirect
measurements is not about the need for mathematics; it's about the need
for additional input from theory.

What I should have said is:

When you calculate the VSWR from measurements of maximum and minimum
voltage on the line, that simple division formula comes directly from
the definition of VSWR. The measurement is direct and completely
self-contained, needing no further input from transmission-line theory.

But you cannot calculate VSWR from a single-point measurement of
reflection coefficient unless without some additional input from
transmission-line theory which connects them together. This dependence
on additional theory is what makes the measurement an indirect one.


Why this keeps on being revisited must be to allow the new lurkers to
observe my correction.

Yes, by all means.


--
73 from Ian G/GM3SEK 'In Practice' columnist for RadCom (RSGB)
http://www.ifwtech.co.uk/g3sek

Reg Edwards wrote:


Let this little anecdote be a friendly warning to they who use meters
with a 0 to infinity SWR scale, or scaled in terms of forward and
reverse power.

This lesson is easier to remember if your first transmitter had an anode
current meter scaled in "Gallons Gone".


--
73 from Ian G/GM3SEK 'In Practice' columnist for RadCom (RSGB)
http://www.ifwtech.co.uk/g3sek

"Cecil Moore" wrote
Reg, I dug up some calculations from sci.physics.electromag
from about a year ago that indicate one foot of 50 ohm coax
on each side of the Bird is enough to make the line real,
i.e. not imaginary, and that's a conservative estimate.

============================================

Cec, you forgot to say sci.physics.electromag were working at 500 MHz
and above. The one and only so-called SWR meter I have stops at 30
MHz. ;o)
----
Reg.

Cecil Moore wrote:
Ian White G/GM3SEK wrote:
All agreed. Along with the math that Cecil has retrieved and quoted
again, everything points towards the distance in question being a
function of coax diameter only; and not wavelength.

Please forgive my previous senior moment.
It was ~2% of a wavelength at 10 MHz for RG-213.
It appears that one foot of coax on each side of
a Bird wattmeter is enough to establish Z0 at
50 ohms which forces Vfor/Ifor=Vref/Iref=50,
the necessary Bird boundary conditions.

The Bird doesn't require any upstream and downstream boundary
conditions. You can insert the instrument between any source impedance
and any load impedance, and what it reports is entirely about the load
impedance, unaffected by the source impedance.

However, it was scaled and calibrated assuming a 50 ohm system reference
impedance, so in order to read correctly, it requires you to agree that
your system reference impedance is 50 ohms too.

The element is trying to sample the voltage and current at a single
point along the instrument's internal line. Because that line is
physically quite long, it is built as an accurate 50-ohm line so that
the instrument will cause minimal disturbance when inserted somewhere
along a 50-ohm cable.


--
73 from Ian G/GM3SEK 'In Practice' columnist for RadCom (RSGB)
http://www.ifwtech.co.uk/g3sek

On Tue, 11 Oct 2005 16:57:51 GMT, Cecil Moore wrote:


****Quote****
Newsgroups: sci.physics.electromag
From: "Kevin G. Rhoads"
Date: Tue, 07 Oct 2003 12:49:14 -0400
Subject: Transmission Line Question
....
So unless almost all the power diverts into an undesireable
mode (by a factor of more than a million to one), one foot
of cable should see pure TEM at the end.
***End Quote***

But Cecil, nowhere in the analysis you quoted does it estimate how
much power is diverted from the dominant mode at the discontinuity.

If the explanation of the discontinuity is that some power is
converted from dominant propagation mode to another mode, and that
those other modes are evanescent, it seems that this analysis of the
impact of the discontinuity considers only estimating the decay rate
of the evanescent modes without estimation of the relative magnitude
of the power diverted to those modes in the first place.

Owen
--

Reg Edwards wrote:
Let this little anecdote be a friendly warning to they who use meters
with a 0 to infinity SWR scale, or scaled in terms of forward and
reverse power.

If the scale were linearly calibrated for |rho| = 0 to 1,
would you still be objecting?

SWR = (1+|rho|)/(1-|rho|) for 0 = |rho| = 1

When SWR = infinity, it doesn't mean infinite current
through the meter. It just means that Vref = Vfor
where |Vref|/|Vfor| = |rho|. My Heathkit HM-15 allows
for full scale to equal Vfor. Then Vref is applied.
The scale is a linear indication for |rho|. The
corresponding SWR scale just follows the above equation.
--
73, Cecil http://www.qsl.net/w5dxp

Reg Edwards wrote:
Cec, you forgot to say sci.physics.electromag were working at 500 MHz
and above.

Not true, Reg. My question was specified using RG-213 at 10 MHz.
--
73, Cecil http://www.qsl.net/w5dxp

Ian White G/GM3SEK wrote:
The Bird doesn't require any upstream and downstream boundary
conditions.

When Bird requires a 50 ohm environment, they are requiring
50 ohm boundary conditions for the reading to be valid. If
you install the Bird in a 450 ohm environment on both sides
of the wattmeter, for instance, it will NOT read a valid forward
power and reflected power. In a matched-line 450 ohm environment
with absolutely zero reflected power, the Bird will indicate an
SWR of 9:1, a |rho| of 0.8 and a ratio of reflected power to
forward power of 0.64 even when the reflected power is zero.
--
73, Cecil http://www.qsl.net/w5dxp

Owen Duffy wrote:
If the explanation of the discontinuity is that some power is
converted from dominant propagation mode to another mode, and that
those other modes are evanescent, it seems that this analysis of the
impact of the discontinuity considers only estimating the decay rate
of the evanescent modes without estimation of the relative magnitude
of the power diverted to those modes in the first place.

I've always had a rule of thumb that 100 times the diameter of
the coax was enough length to 99% establish the TEM mode so
Kevin's explaination made sense to me.

Apparently, you are not satisfied with Kevin's explaination. Kevin
Rhodes email address is available on Google. The reason not to
publish it here is to avoid spaming his email account.
--
73, Cecil http://www.qsl.net/w5dxp

On Tue, 11 Oct 2005 21:15:22 GMT, Cecil Moore wrote:


Not true, Reg. My question was specified using RG-213 at 10 MHz.

True enough, but in the context of the question as to whether the Bird
43 reads sufficiently accurately, the transmission line on which one
is interested in the decay of the evanescent modes is the Bird
Thruline coupler section, not Rg-213 or any other cable that might be
attached to the Bird.

Owen
--

On Tue, 11 Oct 2005 21:25:53 GMT, Cecil Moore wrote:

Ian White G/GM3SEK wrote:
The Bird doesn't require any upstream and downstream boundary
conditions.

When Bird requires a 50 ohm environment, they are requiring
50 ohm boundary conditions for the reading to be valid. If
you install the Bird in a 450 ohm environment on both sides
of the wattmeter, for instance, it will NOT read a valid forward
power and reflected power. In a matched-line 450 ohm environment
with absolutely zero reflected power, the Bird will indicate an
SWR of 9:1, a |rho| of 0.8 and a ratio of reflected power to
forward power of 0.64 even when the reflected power is zero.

(I am assuming your 450 ohm line to be an unbalanced line, impractical
as that is, but the issues of balance to unbalanced transition are
just noise to the discussion.)

Is this about whether the Bird readings are correct for the conditions
on the Bird Thruline, or whether the meter readings are extensible to
the adjacent transmission line without further interpretation /
modelling?

The Bird readings should be correct for the conditions on the Bird
Thruline. You can safely extend those measurements literally to the
adjacent line where the adjacent line is the same as the Bird Thruline
and of negligible loss. In other cases, knowing the line parameters,
you may be able to use the measurements to some extent to calculating
some conditions on the other line.

Though the Bird readings in your example for Forward and Reflect Power
cannot be assumed valid for the adjacent line, the net power should be
correct.

I don't think anyone is suggesting that the Bird could be used in a
general sense to estimate the VSWR on your 450 ohm line.

Owen
--

On Tue, 11 Oct 2005 14:09:03 -0400, "Fred W4JLE"
wrote:

I would then assume you disregard anything written in books as it falls in
the same category.

Hi Fred,

Certainly anything that is third hand and name dropping - but you
already knew that from my previous posting.

73's
Richard Clark, KB7QHC

On Tue, 11 Oct 2005 17:45:51 GMT, Cecil Moore wrote:
Exactly what did you find technically wrong with that third hand quote?
Why would I bother? It reveals any one of several many problems:
1. Incorrect reporting through poor transcription;
2. Presumed ascription of source;
3. No access to the original (as third hand offers no citations, or
if it did, it renders third hand reporting as immaterial);
4. No access to the presumed source of the third hand-off;
5. No access to the original (op. cit.) author;
6. Discussion does not attend the topic (and is of no interest to me,
nor others barring their continuing that line of inquiry - I won't
hold my breath for that);
7. Third hand-off reporting is the lame equivalent of celebrity news.

Cecil Moore wrote:
Ian White G/GM3SEK wrote:
The Bird doesn't require any upstream and downstream boundary
conditions.

When Bird requires a 50 ohm environment, they are requiring
50 ohm boundary conditions for the reading to be valid.

No, they're not. They are requiring a 50-ohm system reference impedance.

What you call the "impedance environment" consists of physical things
like the source impedance, line impedance and load impedance. You're
confusing those with the system reference impedance, which something
completely different.

System reference impedance is purely a matter of definition. The most
common choice is 50 ohms... and by definition, that means 50 ohms
exactly.

Having made that choice, then you obviously design and calibrate your
instruments to give correct readings in an impedance environment that is
as close to your chosen reference impedance as you can practically make
it.

Your example shows the difference between impedance environment and
reference impedance most clearly.

If
you install the Bird in a 450 ohm environment on both sides
of the wattmeter, for instance, it will NOT read a valid forward
power and reflected power. In a matched-line 450 ohm environment
with absolutely zero reflected power, the Bird will indicate an
SWR of 9:1, a |rho| of 0.8 and a ratio of reflected power to
forward power of 0.64 even when the reflected power is zero.

You have changed the impedance environment to 450 ohms, and that's
fine... but all of the Bird's readings are perfectly correct if the
system reference impedance remains defined at 50 ohms. The reason why
say they are incorrect is that you also changed your definition of
system reference impedance to 450 ohms, without acknowledging that you
did it.

It's like doing a financial calculation without mentioning that you
switched into another base currency... darn right the results are not
valid.


--
73 from Ian G/GM3SEK 'In Practice' columnist for RadCom (RSGB)
http://www.ifwtech.co.uk/g3sek

On Tue, 11 Oct 2005 21:11:55 +0100, Ian White G/GM3SEK
wrote:

Thank you, you're right. The key difference between direct and indirect
measurements is not about the need for mathematics; it's about the need
for additional input from theory.

What I should have said is:

Hi Ian,

Remarkable touch of admission - especially to my over-arching method
of criticism.

Another point needs to be attended; the discussion of the measurement
of SWR seems quite, and absurdly, drawn in kindergarten terms of
mathematics as if the determination were king and the numbers simply
fell out be virtue of cranking the equation.

Nothing could be further from the truth.

As much as Reg pines away about what a SWR does not measure,
absolutely the same could be said for his unspoken inference that
probe measurement along a line does measure it.

True, probing the line reveals a pattern that in the mind conforms to
the expectation of standing waves, but this is simply trying to
measure your own shadow when each time you stretch out the rule your
shadow moves further out. What size is your shadow - when?

Using the formula everyone here leans upon as the archetypal equation
for SWR, and claiming they've measured at the appropriate points along
the line (undoubtedly only in their imagination) easily leads to
errors above 20%. Worse yet is that this same formula fails utterly
at the bench (am I embarrassing anyone?) for any but the most
pedestrian of SWRs which are easily resolved by a SWR meter in the
first place.

Beyond this issue of accuracy (certainly no one is interested in that
are they?) stands the fictions of requiring slightly more than a
quarterwave length, or access along the line to both the trough and
the peak. Anyone so hamstrung to NEED these criteria, hasn't ever
really faced the problem of measuring SWR on an open line in the first
place. The double-minima method offers an exceptional accuracy for
high SWRs and occupies a smaller region of line than otherwise
demanded. Measuring at the minima also reduces the error introduced
in the very act of measuring SWR.

To this last, how many here can guarantee their probes will approach
the line with the same offset? There's a very good reason why SWR
probes are mounted on vernier carriages. How many would recognize
when the probes were too deep, or not deep enough to justify the
measurement? To imagine any approaching the line with hand held leads
raises the prospect of shooting marbles without thumbs.

As to these meters that everyone is rushing to use - Square Law or
Linear? Don't know the difference? You don't know accuracy or how to
obtain it when you have no choice. How do you render a Square Law
detector linear? How do you linearize a Square Law detector
measurement? No concern? You aren't measuring SWR then either. The
method of measurement for low, medium, and high SWRs is not the same.
One size does not fit all as the discussion in this group might imply
(from that same lack of actually having done it). Even the math is
different - and if any argue that this observation flies in the face
of simple transmission line equations, then these casual tourists are
comfortably remote from actually measuring the rough terrain of SWR.

However, none of this practicality is going to disturb the armchair
SWR analyzer. It comes to their great fortune that one simple
instrument will probably offer far more accuracy than they could ever
obtain by trying to be literal about SWR "on the line."

73's
Richard Clark, KB7QHC

Owen Duffy wrote:
Cecil Moore wrote:
Not true, Reg. My question was specified using RG-213 at 10 MHz.

True enough, but in the context of the question as to whether the Bird
43 reads sufficiently accurately, the transmission line on which one
is interested in the decay of the evanescent modes is the Bird
Thruline coupler section, not Rg-213 or any other cable that might be
attached to the Bird.

I didn't read it that way, Owen. IMO, the real question is:
What length of 50 ohm coax needs to be attached to the Bird
input and output to ensure that a 50 ohm environment is
present?
--
73, Cecil http://www.qsl.net/w5dxp

Owen Duffy wrote:
I don't think anyone is suggesting that the Bird could be used in a
general sense to estimate the VSWR on your 450 ohm line.

I thought that was the subject of the discussion.
--
73, Cecil http://www.qsl.net/w5dxp

Richard Clark wrote:
Certainly anything that is third hand and name dropping - but you
already knew that from my previous posting.

It's obvious that you cannot bring yourself to believe that e=mc^2.
--
73, Cecil http://www.qsl.net/w5dxp

On Wed, 12 Oct 2005 02:04:52 GMT, Cecil Moore wrote:

Owen Duffy wrote:
I don't think anyone is suggesting that the Bird could be used in a
general sense to estimate the VSWR on your 450 ohm line.

I thought that was the subject of the discussion.

From an earlier post:

In the case of the Bird 43, I suggest that if had, say, at 1MHz, 75
ohm line and a 75 ohm load on the load side, that the V/I raio for the
travelling waves in the region of the sampling element would be so
close to 50 ohms as to not materially affect the accuracy of
measurements on the 50 ohms coupler section, irrespective of the fact
that the sampling element has only 0.02% of a wavelength of 50 ohm
line on its load side.

(For avoidance of doubt, nothing in the foregoing is to imply the Bird
43 would be directly measuring or indicating the conditions on the 75
ohm line.)

Owen
--

Ian White G/GM3SEK wrote:

You have changed the impedance environment to 450 ohms, and that's
fine... but all of the Bird's readings are perfectly correct if the
system reference impedance remains defined at 50 ohms.

I have changed the system reference impedance to 450 ohms. Assuming
a tube PA with a pi-net output, 50 ohms doesn't exist anywhere anymore.
The system reference impedance is no longer 50 ohms so the Bird wattmeter
is being abused and misused. You can do the same thing by using a DC
voltmeter on an RF voltage or by using a hammer on a screw. If you
want to know the SWR on 450 ohm line, use a 450 ohm SWR meter.
--
73, Cecil http://www.qsl.net/w5dxp

Owen Duffy wrote:
In the case of the Bird 43, I suggest that if had, say, at 1MHz, 75
ohm line and a 75 ohm load on the load side, that the V/I raio for the
travelling waves in the region of the sampling element would be so
close to 50 ohms as to not materially affect the accuracy of
measurements on the 50 ohms coupler section, irrespective of the fact
that the sampling element has only 0.02% of a wavelength of 50 ohm
line on its load side.

If there is 75 ohm coax on the input of the Bird, the reflected
power reported by the Bird on the coax will be off by an infinite
percent. That's pretty inaccurate.
--
73, Cecil http://www.qsl.net/w5dxp

On Wed, 12 Oct 2005 02:06:22 GMT, Cecil Moore wrote:
It's obvious that you cannot bring yourself to believe that e=mc^2.
If "that is obvious" to you, then such are your problems of a third
hand education resulting in uncontrolled topic inflation.

On Wed, 12 Oct 2005 02:03:10 GMT, Cecil Moore wrote:
What length of 50 ohm coax needs to be attached to the Bird
input and output to ensure that a 50 ohm environment is
present?
None.

On Wed, 12 Oct 2005 02:25:33 GMT, Cecil Moore wrote:

Owen Duffy wrote:
In the case of the Bird 43, I suggest that if had, say, at 1MHz, 75
ohm line and a 75 ohm load on the load side, that the V/I raio for the
travelling waves in the region of the sampling element would be so
close to 50 ohms as to not materially affect the accuracy of
measurements on the 50 ohms coupler section, irrespective of the fact
that the sampling element has only 0.02% of a wavelength of 50 ohm
line on its load side.

If there is 75 ohm coax on the input of the Bird, the reflected
power reported by the Bird on the coax will be off by an infinite
percent. That's pretty inaccurate.

The Bird does not measure or report the conditions on the coax, it
measures and reports the conditions in the immediate region of the
sampling element which is some 40mm inside the Thruline coupler
section.

Owen
--

Owen Duffy wrote:
The Bird does not measure or report the conditions on the coax, it
measures and reports the conditions in the immediate region of the
sampling element which is some 40mm inside the Thruline coupler
section.

I don't know what this argument is all about. Consider the
following:

XMTR---75 ohm coax---Bird---75 ohm load

Are you saying the Bird's placement will result in a reflection
coefficient of 0.2? I seriously doubt that is true.
--
73, Cecil http://www.qsl.net/w5dxp

On Wed, 12 Oct 2005 04:06:07 GMT, Cecil Moore wrote:
I seriously doubt that is true.
We are genuinely grateful that you don't list all the combinations and
permutations of those topics of invention you "seriously doubt" being
true...
.... or false.

On Wed, 12 Oct 2005 04:06:07 GMT, Cecil Moore wrote:

Owen Duffy wrote:
The Bird does not measure or report the conditions on the coax, it
measures and reports the conditions in the immediate region of the
sampling element which is some 40mm inside the Thruline coupler
section.

I don't know what this argument is all about. Consider the
following:

XMTR---75 ohm coax---Bird---75 ohm load

Are you saying the Bird's placement will result in a reflection
coefficient of 0.2? I seriously doubt that is true.

I don't have the equipment at hand to do that experiment, but I have
done another experiment.


XMTR -- 2m of RG58 -- Bird43 -- 1.2m of RG213 Bird43 -- 20m RG6 (75
ohms) -- antenna.

1.2m of 50 ohm coax between the Birds is 4.2% of an electrical
(wavelength.)

I have made measurements with only one 100W slug which is moved from
instrument to instrument.

The tx was adjusted to 100W forward on the first instrument.

Both instruments read 100W forward.

Both instruments read 2W reflected.

When I swap the instruments around, I get the same results. It is only
a simple test, but I am not convinced that measurements from one
position are signficantly different to the other position, despite the
transmission line "environment" being different.

I am not surprised that both instruments read similarly, despite the
fact that one doesn't have any 50 ohm coax on the load side of itself,
whereas the other one does.

Owen
--

Owen Duffy wrote:
XMTR -- 2m of RG58 -- Bird43 -- 1.2m of RG213 Bird43 -- 20m RG6 (75 ohms) -- antenna.

I don't have any argument with your results. Try this instead.

XMTR -- 2m of RG58 -- Bird43 -- 1.2m of RG6 Bird 43 -- 20m RG6 -- antenna

The second Bird will NOT indicate the actual SWR on the RG6.
--
73, Cecil http://www.qsl.net/w5dxp

On Wed, 12 Oct 2005 05:36:38 GMT, Cecil Moore wrote:

Owen Duffy wrote:
XMTR -- 2m of RG58 -- Bird43 -- 1.2m of RG213 Bird43 -- 20m RG6 (75 ohms) -- antenna.

I don't have any argument with your results. Try this instead.

XMTR -- 2m of RG58 -- Bird43 -- 1.2m of RG6 Bird 43 -- 20m RG6 -- antenna


I won't waste the time, I can predict that they will most likely be
different. Transmission line theory tells me that the Z in the region
of each Bird will be different, and that will probably result in a
different indicated VSWR.

It is a sidetrack, just noise.

The measurements I reported were identical, although one Bird was
surrounded by 50 ohm line, and the other had 75 ohm line on one side
of it. The 75 ohm line did not cause a measureable difference in meter
readings in that simple trial.

Owen
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Owen Duffy wrote:
It is a sidetrack, just noise.

The main point of the discussion has been: An SWR meter
calibrated for Z0=50 ohms will not accurately report the
actual SWR on a line that is not Z0=50 ohms.

Of the four following configurations, in a lossless situation,
which one accurately reports the SWR on both sides of the SWR
meter?

1. XMTR---50 ohm coax---SWR meter---50 ohm coax---load

2. XMTR---50 ohm coax---SWR meter---75 ohm coax---load

3. XMTR---75 ohm coax---SWR meter---50 ohm coax---load

4. XMTR---75 ohm coax---SWR meter---75 ohm coax---load
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73, Cecil http://www.qsl.net/w5dxp

Cecil Moore wrote:


If there is 75 ohm coax on the input of the Bird, the reflected
power reported by the Bird on the coax will be off by an infinite
percent. That's pretty inaccurate.

Why wouldn't the meter correctly indicate the reflection resulting from
the mismatch between the 50 ohm wattmeter and the 75 ohm transmission line?

ac6xg

On Wed, 12 Oct 2005 11:56:44 GMT, Cecil Moore wrote:

Owen Duffy wrote:
It is a sidetrack, just noise.

The main point of the discussion has been: An SWR meter
calibrated for Z0=50 ohms will not accurately report the
actual SWR on a line that is not Z0=50 ohms.


No Cecil, the thread subject and the quote of your text in the first
message of the post was the main point of discussion:

The transmission line length must only be long enough such that
the V/I ratio is forced to the Z0 value. According to some pretty
smart guys I asked, that's about 2% of a wavelength.

It seems that the statement you have quoted is not born out in
practice, though I note that you "seriously doubt that is true".

I have found out what I needed to know, thanks... Owen
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