On Wed, 29 Jun 2005 23:00:17 -0500, Cecil Moore
wrote:
Optics engineers figured it out a long time ago.
And you have consistently failed in its demonstration - so what?

Depends on what you mean by error. Is it a linear system? If the near
field strength increases with no change in the radiating structure
itself (and propagation is stable, etc.), does the far field not
increase by the same ratio? But of course, with a repositionable
(rotatable) directional antenna, it's pretty hard to calibrate the FSM
in a meaningful way since the antenna system changes (quite a lot, with
respect to the FSM) as you rotate it, so you don't know from one time
to the next that you have the RIGHT field strength. I'd (ideally) like
to know that the transmitter is properly adjusted to output a clean
signal, and that the antenna system presents the proper load to the
transmitter, AND that the antenna system is radiating like I'd like it
to. The "SWR meter" is one component that helps me, but with only one
of those tasks. (And yes, it's fine with me if you care also about the
SWR on your 450 ohm balanced line...there may also be good reason for
wanting to know that.)

Cheers,
Tom

PS--Frank, if you look back in the archives from this group, you'll
find directional couplers (of the sort that measure the line at a
single point) explained in great detail with four-part harmony and the
whole nine yards. Go study them and you may understand why calibration
is important.

Roy Lewallen wrote:
There are several
examples at Food for thought.txt available at
http://eznec.com/misc/food_for_thought/.

Regarding errors in the first food_for_thought:

A 100w source equipped with a circulator and load while
looking into an open line, will generate 100w and dissipate
100w in the circulator load. That 100w is definitely not free
power. It can be demonstrated to have made a round trip to
the open end of the feedline and then back to the circulator
load.

The error in your thinking is that the source would see an open
circuit when it is equipped with a circulator and load. It won't.
It will *always* see the Z0 of the feedline as its load (assuming
the circulator load equals Z0). That's the purpose of using
the circulator and load - to allow the source to see a fixed
load equal to Z0.
--
73, Cecil http://www.qsl.net/w5dxp

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Richard Clark wrote:

Cecil Moore wrote:
Optics engineers figured it out a long time ago.

And you have consistently failed in its demonstration - so what?

I can lead you to water but I can't make you drink.
--
73, Cecil http://www.qsl.net/w5dxp

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On Wed, 29 Jun 2005 22:39:10 -0500, Cecil Moore
wrote in :

Frank Gilliland wrote:
There lies our misperceptions; I was not referring to using an HF SWR
meter designed for coax and plugging it into 450 ohm ladder line.

But I specifically stated above the Z0 environment was different
from 50 ohms. The same type of error happens when one uses a
50 ohm SWR meter in a 75 ohm coaxial line.


If that were true then the mere existence of standing waves could
render any measurements worthless. Regardless, I did the experiment a
long time ago -- take a 50 ohm SWR meter and plug it into a 75 ohm
line -- it gives you almost the same measurement (in fact, I didn't
see -any- difference at all). Any small error you might see is, as I
said before, insignificant, especially considering the reason you are
measuring SWR in the first place. The objective is simply to get the
reading as low as practially possible. If you feel the need to quibble
about a couple tenths of a point on a ratio then maybe you're spending
a little too much time playing with the calculator instead of the
antenna.





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On Wed, 29 Jun 2005 23:10:25 -0500, Cecil Moore
wrote in :

Frank Gilliland wrote:

Wes Stewart wrote:
In a word, baloney. The error is independent of length. A zero length
bridge calibrated at 75 ohm is in error when measuring in a 50 ohm
system. Period.

Prove it.

A 75 ohm bridge is expecting the ratio of voltage to current
to be 75 for a matched system. In a 50 ohm matched system, the
ratio of voltage to current will be 50. Therefore, the 75 ohm
bridge won't be balanced. A 50 ohm bridge would be balanced.


The bridge is calibrated to the impedance of the directional coupler
(which is usually built to match the expected line impedance, but
cannot be "zero length" in the present state of reality). If the
impedance of the signal is different than what is expected by the
bridge then your power measurements will probably be wrong (to what
extent they are wrong may or may not be important). But if that's the
case then any error will be the same by percentage and sign for both
forward =AND= reflected power because the impedance of the signal is
the same for both forward and reflected power. IOW, the ratio is the
same -despite- the impedance.

If you don't believe me, try it yourself.







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I'm not quite sure what you are trying to say Frank.

Frank Gilliland wrote:
On Wed, 29 Jun 2005 23:10:25 -0500, Cecil Moore
wrote in :

The bridge is calibrated to the impedance of the directional coupler
(which is usually built to match the expected line impedance, but
cannot be "zero length" in the present state of reality).

The direction coupler samples voltage across and current through a
given point. There is always a current transformer of some type and a
voltage sample through some type of divider. The "voltages"
representing E and I are summed before detection (conversion to dc).

The "directivity" comes because the current phase sample is reversed
180 degrees from the summing phase, causing voltages to subtract.

This means the directional coupler is calibrated for a certain ratio of
voltage and current, so when they exist you have twice the voltage in
the direction where E and I add, and zero voltage where they subtract.


If the
impedance of the signal is different than what is expected by the
bridge then your power measurements will probably be wrong (to what
extent they are wrong may or may not be important). But if that's the
case then any error will be the same by percentage and sign for both
forward =AND= reflected power because the impedance of the signal is
the same for both forward and reflected power. IOW, the ratio is the
same -despite- the impedance.

?What does that mean?

If the directional coupler is calibrated at 50 ohms and you use it in a
75 ohm system you won't get a total reflected null even if the 75 ohm
line has a 1:1 SWR. But if you subtract reflected power from forward
power readings you will get the correct power, within linearity and
calibration limits of the "meter system". This has nothing to do with
standing waves. It has only to do with the relationship between current
and voltage at the point where the directional coupler is inserted.

I'm not sure if you are saying that or not.

73 Tom

Frank Gilliland wrote:

Cecil Moore wrote:
But I specifically stated above the Z0 environment was different
from 50 ohms. The same type of error happens when one uses a
50 ohm SWR meter in a 75 ohm coaxial line.

If that were true then the mere existence of standing waves could
render any measurements worthless. Regardless, I did the experiment a
long time ago -- take a 50 ohm SWR meter and plug it into a 75 ohm
line -- it gives you almost the same measurement (in fact, I didn't
see -any- difference at all).

Please run it again in the following configuration:

Xmtr--1/4WL 75 ohm line--SWR meter--1/4WL 75 ohm line--50 ohm load

The SWR meter will read 2.25:1 when the actual SWR is 1.5:1

Xmtr--1/2WL 75 ohm line--SWR meter--1/2WL 75 ohm line--50 ohm load

The SWR meter will read 1:1 when the actual SWR is 1.5:1

Any small error you might see is, as I
said before, insignificant, especially considering the reason you are
measuring SWR in the first place.

A 50% error in SWR reading is NOT insignificant.

The objective is simply to get the
reading as low as practially possible. If you feel the need to quibble
about a couple tenths of a point on a ratio then maybe you're spending
a little too much time playing with the calculator instead of the
antenna.

A 50% error in SWR is NOT a couple tenths of a point.
--
73, Cecil http://www.qsl.net/w5dxp

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Frank Gilliland wrote:

Cecil Moore wrote:
A 75 ohm bridge is expecting the ratio of voltage to current
to be 75 for a matched system. In a 50 ohm matched system, the
ratio of voltage to current will be 50. Therefore, the 75 ohm
bridge won't be balanced. A 50 ohm bridge would be balanced.

The bridge is calibrated to the impedance of the directional coupler
(which is usually built to match the expected line impedance, but
cannot be "zero length" in the present state of reality). If the
impedance of the signal is different than what is expected by the
bridge then your power measurements will probably be wrong (to what
extent they are wrong may or may not be important). But if that's the
case then any error will be the same by percentage and sign for both
forward =AND= reflected power because the impedance of the signal is
the same for both forward and reflected power. IOW, the ratio is the
same -despite- the impedance.

The error is NOT the same percentage. In a matched 50 ohm system,
the 75 ohm bridge reflected power reading will be off by an
infinite percentage, i.e. division by zero.

If you don't believe me, try it yourself.

I have tried it and you are wrong. Maybe you should try it.
--
73, Cecil http://www.qsl.net/w5dxp

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Much noise has been radiated. I speculate that a reminder about what
linearity means might get things back on track. In a linear network (lumped
or distributed) superposition (of linear signals) produces correct results.

The last statement works in both directions. (The degree to which a
network is linear is the same as the degree to which superposition is
valid.) (If one supplies a large enough signal to any network, it will
become non-linear - as in letting-out-the-smoke-put-in-at-the-factory.)

The catch in all of the above is that superposition only applies to
linear signals and power (however indicated) is not a linear signal.
Power, which could be complex power S = V*I* (the phasor voltage time the
conjugate of the phasor current) or the magnitude of S (apparent power) or
the real part of S ("real" power), simply does not obey superposition even
in a network that is linear.

Bottom line: assuming the use of networks (lumped or distributed) that
are essentially linear, one is only allowed to combine phasor voltages or
phasor currents (but not their product nor the square of such linear
signals). Once combined, the resultant voltage and the resultant current
may be used to find a measure of power. (The "combined" mentioned must be
a linear, additive process.)

It seems to me that Roy, and others, have plowed this ground many times.

73 Mac N8TT

--
J. Mc Laughlin; Michigan U.S.A.
Home:

"Roy Lewallen" wrote in message
snip

I've posted many, many times on this topic and have shown a number of
cases where the load is perfectly matched but the power dissipated in
the source resistor is less than or greater than the "reverse power",
clearly demonstrating that this concept is incorrect. There are several
examples at Food for thought.txt available at
http://eznec.com/misc/food_for_thought/.

Because I've posted so much on the topic I won't do it all again. But I
know at least one person on this newsgroup would be glad to have an
opportunity to express his views once again. I'll leave this discussion
to those who want to revisit it; I don't. But I do want to caution
readers that this view of "reflected power" is demonstrably incorrect.

Roy Lewallen, W7EL