On Wed, 29 Jun 2005 16:37:10 -0500, Cecil Moore
wrote in :

Frank Gilliland wrote:
Point is that they are usually calibrated for Z0=50 ohms
and are in error when used in Z0 environments differing
from Z0=50 ohms, e.g. Z0=75 ohms.

The point is that the error is insignificant when the directional
coupler is much shorter than the wavelength.

Nope, that's not the point at all. It is true that a 50 ohm
SWR meter designed for HF may not work on 70 cm but the error
I'm talking about is the calibration error in a 50 ohm SWR meter
designed for HF and used on HF in, for instance, a Z0 = 450 ohm
environment instead of its calibrated-for 50 ohm environment....


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.






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On Wed, 29 Jun 2005 22:58:51 GMT, "Tom Donaly"
wrote:

james wrote:

On Wed, 29 Jun 2005 16:42:49 -0500, Cecil Moore
wrote:


Sometimes yes, sometimes no. If the reflected current arrives out
of phase with the forward current, then the final dissipation can
actually be *reduced* by the mismatch.

*****

Power is power. Phase is not a problem. Take the mafnitude of the
transmitted power and teh magnitude of the reflected power. The
results are phaseless. The magnitudes add linearly.

QED

james


Cecil was talking about current, not power. You can't add
power the way you can voltage and current. If you could, you
could build a very nice perpetual motion machine just by using the
reflections in a transmission line to add power so that the output
was greater than the input.
73,
Tom Donaly, KA6RUH
******

Tom

The problem is that current is not reflected back from the load, power
is. Thus the you can add magnitudes of power.


james

On Wed, 29 Jun 2005 16:29:26 -0700, Roy Lewallen
wrote:

The rig has no way of detecting any alleged "reflected power". It can't
tell the difference between a feedline with a lot of "reflected power",
a feedline with no "reflected power", and a plain resistor. It behaves
exactly the same in all cases, provided only that the impedance that
each provides to it is the same.

*****

Agreed that a rig cannot detect the difference between forward and
reflected power. If the reflection coeffiecient of the source is zero
then final stage of a transmiter will look purely resistive to any
power reflected by the load. Thereby that refelcted power is
dissapated as heat. Other reflection coefficients at the source will
yield lesser amounts of reflected power from the load as heat.

james

Anyone not convinced of this should put a couple or more dummy loads in
series or parallel, make up a few lengths of transmission line of
various impedances, and see for himself.

Roy Lewallen, W7EL

james wrote:
On Wed, 29 Jun 2005 16:42:49 -0500, Cecil Moore
wrote:


Sometimes yes, sometimes no. If the reflected current arrives out
of phase with the forward current, then the final dissipation can
actually be *reduced* by the mismatch.

*****

Power is power. Phase is not a problem. Take the mafnitude of the
transmitted power and teh magnitude of the reflected power. The
results are phaseless. The magnitudes add linearly.

QED

james

On 29 Jun 2005 14:53:29 -0700, "K7ITM" wrote in
.com:

Frank Gilliland wrote, among other things, "The point is that the error
is insignificant when the directional coupler is much shorter than the
wavelength."

Certainly "directional couplers" for HF may be built at essentially
zero length, and ideally would have exactly zero length, monitoring the
current and voltage at a single point on a line. Then SWR or
reflection coefficient magnitude or even complex reflection coefficient
may be calculated under the assumption we know the desired reference
impedance. But if the equipment combines the voltage and current
samples in the wrong ratio, you will get the WRONG answer. Even if the
coupler looks like a perfect 50 ohms impedance section of transmission
line (with some attenuation), the error _in_measurement_output_ can be
significant indeed. Just because the coupler looks like a 50 ohm line
to the line it's hooked to doesn't mean it will read zero reflection
when IT's presented with a 50 ohm load.


SWR is a ratio, not an absolute value. It doesn't matter if the meter
reads a forward power that's off by 1 or 1000 watts just as long as
the reflected power is in error by the same percent, which will be the
case unless you are using two different meters for forward and
reflected power. Calibrated or not, SWR is the same.


And by the way, not everyone who measures and cares very much about SWR
(or reflection coefficient) cares a whit about field strength. Not all
loads are antennas.


That point might be relevant if this thread were cross-posted to
alt.heaters.induction or rec.dummy-loads.


Indeed, as Reg says, we might do better in amateur applications to
consider the SWR meter as an indicator of the degree to which we're
presenting a transmitter with the desired load.


I agree 100%.


That's really what
we're using it for, most of the time.


Unfortunately, it's that "most of the time" part that starts threads
like this. Some radio operators mistakenly think SWR is a measure of
antenna efficiency.


It may ALSO be interesting to
know the field strength, but please be aware that a transmitter's
distortion products may be significantly higher if it's presented the
wrong load impedance, even though the power output may be increased.
Field strength alone is not acceptable to me as a means to adjust an
antenna load to a transmitter, or as a way to adjust the operating
point of the transmitter.


True story. It's certainly better to use a tunable FSM if one is
available. And such meters are readily available -- any receiver with
a good S-meter.






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james wrote:
On Wed, 29 Jun 2005 22:58:51 GMT, "Tom Donaly"
wrote:


james wrote:


On Wed, 29 Jun 2005 16:42:49 -0500, Cecil Moore
wrote:



Sometimes yes, sometimes no. If the reflected current arrives out
of phase with the forward current, then the final dissipation can
actually be *reduced* by the mismatch.

*****

Power is power. Phase is not a problem. Take the mafnitude of the
transmitted power and teh magnitude of the reflected power. The
results are phaseless. The magnitudes add linearly.

QED

james


Cecil was talking about current, not power. You can't add
power the way you can voltage and current. If you could, you
could build a very nice perpetual motion machine just by using the
reflections in a transmission line to add power so that the output
was greater than the input.
73,
Tom Donaly, KA6RUH

******

Tom

The problem is that current is not reflected back from the load, power
is. Thus the you can add magnitudes of power.


james

Nope. You need a course in electromagnetics. Who put all these
ideas into your head, anyway?
73,
Tom

james wrote:

On Wed, 29 Jun 2005 16:29:26 -0700, Roy Lewallen
wrote:


The rig has no way of detecting any alleged "reflected power". It can't
tell the difference between a feedline with a lot of "reflected power",
a feedline with no "reflected power", and a plain resistor. It behaves
exactly the same in all cases, provided only that the impedance that
each provides to it is the same.


*****

Agreed that a rig cannot detect the difference between forward and
reflected power. If the reflection coeffiecient of the source is zero
then final stage of a transmiter will look purely resistive to any
power reflected by the load. Thereby that refelcted power is
dissapated as heat. Other reflection coefficients at the source will
yield lesser amounts of reflected power from the load as heat.

james


Anyone not convinced of this should put a couple or more dummy loads in
series or parallel, make up a few lengths of transmission line of
various impedances, and see for himself.

Roy Lewallen, W7EL

james wrote:

On Wed, 29 Jun 2005 16:42:49 -0500, Cecil Moore
wrote:



Sometimes yes, sometimes no. If the reflected current arrives out
of phase with the forward current, then the final dissipation can
actually be *reduced* by the mismatch.

*****

Power is power. Phase is not a problem. Take the mafnitude of the
transmitted power and teh magnitude of the reflected power. The
results are phaseless. The magnitudes add linearly.

QED

james




You need to read _Reflections II, Transmission Lines and Antennas_
by M. Walter Maxwell, W2DU. Even better, get a book on electromagnetics.
You might be able to puzzle some of it out although much of
the math might be too esoteric for you.
73,
Tom Donaly, KA6RUH

On Wed, 29 Jun 2005 22:45:03 GMT, "Tom Donaly"
wrote:

james wrote:
On Wed, 29 Jun 2005 11:07:17 -0400, "Fred W4JLE"
wrote:


What is the reason a 2:1 SWR can cause such havoc?

How can I avoid this catastrophic condition?

I feed my dipoles with 450 Ohm ladder line, but the last 20 feet or so is 50
Ohm coax, I guess that makes it work ok. I haven't blown up my finals yet.

Lions and tigers and bears Oh my...

*****

Actually can happen if you push the finals to where there is
insufficeint margin to the maximum heat dissapation. Tubes are a bit
more forgiving. Transistor inadequately heatsinked and overdriven,
typical CB usage, often have little of no margin for heat dissapation.

If the transmitter has a refelction coefficient of zero and the load
say .3, then that reflected power from the load is dissapated as heat
in the output circuits and any final transistors or tubes. Now if the
radio has a reflection coefficient other than zero that will lessen
the heat dissapation on the transimiiter. Now you get load and source
reflections convoluting within the transmission line.

You ought to model a 400 Mhz square wave with source and load
refelctions coefficients other than zero. It can get ugly


james



Consider the MRF 140, a 150 Watt 2.0 - 150.0 Mhz N-Channel
linear RF power fet. From the technical data sheet: "100% Tested
For Load Mismatch At All Phase Angles With 30:1 VSWR." You'd have
a tough time damaging this device with a mere 2:1 VSWR.
How do load and source reflections convolute within the
transmission line? That's a new one on me. My old dictionary
defines 'convolute' as "Rolled or folded together with one part
over another; twisted; coiled." The rest of the post is pretty
fanciful, too. A trip to the library would do wonders.
73,
Tom Donaly, KA6RUH
******

In electrical engineering it is the instantaneous power density of two
signals passing at the same spot from two directions. That is called
Convolution. It also is a nice mathematical means of modeling SWR at
any point on a transmisison line at a particular time.

Well if you knew CBers, they are not satidied getting 150 watts from
a transistor rated for 150 watts. But in my first paragraph I thought
I made it clear but evidently I did not. I guess I must strive to
better explain myslef.

james

On Thu, 30 Jun 2005 00:12:31 GMT, "Tom Donaly"
wrote:

Nope. You need a course in electromagnetics. Who put all these
ideas into your head, anyway?
73,
Tom
Electromagnetics

james

On Wed, 29 Jun 2005 22:45:03 GMT, "Tom Donaly"
wrote in
:

snip
..... A trip to the library would do wonders.
73,
Tom Donaly, KA6RUH


I keep trying to stress that fact, but some people persist under the
delusion that they can learn everything they need from the internet.






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On Thu, 30 Jun 2005 00:22:48 GMT, james wrote:

In electrical engineering it is the instantaneous power density of two
signals passing at the same spot from two directions. That is called
Convolution.

Hi James,

No, it is called Superposition, and that is done only with voltage or
current. What you are describing may be associated with the Fourier
convolution of power series - an entirely different field (and not
even additive).

73's
Richard Clark, KB7QHC

On Thu, 30 Jun 2005 00:19:22 GMT, "Tom Donaly"
wrote:

You need to read _Reflections II, Transmission Lines and Antennas_
by M. Walter Maxwell, W2DU. Even better, get a book on electromagnetics.
You might be able to puzzle some of it out although much of
the math might be too esoteric for you.
73,
Tom Donaly, KA6RUH
*****

Whats wrong with stating that power is reflected by the load? Isn't
power delivered to the load from the source? Elementary electronics
states that power is voltage time current.

Currents in a transmission line are induced currents. They are induced
from the E and H fields of the TEM wave. I hope that you don't think
that current races up and down the coax millions a times per second?

james

On Thu, 30 Jun 2005 00:22:48 GMT, james wrote
in :

On Wed, 29 Jun 2005 22:45:03 GMT, "Tom Donaly"
wrote:

james wrote:
On Wed, 29 Jun 2005 11:07:17 -0400, "Fred W4JLE"
wrote:


What is the reason a 2:1 SWR can cause such havoc?

How can I avoid this catastrophic condition?

I feed my dipoles with 450 Ohm ladder line, but the last 20 feet or so is 50
Ohm coax, I guess that makes it work ok. I haven't blown up my finals yet.

Lions and tigers and bears Oh my...

*****

Actually can happen if you push the finals to where there is
insufficeint margin to the maximum heat dissapation. Tubes are a bit
more forgiving. Transistor inadequately heatsinked and overdriven,
typical CB usage, often have little of no margin for heat dissapation.

If the transmitter has a refelction coefficient of zero and the load
say .3, then that reflected power from the load is dissapated as heat
in the output circuits and any final transistors or tubes. Now if the
radio has a reflection coefficient other than zero that will lessen
the heat dissapation on the transimiiter. Now you get load and source
reflections convoluting within the transmission line.

You ought to model a 400 Mhz square wave with source and load
refelctions coefficients other than zero. It can get ugly


james



Consider the MRF 140, a 150 Watt 2.0 - 150.0 Mhz N-Channel
linear RF power fet. From the technical data sheet: "100% Tested
For Load Mismatch At All Phase Angles With 30:1 VSWR." You'd have
a tough time damaging this device with a mere 2:1 VSWR.
How do load and source reflections convolute within the
transmission line? That's a new one on me. My old dictionary
defines 'convolute' as "Rolled or folded together with one part
over another; twisted; coiled." The rest of the post is pretty
fanciful, too. A trip to the library would do wonders.
73,
Tom Donaly, KA6RUH
******

In electrical engineering it is the instantaneous power density of two
signals passing at the same spot from two directions. That is called
Convolution.


I'm an EE and I've NEVER heard the term used in reference to
electronics, let alone used to describe standing waves.


It also is a nice mathematical means of modeling SWR at
any point on a transmisison line at a particular time.


I know how to model standing waves, but how do you model a standing
wave ratio?


Well if you knew CBers, they are not satidied getting 150 watts from
a transistor rated for 150 watts. But in my first paragraph I thought
I made it clear but evidently I did not. I guess I must strive to
better explain myslef.


Agreed.







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On Thu, 30 Jun 2005 00:25:23 GMT, james wrote
in :

On Thu, 30 Jun 2005 00:12:31 GMT, "Tom Donaly"
wrote:

Nope. You need a course in electromagnetics. Who put all these
ideas into your head, anyway?
73,
Tom
Electromagnetics


Did "Electromagnetics" teach you that power = voltage * current?






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On Wed, 29 Jun 2005 17:31:36 -0700, Richard Clark
wrote:

On Thu, 30 Jun 2005 00:22:48 GMT, james wrote:

In electrical engineering it is the instantaneous power density of two
signals passing at the same spot from two directions. That is called
Convolution.

Hi James,

No, it is called Superposition, and that is done only with voltage or
current. What you are describing may be associated with the Fourier
convolution of power series - an entirely different field (and not
even additive).

73's
Richard Clark, KB7QHC
*****

Okay maybe I am not expressing my self correctly and right now I don't
realy have the time or patience to look back through my old text
books. It has been several years since I have done a lot of RF work
and some things are not as fresh in my mind. It does seem like the
less you use the more you forget or have trouble explaining what you
think.

Most of my work lately has been away from RF.

james

james wrote:

Agreed that a rig cannot detect the difference between forward and
reflected power. If the reflection coeffiecient of the source is zero
then final stage of a transmiter will look purely resistive to any
power reflected by the load. Thereby that refelcted power is
dissapated as heat. Other reflection coefficients at the source will
yield lesser amounts of reflected power from the load as heat.

james

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

Richard Clark wrote:
On Thu, 30 Jun 2005 00:22:48 GMT, james wrote:


In electrical engineering it is the instantaneous power density of two
signals passing at the same spot from two directions. That is called
Convolution.


Hi James,

No, it is called Superposition, and that is done only with voltage or
current. What you are describing may be associated with the Fourier
convolution of power series - an entirely different field (and not
even additive).

73's
Richard Clark, KB7QHC

Convolution is a mathematical stunt you can perform with
two functions: f(x)* g(x) = (integral from 0 to x) f(t)g(x-t) dt.
At least that's how it's explained in Schaum's Outline book
_Differential Equations_. It's pretty tough to see how it relates
to power in a transmission line. Maybe someone has a use for it
there.
73,
Tom Donaly, KA6RUH

On Thu, 30 Jun 2005 01:02:06 GMT, james wrote:



Most of my work lately has been away from RF.

Uh huh. And do you drive a Kenworth or a Volvo?

On Wed, 29 Jun 2005 14:07:26 -0700, Frank Gilliland
wrote:

On Tue, 28 Jun 2005 23:17:15 -0500, Cecil Moore
wrote in :

Frank Gilliland wrote:
Impedance matching of an SWR meter is generally unimportant since most
SWR meters used for HF have a directional coupler that is much shorter
than the operating wavelength.

Point is that they are usually calibrated for Z0=50 ohms
and are in error when used in Z0 environments differing
from Z0=50 ohms, e.g. Z0=75 ohms.


The point is that the error is insignificant when the directional
coupler is much shorter than the wavelength.

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.


The error is even more
insignificant when there are a host of variables and confounds between
the SWR meter and the transmitted field that can (and frequently do)
affect the objective -- field strength.

Often, field strength is of zero importance. What do you do when the
device under test isn't supposed to radiate? The simplest example of
this would be a CATV system, yet VSWR is *extremely* important in
cascaded networks.


It's much simpler (and just
plain logical) to measure the field strength directly instead of
measuring an abstract value halfway towards the objective and relying
on nothing more than speculation that the rest is working according as
expected.

More baloney and it isn't even sliced.

On Wed, 29 Jun 2005 19:06:14 -0700, Wes Stewart
wrote in :

On Wed, 29 Jun 2005 14:07:26 -0700, Frank Gilliland
wrote:

On Tue, 28 Jun 2005 23:17:15 -0500, Cecil Moore
wrote in :

Frank Gilliland wrote:
Impedance matching of an SWR meter is generally unimportant since most
SWR meters used for HF have a directional coupler that is much shorter
than the operating wavelength.

Point is that they are usually calibrated for Z0=50 ohms
and are in error when used in Z0 environments differing
from Z0=50 ohms, e.g. Z0=75 ohms.


The point is that the error is insignificant when the directional
coupler is much shorter than the wavelength.

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.


The error is even more
insignificant when there are a host of variables and confounds between
the SWR meter and the transmitted field that can (and frequently do)
affect the objective -- field strength.

Often, field strength is of zero importance. What do you do when the
device under test isn't supposed to radiate?


That device probably wouldn't make a very good radio, would it?


The simplest example of
this would be a CATV system, yet VSWR is *extremely* important in
cascaded networks.


Thank you for making my point.


It's much simpler (and just
plain logical) to measure the field strength directly instead of
measuring an abstract value halfway towards the objective and relying
on nothing more than speculation that the rest is working according as
expected.

More baloney and it isn't even sliced.


The word is "blarney". And although the syntax of my statement was
somewhat 'convoluted', the logic is sound -- you can dyno your engine
all day, but the only way to know for sure how fast you can get down
the quarter mile is to run the race.





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james wrote:
Power is power. Phase is not a problem. Take the mafnitude of the
transmitted power and teh magnitude of the reflected power. The
results are phaseless. The magnitudes add linearly.

Strangely enough, the results are not phaseless. The equation
for reflected power at an impedance discontinuity is:

Pref = P3 + P4 - SQRT(P3*P4)cos(theta)

Where theta is the phase angle between V3 and V4, the
associated reflected interferring voltages.

Reference "Optics", by Hecht, Chapter 9 - Interference

The last term in the equation above is known as the "interference"
term. Unless you take the interference term into account, you
don't have a ghost of a chance of ascertaining where the power
goes.
--
73, Cecil http://www.qsl.net/w5dxp

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K7ITM wrote:
Field strength alone is not acceptable to me as a means to adjust an
antenna load to a transmitter, ...

Doesn't being located in the near field introduce
a measurement error?
--
73, Cecil http://www.qsl.net/w5dxp

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Tom Donaly wrote:
Cecil was talking about current, not power. You can't add
power the way you can voltage and current.

That's absolutely correct. When one adds powers, one must
include the interference term which takes care of conservation
of energy. The equation is:

Ptot = P1 + P2 + SQRT(P1*P2)cos(theta)

where theta is the phase angle between V1 (associated with
P1) and V2 (associated with P2). The last term is labeled
the "interference term" and is absolutely necessary when
adding powers. If the interference term is positive, the
interference is constructive. If the interference term is
negative, the interference is destructive.

The best reference on interference and the adding of powers
that I have found is Chapter 9 in "Optics", by Hecht.
--
73, Cecil http://www.qsl.net/w5dxp

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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.
--
73, Cecil http://www.qsl.net/w5dxp

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james wrote:
The problem is that current is not reflected back from the load, power
is. Thus the you can add magnitudes of power.

If power (ExH) is reflected then, of course current is reflected.
Powers can be added but you *must* include the interference term.
--
73, Cecil http://www.qsl.net/w5dxp

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james wrote:
Agreed that a rig cannot detect the difference between forward and
reflected power. If the reflection coeffiecient of the source is zero
then final stage of a transmiter will look purely resistive to any
power reflected by the load. Thereby that refelcted power is
dissapated as heat. Other reflection coefficients at the source will
yield lesser amounts of reflected power from the load as heat.

This is true for a source equipped with a circulator and load but
most ham transmitters are not equipped with a circulator and load.
You must take the phase of the voltages or currents into account
in order to calculate the interference power term. Only then will
you be able to tell where the power goes.
--
73, Cecil http://www.qsl.net/w5dxp

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Roy Lewallen wrote:
But I do want to caution
readers that this view of "reflected power" is demonstrably incorrect.

So is your concept of "sloshing" energy. Reflected energy
waves are demonstrably real. One can find out exactly where
the reflected power goes by taking the interference power
terms into account. Optics engineers figured it out a long
time ago.
--
73, Cecil http://www.qsl.net/w5dxp

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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.
--
73, Cecil http://www.qsl.net/w5dxp

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james wrote:

snip

*****

Agreed that a rig cannot detect the difference between forward and
reflected power. If the reflection coeffiecient of the source is zero
then final stage of a transmiter will look purely resistive to any
power reflected by the load. Thereby that refelcted power is
dissapated as heat. Other reflection coefficients at the source will
yield lesser amounts of reflected power from the load as heat.

james

So if I have a perfect voltage source in series with a 50 ohm resistor,
and I set the voltage source for 100Vrms (50W to the load, 50W to the
source resistor) and I leave the output terminals open (100% reflected
power) then the resistor will dissipate 100W? With no current flowing
through it?

Wow. I gotta review my basic electronics.

-------------------------------------------
Tim Wescott
Wescott Design Services
http://www.wescottdesign.com

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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On Wed, 29 Jun 2005 22:53:24 -0700, Frank Gilliland
wrote:


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.

Lets make an assumption that we are talking about lossless lines. (If
you are not, then the reflectometer does not provide an accurate
indication of forward and reverse power.)

If you use an ideal 50 ohm reflectometer (that means it is a
negligibly short 50 ohm through line and it is nulled to show zero
reflected power when connected to a 50+j0 load) to measure conditions
in a line, the power flow at that point is the indicated Pf-Pr. If you
had placed an ideal 75 ohm instrument in that spot, the readings are
not necessarily in the same ratio (they are unlikely to be so), but
the difference between Pf and Pr will be the same.

The only other inference you can make from one instrument with regard
to the other will be if one of the instruments shows zero reflected
power, then you know the VSWR that the other instrument will indicate.

Real instruments aren't of zero length, but some types of design are
so close to it at low HF frequencies, you will not detect the error
that is introduced.

Owen
--

Updated:

On Wed, 29 Jun 2005 22:53:24 -0700, Frank Gilliland
wrote:


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.

Lets make an assumption that we are talking about distortionless
lines. (If you are not, then the reflectometer does not provide an
accurate indication of forward and reverse power.)

If you use an ideal 50 ohm reflectometer (that means it is a
negligibly short 50 ohm through line and it is nulled to show zero
reflected power when connected to a 50+j0 load) to measure conditions
in a line, the power flow at that point is the indicated Pf-Pr. If you
had placed an ideal 75 ohm instrument in that spot, the readings are
not necessarily in the same ratio (they are unlikely to be so), but
the difference between Pf and Pr will be the same.

The only other inference you can make from one instrument with regard
to the other will be if one of the instruments shows zero reflected
power, then you know the VSWR that the other instrument will indicate.

Real instruments aren't of zero length, but some types of design are
so close to it at low HF frequencies, you will not detect the error
that is introduced.

Owen
--

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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Owen wrote:
Real instruments aren't of zero length, but some types of design are
so close to it at low HF frequencies, you will not detect the error
that is introduced.

The error that we are talking about has nothing to do with
the length of the directional coupler.

The error that we are talking about has everything to do
with an infinite error in the measurement of reflected power.
Infinite errors are hard to sweep under the rug.
--
73, Cecil http://www.qsl.net/w5dxp

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