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 Wed, 29 Jun 2005 17:17:53 -0700, Frank Gilliland
wrote:

That's basically what Lance said, just in different words. So what's
the problem?
*****

I have problems when people state "current flows". That is not
actually true. Power in the electromagnetic wave that flows down a
transmission line and currents on the center conductor and outer
conductor are dependant on the E and H fields of the wave within the
transmission line at any point on the transmission line.

outside that I think we are primarily in agreement.

james

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 Thu, 30 Jun 2005 00:45:25 GMT, james wrote
in :

On Wed, 29 Jun 2005 17:17:53 -0700, Frank Gilliland
wrote:

That's basically what Lance said, just in different words. So what's
the problem?
*****

I have problems when people state "current flows". That is not
actually true.


So current -doesn't- flow?


Power in the electromagnetic wave that flows down a
transmission line and currents on the center conductor and outer
conductor are dependant on the E and H fields of the wave within the
transmission line at any point on the transmission line.


Those E and H fields are created by voltage and current introduced
onto the conductors. A transmission line simply establishes a kind of
self-propogating 'symbiosis' between the fields and the current &
voltage on the conductors. Current does indeed flow on those
conductors -- if it didn't they wouldn't be necessary.


outside that I think we are primarily in agreement.


What you are missing is that different things happen on the outside of
the coax than what happens on the inside. And what Lance was pointing
out (if I may be so bold as to speak for him here) was that currents
on the outside of the coax can screw up things when you try to measure
what's happening on the inside.







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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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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

On Wed, 29 Jun 2005 19:42:53 -0700, Frank Gilliland
wrote:

[snipped in the interest of brevity]

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?

My "SWR Meter" is one of these:

http://users.adelphia.net/~n2pk/VNA/VNAarch.html

I have 1 mW to radiate. What kind of FSM should I use?



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


Thank you for making my point.

Not even you have made your 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".

My Webster's says:

baloney n (bologna) : pretentious nonsense : BUNKUM --- often used as
a generalized expression of disagreement....

I could not be more accurate.


And although the syntax of my statement was
somewhat 'convoluted',

A ray of hope

the logic is sound

smashed

-- 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.

Uh huh. By this convoluted "logic" I guess you would avoid any dyno
testing at all and just go do hit-and-miss tuning at the drag strip.

Reading the mail appearing in this thread is more fun than watching Saturday
Night Live!

Walt, W2DU

J. Mc Laughlin wrote:
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).

Power can certainly be combined, just not superposed. Here is
the irradiance equation from _Optics_, by Hecht.

Itot = I1 + I2 + Sqrt(I1*I2)cos(theta)

Irradiance is power/unit-area. The last term is known as the
"interference" term. Hecht provides separate chapters for
interference and superposition, the best treatment of those
two subjects of which I am aware.

Here is the transmission line forward power equation from Dr.
Best's QST transmission line article.

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

It is virtually identical to the irradiance equation above.
The last term is known to be the "interference" term and
for a Z0-matched system, THETA EQUALS ZERO, so for a Z0-
matched system, a complete analysis can be done using only
the forward and reflected power magnitudes. This is something
I and others have been saying for years and it has been
called "gobbledegook" (sic) by Roy (and worse by others) even
though Roy admits that he doesn't care to understand where the
power goes.

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

Yes, and they are still not 100% correct. Roy has said:
'I personally don't have a compulsion to understand where
this power "goes".' Too bad he doesn't have that compulsion
because, with a small amount of mental effort, he surely would
have figured it out before me - if by no other means, simply by
referencing the chapter on EM wave interference in _Optics_.
--
73, Cecil, http://www.qsl.net/w5dxp

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Frank wrote, "Prove it."

OK, here I am at the track (the bench). I have an SWR meter that I've
verified with my HP8653 to behave like a short section of 50 ohm line
at the frequency of interest. I put a load on its output that I've
also verified to be 50 ohms at the frequency of interest. I've applied
power to the load through the SWR meter. The indicated SWR is 1.23:1.

I took the SWR meter apart, and located a particular resistor. I
changed its value slightly. I re-verified that the meter still looks
like a short section of 50 ohm line. I re-ran the experiment of
applying power through the meter to the load. The indicated SWR is now
1.05:1.

Yes, I really have done that!

This particular meter is built, as very many of them are, to sample
current and voltage at a point of essentially zero length on the line.
The current sample (through a current transformer: line center passes
through a toroid; secondary is several turns, loaded by that
calibration resistor) is converted to a voltage by dropping it through
a resistance, and by changing that resistance, I can change the
relative amount the current contributes to the measurement. In other
words, if the voltage sample is v(samp)=k*v(line), I want to adjust the
current sampling so v(i(samp)) = k*Zo*i(line), where Zo is the
impedance to which the meter is calibrated to measure SWR. In some
meters, there is a means to adjust the voltage sampling ratio easily
with a variable trimmer capacitor. Either way works. The adjustment
DOES have a TINY effect on the impedance the meter presents to the line
it's in, but that is very minor, compared with the range of adjustment
of the impedance calibration value.

Yes, I really have adjusted a meter which uses the variable capacitor,
too.

Cheers,
Tom

On Thu, 30 Jun 2005 09:41:05 -0500, Cecil Moore
wrote:
Power can certainly be combined, just not superposed. Here is
the irradiance equation from _Optics_, by Hecht.
Itot = I1 + I2 + Sqrt(I1*I2)cos(theta)
According to this formula, for two laser beams of 100W each, without
any phase difference, we can illuminate a target with
Itot = 100 + 100+ Sqrt(100*100)cos(0)
Itot = 300W

Every CBer's dream....

Hecht should sue you for copyright Unfair Use.

SWR=Strewn With Rumor

"Walter Maxwell" wrote in message
...
Reading the mail appearing in this thread is more fun than watching
Saturday
Night Live!

Walt, W2DU

K7ITM wrote:
Yes, I really have adjusted a meter which uses the variable capacitor,
too.

In the old Heathkit SWR meter were instructions to
install either two 50 ohm resistors for a 50 ohm SWR
meter or two 75 ohm resistors for a 75 ohm SWR meter.
We used a lot of RG-11 back in those days.
--
73, Cecil, http://www.qsl.net/w5dxp

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On Thu, 30 Jun 2005 01:30:39 GMT, "Tom Donaly"
wrote:


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
****

Yes Tom

Convultion was the wrong term to use. I made a mistake because i type
as i think and on occasion hit send before i reread what i have
written.

I still contend that a sinusoidal wave travelling down a coax is
comprised of perpendicular(orthogonal) E and H fields. The these
vector fields that induce sinusodial current and voltage potential
vectors in and between the shield and center conductors as the wave
travels. Both the source and reflected waves are comprised of two
vector fields, E and H. Granted this is true only when the load
reflection coefficient is not zero. In that case of zero, then there
is no reflected power.

It is possible to derive from the vector current and vector voltage a
magnitude of those vectors and thus a produce two scalar quantities
that can be pluged into Ohm's Law and derive an instantaineous power
at a given time and position on the coax. That both source and
reflected sinusoidal current and voltage can have derived scalar
values. These values can be directly added.

This all started from an SWR question. I contend that the
instantaineous power at any given time and position of the coax can be
expressed as the sum of the magnitudes or scalar quantities of the
source and reflected powers. If you are wanting just the magnitudes of
the power, then this should work.

james

Richard Clark wrote:

On Thu, 30 Jun 2005 09:41:05 -0500, Cecil Moore
wrote:

Power can certainly be combined, just not superposed. Here is
the irradiance equation from _Optics_, by Hecht.
Itot = I1 + I2 + Sqrt(I1*I2)cos(theta)

According to this formula, for two laser beams of 100W each, without
any phase difference, we can illuminate a target with
Itot = 100 + 100+ Sqrt(100*100)cos(0)
Itot = 300W

Sorry, I made a mistake in the equation. Please forgive
my omission. Here's the correct equation:

Itot = I1 + I2 + 2*Sqrt(I1*I2)cos(theta)

So if theta equals zero, Itot would be 400w, not 300w.

This is the power of superposition of coherent waves. When
you superpose two 100w coherent laser beams, the resultant
power is indeed 400w and must be supplied by the sources
or supplied by destructive interference from somewhere else.
This is all explained in _Optics_, by Hecht. Hows about
reading it so I won't have to explain superposition to you?
--
73, Cecil, http://www.qsl.net/w5dxp

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Cecil Moore wrote:
Itot = I1 + I2 + Sqrt(I1*I2)cos(theta)

Sorry, should be:
Itot = I1 + I2 +2*SQRT(I1*I2)cos(theta)

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

Sorry should be:
Ptot = P1 + P2 + 2*Sqrt(P1*P2)cos(theta)
--
73, Cecil, http://www.qsl.net/w5dxp

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