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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