Richard Harrison wrote:

Tor, N4OGW wrote:
"You CANNOT superimpose POWERS, or even talk about the "power" of
various reflections in the same media."

It is done all the time. P=Esquared / R = Isquared x R.


On page 99 of the 1955 edition of "Electronic and Radio Engineering"
Terman writes:

"The standing-wave ratio S is one means of expressing the magnitude of
the reflectiom coefficient; the exact relation between the two is:

S=1+absolute value of the reflection coefficient / 1-absolute value of
the reflection coefficient

or

absolute value of the reflection coefficient= S-1/S+1

Standing-wave ratio=S=Emax/Emin

This definition of standing-wave ratio is sometimes called voltage
standing-wave ratio (VSWR) to distinguish it from the standing-wave
ratio expressed as a power ratio, which is (Emax/Emin)squared."

I think he means that it does not occur in nature, and it therefore
shouldn't be done mathematically. That's because power doesn't
propagate, and
I hasten to add - neither do Poynting vectors. Nor does power reflect,
refract, or diffract. Power is a rate at which energy is transferred,
absorbed, or dissipated. It's not a wave which propagates. It's a
mathematical product of two of the characteristics of a wave that
propagates. Numbers on the other hand can indeed be multiplied, divided,
squared, added, and subtracted. We can often even get the right answer
when we do that with power numbers. But that fact can't necessarily be
extrapolated to mean that the given mathematical operation also takes
place in a transmission line. An that is what the author in question
is attempting to have us believe.

73, ac6xg

Richard Harrison wrote:
. . .
Standing-wave ratio=S=Emax/Emin

This definition of standing-wave ratio is sometimes called voltage
standing-wave ratio (VSWR) to distinguish it from the standing-wave
ratio expressed as a power ratio, which is (Emax/Emin)squared."

Now, that's an interesting concept.

The VSWR is the ratio of the maximum voltage anywhere along the line to
the minimum voltage anywhere along the line (presuming the line is
sufficiently long for a full maximum and minimum to occur). We can use a
voltage probe to actually measure this voltage in a slotted line.

If the voltage is sinusoidal, so is the power, just at twice the
frequency. (I'm sure you can find that in Terman somewhere, but it's
simple to derive with a little trigonometry.) Since you say that power
superposes, we should expect power waves to add and cancel just like
voltage waves. And this should result in dips and peaks in the power as
we move along the line, just like voltage, right? I'd expect the dips
and peaks to be twice as closely spaced, however, because the wavelength
of the power waves would be half the wavelength of the voltage waves.

So is the ratio of the maximum to minimum power along the line due to
this superposition of power waves equal to the PSWR?

Roy Lewallen, W7EL

On Fri, 29 Jul 2005 13:42:49 -0500, (Richard
Harrison) wrote:

Tor, N4OGW wrote:
"You CANNOT superimpose POWERS, or even talk about the "power" of
various reflections in the same media."

Hi Richard,

Given the original subject line has been munged up, it is difficult to
know if you were responding to Tor directly, or through another
posting. Either way, it would be instructive to complete his quote,
as follows:

However, your understanding of superposition is wrong. You
CANNOT superimpose POWERS, or even talk about the "power" of various
reflections in the same media. You can only add the wave amplitudes
(electric fields). THEN you take the total amplitude and square that
to get the power.

which seems to "square" with where you take this:

It is done all the time. P=Esquared / R = Isquared x R.

Of course the topic has drifted into this problem of superposition
which has been approached before along similar lines. The truth of
the matter in the thin film discussion is that it doesn't require a
complete solution of either electric fields or power to realize that
"total cancellation" of the waves described in the example is
impossible.

It is quite evident that through the actions of the first interface,
that there is less energy incident upon the second interface.
Further, given that both interfaces operate with identical reflective
and transmissive properties, it follows the second interface could not
reflect enough to totally negate the reflections of the first.

Torr had a question in this regard, but as he is a casual
correspondent we have not seen any further comment from him - nor
would I expect him to have amplified on your observation above.

73's
Richard Clark, KB7QHC

Richard Clark wrote:

It is quite evident that through the actions of the first interface,
that there is less energy incident upon the second interface.
Further, given that both interfaces operate with identical reflective
and transmissive properties, it follows the second interface could not
reflect enough to totally negate the reflections of the first.

True for any one reflection. But as an optical engineer I'm sure you're
aware that, even in a lossy medium, a given wave reflects back and forth
multiple times before it's amplitude is reduced to insignificance. As
you know, the measured amplitude at a surface would then be the
superposition of multiple successively reflected waves.

So says JM Vaughn in his book "The Fabry-Perot Interfermeter". He cites
Born and Wolf a lot. Also Kuhn, Steel, Liddell, Mcleod, Meissner,
Tolansky, Jacquinot, and of course, Fabry and Perot. Strangely, no
mention of Hecht.

For those who are interested, a Fabry-Perot interferometer is an optical
instrument comprising two partially reflective parallel mirror surfaces
separated by some fixed or variable distance. It's basically a
narrow-bandpass filter at light wavelengths.

73, ac6xg

Roy Lewallen wrote:
Since you say that power
superposes, we should expect power waves to add and cancel just like
voltage waves.

Strawman Alert!!! Richard H. did NOT say power superposes.
It is obvious that what he was disagreeing with was the
statement: "You CANNOT ... even talk about the "power"
of various reflections in the same media."
--
73, Cecil http://www.qsl.net/w5dxp

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Richard Clark wrote:
It is quite evident that through the actions of the first interface,
that there is less energy incident upon the second interface.

On the contrary, it is quite evident that through the actions of the first
interface, namely reflection and wave cancellation, that there is *MORE*
energy incident upon the second interface. Here's the example from my
earlier posting. Those who can, please calculate the forward power in the
1/4WL 100 ohm section. Who believes, like Richard C., that it cannot
possibly be 100 watts or more?

100W 1/4WL
XMTR---50 ohm---+--------100 ohm---------+---200 ohm---200 load
feedline A feedline B feedline

Torr had a question in this regard, but as he is a casual
correspondent we have not seen any further comment from him - nor
would I expect him to have amplified on your observation above.

Why don't you send Tor an email and allow him to teach you how to
handle problems like these?
--
73, Cecil http://www.qsl.net/w5dxp

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Jim Kelley wrote:
That's because power doesn't propagate, and
I hasten to add - neither do Poynting vectors. Nor does power reflect,
refract, or diffract. Power is a rate at which energy is transferred,
absorbed, or dissipated. It's not a wave which propagates. It's a
mathematical product of two of the characteristics of a wave that
propagates.

I'm not arguing with you but as with all words, it depends upon how
one defines "power". You have your own personal narrow definition.
The power industry has a different definition.
Even the field of RF engineering has a definition of power
different from yours. The IEEE Dictionary has 11 pages of definitions
dealing with power including the "power reflection coefficient",
"power density of a traveling wave", "radiated power", "power-flow
vector", "power transfer", "power 'carried' by a waveguide" ...

Why don't you visit your local power company, call a meeting of their
engineers, and inform them that there is no power flowing in their
transmission lines? :-)
--
73, Cecil http://www.qsl.net/w5dxp

Jim Kelley wrote:
"That`s because power doesn`t propagate, and hasten to add - neither do
Poynting vectors."

Words are not objects. They merely represent objects. We use
abstractions for brevity and clarity. Even the best say: "--the Poynting
vector or power density (watts per square meter)." See Kraus` 3rd
edition of "Antennas" page 73, under "Power Patterns".

Best regards, Richard Harrison, KB5WZI

On Fri, 29 Jul 2005 16:58:29 -0700, Jim Kelley
wrote:
It is quite evident that through the actions of the first interface,
that there is less energy incident upon the second interface.
Further, given that both interfaces operate with identical reflective
and transmissive properties, it follows the second interface could not
reflect enough to totally negate the reflections of the first.

True for any one reflection.

Hi Jim,

And true for ALL accumulated reflections there after. Reflections do
not add any energy to the cup when the first interface is draining it
more quickly.

But as an optical engineer I'm sure you're
aware that, even in a lossy medium, a given wave reflects back and forth
multiple times before it's amplitude is reduced to insignificance. As
you know, the measured amplitude at a surface would then be the
superposition of multiple successively reflected waves.

My analysis allowed ALL of the energy in the reflection from the
second interface ( 0.098X) to combine with the first reflection
(0.11X). This total superposition was both more than generous, and at
the same time very unlikely; and yet with this generous allowance
there is still excess reflection from the first interface. Hence for
something less than total superposition of ALL energies, it hardly
bodes a better yield in total cancellation - the energy just isn't
there in the first place. 0.098X 0.11X is the simple economics of
the balance.

As an optical engineer, I've dealt with the harsh reality of this myth
of total reflection cancellation. I've designed systems with 9 orders
of dynamic range and the couple of percent dashed off as being
invisible by academics was distinctly and overwhelmingly present.
Basically I had the advantage in fluorescence measurement in being
able to turn off my detector during the initial flash to suppress the
reflection products in time rather than in these shenanigans (I did
them too - and certainly much more - because even the detector can be
blinded in its "off" state).

Basically these claims are for first year students where demanding too
much inquiry would push them into switching majors to Business school.
Simple optics with simple, ordinary glasses exhibit quite useful
results, but they do not embody a proof. To anyone following the math
of my presentation, it is quite obvious what WOULD tend towards a more
complete cancellation - and such a subtle shift in the formula
diverges only slightly from the choir book hymn. It's not that hard
when the interface ratios drive the answer.

73's
Richard Clark, KB7QHC

Richard Harrison wrote:

Jim Kelley wrote:
"That`s because power doesn`t propagate, and hasten to add - neither do
Poynting vectors."

Words are not objects. They merely represent objects. We use
abstractions for brevity and clarity. Even the best say: "--the Poynting
vector or power density (watts per square meter)." See Kraus` 3rd
edition of "Antennas" page 73, under "Power Patterns".

If we send a pulse of energy down a transmission line, one
wonders where the power in the pulse is if not in the pulse.
--
73, Cecil http://www.qsl.net/w5dxp

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