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Old May 25th 10, 03:38 PM posted to rec.radio.amateur.antenna
Cecil Moore Cecil Moore is offline
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First recorded activity by RadioBanter: Aug 2006
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Default Question about "Another look at reflections" article.

On May 25, 2:56*am, Roy Lewallen wrote:
The chart and discussion in the "Forward and reverse power" section show
that the concept of "reflected power" being absorbed in or dissipated by
the source is incorrect.


I'm sorry, Roy, but that is a very misleading statement. There are
THREE things that can possibly happen to the reflected energy.

1. It can indeed, be absorbed/dissipated by the source but it
certainly doesn't have to be. The conditions governing absorption,
reflection, and redistribution of EM wave energy are well understood
in the field of optics. RF gurus seem to be myopic about interference
effects.

2. It can obey the laws of the reflection model for EM waves, e.g. if
the source impedance is different from the Z0 of the transmission
line, some reflected energy will be re-reflected. (I know that doesn't
apply to your "food for thought" examples because the source resistor
is equal to the Z0 of the feedline.)

3. The interference phenomenon that you completely ignored in your
discussion, presumably through ignorance. This third possibility for
the reflected energy redistribution is associated with constructive
and destructive interference. It is the same phenomenon that is in
operation with the 1/4WL coating on non-reflective glass. Here's a
quote from a Florida State University web page with capitals added by
me for emphasis:

micro.magnet.fsu.edu/primer/java/scienceopticsu/interference/
waveinteractions/index.html

"... when two waves of equal amplitude and wavelength that are 180-
degrees ... out of phase with each other meet, they are not actually
annihilated, ... ALL OF THE PHOTON ENERGY present in these waves must
somehow be recovered or REDISTRIBUTED IN A NEW DIRECTION, according to
the law of energy conservation ... Instead, upon meeting, the photons
are REDISTRIBUTED TO REGIONS THAT PERMIT CONSTRUCTIVE INTERFERENCE, so
the effect should be considered as a REDISTRIBUTION OF light (and RF)
waves and PHOTON ENERGY rather than the spontaneous construction or
destruction of light."

This is the concept that you have been missing for eight years.
"Redistribution of energy" in a transmission line is simply a reversal
of energy flow. Each of your examples have different magnitudes of
interference (depending on the phasing between the forward energy and
the reflected energy). The phasing is a variable for all of your
examples. Why did you completely ignore the interference? Why would
you expect a CONSTANT magnitude of reflected energy to be dissipated
in examples with VARIABLE phasing and VARIABLE levels of
interference??? Here are the possibilities:

1. No Interference at the source resistor - this happens when the
forward wave is 90 degrees out of phase with the reflected wave at the
source resistor. In this case, the reflected energy is indeed
dissipated in the source resistor. I have written a short article on
the no interference case at:

http://www.w5dxp.com/nointfr.htm

2. Constructive Interference at the source resistor - This happens
when the phase angle between the forward wave and reflected wave is
less than 90 degrees at the source resistor. The voltage across the
source resistor increases and therefore more power is dissipated in
the source resistor. In fact, for total constructive interference, all
of the forward power and all of the reflected power is dissipated in
the source resistor. This happens when the SWR is infinite and the
reflected wave arrives at the source resistor in phase with the
forward wave from the source.

3. Destructive interference at the source resistor - This happens when
the angle between the forward wave and reflected wave is between 90+
degrees and 180 degrees. In this case, reflected energy is
redistributed back toward the load and less is dissipated in the
source resistor. In fact, for total destructive interference, zero
reflected power is dissipated in the source resistor and all of the
reflected energy is redistributed back toward the load. This happens
when the SWR is infinite and the reflected wave arrives at the source
resistor 180 degrees out of phase with the forward wave from the
source.

I explained all of this in my "WorldRadio" article way back in 2005.
Presumably, you have never read it. In order to understand the role
that interference plays during the superposition of EM waves, you
might want to read it and take a look at the references. The question
of where reflected energy goes was answered long ago by optical
physicists. Too bad that RF gurus remain so ignorant of that
knowledge. Here's that five year old "WorldRadio" article.

http://www.w5dxp.com/energy.htm

You obviously know how to add voltage phasors but you are obviously
ignorant as to what happens to the ExH power in each of those waves.
In any optics reference book, you will find the following irradiance
equation. Since irradiance uses the same units as the Poynting vector,
I have changed I (irradiance) to P (power density) in the irradiance
equation. I first saw this equation in Dr. Best's QEX article[1].

Ptot = P1 + P2 + 2*SQRT(P1*P2)cos(theta)

When any two EM waves are superposed, whether light waves or RF waves,
this is what happens to the component powers. The phase angle, theta,
is the relative phase angle between the two electric fields before
superposition. The last term in the equation is called the
"interference term" and it is easy to see how it modifies the total
power in the wave after superposition. Energy must be conserved. If
destructive interference occurs between two waves in a transmission
line, constructive interference must occur in the opposite direction -
and vice versa.

By completely ignoring the destructive and constructive interference
occurring in your "food for thought" examples, you have ignored the
accepted laws of physics (available in any physics optics reference)
and arrived at completely false conclusions. Roy, every one of your
examples can be explained simply by using the above equation from the
field of EM wave optics. Differing types/levels of interference
explains every one of your examples perfectly.

[1] Best, Steven R., "Wave Mechanics of Transmission Lines, Part 3",
QEX, Nov/Dec 2001
--
73, Cecil, w5dxp.com