On May 27, 9:50*am, lu6etj wrote:
I am reading this newsgroup through Google groups web page and I just
realized that later replies to previous post are intercalated in the
thread, while I expected to see it always at the end of it, for that
reason I did not ACK before to it. (I hope yours be the only one, I
will review all thread tho chek for others).

I am also using Google since ATT dropped Usenet. I liked Thunderbird a
lot better than Google's usenet interface but I am adapting. The above
information is good to know. Thunderbird has a way to keep up with
unread vs read postings but Google doesn't seem to - at least I don't
know how to do it on Google.

In a early post I wrote = "of course if we insert a circulator to
separate both powers, generator now would see 1 ohm load, could
develope 1 W incident, 0 W reflected (Pn=1W) on circulator input, 0.36
W would be outputting on the other port to render 0.64 W (Pn) to the
load with 1 W Pf and 0,36 W Pr again"
Is this result OK for you?.

The SGCR source is usually designed for 50 ohms, i.e. the signal
generator always "sees" a 50 ohm load because it does not "see" any
reflected energy. The ideal circulator is usually designed with 50 ohm
line and a 50 ohm load resistor. If we could stick with that
particular configuration for the SGCR source, it would aid in my
understanding what is the actual system configuration, i.e. not your
fault but I am confused by your above posting.

I am interested in your optic analogy, I can imagine the load as a
partially reflecting surface, real part of it as absorbance
(transmittance if it was a radiator). line as a unidimensional medium
and reflection as the form of "redistribute energy" (is it OK?) and a
coherent light source for the voltage source, but I am still trying to
visualze the optical equivalent of source resistance and its job to be
a good analog, Also I am interested in check other values and
conditions in your other article (first part) with 45 degree line.

I don't think a laser source handles reflected energy like an RF amp
does. So, to start with, let's avoid reflected energy being incident
upon the laser source. Here is a good example to start with, a 1/4WL
non-reflective coating on glass.

Laser-----air-------|--1/4WL thin-film, r = 1.2222---|---Glass, r =
1.4938---...

The 1/4WL thin-film coating on the glass acts exactly like a 1/4WL
matching section of transmission line. Reflections at the air to thin-
film interface are eliminated by wave cancellation just as the FSU web
page says,

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 waves and
photon energy rather than the spontaneous construction or destruction
of light."

Note that the reflection coefficient, r, is 1.0 for air. Thus the
SQRT[(1.0)(1.4938)] = 1.2222 ensures that reflections are eliminated
by the r = 1.2222 thin-film coating.

The same thing happens at the '+' Z0-match in the following RF system.

XMTR---50 ohm coax---+---1/4WL 300 ohm feedline---1800 ohm load

Note that SQRT[(50)(1800)] = 300 ensuring that reflections are
eliminated.
--
73, Cecil, w5dxp.com