On Jun 21, 1:03*am, Keith Dysart wrote:
On Jun 18, 8:36*pm, K1TTT wrote:



On Jun 19, 12:22*am, Keith Dysart wrote:

On Jun 17, 8:38*pm, Cecil Moore wrote:

On Jun 17, 7:00*pm, Keith Dysart wrote:

I will caution again, do not assign too much reality to these
forward and reflected waves; very useful for solving problems,
but trouble if carried too far.

I'm afraid you are too late with that advice. Optical physicists have
assigned reality to forward and reflected waves inside an
interferometer and have tracked all of the energy including the energy
in canceled wavefronts.

http://www.teachspin.com/instruments...eriments.shtml

Scroll down to, "Using Dielectric Beamsplitters to find the "missing
energy" in destructive interference" It says: "... when interference
is destructive at the standard output, it is constructive at the non-
standard output."

In RF terms, when interference is destructive at a Z0-match, it is
constructive in the direction of the load.
--
73, Cecil, w5dxp.com

You have fallen in to the same trap that you did with circulators.
When you introduce a circulator in to the transmission line, you
change the experiment and you get certain results. The circulator
provides some numerology that appears to support the hypothesis
of 'reflected energy' being real. The same with looking for
"missing energy" (note they even quote it) with a beamsplitter.
The beamsplitter reveals what is happening at the beamsplitter,
not what is happening at points between the two beamsplitters.

Just like a circulator.

...Keith- Hide quoted text -

- Show quoted text -

he is even more hopeless than most, first he determines there are
forward and reflected waves when a dc source has charged the line,
then he rejects circulators and beamsplitters that obviously work
because of the waves. *so dc waves exist, but rf and optical waves
don't. *thats a unique viewpoint.- Hide quoted text -

- Show quoted text -

You might consider the following experiment:

- a sinusoidal generator that will produce 50W in to 50 ohms
- attach one half wavelength of transmission line
- leave the far end of the transmission line open

After the line settles, a directional "wattmeter" anywhere on
the line will indicate 50W forward and 50W reflected.
The generator will not be putting any energy in to the line
since the line input appears as an open circuit at the
generator output.

Insert a circulator (with the reflected port terminated in
50 ohms) between the generator and the line. The circulator
termination will be dissipating 50W and the generator will
now be delivering 50W in to the circulator.

1. Please explain how inserting the circulator did not change
* *the circuit conditions. The generator went from delivering
* *0W to delivering 50W.

2. Where do you think the 50W being dissipated in the circulator
* *termination resistor is coming from? The line? Or the
* *generator (which is now outputting 50W)?

well, its coming from the generator via the far end of the line of
course.


Just for fun, you may be interested in a DC coupled circulator
that works all the way down to 0 Hz:http://www.techlib.com/files/RFDesign3.pdf

Answer question 2 for this circulator design.

...Keith

the result is no current, but 50 v, as would be expected of course.