On Jun 20, 8:03*pm, Keith Dysart wrote:
1. Please explain how inserting the circulator did not change
* *the circuit conditions. The generator went from delivering
* *0W to delivering 50W.

The generator went from experiencing total destructive interference to
experiencing zero interference. Of course, the entire purpose of
inserting the circulator is to cause the generator to see 50 ohms as a
load impedance.

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)?

TV ghosting experiments will verify that the signal being dissipated
in the circulator load resistor has made a round trip from the
generator to the end of the N(1/2WL) stub and back to the circulator.
Please don't insult our intelligence by arguing that is not a steady-
state condition.
--
73, Cecil, w5dxp.com

On Jun 21, 9:04*am, Cecil Moore wrote:
On Jun 20, 8:03*pm, Keith Dysart wrote:

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

The generator went from experiencing total destructive interference to
experiencing zero interference. Of course, the entire purpose of
inserting the circulator is to cause the generator to see 50 ohms as a
load impedance.

Along with the other use for circulators: To present an impedance
match to the reflected wave so that it will not be re-reflected.

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)?

TV ghosting experiments will verify that the signal being dissipated
in the circulator load resistor has made a round trip from the
generator to the end of the N(1/2WL) stub and back to the circulator.

It is not clear why you think this verifies something.

Enhancing the details slightly of the generator in the original
example: The generator is constructed using the Thevenin model
of a source followed by a 50ohm resistor.

With the circulator, there is no re-reflection at the driven
end of the line, 50W is dissipated in the circulator
resistor and 50W is dissipated in the source resistor.

Without the circulator, there is no re-reflection and nothing is
dissipated in the source resistor (and there is no circulator
resistor).

The dissipation seems to correlate much more strongly with the
circuit design than it does with the magnitude of the "reflected
power".

Please don't insult our intelligence by arguing that is not a steady-
state condition.

Well you are right, it is not steady-state when there is modulation,
but modulation does not affect the examples I have provided.

....Keith

PS: For further understanding, substitute a Norton style generator,
then do it again for both kinds of generators but with the end
of the line shorted yielding more examples where the "reflected
power" does not correlate with the dissipation.

On Jun 21, 7:25*pm, Keith Dysart wrote:
The dissipation seems to correlate much more strongly with the
circuit design than it does with the magnitude of the "reflected
power".

Given a 50 ohm source resistor resulting in zero re-reflections, *the
dissipation correlates perfectly with the level and type of
interference*. I am amazed at the level of ignorance concerning
interference and the conservation of energy principle. I studied
interference in detail in my 1950's college physics courses including
what happens to the energy during interference events. Although Dr.
Best didn't realize it, in his QEX Nov/Dec 2001 article on
transmission lines, he presented the equations governing energy
redistribution associated with interference. What didn't help was his
equation of the form:

Ptot = 75w + 8.33w = 133.33w

which is obviously false. What he left out was the constructive
interference term of:

2*SQRT(75w*8.33w) = 50w

Given two coherent waves, each associated with a Poynting Vector, when
those two waves are superposed, interference will result. I'm sure
everyone has seen interference rings and pictures of interference
rings where the energy in the interference rings has obviously been
redistributed away from a homogeneous power density. The same thing
happens in a transmission line and/or at a source, just at longer
wavelengths where we cannot "see" them with our eyes.

Constructive interference requires "extra" energy. Superpose two 50
watt coherent waves in phase and the result is a 200 watt wave.
Destructive interference has energy left over. Superpose two 50 watt
coherent waves 180 degrees out of phase and the result is a zero watt
wave. At a source, the source is capable of supplying extra power or
throttling back on its power output depending on the level and kind of
interference. If one simply takes the time to understand interference,
one knows where the energy components go even when there is zero re-
reflection.

Given a single traveling wave in a lossless transmission line with a
purely resistive characteristic impedance of Z0, the power (joules/
second) measured at a point on the line is P=V^2/Z0=I^2*Z0=V*I. This
is classic transmission line math no matter which direction the wave
is traveling. Superposing two coherent traveling waves cannot create
or destroy the energy pre-existing in the two waves, but it can (and
often does) redistribute the energy according to the conservation of
energy principle. If the total energy in the two superposed waves is
different from the sum of the two energy components before
superposition, then interference exists and some interference energy
has been redistributed in the system. That some people are ignorant of
where the constructive interference energy came from or where the
destructive interference energy went is not a good reason to deny its
existence (which violates the accepted laws of physics).

The thing that makes an interference event so easy to analyze in a
transmission line is the fact that the transmission line is
essentially one-dimensional. At an impedance discontinuity in a
transmission line (away from any active source) if interference
exists, the destructive interference in one direction must equal the
constructive interference in the other direction. Any other outcome
would violate the conservation of energy principle and that is a
common occurence on this newsgroup.

In my earlier example, the two results of *(1) reflection* are
traveling waves containing ExH power densities:

Pfor1(rho^2) toward the source and

Pref2(rho^2) toward the load

The complimentary transmissions a

Pfor1(1-rho^2) toward the load and

Pref2(1-rho^2) toward the source

Those two reflections *(2) superpose* with the two complimentary
transmissions in each same direction according to the power density
equation:

0 = Pref1 = Pfor1(rho^2) + Pref2(1-rho^2) minus total destructive
interference toward the source

Pfor2 = Pfor1(1-rho^2) + Pref2(rho^2) plus constructive interference
toward the load

When one assumes that reflection is the only phenomenon that can
redistribute wave energy, one is ignoring the superposition process
which is also known to be perfectly capable of redistributing wave
energy. In Roy's food-for-thought example, he designed it so
reflections would not exist at the source. That leaves superposition
as the phenomenon that is accomplishing the obvious redistribution of
energy. It certainly does NOT mean that the laws of physics have been
suspended due to the ignorance of the writer.

Here's a simple exercise in phasor superposition. Given two 50w
(joules/sec) coherent EM waves traveling in the same direction in
Z0=50 ohm coax:

1. What is the magnitude of the two voltages? ________

2. If one superposes those two voltages in phase in the Z0=50 ohm
transmission line, what is the resulting power (joules/sec) in the
total wave? ___________

3. If one superposes those two voltages 180 degrees out of phase in
the Z0=50 ohm transmission line, what is the resulting power (joules/
sec) in the total wave? ____________

In #2, if there is no source around, where does the "extra" wave
energy come from?

In #3, if there is no source or sink around, where does the "excess"
wave energy go?
--
73, Cecil, w5dxp.com