David Gilbert wrote:
> Unless I'm using the term incorrectly (certainly a possibility), a
> directional coupler routes signals leaving Port A to Port B, whereas
> signals leaving Port B in the reverse direction end up at Port C ... not
> Port A. As far as Ports A and B are concerned, I don't think we have
> reciprocity.
That's a circulator...
A directional coupler typically has 4 ports (with one terminated)
The main line goes from A to B (or B to A)
Some fraction (e.g. 1/10th for a 10dB coupler) of the power going from
A to B winds up at port C.
Some fraction of the power going from B to A winds up at port D (often
terminated).
If you put power into port C, some fraction winds up at port A, the rest
winds up at port D (absorbed in the load).
The attenuation from port A to port C is the same as the attenuation
from port C to port A, so the device is reciprocal.
>
> You say that non-reciprocal propagation (recognizing that neither of us
> is referring to the ionosphere here) is impossible even taking into
> account diffraction. If diffraction happens to cause energy to take a
> different path in the return direction than it took in the outgoing
> direction, why is non-reciprocal propagation not possible? Please
> explain without simply saying it is impossible.
Tricky to come up with a concise example here. I'm thinking about it,
because it IS sort of non-intuitive.
>
> I myself have tried to rely on an optical analogy in attempting to
> reason this through, but we're not talking about light passing through a
> uniform medium here ... we're talking light diffracting around an edge.
> The mechanisms are totally different. An edge disperses single
> wavelength energy ... a prism does not.
For a monochromatic signal (which your radio signal is, for practical
purposes), the prism and the edge work the same (granted, the "pattern"
is different, but for any single ray....)
An edge can look entirely
> different from one side than from the other (picture a half-cylinder),
> and light passing across that edge will refract and disperse in a manner
> that is a function of the radius of curvature seen by the wave.
Not exactly.. your analogy has lots of rays, and, indeed, a fan of rays
from one side will not produce an identical distribution on the other
side as the reverse. BUT, for any one ray going from A to B, the path
is reciprocal.
>
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