I don't think it works that way. We're not talking diffraction through
a prism here at all. A prism is a uniform medium that refracts a wave
front uniformly. Edge diffraction is a different mechanism ... wave
energy passing over an edge (sharp, blunt, whatever) bends in a
dispersive manner that isn't uniform at all. As K8RI just noted, to
recreate the original wave you'd have to generate and sum all the
various rays from their various angles that had been dispersed by the
edge. The calculation to figure the bending caused by a prism is
trivial ... the calculations necessary to determine the angular density
of a wave passing over a uniform edge, much less a complex shape, are
horrendous ... which is why we have HFTA.
I remember optics experiments in high school that demonstrated the
difference between a wave passing over a rounded edge versus a wave
passing over a sharp edge (radii dimensions referenced to a wavelength)
... a knife edge creates more dispersion and bends at least some of the
rays more than a blunt edge does. The principle is well-defined. HFTA
uses ray tracing calculations for edge diffractions, surface
reflections, and combinations thereof to compile a net elevation profile
of the energy passing over the terrain after leaving the antenna. To
say that a single ray coming from the far end of the path and "seeing"
an entirely different terrain profile from the return side (assuming
asymmetric features) would follow the same path to the antenna just
doesn't make sense. Even HFTA says (per the very simple test profiles I
generated) that an asymmetrical terrain feature results in asymmetrical
diffraction when viewed from both sides. Unless you want to challenge
the algorithms and calculations built into HFTA, I don't see how how you
and N6RK can challenge that result.
It does NOT require nonlinearity to achieve non-reciprocity ....
non-equivalency in the paths does the job very nicely.
Let's try this a different way. It is a given that a wave passing over
a rounded edge diffracts and disperses differently than a wave passing
over a sharp edge. So ... assume we have an edge with a sharp radius on
one side and a more blunt radius on the other side. If we orient that
edge horizontally and shine light past it from the sharp side, we can
cast a vertical spread beam pattern on a wall beyond it. Now pick any
spot on the wall that has been illuminated and shine a light back toward
the edge ... which of course would now be from the blunt side. Any
light passing across that blunt edge is going to bend less and disperse
less, and there is no guarantee that any of it will land on the original
light source since the path conditions are not the same. The entire
system is fully linear, but the paths are not equivalent since one
passes across a sharp edge and the other passes across a blunt edge.
I can't see how signals going out and coming back over asymmetric
terrain is any different, but I'd love to have someone clearly explain
it to me if I am wrong.
73,
Dave AB7E
Jim Lux wrote:
>>> It seems to me that a good portion of Station B's signal arriving at the
>>> hill could overshoot Station A because it wasn't diffracted enough. The
>>> path taken by the signal from B to A would be identical to the path
>>> taken by the signal from A to B EXCEPT for the portion between Station A
>>> and the hill. Think directional coupler.
>>>
>>>
>
> Uhh, nope.. even though it sort of seems that way at first glance..
>
>
> Imagine that the top of the hill is a big lens or prism that bends the
> ray (which is what diffraction is, in one sense)..
>
> Launch a ray from A towards B and it follows a certain path, apparently
> bending over the top of the ridge. Some distance after the hill, stop
> the ray and send it back exactly as it came. When it gets to the
> prism/lens it will bend it just like it did on the outbound trip and
> wind up at A.
>
> As long as there no nonlinearity in the system, it works.
>
> Where there IS non-reciprocal propagation you need something else. here
> are some examples:
> a) propagation through the ionosphere, which is anisotropic. The
> polarization is rotated by the Faraday effect, but the same rotation
> regardless of direction, so, if you have polarized transmit and receive
> antennas, oriented, say, 45 degrees apart, and the rotation is 45
> degrees, A signal from A to B winds up perfectly lined up, but a signal
> from B to A is cross polarized.
>
> b) the ordinary and extraordinary ray (polarization dependent) follow
> different paths, so when they combine the net effect is different
> depending on the direction (actually another case of the phenomenon in A)
>
> c) sound propagating in a medium with a velocity gradient. This causes
> sound to bend down when going downwind, and bend up when going upwind.
>
> d) various and sundry light propagation through various crystal
> phenomena (e.g. polarizers, birefringent materials, etc.)
>
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