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.
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.
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. 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.
By the way, I've tried to contact N6BV to see what his take on all of
this is but so far I haven't gotten a reply. The "manual" he wrote to
accompany HFTA, however, includes some fairly intricate diagrams of the
ray combinations that are included in the program calculations and it is
very hard to believe that energy in the reverse direction would sum at
the antenna in the same manner it left when taking into account
asymmetric terrain. In any case, I would certainly defer to his
expertise on the subject if we are lucky enough to hear from him.
Dave AB7E
Richard (Rick) Karlquist wrote:
> Directional couplers are reciprocal. Are you implying
> that they might not be?
>
> An example of a non reciprocal circuit is a circulator.
> This requires a non-reciprocal medium (ie ferrite).
> All electromagnetic structures with reciprocal media such
> as air as reciprocal. If there were a way to make a
> non reciprocal microwave structure using only air, it would
> have been discovered and exploited long ago. It would
> be worth a lot of money.
>
> I understand what you are saying below. What you are
> saying is that non-reciprocal propagation is possible
> if diffraction is involved. What I am trying to say
> is that non-reciprocal propagation in general is impossible,
> including diffraction effects. It is not necessary to
> specifically prove diffraction is reciprocal.
>
> The reason why I mentioned the other concepts was because
> you brought up HFTA. It would address those other
> concepts (except noise) but not reciprocity.
> Again, if there were non-reciprocity, HFTA
> would ask you to specify the direction of propagation.
>
> Perhaps an optical analogy would be easier to understand.
> Show me an example of where I can shine a laser beam at
> a corner cube mirror and have the beam reach the mirror
> but not be able to reflect back to the laser source. You
> can include diffraction effects in the path.
>
>
> Rick N6RK
>
> David Gilbert wrote:
>> Rick,
>>
>> I don't see how your 2-port S-parameter example is analogous at all.
>> If anything, a directional coupler might be a closer analogy to this
>> discussion.
>> Also, I know that reciprocity does not require the takeoff angle of
>> the transmitted signal to be the same as the arrival angle of the
>> receiving station at the far end, and I did not say or imply that it
>> was. I'm also not implying differences in noise levels on the two
>> ends of a path, I'm not implying differences in path loss for the two
>> directions, and I'm for sure not referring to station A working Asia
>> from the east side of a hill while station B works Africa from the
>> west side of the same hill.
>>
>> I'll try to simplify the hypothesis. Assume Station A in Colorado
>> has a yagi pointed east and it is located a few thousand feet west of
>> an asymmetrical hill (steep on the west side, blunt on the east side)
>> high enough to cause diffraction products that bend his signal lower
>> than it would be without the hill. Let's say the partially
>> diffracted signal has a takeoff angle of 4 degrees. Station A's
>> signal travels from the hill upward at that 4 degree angle where it
>> interacts with the ionosphere and eventually returns to earth (at
>> some arbitrary angle depending upon propagation) at Station B in
>> Spain. Station B is identical to Station A in terms of power,
>> antenna gain and efficiency, ground conductivity, and whatever other
>> factors may normally affect signal strength. Station B receives the
>> signal from Station A and sends a reply. That reply travels back
>> toward Station A along the same path, suffering the same path losses,
>> and presumably arrives at the asymmetrical hill at the same 4 degree
>> angle that Station A's signal left it. However .... the signal from
>> Station B is not "processed" by the asymmetrical hill the same way
>> that the signal from Station A was. The asymmetrical hill (blunt on
>> the east side) does not bend Station B's signal downward toward
>> Station A's antenna the same amount as the signal going from A to B
>> was bent.
>> 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.
>>
>> So where am I wrong?
> Di
>> Dave AB7E
>>
>>
>
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