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Re: [TowerTalk] Takeoff Angles and Non-Reciprocal Propagation

To: towertalk@contesting.com
Subject: Re: [TowerTalk] Takeoff Angles and Non-Reciprocal Propagation
From: K4SAV <RadioIR@charter.net>
Date: Sat, 03 Nov 2007 22:42:13 -0500
List-post: <mailto:towertalk@contesting.com>
I don't know the answers here, but I have been following this discussion 
trying to make sense of it.  I am beginning to get a dim view of how 
this may work.  The definitions of reciprocity given, or implied, so far 
have been causing me some confusion.  I have seen two or three 
definitions so far along with some implied definitions and they are not 
the same, however all of these definitions seem to have something in common.

For example K1MK wrote:

.."Suppose I send a plane wave at some complicated diffractive surface. 
Let's say this plane wave was initially moving from left-to-right in one 
direction with a uniform amplitude. Upon encountering the surface, the 
wave gets diffracted and ends up going off to the right in a bunch of 
different directions with differing amplitudes and phases.

What reciprocity says is if I send in a bunch of waves from the right 
from different directions with differing amplitudes and phases such that 
they replicate the output seen above, I'll get a copy of the original 
wave moving to the left."...

Assuming no energy dissipated in the object, that kinda makes sense, 
however this doesn't completely explain the question at hand.  If you 
apply that to a radio example you also don't have the complete set of 
incoming rays to do the combining.

 N3OX wrote:

..."The demonstration of reciprocity ends in equation 9, showing that 
the wave coming in from the left (1) and exiting the right (t) gives the 
same fraction of transmission across the barrier the as the wave 
entering the right (1) and leaving the left (t').  If you transmit 1 and 
the DX station gets t, reciprocity says that if the DX station transmits 
1, you get t.  You transmit 100W they get, say, 20 microwatts... they 
transmit 100W, you get 20 microwatts."...

This is a little different from the first statement, but it doesn't 
contradict it.

K1MK previously made a statement that agrees with the previous one by N3OX:

..." All reciprocity says is that for any one of the multiple diffracted 
rays, the amplitude of that diffracted ray depends only on the included 
angle between the incident ray and the diffracted ray and not on the 
direction of the incident ray."....

Hmmm. That would imply that the shape of the object that is doing the 
diffracting is not important.  Maybe this statement is too simplified or 
maybe I'm not intrepreting it properly.

However from these comments, the one thing in common is that the 
transmitted path loss, in either direction,  has to be the same for any 
angle pairs you pick regardless of the shape of the object.  At first I 
though that required an identical diffraction pattern for both 
directions, but maybe not. I had to draw a few diagrams to convince 
myself that this was possible.  I drew an incident received waveform and 
split in into several diffracted waves.  Then from the transmit side I 
looked at a wave hitting the object at one of these same angles.  As 
long as you make the angle pair (one side to the other) the same loss, 
the rest of the pattern can be anything.

Let me do a bit of rambling and maybe someone will tell me how far off 
base I am.  For an incident wave, there would be a completely different 
diffraction pattern for each angle and direction, but for any particular 
angle pair, the path loss in both directions would have to be the same 
(according to these definitions).  A practical radio example, of course, 
would consist of more than one wave.  For a radio wave (ignoring 
ionosphere effects) looking at an object near a source, there will be an 
infinite number of incident waves hitting the object and these will 
produce an infinite number of diffraction patterns, but they will 
combine with some attenuation to produce a signal at a desired angle on 
the other side.  To simplify, let's assume only one angle is used for 
this path and all signal for the path came from diffraction by the 
object.  In the reverse direction, a single incoming wave at the path 
angle, will produce a diffraction pattern consisting of the same number 
of waves that were used to produce the signal from the other direction 
(excluding any waves which did not hit the object and those that 
contributed zero to the path).

I'm not sure I understand exactly why this happens, but it does seem 
possible, and it agrees with the definitions presented so far, unless I 
have gone astray somewhere.  It also doesn't contradict HFTA, which says 
that the attenuation produced by a path object is different for 
different shapes.  Unfortunately it is not possible to turn the path 
around in HFTA to prove the point.  You would need to be able to 
generate a single wave of selectable angle to do it.  Turning the object 
and using the same source isn't the same thing.

It was also interesting to see some examples listed where reciprocity 
does not apply, but these are due to some special cases, mostly 
associated with polarization.

Comments, either way, appreciated.

Jerry, K4SAV

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