Diffraction Across a Knife Edge

[skeptic2]

I was having a discussion with someone far more expert than I and for whom I have a lot of respect. He was telling me that the path loss of a radio signal from A to B is not necessarily the same as it is from B to A. One of the reasons he mentioned is diffraction over a knife edge which in radio could be the edge created where a highway cuts through a hill. In radio as well as optics, the diffraction produces a series of light and dark lines on the far side of the edge.

My position is that the path loss is symmetrical and all the effects that affect path loss such as diffraction, absorption, multipath, the Fresnel Effect, which is really the same thing as multipath, and shadowing are all symmetrical with respect to the direction of propagation. With respect to diffraction, if the transmitter is a A and the receiver is at B, with the knife edge closer to B, B may be in a bright line or dark line but either way the path loss will be the same in both directions.

Can anyone think of any reason this should not be so?

 

[Born2bwire]

The theory of reciprocity supports your assertion. For most cases, the performance of a link is going to be the same when you switch the locations of the transmitter and receiver (and obviously correct the direction of transmission and receiving). There are exceptions to this but they generally do not come up with most line of sight and multipath problems. The one exception I know of off hand is the reflection off of the ionosphere because the ionosphere is a plasma. A plasma is not a reciprocal material and this means that when bouncing shortwave off of the ionosphere, you are not guaranteed that you can transmit back to a station that you receive. I don't see any reason by diffraction would change reciprocity as it is still a basic scattering problem. For your own edification, you could easily calculate it out yourself. Constantine Balanis' "Advanced Electromagnetics" text discusses how to calculate the diffraction from a wedge and even has Fortran 77 code in the text to do the calculation. You could easily work out the received power at a given position relative to a source position and then swap them to see what happens.

 

[skeptic2]

Thank you very much Born2bwire. I really appreciate your answer. As I said this came out of a discussion with someone who knows a lot more than me. I didn't want to get into an argument with him, but I never did understand why both paths would not be the same. In my application ozone refraction is not a consideration.

 

[Born2bwire]

Thank you very much Born2bwire. I really appreciate your answer. As I said this came out of a discussion with someone who knows a lot more than me. I didn't want to get into an argument with him, but I never did understand why both paths would not be the same. In my application ozone refraction is not a consideration.

There are simplified ways at looking at it but reciprocity can be proven formally, I recall that Weng Cho Chew's "Waves and Fields in Inhomogeneous Media" has a derivation proving reciprocity for isotropic media. Looking at the Wikipedia article, their derivations are reminiscent of Chew's I believe.

 

[Antiphon]

To expand on this, the requirement for reciprocity to hold is that the media must be linear. It does not have to isotropic however.

The most direct example of a non-reciprocal system would be a three-port circulator with one port impedance-matched in a resistive termination. The circulator relies on a non-linear ferromagnetic element to do its job.