Basic knife-edge diffraction

Question 1

How is the Fresnel radius defined?

How does it relate to diffraction?

Answer 1

r_f = sqrt(lambdad1d2 / d1+d2)

The radius forms a 3d elliptical zone between links in a communication channel. objects that fall within 0.6 * r_f will cause diffraction, regardless of them being in the LoS or not. As such a diffraction loss needs to be modelled into the propagation.

Question 2

How is the knife-edge diffraction loss related to the electric fields involved in this model?

Answer 2

L_diff = | E_diff / E_Tx |

Question 3

What is the equation for the Fresnel diffraction parameter?

What does the delta h correspond to in the equation and how does this impact the diffraction loss?

Answer 3

v = deltah * sqrt (2 * (d1 + d2) / (lambda * d1 * d2))

Question 4

How is the diffraction loss calculated from the Fresnel parameter?

Answer 4

L_diff = 1/ ( pi * v * sqrt (2) )

Question 5

Above what threshold loss does the approximated diffraction loss agree with the actual loss?

What consequence does this have on how the approximation is used?

Answer 5

threshold is about 20 dB as per the graph in the notes. losses lower than this see a divergence between the two models and thus the approximation should not be used in these cases.