# Modelling electromagnetic wave reflection

1. Aug 28, 2009

### niko2000

Hi,

I am trying to calculate the power of a reflected electromagnetic field and can't find a physical explanation for a given solution.

I've noticed the following example:

The Plane has a radar for altitude measurement which emmits the power Ptx at the frequency ftx. Calculate the power of the received waves when the plane altitude is h=60m and gamma (reflection) factor of the ground is 0.1.

The solution is Prx = Gtx*Grx((c/ftx)/(4*Pi*2*h))^2*gamma^2

What I don't understand here is why is it appropriate to take 2*h as a radius.

c/fth is the wave length
Gtx and Grx are the gains of the receiver and transmitter
and the factor (4*Pi*r)^2 in the denominator comes from spherical geometry (the sphere with radius r has a surface 4*Pi*r^2 - another 4*Pi comes from gain formula derivation)

The factor gamma is squared because the Poynting vector is a product of Electric and magnetic force and the ground reflects both thus gamma^2.

What I don't understand here is why the radius is set as 2*h.

I tried to solve the problem the following way:

The wave power at the point of collision is proportional to 1/(4*Pi*h)^2. At that point we consider the ground reflets 1/gamma^2 of the power and the reflected power at the receiving point is proportional to 1/(4*Pi*h^2).

I would appreciate if anyone could correct my point of view and explain what's the correct way to model a reflected electromagnetic wave.

Thank you!

Niko

2. Aug 28, 2009

### Born2bwire

That isn't the radius, the \lambda / (4\pi r^2) factor is called the space loss factor. In the far field, the source of an electromagnetic wave appears to be like a point source, thus, there is an inherent spreading of the wave over 4\pi r^2 in the same sense as a spherical surface is spreading. r, the distance between source and receiver is taken as 2h because the wave travels h to the ground, and then another distance h back up to the receiver on the plane.

The given equation is the antenna link equation for the far-field, you will find this in any antenna textbook.

Last edited: Aug 28, 2009