Modelling electromagnetic wave reflection

[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

 

[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.

 

[astrokreng]

Hi Niko,

Thank you for sharing your thoughts and concerns about modelling electromagnetic wave reflection. The solution provided in the example is based on the principles of electromagnetic wave propagation and reflection. Let me explain it in more detail.

When an electromagnetic wave is emitted from a source, it propagates in all directions as a spherical wave. As the wave travels, it spreads out and its power is distributed over a larger and larger area. When the wave hits a reflecting surface, a part of the wave is reflected back towards the source, while the rest continues to propagate in the original direction. The reflected wave has a smaller power compared to the incident wave due to the reflection coefficient, which in this case is represented by the factor gamma.

Now, in order to calculate the power of the reflected wave at a certain point, we need to consider the distance between the source and the reflecting surface, as well as the distance between the reflecting surface and the receiving point. In this example, the altitude of the plane (h) is the distance between the source (radar) and the reflecting surface (ground). The factor of 2 is included because the wave has to travel from the source to the ground and then back to the receiving point, thus covering a total distance of 2h.

Furthermore, the formula also takes into account the gains of the receiver and transmitter, as well as the wavelength of the emitted wave. These factors play a crucial role in determining the power of the received wave.

I hope this explanation helps in understanding the reasoning behind the given solution. It is important to note that modelling electromagnetic wave reflection can be complex and there are various factors that need to be considered. I would suggest consulting with a more experienced physicist or conducting further research for a more comprehensive understanding.

Best regards,