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Issue with polarization in the ''raypl'' function

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Masoud
Masoud on 20 Jun 2022
Answered: Varun on 29 Nov 2023
Consider the attached figure. The red point is the transmitter and the blue one is the receiver on the same z-plane. There are two paths with equal distance and angle of incidence. The pathloss is the same for these two paths when polarization is not considered. But when using the ''raypl'' function with vertical polarization, the pathloss is no longer the same (the difference is almost 1.5 dB). How is this possible?

Answers (1)

Varun
Varun on 29 Nov 2023
Hi Masoud,
Looks like the pathloss is not same with the difference of almost 1.5 dB when using “raypl” function with vertical polarization.
The difference in pathloss with vertical polarization is due to the impact of polarization on radio wave propagation. Polarization refers to the orientation of the electric field vector in an electromagnetic wave, and it significantly influences signal reception in wireless communication. When the polarization of the antennas is not considered, the assumption is often made that the antennas are operating with a vertical and horizontal orientation (or any other orthogonal polarization). In such cases, the pathloss is calculated without considering the impact of the actual polarization of the antennas.
However, when you use the “raypl” function with vertical polarization, it means that the antennas are now polarized in the vertical direction. This introduces a polarization mismatch, affecting signal reception. Despite equal distance and angle of incidence, a lower effective received power may occur due to this mismatch.
You can polarize the antennas by specifying the “TransmitterPolarization” and “ReceiverPolarization” name-value arguments.
For example:
raypl(ray,"TransmitterPolarization","V","ReceiverPolarization","V");
Please refer to the following documentation to learn more about using “raypl”:
Hope this helps.

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