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How correctly choose a channel model: Rician and Rayleigh channel model?

The Rician channel assumes dominant line of sight (LoS) paths from the transmitter to the receiver and other scattering paths. Rayleigh channel consists of scattering channels from the transmitter to the receiver, i.e., there is no dominant LOS propagation between the transmitter and receiver.

In my model, one link (A to B) is the dominant LOS path, and another link ( B to C) can have an NLOS path, which can be dominant. Can I assume in my design 2 channel model: Rician and Rayleigh or should I choose only one?

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  • $\begingroup$ What do you mean by one link in your model? Do you have two transmitter-receiver pairs? Rayleigh can be used as a worst case scenario. $\endgroup$ May 23, 2023 at 11:52
  • $\begingroup$ @Math_Novice link, path between a transmitter and a receiver $\endgroup$
    – Aid22
    May 24, 2023 at 13:25

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Different links can have different channel characteristics.

If your (A) is very close to (B) and any scatterers are far away then the LOS component dominates which could be modeled using a Rician model with high k-factor.

If your (B) is far from (C) or has many scatterers between then that link might be most accurately modeled by a lower k-factor Rician model or Rayleigh in the extreme (Rician w/ zero k-factor = Rayleigh)

Short article which touches on k-factor: http://www.wirelesscommunication.nl/reference/chaptr03/ricepdf/rice.htm

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  • $\begingroup$ can I assume Rician channel model for a link between a satellite and a receiver for example an aeroplane? In this example I can have NLOS link; obstacles other airborne platforms and then the best way is Rayleig model isn't? $\endgroup$
    – Aid22
    May 24, 2023 at 13:26
  • $\begingroup$ you can choose an appropriate k-factor to set how strong the NLOS paths are compared to the LOS path. For example, if you want the LOS path to be 10x stronger than the NLOS paths then k-factor = 10. If there is no LOS path, then k-factor = 0 (Rayleigh) would likely make more sense to use $\endgroup$
    – Engineer
    May 25, 2023 at 10:15
  • $\begingroup$ Can a link between space or air and a ground user be simulated by Rayleigh? $\endgroup$
    – Aid22
    Jun 6, 2023 at 9:52

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