I have studied Rayleigh fading as a channel model for fast fading, capturing the received signal power at any time $t$ of the sum of all multipath components as Rayleigh distributed. Currently, I am particularly interested in how Rayleigh fading evolves over a period of time. Under Rayleigh fading model, is the channel gain at time $t+c$ independent from $t$ for arbitrary $c$? What if $c$ is really small? I am aware in some discretized cases, the channel gain is assumed to be static in one coherence time. What about continuous scenarios?
1 Answer
$\begingroup$ $\endgroup$
3 Rayleigh fading itself doesn't specify the temporal characteristics of the channel. This is why a Rayleigh fading channel can be either "fast fading" or "slow fading". To be fast fading, the transmitter, receiver or channel elements must be changing such that the received signal (as derived from a Rayleigh statistic) is changing significantly faster than the symbol rate, while with slow fading the elements involved are changing very slowly such that the received signal level (as also derived from a Rayleigh statistic) stays at that level for several symbols (long coherence time). The rate of change of a channel is often captured in the "Doppler Spread", with some nice comparison plots given at this Wikipedia link.
Please see this related post for further suggestions on generating a Rayleigh fading model.
- $\begingroup$ Thanks! I am clear now. Is there any models capture the temporal characteristics of the channel gain then? $\endgroup$LOREY CHU– LOREY CHU2025-01-07 07:08:22 +00:00Commented Jan 7 at 7:08
- $\begingroup$ Yes, see the bottom of the Wikipedia link summarizing the common approaches to doing this: Jakes's model, and filtered white noise. I like the filter approach due to its simplicity. What I essentially do in that case is generate a magnitude and phase from a Rayleigh distribution (which is the mag and phase of a complex white noise) for each received sample, but then filter those values first with a response consistent with the expected bathtub PSD (which you can generate most easily using the approach described in the last paragraph in this post: dsp.stackexchange.com/a/93746/21048 ). $\endgroup$Dan Boschen– Dan Boschen2025-01-07 13:48:57 +00:00Commented Jan 7 at 13:48
- $\begingroup$ the filtered stream of channel gain and phase values is then multiplied with the received signal. This is all done at complex baseband to be simplest. The frequency response you want for the resulting filter is given by the $S(v)$ in the wikipedia link under "Doppler power spectral density". (if using the FFT approach, the FFT values would be the square root of the PSD) $\endgroup$Dan Boschen– Dan Boschen2025-01-07 13:55:38 +00:00Commented Jan 7 at 13:55