If i have distances of users in vector d=[50 100 150 200 250 300 ] ;
h=gain.path loss
h=g . pl^-1(d)
where's g is complex random variable with zero mean and variance (0.5) g~CN(0,0.5) and it has CDF and PDF
so how i can simulate the the Rayleigh fading channel h for each user and eventually i am confused am i getting the gain results as scale or matrix ? for each user .
Thanks in advance
Hi Abdalrhman Abdallah
I will suggest you follow this. This is also a way to generate Rayleigh fading channels with a given path loss model in MATLAB.
First, you evaluate the path loss in dB for the considered path loss model based on the distance between the users, since you have mentioned the vector of distances.
e.g. path_lossdB = reference path loss in dB + (10*path loss exponent*log(distance in Km) ).
Then,
variance_of_channeldB = - path_lossdB
(i.e. minus of channel attenuation variance in dB evaluated from considered path loss model).
Thus,
variance_of_channel = 10^(0.1*variance_of_channeldB).
e.g. denote hsd as a Rayleigh fading channel between the users with scale parameter sigma, then
variance [hsd] = ((4-pi)/2)*sigma2 = 10^(0.1*variance_of_channeldB).
Thus, sigma = sqrt(2/(4-pi)) * sqrt (10^(0.1*variance_of_channeldB)).
Now simply generate Rayleigh distributed random number (or numbers in a case for large samples) with this evaluated value of scale parameter sigma as
hsd = random( 'rayl', sigma );
I hope this will be helpful to you.
Cheers!
Hello,
yes, you seems to have a good idea of the problem: any propagation channel is usually composed of both deterministic and random phenomena 1/ the path loss is actually function of the distance and provide the average energy of the path 2/ the random component can be modelized through Rayleigh distribution. Why does Rayleigh channel means? A channel coefficient h is said to be Rayleigh distributed if h is a zero-mean complex Gaussian variable with variance 0.5 for both real and imaginary parts. In that case the module |h| actually obeys a Rayleigh distribution.
best regards,
Vincent
OK, the path loss gives you the average gain of the path, depending on the distance let it note PL(d).
Written in matlab, each channel gain is then: h(d) = sqrt(PL(d)/2)*(randn(1) +1i*randn(1));
It results that the channel coefficient has a Rayleigh distribution with variance PL(d). However, note this does not take into account the possible multipath for each distance, do you consider multipath channel or singlepath?
Vincent
Hi Abdalrhman Abdallah
I will suggest you follow this. This is also a way to generate Rayleigh fading channels with a given path loss model in MATLAB.
First, you evaluate the path loss in dB for the considered path loss model based on the distance between the users, since you have mentioned the vector of distances.
e.g. path_lossdB = reference path loss in dB + (10*path loss exponent*log(distance in Km) ).
Then,
variance_of_channeldB = - path_lossdB
(i.e. minus of channel attenuation variance in dB evaluated from considered path loss model).
Thus,
variance_of_channel = 10^(0.1*variance_of_channeldB).
e.g. denote hsd as a Rayleigh fading channel between the users with scale parameter sigma, then
variance [hsd] = ((4-pi)/2)*sigma2 = 10^(0.1*variance_of_channeldB).
Thus, sigma = sqrt(2/(4-pi)) * sqrt (10^(0.1*variance_of_channeldB)).
Now simply generate Rayleigh distributed random number (or numbers in a case for large samples) with this evaluated value of scale parameter sigma as
hsd = random( 'rayl', sigma );
I hope this will be helpful to you.
Cheers!
Prasanna Raut Realy you did a great job and i do appreciate it, many thanks to you
Check out this link
https://ecewireless.blogspot.com/2020/06/ber-of-3-user-non-orthogonal-multiple.html
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