generating a weibull channel on Matlab

Hi,
I am new here.. I just have a doubt on how to create a Weibull channel on Matlab. For Rayleigh, I know it can be given as: h = 1/sqrt(2)*[randn(nRx(jj),N) + j*randn(nRx(jj),N)] where, nRx is the length i assumed and N is the number of bits in this case.
But how can I generate Weibull channel in the same manner ?
Please help me in this regards. That will be much appreciated.
Thanks and regards,
Raj.

 採用された回答

bym
bym 2011 年 11 月 27 日

2 投票

hw = lamda.*(-log(rand(nRx(jj),N)).^(1/beta))
where lamda & beta are your Weibull parameters. BTW you can generate your Rayleigh channel by using beta = 2

4 件のコメント

raj0689
raj0689 2011 年 11 月 27 日
Thanks a lot.. :)
now i can proceed with my project with your help.. :)
yogesh chinnaraja sundararajan
yogesh chinnaraja sundararajan 2013 年 4 月 4 日
can any one help me what values can be substituted for Weibull parameters lamda and beta in the above equation hw= lamda.*(-log(rand(nRx(jj),N)).^(1/beta))
can
can 2017 年 5 月 22 日
What about Rician Channel? How can I generate Rician channel in the same manner ? Please help me, Thanks and regards.
senem cönger
senem cönger 2022 年 4 月 27 日
lamba ve beta kaç alınmalı burada?

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