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lab3

于 2012-05-17 发布 文件大小:2KB
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  lear N = 10^5 EB_NO = [-1:30] liu1 = zeros(1,N) liu2 = zeros(1,N) for ii = 1:length(EB_NO) m= (2*(rand(1,N)>0.5)-1) + j*(2*(rand(1,N)>0.5)-1) s = (1/sqrt(2))*m normalization of energy to 1 n = 1/sqrt(2)*[randn(1,N) + j*randn(1,N)]

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