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  • 大小: 3KB
    文件類型: .m
    金幣: 1
    下載: 0 次
    發布日期: 2021-06-17
  • 語言: Matlab
  • 標簽: 中繼選擇??

資源簡介

進行了中繼選擇的描述和仿真,和傳統的中繼選擇算法進行比較

資源截圖

代碼片段和文件信息

clear?all;
echo?off;

%?Performance?of?Onat_Threshold?Scheme

%?Variable?Parameters

R=3;

snrdB=[8101214161820222426];

N1=1;
N2=1;
N_l=1;

be=1;

NI=10^6;

%?Model?

prb_O=zeros(1length(snrdB));

numerr_O=zeros(1length(snrdB));


r=zeros(1R);
y=zeros(1R);

for?i=1:10
?????i
?????SNR=exp((snrdB(i)/10)*log(10))/(R+1);
?????Th=log(SNR).*R/be;
?????%Th=log(SNR+1)./2;
????
?????for?ii=1:NI
????????
??????????s=zeros(1R);
????????????
??????????f=sqrt(N1./2).*(randn(1R)+j.*randn(1R));??
??????????g=sqrt(N2./2).*(randn(1R)+j.*randn(1R));
??????????v=sqrt(1./2).*(randn(1R)+j.*randn(1R));
??????????w=sqrt(1./2).*(randn(1R)+j.*randn(1R));
??????????
??????????f_l=sqrt(N_l./2).*(randn(11)+j.*randn(11));
??????????v_l=sqrt(1./2).*(randn(11)+j.*randn(11));
?
?????????
??????????
??????????????Mg=(f_l*f_l‘)*SNR;
??????????????y_l=SNR^(1/2)*f_l+v_l;?
??????????????dy1=(y_l-SNR^(1/2)*f_l)*(y_l-SNR^(1/2)*f_l)‘;
??????????????dy2=(y_l+SNR^(1/2)*f_l)*(y_l+SNR^(1/2)*f_l)‘;
??
???????????????if?(dy1>dy2)
???????????
??????????????????s_m=-1;
???????????
???????????????else
???????????????
??????????????????s_m=+1;
??????????????end

???????????
??????????for?iii=1:R
??????????????
????????????Nf=((f(iii))*(f(iii))‘)*SNR;
????????????Ng=((

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