資源簡介
多用戶MIMO預編碼技術的對比,bd預編碼,zf預編碼,mmse預編碼等,對學習多用戶mimo系統很有幫助。
代碼片段和文件信息
function?[WM]=BD(NtNriSKH)
W=zeros(NtK*S);
M=zeros(K*SNri);
gama=zeros(1K*S);
for?i_user=1:K
????H_others=zeros((K-1)*NriNt);
????i2=1;
????for?i1=1:K
????????if?i1~=i_user
????????????H_others((i2-1)*Nri+1:i2*Nri:)=H((i1-1)*Nri+1:i1*Nri:);
????????????i2=i2+1;
????????end
????end
?????
????U=zeros((K-1)*Nri(K-1)*Nri);
????D=zeros((K-1)*NriNt);
????V=zeros(NtNt);
????[UDV]=svd(H_others);
????rank=(K-1)*Nri;
????for?i3=(K-1)*Nri:(-1):1
????????if?D(i3i3)==0
????????????rank=i3-1;
????????end
????end
????T=V(:rank+1:Nt);
????Hi=H((i_user-1)*Nri+1:i_user*Nri:);
????U=zeros(NriNri);
????D=zeros(NriNt-rank);
????V=zeros(Nt-rankNt-rank);
????[UDV]=svd(Hi*T);
?????
????W(:(i_user-1)*S+1:i_user*S)=T*V(:1:S);%S ????M((i1-1)*S
?屬性????????????大小?????日期????時間???名稱
-----------?---------??----------?-----??----
?????目錄???????????0??2017-02-22?15:30??MU_MIMO?預編碼對比\
?????文件?????????961??2017-02-22?15:20??MU_MIMO?預編碼對比\BD.m
?????文件????????1001??2017-02-22?15:23??MU_MIMO?預編碼對比\BD2.m
?????文件????????4757??2017-02-22?15:24??MU_MIMO?預編碼對比\main1.m
?????文件????????4606??2017-02-22?15:25??MU_MIMO?預編碼對比\main2.m
?????文件??????????82??2017-02-22?15:25??MU_MIMO?預編碼對比\MF.m
?????文件?????????755??2017-02-22?15:26??MU_MIMO?預編碼對比\MMSE.m
?????文件?????????476??2017-02-22?15:26??MU_MIMO?預編碼對比\QPSK_mapper.m
?????文件?????????565??2017-02-22?15:27??MU_MIMO?預編碼對比\receiver.m
?????文件?????????531??2017-02-22?15:27??MU_MIMO?預編碼對比\SLNR.m
?????文件?????????464??2017-02-22?15:28??MU_MIMO?預編碼對比\SLNR2.m
?????文件????????1405??2017-02-22?15:29??MU_MIMO?預編碼對比\sumrate.m
?????文件?????????135??2017-02-22?15:29??MU_MIMO?預編碼對比\svdprecoding.m
?????文件?????????317??2017-02-22?15:30??MU_MIMO?預編碼對比\waterfilling.m
?????文件??????????92??2017-02-22?15:30??MU_MIMO?預編碼對比\ZF.m
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