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    發布日期: 2023-07-22
  • 語言: 其他
  • 標簽: matlab??

資源簡介

關于用戶分布的3D波束賦形方案的研究,基于3GPP標準,編譯可過

資源截圖

代碼片段和文件信息

function[maxrsmaxrs14maxrsdismaxrt]?=?angle_chose
%?maxrs:基站協作選擇傾角,使總吞吐量最大(最佳)
%?maxrs14:取固定角14度時的總速率
%?maxrsdis:垂直波峰對準用戶時的總速率
%?maxrt:每個基站自己選擇傾角時的總吞吐量(其次)
global?UE
global?BS
BS.angle_num?=?60;??%傾角變化范圍
[PtNangle_el_du]?=?PL_FS_pattern;%路損和小陰影衰落,瑞利。
a?=?round(angle_el_du(11));???%基站1和用戶1之間的傾角,user-specific?tilting,四舍五入
b?=?round(angle_el_du(22));
c?=?round(angle_el_du(33));
for?P=1:8??%cheng對應小區邊緣8個發射功率,(-5~30dBm)
????for?S=1:3??????????%基站
????????for?t=1:3????%用戶
????????????for?ll?=?1:60??%角度變化
????????????UE.large_dBm(PStll)?=?Pt(P)?+?UE.large_scale_dB(Stll);??????%SNR為-5到30dB時的信道增益單位dBm
????????????UE.large(PStll)?=?10^(UE.large_dBm(PStll)/10)/N;???????????%dBm轉換成mW并歸一化使N為1按N歸一
????????????end
????????end
????end
end
for?P=1:8
????maxrt1(P)?=?UE.large(P111)?/?(UE.large(P211)+UE.large(P311));??%基站1對用戶1的信號/對用戶2、3的信號,分基站選最優角;
????maxrt2(P)?=?UE.large(P221)?/?(UE.large(P121)+UE.large(P321));
????maxrt3(P)?=?UE.large(P331)?/?(UE.large(P231)+UE.large(P131));
????
????maxrs1(P)?=?log2(1?+?UE.large(P111)/(1+UE.large(P121)+UE.large(P131)));??%用戶1的吞吐量log2(?1+u1/[1+u2+u3]?)
????maxrs2(P)?=?log2(1?+?UE.large(P221)/(1+UE.large(P211)+UE.large(P231)));
????maxrs3(P)?=?log2(1?+?UE.large(P331)/(1+UE.large(P311)+UE.large(P321)));
????maxrs(P)?=?maxrs1(P)?+?maxrs2(P)?+?maxrs3(P);????%總吞吐量
end

for?P=1:8
????for?mm=2:BS.angle_num
????????rt(P1mm)?=?UE.large(P11mm)?/?(UE.large(P21mm)+UE.large(P31mm));
????????if((rt(P1mm)?>?maxrt1(P))?&&?(abs(mm-a)?????????????maxrt1(P)?=?rt(P1mm);??????????????????????%最優傾角在“基站與服務用戶之間的傾角”正負5度范圍內?????,基站1
????????????angle_distr(P1)?=?mm;%保存邊緣功率為5*(2-P)時的最佳下傾角(單基站,有用/干擾最小);---------------基站1
????????end
????????rt(P2mm)?=?UE.large(P22mm)?/?(UE.large(P12mm)+UE.large(P32mm));
????????if((rt(P2mm)?>?maxrt2(P))?&&?(abs(mm-b)?????????????maxrt2(P)?=?rt(P2mm);
????????????angle_distr(P2)?=?mm;
????????end????????
????????rt(P3mm)?=?UE.large(P33mm)?/?(UE.large(P23mm)+UE.large(P13mm));
????????if((rt(P3mm)?>?maxrt3(P))?&&?(abs(mm-c)?????????????maxrt3(P)?=?rt(P3mm);
????????????angle_distr(P3)?=?mm;
????????end
????end
end

for?P=1:8
????for?ii=2:BS.angle_num%三個基站都遍歷一遍下傾角,求得三個基站下傾角的最佳組合
????????for?jj=2:BS.angle_num
????????????for?kk=2:BS.angle_num
????????????????rs(P1iijjkk)?=?log2(1?+?UE.large(P11ii)/(1+UE.large(P12jj)+UE.large(P13kk)));
????????????????rs(P2iijjkk)?=?log2(1?+?UE.large(P22jj)/(1+UE.large(P21ii)+UE.large(P23kk)));
????????????????rs(P3iijjkk)?=?log2(1?+?UE.large(P33kk)/(1+UE.large(P32jj)+UE.large(P31ii)));
????????????????rstot(Piijjkk)?=?rs(P1iijjkk)?+?rs(P2iijjkk)?+?rs(P3iijjkk);%三個用戶的吞吐量和??????
????????????????if(rstot(Piijjkk)?>?maxrs(P))
????????????????????maxrs(P)?=?rstot(Piijjk

?屬性????????????大小?????日期????時間???名稱
-----------?---------??----------?-----??----
?????目錄???????????0??2015-03-17?20:02??code_el_az\
?????文件????????3906??2015-03-13?16:39??code_el_az\angle_chose.m
?????文件????????4596??2015-03-05?11:21??code_el_az\angle_el_az.m
?????文件?????????484??2013-07-08?10:39??code_el_az\BS_creation.m
?????文件?????????622??2014-12-10?20:58??code_el_az\huatu10.m
?????文件????15251308??2015-03-11?19:50??code_el_az\matlab.mat
?????文件?????????840??2013-11-12?11:04??code_el_az\pattern.m
?????文件????????4705??2015-03-14?23:22??code_el_az\PL_FS_pattern.m
?????文件????????4182??2014-12-10?21:01??code_el_az\PL_FS_pattern_az.m
?????文件????????3458??2015-03-12?14:31??code_el_az\run.m
?????文件????????1017??2014-03-02?13:50??code_el_az\sum?and?edge10.mat
?????文件???????28834??2015-03-06?10:59??code_el_az\sum?and?edge_201536_10_58.mat
?????文件????????1194??2015-03-14?15:37??code_el_az\UE_creation.m

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