資源簡介
在MATLAB上實現(xiàn)的4*4的空時編碼的仿真,包括解碼。形成一個完整的通信系統(tǒng)。
代碼片段和文件信息
function?[mod_symbolssym_tableM]?=?modulator(bitseqb)
%MIMO-OFDM?Wireless?Communications?with?MATLAB㈢???Yong?Soo?Cho?Jaekwon?Kim?Won?Young?Yang?and?Chung?G.?Kang
%?2010?John?Wiley?&?Sons?(Asia)?Pte?Ltd
N_bits?=?length(bitseq);
if?b==1??????%?BPSK?modulation
???sym_table=exp(j*[0?-pi]);??sym_table=sym_table([1?0]+1);
???inp=bitseq;???mod_symbols=sym_table(inp+1);???M=2;
?elseif?b==2????%?QPSK?modulation
???sym_table?=?exp(j*pi/4*[-3?3?1?-1]);?sym_table=sym_table([0?1?3?2]+1);
???inp=reshape(bitseqbN_bits/b);
???mod_symbols=sym_table([2?1]*inp+1);???M=4;
?elseif?b==3????%?generates?8PSK?symbols
???sym_table=exp(j*pi/4*[0:7]);?sym_table=sym_table([0?1?3?2?6?7?5?4]+1);
???inp=reshape(bitseqbN_bits/b);?mod_symbols=sym_table([4?2?1]*inp+1);
???M=8;???
?elseif?b==4????%?16-QAM?modulation
???m=0;??sq10=sqrt(10);
???for?k=-3:2:3
??????for?l=-3:2:3
?????????m=m+1;?sym_table(m)?=?(k+j*l)/sq10;?%?power?normalization
??????end
???end
???sym_table?=?sym_table([0?1?3?2?4?5?7?6?12?13?15?14?8?9?11?10]+1);?%?Gray?code?mapping?pattern?for?8-PSK?symbols
???inp?=?reshape(bitseqbN_bits/b);
???mod_symbols?=?sym_table([8?4?2?1]*inp+1);??%?maps?transmitted?bits?into?16QAM?symbols
???M=16;?%16?constellation?points
?else
???error(‘Unimplemented?modulation‘);
end
?屬性????????????大小?????日期????時間???名稱
-----------?---------??----------?-----??----
?????目錄???????????0??2019-07-04?18:41??STBC\
?????文件??????247644??2019-05-23?18:27??STBC\1,Space–Time?Block?Codes?from?Orthogonal?Designs.pdf
?????文件??????112641??2019-05-26?22:00??STBC\4X4_STBC.mat
?????文件????????1305??2010-08-25?02:06??STBC\modulator.m
?????文件????????3384??2019-07-05?09:30??STBC\STBC_LM.m
?????文件????????2787??2019-07-03?20:57??STBC\unti
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