資源簡介
用MATLAB程序產(chǎn)生16QAM調(diào)制信號,并對該信號進行上變頻操作
代碼片段和文件信息
%16QAM中頻調(diào)制
%升余弦窗成形,滾降系數(shù)0.35,符號率1MSybol/s中頻頻率21MHz
%給出各級濾波器的系數(shù),時域/頻域響應(yīng)以及信號經(jīng)過各濾波器的時域/頻域圖
%*************************產(chǎn)生QAM基帶信號*************************%
%產(chǎn)生偽隨機序列PN
N=500;?%二進制數(shù)據(jù)長度,長度不足會造成星座點缺失
x=randint(1N2);?
%數(shù)據(jù)分組串并變換
x1=x(1:2);
x2=x(3:4);?%完成第一組轉(zhuǎn)換
for?i=1:(N/4-1)??%完成所有點的轉(zhuǎn)換??
????x1=[x1?(x(i*4+1:i*4+2))];
????x2=[x2?(x(i*4+3:i*4+4))];
end
%二-十進制轉(zhuǎn)換(00-001-110-211-3)
xi=x1(1)*2+x1(2);
xq=x2(1)*2+x2(2);?%完成第一組轉(zhuǎn)換
n=length(x1);
for?i=1:n/2-1????%完成所有點的轉(zhuǎn)換??
????xi=[xi?(x1(i*2+1)*2+x1(i*2+2))];
????xq=[xq?(x2(i*2+1)*2+x2(i*2+2))];
end
%信號映射
for?i=1:n/2
????switch(xi(i))
????????case?0
????????????xi(i)=-3;
????????case?1
????????????xi(i)=-1;
????????case?2
????????????xi(i)=1;
????????case?3
????????????xi(i)=3;
????end
????
????switch(xq(i))
?屬性????????????大小?????日期????時間???名稱
-----------?---------??----------?-----??----
?????文件???????4641??2012-05-08?11:17??11603454QAM\QAM.asv
?????文件???????4641??2012-05-08?11:19??11603454QAM\QAM.m
?????文件????????218??2007-06-05?03:14??11603454QAM\www.pudn.com.txt
?????目錄??????????0??2012-05-22?15:38??11603454QAM
-----------?---------??----------?-----??----
?????????????????9500????????????????????4
評論
共有 條評論