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資源簡介

16qam-64qam誤碼率對比的matlab仿真,其中qam16為主函數,modqam和demodqam分別是調制和解調

資源截圖

代碼片段和文件信息

function?[redata16]=demodqam16(receive16rreceive16iN)
redata16=zeros(N4);
for?a=1:N
????if?receive16r(a)>2/sqrt(10)
????????redata16(a3)=1;
????????redata16(a4)=0;
????elseif?receive16r(a)>0&&receive16r(a)<2/sqrt(10)
????????redata16(a3)=1;
????????redata16(a4)=1;
????elseif?receive16r(a)>-2/sqrt(10)&&receive16r(a)<0
????????redata16(a3)=0;
????????redata16(a4)=1;
????elseif?receive16r(a)<-2/sqrt(10)
????????redata16(a3)=0;
????????redata16(a4)=0;
????end
????
????if?receive16i(a)>2/sqrt(10)
????????????redata16(a1)=1;
????????????redata16(a2)=0;
????elseif?receive16i(a)>0&&receive16i(a)<2/sqrt(10)
????????????redata16(a1)=1;
????????????redata16(a2)=1;
????elseif?receive16i(a)>-2/sqrt(10)&&receive16i(a)<0
????????????redata16(a1)=0;
????????????red

?屬性????????????大小?????日期????時間???名稱
-----------?---------??----------?-----??----

?????文件????????927??2019-10-31?15:25??demodqam16.m

?????文件???????2289??2019-10-31?15:40??demodqam64.m

?????文件????????737??2019-10-30?21:15??modqam16.m

?????文件???????1573??2019-10-31?15:13??modqam64.m

?????文件???????1202??2019-10-31?14:52??qam16.m

-----------?---------??----------?-----??----

?????????????????6728????????????????????5


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