資源簡介
之前通信仿真課寫的,曲線還蠻平滑。MATLAB仿真AWGN信道中,16QAM和64QAM在不同信噪比下的誤碼率比較。

代碼片段和文件信息
function?[redata1]=demoduQ16(Receive16RealReceive16ImagN)%demodulation?函數
????data1demod1?=?zeros(1N);
????data1demod2?=?zeros(1N);
????data1demod3?=?zeros(1N);
????data1demod4?=?zeros(1N);
????redata1=zeros(14*N);
????for?k?=?1:N
????????if?Receive16Real(k)>2?
????????????data1demod1(k)?=?0;
????????????data1demod2(k)?=?0;
????????????if?Receive16Imag(k)>2
????????????????data1demod3(k)?=?0;
????????????????data1demod4(k)?=?0;
????????????elseif?Receive16Imag(k)>0?&&?Receive16Imag(k)<2
????????????????data1demod3(k)?=?0;
????????????????data1demod4(k)?=?1;
????????????elseif?Receive16Imag(k)>-2?&&?Receive16Imag(k)<0
????????????????data1demod3(k)?=?1;
????????????????data1demod4(k)?=?1;
????????????elseif?Receive16Imag(k)<-2
????????????????data1demod3(k)?=?1;
????????????????data1demod4(k)?=?0;
????????????end
????????elseif?Receive16Real(k)>0?&&?Receive16Real(k)<2
????????????data1demod1(k)?=?0;
????????????data1demod2(k)?=?1;
????????????if?Receive16Imag(k)>2
????????????????data1demod3(k)?=?0;
????????????????data1demod4(k)?=?0;
????????????elseif?Receive16Imag(k)>0?&&?Receive16Imag(k)<2
????????????????data1demod3(k)?=?0;
????????????????data1demod4(k)?=?1;
????????????elseif?Receive16Imag(k)>-2?&&Receive16Imag(k)<0
????????????????data1demod3(k)?=?1;
????????????????data1demod4(k)?=?1;
????????????elseif?Receive16Imag(k)<-2
????????????????data1demod3(k)?=?1;
????????????????data1demod4(k)?=?0;
????????????end
????????elseif?Receive16Real(k)<0?&&?Receive16Real(k)>-2
????????????data1demod1(k)?=?1;
????????????data1demod2(k)?=?1;
????????????if?Receive16Imag(k)>2
????????????????data1demod3(k)?=?0;
????????????????data1demod4(k)?=?0;
????????????elseif?Receive16Imag(k)>0?&&?Receive16Imag(k)<2
????????????????data1demod3(k)?=?0;
????????????????data1demod4(k)?=?1;
????????????elseif?Receive16Imag(k)>-2?&&?Receive16Imag(k)<0
????????????????data1demod3(k)?=?1;
????????????????data1demod4(k)?=?1;
????????????elseif?Receive16Imag(k)<-2
????????????????data1demod3(k)?=?1;
????????????????data1demod4(k)?=?0;
????????????end
????????elseif?Receive16Real(k)<-2
????????????data1demod1(k)?=?1;
????????????data1demod2(k)?=?0;
????????????if?Receive16Imag(k)>2
????????????????data1demod3(k)?=?0;
????????????????data1demod4(k)?=?0;
????????????elseif?Receive16Imag(k)>0?&&?Receive16Imag(k)<2
????????????????data1demod3(k)?=?0;
????????????????data1demod4(k)?=?1;
????????????elseif?Receive16Imag(k)>-2?&&?Receive16Imag(k)<0
????????????????data1demod3(k)?=?1;
????????????????data1demod4(k)?=?1;
????????????elseif?Receive16Imag(k)<-2
????????????????data1demod3(k)?=?1;
????????????????data1demod4(k)?=?0;
????????????end
????????end
????????redata1(4*k-3)=data1demod1(k);
????????redata1(4*k-2)=data1demod2(k);
????????redata1(4*k-1)=data1demod3(k);
????????redata1(4*k)=data1demod4(k);
????end
?屬性????????????大小?????日期????時間???名稱
-----------?---------??----------?-----??----
?????文件????????1223??2019-01-12?18:29??QAM\Qam1.m
?????文件????????2915??2018-12-01?00:14??QAM\demoduQ16.m
?????文件???????12741??2018-12-01?00:14??QAM\demoduQ64.m
?????文件????????2117??2018-11-30?23:47??QAM\moduQ16.m
?????文件????????7253??2018-11-30?23:47??QAM\moduQ64.m
?????目錄???????????0??2019-04-19?22:34??QAM\
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