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  • 大小: 3KB
    文件類型: .zip
    金幣: 2
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    發(fā)布日期: 2021-06-04
  • 語言: Matlab
  • 標(biāo)簽: QAM??

資源簡介

模擬實(shí)現(xiàn)一個16-QAM的調(diào)制器和解調(diào)器,同時信道采用加性高斯白噪聲信道

資源截圖

代碼片段和文件信息




????????????????????????%%%?16-QAM?Modulator?and?Demodulator?Implementation?with?an?AWGN?Channel?%%%
???????????????????????

clc;
clear;
close?all;


????%%%%?Input?Signal?%%%%
????%?Input?is?a?stream?of?binary?digits?of?integers?0-15
dec=[0?0?0?0?0?0?0?1?0?0?1?0?0?0?1?1?0?1?0?0?0?1?0?1?0?1?1?0?0?1?1?1?1?0?0?0?1?0?0?1?1?0?1?0?1?0?1?1?1?1?0?0?1?1?0?1?1?1?1?0?1?1?1?1];?
dl=length(dec);




%%%%?Serial?To?Parallel?and?2-to-4?Level?Converter?%%%%
????sp=[];
????o1=[];
????o2=[];
????clear?i;
????for?i=1:4:64;
????????sp=[dec(1i:i+3)];??%?Serial?to?Parallel?4-bit?Register
????????I=sp(11);??????????%?Separation?of?I?and?Q?bits
????????Id=sp(12);
????????Q=sp(13);
????????Qd=sp(14);
????
????????if?I==0?????????????%?Assigning?Amplitude?levels?for?I-channel
????????????if?Id==0
????????????????o1=[o1?-3];?%?if?input?is?00?output?is?-3
????????????else
????????????????o1=[o1?-1];?%?if?input?is?01?output?is?-1
????????????end
????????
????????else
????????????if?Id==0
????????????????o1=[o1?1];?%?if?input?is?10?output?is?1
????????????else
????????????????o1=[o1?3];?%?if?input?is?11?output?is?3
????????????end
????????end
????
????????if?Q==0?????????????%?Assigning?Amplitude?levels?for?Q-channel
????????????if?Qd==0
????????????????o2=[o2?-3];?%?if?input?is?00?output?is?-3
????????????else
????????????????o2=[o2?-1];?%?if?input?is?01?output?is?-1
????????????end
????????
????????else
????????????if?Qd==0
????????????????o2=[o2?1];?%?if?input?is?10?output?is?1
????????????else
????????????????o2=[o2?3];?%?if?input?is?11?output?is?3
????????????end
????????end
????
????clear?sp?clear?I?clear?Id?clear?Q?clear?Qd;?
????end


????
????%%%%?Oscillator?and?Balanced?Modulator?%%%%
????fc=500;??????????????%?Carrier?Frequency
????fs=20000;????????????%?Sampling?Frequency
????t=1:100;?????????????%?Duration?of?Signal
????s=[];
????clear?i;
????for?i=1:1:16;????????%?Modulation?(multiplication?with?carrier?signals?cos?and?sin)
????????Ac=o1(i);
????????As=o2(i);
????????s1=Ac*cos(2*pi*(fc/fs)*t);
????????s2=As*sin(2*pi*(fc/fs)*t);
????????s=[s?(s1+s2)];
????end
????figure(1)
????plot(s)?????%?‘s‘?demotes?the?transmitted?signal
????title(‘16-QAM?Modulated?Signal‘‘FontSize‘14)
????Xlim([0?1610])
????
???????
????
????
????
????%%%%?AWGN?Addition?While?Passing?Through?The?Channel?%%%%
????r=awgn(s10);???????%?Transmitted?Signal?passes?through?AWGN?Channel
????figure(2)???????????%?Received?Signal?SNR?is?10?db
????plot(r)?????????????%?received?signal?is?denoted?by?‘r‘
????title(‘Transmitted?Signal?With?Additive?Noise?After?Passing?Through?The?Channel‘‘FontSize‘14);
????
????
????
????%%%%?COHERENT?DEMODULATION?%%%%
????ss1=[];
????ss2=[];
????rs1=2*cos(2*pi*(fc/fs)*t);??%?Sin?and?Cos?generation?by?Local?Oscillator
????rs2=2*sin(2*pi*(fc/fs)*t);
????clear?i;
????for?i=1:100:1600;???????????%?Demodulation?of?Received?Signal
????ss1=[ss1?rs1.

?屬性????????????大小?????日期????時間???名稱
-----------?---------??----------?-----??----
?????文件????????9048??2012-12-25?11:25??qam.m
?????文件????????1312??2014-02-12?14:14??license.txt

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