資源簡介
matlab的QPSK系統(tǒng)仿真報(bào)告及源代碼

代碼片段和文件信息
function?[pbps]=cm_sm32(snr_in_dB)
%?[pbps]=cm_sm32(snr_in_dB)
%????????????????CM_SM3?finds?the?probability?of?bit?error?and?symbol?error?for?
%????????????????the?given?value?of?snr_in_dB?signal?to?noise?ratio?in?dB.
N=100;
E=1;????????????????????????????????????????????%?energy?per?symbol
numofsymbolerror=0;
numofbiterror=0;
counter=0;
snr=10^(snr_in_dB/10);??????????????????????????%?signal?to?noise?ratio
sgma=sqrt(E/snr)/2;?????????????????????????????%?noise?variance
s00=[1?0];?s01=[0?1];?s11=[-1?0];?s10=[0?-1];???%?signal?mapping
%?generation?of?the?data?source
while(numofbiterror<100)
for?i=1:N
????temp=rand;??????????????????????????????????%?a?uniform?random?variable?between?0?and?1
????if?(temp<0.25)?????????????????????????????%?with?probability?1/4?source?output?is?“00“
????????dsource1(i)=0;?dsource2(i)=0;
????elseif?(temp<0.5)??????????????????????????%?with?probability?1/4?source?output?is?“01“
????????dsource1(i)=0;?dsource2(i)=1;
????elseif?(temp<0.75)?????????????????????????%?with?probability?1/4?source?output?is?“10“
????????dsource1(i)=1;?dsource2(i)=0;
????else????????????????????????????????????????%?with?probability?1/4?source?output?is?“11“
????????dsource1(i)=1;?dsource2(i)=1;
????end;
end;
%?detection?and?the?probability?of?error?calculation
for?i=1:N
????ray=raylrnd(0.8);
????n=sgma*randn(12);??????????????????????????%?2?normal?distributed?r.v?with?0?variance?sgma
????if?((dsource1(i)==0)?&?(dsource2(i)==0))?
????????r=ray*s00+n;
????elseif?((dsource1(i)==0)?&?(dsource2(i)==1))?
????????r=ray*s01+n;
????elseif?((dsource1(i)==1)?&?(dsource2(i)==0))?
????????r=s10*ray+n;
????else?
????????r=s11*ray+n;
????end;
????%?The?correlation?metrics?are?computed?below
????c00=dot(rs00);?c01=dot(rs01);?c10=dot(rs10);?c11=dot(rs11);
????%?The?decision?on?the?ith?symbol?is?made?next
????c_max=max([c00c01c10c11]);
????if?(c00==c_max)?decis1=0;?decis2=0;
????elseif?(c01==c_max)?decis1=0;?decis2=1;
????elseif?(c10==c_max)?decis1=1;?decis2=0;
????else?decis1=1;?decis2=1;
????end;
????%?Increment?the?error?counter?if?the?decision?is?not?correct
????symbolerror=0;
????if?(decis1~=dsource1(i))?numofbiterror=numofbiterror+1;?symbolerror=1;?
????end;
????if?(decis2~=dsource2(i))?numofbiterror=numofbiterror+1;?symbolerror=1;?
????end;
????if?(symbolerror==1)?numofsymbolerror=numofsymbolerror+1;?
????end;
????
end
counter=counter+1;
end
ps=numofsymbolerror/(N*counter);??????????????????????????%?since?there?are?totally?N?symbols
pb=numofbiterror/(2*N*counter);?????????????????????????%?since?2N?bits?are?transmitted?
?屬性????????????大小?????日期????時(shí)間???名稱
-----------?---------??----------?-----??----
?????文件???????2666??2013-06-17?08:58??基于matlab的QPSK系統(tǒng)仿真\cm_sm32.m
?????文件???????2695??2013-06-17?09:10??基于matlab的QPSK系統(tǒng)仿真\cm_sm33.m
?????文件???????9282??2013-06-17?08:56??基于matlab的QPSK系統(tǒng)仿真\qpsk.m
?????文件?????211770??2013-09-25?23:22??基于matlab的QPSK系統(tǒng)仿真\基于matlab的QPSK系統(tǒng)仿真.doc
?????目錄??????????0??2013-09-25?23:22??基于matlab的QPSK系統(tǒng)仿真
-----------?---------??----------?-----??----
???????????????226413????????????????????5
- 上一篇:波長與RGB的轉(zhuǎn)換
- 下一篇:圖像的簡單處理
評(píng)論
共有 條評(píng)論