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    發布日期: 2021-05-16
  • 語言: 其他
  • 標簽: PTS??PAPR??CCDF??

資源簡介

使用PTS方法降低PAPR,子載波數為128,4倍過采樣,并將原始信號與PTS后的CCDF曲線進行比較。

資源截圖

代碼片段和文件信息

%??PTS方法的CCDF曲線?pts_ccdf.m??
clear?all;
close?all;
n1=10;
c1=128;
v=8;
Fs=10^3;
NN=0:.1:13;??%??CCDF的門限值
ccdf0=zeros(1131);
ccdf2=ccdf0;
for?m=1:n1;
????%??generate?QAM?signal?
????x(:1)=randsrc(c11[1-13-3]);
????x(:2)=randsrc(c11[1-13-3]);
?????y=squeeze(x);
?????x1=amodce(yFs‘qam‘);
?????y1=x1.‘;
?????%??oversample?4
?????z1=[y1(1:c1/2)zeros(13*c1)y1(c1/2+1:c1)];
?????z2=ifft(z1)*c1;
?????x0=abs(z2).^2;
?????v3=max(x0);
?????m3=mean(x0);
?????papr3(m)=10*log10(v3/m3);
?????l1=length(z1);
?????%??divided?into?v?sets?
?????x2=zeros(1l1);
?????x4=x2;
?????x6=x4;?
?????b=ones(1v);
?????for?k=1:v;
?????????z(k:)=[zeros(1(k-1)*l1/v)z1((k-1)*l1/v+1:k*l1/v)zeros(1(v-k)*l1/v)];
?????????iz(k:)=ifft(z(k:));??
?????????iz(k:)=iz(k:)*c1;
?????????x2(1:)=x2(1:)+b(k)*iz(k:);
?????end
?????????x3=abs(x2).^2;
?????????m0=mean(x3);
?????????v0=max(x3);
?????????papr0(m)=10*log10(v0/m0);
?????????index=1;?
????while?index?????????b(index)=-1;
?????????for?n=1:v;
????????????x4(1:)=x4(1:)+b(n)*iz(n:);
?????????end
?????????x5=abs(x4).^2;
?????????m1=mean(x5);
?????????v1=max(x5);
????????????papr1(m)=10*log10(v1/m1);
?????????if?papr1(m)????????????papr0(m)=papr1(m);
?????????else
?????????????b(index)=1;
?????????end
?????????index=index+1;
????end
????for?l=1:131;
????if?papr3(m)>NN(l);
???????ccdf0(l)=ccdf0(l)+1;
????end??
????if?papr0(m)>NN(l);
???????ccdf2(l)=ccdf2(l)+1;
????end
end
end
ccdf1=ccdf0./n1;
ccdf3=ccdf2./n1;
%?NN=4:.5:10;
semilogy(NNccdf1NNccdf3‘b-*‘);
xlabel(‘PAPR門限值/dB‘);?ylabel(‘CCDF‘);
legend(‘原始信號‘‘PTS后‘0);
title(‘PTS方法的CCDF曲線‘);



????

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

?????文件???????1729??2010-05-07?21:25??pts_ccdf.m

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

?????????????????1729????????????????????1


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