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信號分選PRI變換的matlab程序,三源抖動
代碼片段和文件信息
%改進處理3源10%抖動
clc;
clear;
N=1000;
taumin=0;
taumax=10;
K=201;
epsilon1=0.05;?
epsilon2=0.05*sqrt(2);
epsilon3=0.05*sqrt(5);
p1=1;
p2=sqrt(2);
p3=sqrt(5);
zeta0=0.03;
i=sqrt(-1);
alpha=0.0375;
beta=0.1;
gamma=3;
T=465;
N1=465;
N2=328;
N3=207;
b=(taumax-taumin)/K;
for?k=1:K
????tau1(k)=(k-0.5)*b+taumin;
????b11(k)=max([b2*epsilon1*tau1(k)]);?
????b12(k)=max([b2*epsilon2*tau1(k)]);?
????b13(k)=max([b2*epsilon3*tau1(k)]);?
end
D1=zeros(size(tau1));
C1=zeros(size(tau1));???????????
O1=zeros(size(tau1));
first1=zeros(size(tau1));
D2=zeros(size(tau1));
C2=zeros(size(tau1));???????????
O2=zeros(size(tau1));
first2=zeros(size(tau1));
D3=zeros(size(tau1));
C3=zeros(size(tau1));???????????
O3=zeros(size(tau1));
first3=zeros(size(tau1));
threshold=zeros(size(tau1));?????????????
for?k1=1:N1
????x1(k1)=k1*p1-epsilon1+(2*epsilon1).*rand(1);
end
for?k2=1:N2
????x2(k2)=k2*p2-epsilon2+(2*epsilon2).*rand(1);
end
for?k3=1:N3
????x3(k3)=k3*p3-epsilon3+(2*epsilon3).*rand(1);
end
for?n=2:N1
????for?m=1:n-1
????????delta_tau1=x1(n)-x1(m);
????????if?((delta_tau1>(1-epsilon1)*taumin)&&(delta_tau1<(1+epsilon1)*taumax))????
???????????for?k6=1:K???????
???????????????if?((delta_tau1>tau1(k6)-b11(k6)/2)&&(delta_tau1 ??????????????????C1(k6)=C1(k6)+1;???????????????????
??????????????????if?first1(k6)==0
?????????????????????O1(k6)=x1(n);%到達時間
?????????????????????first1(k6)=1;?????????%v????????????
??????????????????end??????
??????????????????eta01=(x1(n)-O1(k6))/tau1(k6);?%初始相位????????
??????????????????nu1=floor(eta01+0.4999);??????%v的迭代公式??
??????????????????if?nu1>0??????
?????????????????????if?(((nu1==1)&&(x1(m)==O1(k6)))||((nu1>=2)&&(abs(eta01/nu1-1)<=zeta0)))???????
????????????????????????O1(k6)=x1(n);????
?????????????????????end?????????????????????????????????????
??????????????????end??????????
??????????????????eta1=(x1(n)-O1(k6))/tau1(k6);????%初始相位???????
??????????????????D1(k6)=D1(k6)+exp(2*pi*i*eta1);??????????
???????????????end?????????????
???????????end?
????????end
????end
end?
for?n=2:N2
????for?m=1:n-1
????????delta_tau2=x2(n)-x2(m);
?????????if?((delta_tau2>(1-epsilon2)*taumin)&&(delta_tau2<(1+epsilon2)*taumax))????
???????????for?k7=1:K???????
???????????????if?((delta_tau2>tau1(k7)-b12(k7)/2)&&(delta_tau2 ??????????????????C2(k7)=C2(k7)+1;???????????????????
??????????????????if?first2(k7)==0
?
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