資源簡介
認知無線電頻譜分配仿真,并對結果進行分析和比較,得出結論

代碼片段和文件信息
clc
clear?all
B=15;
BER=10^(-4);
K=1.5/(logm(0.2/BER));
SNR=[810];
k=log2(1+K.*SNR);
r1=10;
r2=12;
x=0;
y=1;
z=1;
%?b=[00];
%?c=x+y*(sum(b));
%?p=r.*k.*b-b.*c;
a(1)=0.001;
for?i=1:300
????a(2)=0.001;
????for?j=1:300
????????eigenvalue(1)=1-10/3*(2*a(1)*k(1)-a(1)*k(2)+2*a(2)*k(2)-a(2)*k(1))+10/3*((a(1)*(k(2)-2*k(1))*a(2)*(k(1)-2*k(2))+(2*a(2)*k(2)-a(2)*k(1)-2*a(1)*k(1)+a(1)*k(2))^2)^(0.5));
????????eigenvalue(2)=1-10/3*(2*a(1)*k(1)-a(1)*k(2)+2*a(2)*k(2)-a(2)*k(1))-10/3*((a(1)*(k(2)-2*k(1))*a(2)*(k(1)-2*k(2))+(2*a(2)*k(2)-a(2)*k(1)-2*a(1)*k(1)+a(1)*k(2))^2)^(0.5));
????????if?(eigenvalue(1)<1)&&(eigenvalue(1)>-1)&&(eigenvalue(2)<1)&&(eigenvalue(2)>-1)
????????????a(2)=a(2)+0.002;
????????end
????end
????if?a(2)>0
????????a(2)=a(2)-0.002;
????else
????????a(2)=0;
????end
????f2(i)=a(2);
????f1(i)=a(1);
????a(1)=a(1)+0.002;
end
figure(1)
plot(f1f2‘--r‘)
axis([00.2500.25])
hold?on
B=15;
BER=10^(-4);
K=1.5/(logm(0.2/BER));
SNR=[1010];
k=log2(1+K.*SNR);
r1=10;
r2=12;
x=0;
y=1;
z=1;
%?b=[00];
%?c=x+y*(sum(b));
%?p=r.*k.*b-b.*c;
a(1)=0.001;
for?i=1:300
????a(2)=0.001;
????for?j=1:300
????????eigenvalue(1)=1-10/3*(2*a(1)*k(1)-a(1)*k(2)+2*a(2)*k(2)-a(2)*k(1))+10/3*((a(1)*(k(2)-2*k(1))*a(2)*(k(1)-2*k(2))+(2*a(2)*k(2)-a(2)*k(1)-2*a(1)*k(1)+a(1)*k(2))^2)^(0.5));
????????eigenvalue(2)=1-10/3*(2*a(1)*k(1)-a(1)*k(2)+2*a(2)*k(2)-a(2)*k(1))-10/3*((a(1)*(k(2)-2*k(1))*a(2)*(k(1)-2*k(2))+(2*a(2)*k(2)-a(2)*k(1)-2*a(1)*k(1)+a(1)*k(2))^2)^(0.5));
????????if?(eigenvalue(1)<1)&&(eigenvalue(1)>-1)&&(eigenvalue(2)<1)&&(eigenvalue(2)>-1)
????????????a(2)=a(2)+0.002;
????????end
????end
????if?a(2)>0
????????a(2)=a(2)-0.002;
????else
????????a(2)=0;
????end
????f2(i)=a(2);
????f1(i)=a(1);
????a(1)=a(1)+0.002;
end
figure(1)
plot(f1f2‘-.*b‘)
axis([00.2500.25])
B=15;
BER=10^(-4);
K=1.5/(logm(0.2/BER));
SNR=[108];
k=log2(1+K.*SNR);
r1=10;
r2=12;
x=0;
y=1;
z=1;
%?b=[00];
%?c=x+y*(sum(b));
%?p=r.*k.*b-b.*c;
a(1)=0.001;
for?i=1:300
????a(2)=0.001;
????for?j=1:300
????????eigenvalue(1)=1-10/3*(2*a(1)*k(1)-a(1)*k(2)+2*a(2)*k(2)-a(2)*k(1))+10/3*((a(1)*(k(2)-2*k(1))*a(2)*(k(1)-2*k(2))+(2*a(2)*k(2)-a(2)*k(1)-2*a(1)*k(1)+a(1)*k(2))^2)^(0.5));
????????eigenvalue(2)=1-10/3*(2*a(1)*k(1)-a(1)*k(2)+2*a(2)*k(2)-a(2)*k(1))-10/3*((a(1)*(k(2)-2*k(1))*a(2)*(k(1)-2*k(2))+(2*a(2)*k(2)-a(2)*k(1)-2*a(1)*k(1)+a(1)*k(2))^2)^(0.5));
????????if?(eigenvalue(1)<1)&&(eigenvalue(1)>-1)&&(eigenvalue(2)<1)&&(eigenvalue(2)>-1)
????????????a(2)=a(2)+0.002;
????????end
????end
????if?a(2)>0
????????a(2)=a(2)-0.002;
????else
????????a(2)=0;
????end
????f2(i)=a(2);
????f1(i)=a(1);
????a(1)=a(1)+0.002;
end
figure(1)
plot(f1f2‘-g‘)
axis([00.2500.25])
B=15;
BER=10^(-4);
K=1.5/(logm(0.2/BER));
SNR=[77];
k=log2(1+K.*SNR);
r1=10;
r2=12;
x=0;
y=1;
z=1;
%?b=[00];
%?c=x+y*(sum(b));
%?p=r.*k.*b-b.*c;
a(1)=0.001;
for?i=1:300
????a(2)=0.001;
????for?j=1:300
???
?屬性????????????大小?????日期????時間???名稱
-----------?---------??----------?-----??----
?????文件???????4719??2009-04-03?13:42??Unti
-----------?---------??----------?-----??----
?????????????????4719????????????????????1
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