資源簡介
自己在matlab中寫的3780點FFT的混合基算法。分為63x60.63又分為7x9.60分為3x4x5。

代碼片段和文件信息
N=3780;
n=0:N-1;
xn=0.5*cos(15/200*pi*n)+cos(6/200*pi*n);
%wfta7_gaijin
for?n3=0:59???
????for?n2=0:8
????????for?n1=0:6
????????????add1(n1+1)=60*mod(9*n1+7*n263)+n3+1;%加1因為MATLAB中數組下標是從1開始的
????????end
????????x1=xn(add1);
????????y1(add1)=fun7(x1);
????end
end
%wfta9_gaijin
for?n3=0:59
????for?n2=0:6
????????for?n1=0:8
????????????add2(n1+1)=60*mod(7*n1+9*n263)+n3+1;
????????end
????????x2=y1(add2);
????????y2(add2)=fun9(x2);
????end
end
%乘旋轉因子
for?n2=0:59
????for?k1=0:62
????????add3=(60*k1+n2)+1;
????????y3(add3)=y2(add3)*exp(-i*2*pi/3780*k1*n2);
????end
end
%wfta5_gaijin
for?n3=0:62
????for?n2=0:11
????????for?n1=0:4
????????????add4(n1+1)=60*n3+mod(12*n1+5*n260)+1;
????????end
????????x4=y3(add4);
????????y4(add4)=fun5(x4);
????end
end
%wfta3_gaijin
for?n3=0:62
????for?n2=0:19
????????for?n1=0:2
????????????add5(n1+1)=60*n3+mod(20*n1+3*n260)+1;
????????end
????????x5=y4(add5);
????????y5(add5)=fun3(x5);
????end
end
%wfta4_gaijin
for?n3=0:62
????for?n2=0:14
????????for?n1=0:3
????????????add6(n1+1)=60*n3+mod(15*n1+4*n260)+1;
????????end
????????x6=y5(add6);
????????y6(add6)=fun4(x6);
????end
end
%倒序輸出
add_temp=1;
for?k2=0:59
????for?k1=0:62
????????add_out=60*k1+k2+1;
????????yn(add_temp)=y6(add_out);
????????add_temp=add_temp+1;
????end
end
?屬性????????????大小?????日期????時間???名稱
-----------?---------??----------?-----??----
?????文件????????791??2012-08-27?14:57??fun9.m
?????文件???????1382??2012-08-27?14:59??fft3780.m
?????文件????????210??2012-08-27?15:25??fun3.m
?????文件?????????96??2012-08-27?14:33??fun4.m
?????文件????????303??2012-08-27?14:33??fun5.m
?????文件????????515??2012-08-27?14:37??fun7.m
-----------?---------??----------?-----??----
?????????????????3297????????????????????6
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