DIT和DIF实现快速傅里叶变换的FFT

DIT和DIF实现快速傅里叶变换的FFT

数字信号处理课程老师要求做的一个小程序，在此分享~

%***************基2的DIF&DIT-FFT算法************************
%****************——徐延宾——*****************************
%****************-9171040G0633-*****************************
%*****************分组：第三组******************************
clc;
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%频率f1、f2输入
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
YN='Y';
while(strcmp(YN'Y'))
f1=input('请输入频率f1的值(200<=f1<300):');
while((f1<200) || (f1>=300))
    disp('输入的频率不满足"200<=f1<300"的条件，请重新输入：');
    f1=input('f1:');
end
f2=input('请输入频率f2的值(300<=f2<400f1+f2>600):');
while((f2<300) || (f2>=400))
    disp('输入的频率不满足"300<=f2<400"的条件，请重新输入：');
    f2=input('f2:');
end
while (f1+f2<=600) %判断输入的f1、f2是否满足条件
    disp('输入的频率不满足"f1+f2>600"的条件，请重新输入：');
    f1=input('请输入频率f1的值(200<=f1<300):');
    while((f1<200) || (f1>=300))
    disp('输入的频率不满足"200<=f1<300"的条件，请重新输入：');
    f1=input('f1:');
   end
    f2=input('请输入频率f2的值(300<=f2<400f1+f2>600):');
    while((f2<300) || (f2>=400))
    disp('输入的频率不满足"300<=f2<400"的条件，请重新输入：');
    f2=input('f2:');
    end
end
disp('请选择求解FFT的算法（输入DIT或者DIF），注意输入字母大写');
select=input('你的选择：''s');
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%确定长度N的值
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
fT=1000;
F1=(300-f1);
F2=(f2-300);
if(F1<F2)
    F=F1;
else
    F=F2;
end
N1=(fT/F);
i=3;
while ((2^i)<N1)
   
    i=i+1;
end
N=2^(i);
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%得到离散时序列x[n]
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
x=zeros(1N);
for n=1:N
    x(n)=1.5*cos(2*pi*f1*n/N)+cos(2*pi*300*n/N)+0.5*cos(2*pi*f2*n/N);
end
%画出x[n]的图像
k=1:1:N;
y=1.5*cos(2*pi*f1*k/N)+cos(2*pi*300*k/N)+0.5*cos(2*pi*f2*k/N);
subplot(211);stem(ky'filled');xlabel('n');ylabel('x[n]');title('x[n]的采样图');
m=log2(N);
if(strcmp(select'DIT'))
    %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
    %基2的DIT-FFT算法
    %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
    change=bin2dec(fliplr(dec2bin([1:N]-1m)))+1;
    Y1=x(change);
    for s=1:m
        Nr=2^s;u=1;
        WN=exp(-1j*2*pi/Nr);
        for j=1:Nr/2
            for k=j:Nr:N
                kp=k+Nr/2;
                g=Y1(kp)*u;
                Y1(kp)=Y1(k)-g;
                Y1(k)=Y1(k)+g;
            end
            u=u*WN;
        end
    end
    Y=Y1;
  elseif(strcmp(select'DFT'))
   %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
   %基2的DIF-FFT算法
   %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
   Y2=x;
   WN=exp(-1j*2*pi/N);
   for s=1:m    
       for j=1:2^(m-s+1):N-1
           for k=1:2^(m-s)
               p=j+k;
               q=p+2^(m-s);
               r=2^(s-1)*mod(p2^(m-s+1));
               %g=Y(kp)*u;)
               Y2(p)=Y2(p)+Y2(q);
               Y2(q)=(Y2(p)+Y2(q))*WN^r;
           end
          u=u*WN;
       end
   end
   NI=N/2;
   for I=1:N-1
       if I<NI
           t=Y2(I+1);
           Y2(I+1)=Y2(NI+1);
           Y2(NI+1)=t;
       end
       T=N/2;
       while(NI>=T)
           NI=NI-T;
           T=T/2;
       end
       NI=NI+T;
   end
   
%    change=bin2dec(fliplr(dec2bin([1:N]-1m)))+1;
%    Y=Y2(change);
   Y=Y2;
end
subplot(212);stem(changereal(Y)'filled');xlabel('K');ylabel('x[K]');title('x[K]');
disp('是否要继续进行计算？【输入：Y(是)或者N（否）】')
YN=input('是？否？''s');
end

然后进行验证DIF、DIT两种方法的结果是否相同：

1. 选择DIT法：
   
   运行结果：
   

2.选择DIF法：

运行结果：

由上面的验证可知程序的可行性与正确性。
欢迎交流~