求实现快速傅立叶变换的程序
有个程序需要快速傅立叶变换,各位大哥帮帮忙啊
#include <math.h> #include <stdio.h> #define N 8 void kkfft(double pr[], double pi[], int n, int k, double fr[], double fi[], int l, int il); void main() { double xr[N],xi[N],Yr[N],Yi[N],l=0,il=0; int i,j,n=N,k=3; for(i=0;i<N;i++) { xr[i]=i; xi[i]=0; } printf("------FFT------\n"); l=0; kkfft(xr,xi,n,k,Yr,Yi,l,il); for(i=0;i<N;i++) { printf("%-11lf + j* %-11lf\n",Yr[i],Yi[i]); } printf("-----DFFT-------\n"); l=1; kkfft(Yr,Yi,n,k,xr,xi,l,il); for(i=0;i<N;i++) { printf("%-11lf + j* %-11lf\n",xr[i],xi[i]); } getch(); } void kkfft(double pr[], double pi[], int n, int k, double fr[], double fi[], int l, int il) { int it,m,is,i,j,nv,l0; double p,q,s,vr,vi,poddr,poddi; for (it=0; it<=n-1; it++) { m = it; is = 0; for(i=0; i<=k-1; i++) { j = m/2; is = 2*is+(m-2*j); m = j; } fr[it] = pr[is]; fi[it] = pi[is]; } pr[0] = 1.0; pi[0] = 0.0; p = 6.283185306/(1.0*n); pr[1] = cos(p); pi[1] = -sin(p); if (l!=0) pi[1]=-pi[1]; for (i=2; i<=n-1; i++) { p = pr[i-1]*pr[1]; q = pi[i-1]*pi[1]; s = (pr[i-1]+pi[i-1])*(pr[1]+pi[1]); pr[i] = p-q; pi[i] = s-p-q; } for (it=0; it<=n-2; it=it+2) { vr = fr[it]; vi = fi[it]; fr[it] = vr+fr[it+1]; fi[it] = vi+fi[it+1]; fr[it+1] = vr-fr[it+1]; fi[it+1] = vi-fi[it+1]; } m = n/2; nv = 2; for (l0=k-2; l0>=0; l0--) { m = m/2; nv = 2*nv; for(it=0; it<=(m-1)*nv; it=it+nv) for (j=0; j<=(nv/2)-1; j++) { p = pr[m*j]*fr[it+j+nv/2]; q = pi[m*j]*fi[it+j+nv/2]; s = pr[m*j]+pi[m*j]; s = s*(fr[it+j+nv/2]+fi[it+j+nv/2]); poddr = p-q; poddi = s-p-q; fr[it+j+nv/2] = fr[it+j]-poddr; fi[it+j+nv/2] = fi[it+j]-poddi; fr[it+j] = fr[it+j]+poddr; fi[it+j] = fi[it+j]+poddi; } } /*逆傅立叶变换*/ if(l!=0) { for(i=0; i<=n-1; i++) { fr[i] = fr[i]/(1.0*n); fi[i] = fi[i]/(1.0*n); } } /*是否计算模和相角*/ if(il!=0) { for(i=0; i<=n-1; i++) { pr[i] = sqrt(fr[i]*fr[i]+fi[i]*fi[i]); if(fabs(fr[i])<0.000001*fabs(fi[i])) { if ((fi[i]*fr[i])>0) pi[i] = 90.0; else pi[i] = -90.0; } else pi[i] = atan(fi[i]/fr[i])*360.0/6.283185306; } } return; }