// 函數(shù)名: 快速傅立葉變換（來源《C常用算法集》）
// 本函數(shù)測試OK,可以在TC2.0,VC++6.0,Keil C51測試通過。
// 如果你的MCS51系統(tǒng)有足夠的RAM時(shí),可以驗(yàn)證一下用單片機(jī)處理FFT有多么的慢。
//
// 入口參數(shù)： 
// l: l = 0, 傅立葉變換; l = 1, 逆傅立葉變換
// il: il = 0,不計(jì)算傅立葉變換或逆變換模和幅角；il = 1,計(jì)算模和幅角
// n: 輸入的點(diǎn)數(shù)，為偶數(shù)，一般為32，64，128，...,1024等
// k: 滿足n=2^k(k>0),實(shí)質(zhì)上k是n個(gè)采樣數(shù)據(jù)可以分解為偶次冪和奇次冪的次數(shù)
// pr[]: l=0時(shí)，存放N點(diǎn)采樣數(shù)據(jù)的實(shí)部
// l=1時(shí), 存放傅立葉變換的N個(gè)實(shí)部
// pi[]: l=0時(shí)，存放N點(diǎn)采樣數(shù)據(jù)的虛部 
// l=1時(shí), 存放傅立葉變換的N個(gè)虛部
//
// 出口參數(shù)：
// fr[]: l=0, 返回傅立葉變換的實(shí)部
// l=1, 返回逆傅立葉變換的實(shí)部
// fi[]: l=0, 返回傅立葉變換的虛部
// l=1, 返回逆傅立葉變換的虛部
// pr[]: il = 1,i = 0 時(shí)，返回傅立葉變換的模
// il = 1,i = 1 時(shí)，返回逆傅立葉變換的模
// pi[]: il = 1,i = 0 時(shí)，返回傅立葉變換的輻角
// il = 1,i = 1 時(shí)，返回逆傅立葉變換的輻角
// data: 2005.8.15,Mend Xin Dong

#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;
        }
    }

    /*逆傅立葉變換*/
    if(l!=0)
    {
      for(i=0; i<=n-1; i++)
       { 
        fr[i] = fr[i]/(1.0*n);
        fi[i] = fi[i]/(1.0*n);
       }
    } 
    
    /*是否計(jì)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;
} 



#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;
        }
    }

    /*逆傅立葉變換*/
    if(l!=0)
    {
      for(i=0; i<=n-1; i++)
       { 
        fr[i] = fr[i]/(1.0*n);
        fi[i] = fi[i]/(1.0*n);
       }
    } 
    
    /*是否計(jì)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;
}