c语言图像傅里叶变换（c++实现傅里叶变换）

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本文目录一览：

1、傅里叶变换用C语言程序怎么实现？2、C语言 1024点快速傅里叶变换（FFT）程序，最好经过优化，执行速度快3、求个快速傅里叶变换的C语言程序

傅里叶变换用C语言程序怎么实现？

#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;iN;i++)

{

xr[i]=i;

xi[i]=0;

}

printf(“——FFT——\n”);

l=0;

kkfft(xr,xi,n,k,Yr,Yi,l,il);

for(i=0;iN;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;iN;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;

}

C语言 1024点快速傅里叶变换（FFT）程序，最好经过优化，执行速度快

void fft()

{

int nn,n1,n2,i,j,k,l,m,s,l1;

float ar[1024],ai[1024]; // 实部 虚部

float a[2050];

float t1,t2,x,y;

float w1,w2,u1,u2,z;

float fsin[10]={0.000000,1.000000,0.707107,0.3826834,0.1950903,0.09801713,0.04906767,0.02454123,0.01227154,0.00613588,};// 优化

float fcos[10]={-1.000000,0.000000,0.7071068,0.9238796,0.9807853,0.99518472,0.99879545,0.9996988,0.9999247,0.9999812,};

nn=1024;

s=10;

n1=nn/2; n2=nn-1;

j=1;

for(i=1;i=nn;i++)

{

a[2*i]=ar[i-1];

a[2*i+1]=ai[i-1];

}

for(l=1;ln2;l++)

{

if(lj)

{

t1=a[2*j];

t2=a[2*j+1];

a[2*j]=a[2*l];

a[2*j+1]=a[2*l+1];

a[2*l]=t1;

a[2*l+1]=t2;

}

k=n1;

while (kj)

{

j=j-k;

k=k/2;

}

j=j+k;

}

for(i=1;i=s;i++)

{

u1=1;

u2=0;

m=(1i);

k=m1;

w1=fcos[i-1];

w2=-fsin[i-1];

for(j=1;j=k;j++)

{

for(l=j;lnn;l=l+m)

{

l1=l+k;

t1=a[2*l1]*u1-a[2*l1+1]*u2;

t2=a[2*l1]*u2+a[2*l1+1]*u1;

a[2*l1]=a[2*l]-t1;

a[2*l1+1]=a[2*l+1]-t2;

a[2*l]=a[2*l]+t1;

a[2*l+1]=a[2*l+1]+t2;

}

z=u1*w1-u2*w2;

u2=u1*w2+u2*w1;

u1=z;

}

}

for(i=1;i=nn/2;i++)

{

ar[i]=a[2*i+2]/nn;

ai[i]=-a[2*i+3]/nn;

a[i]=4*sqrt(ar[i]*ar[i]+ai[i]*ai[i]); // 幅值

}

}

求个快速傅里叶变换的C语言程序

void  fft()

{

      int nn,n1,n2,i,j,k,l,m,s,l1;

      float ar[1024],ai[1024]; // 实部 虚部

      float a[2050];

      float t1,t2,x,y;

      float w1,w2,u1,u2,z;

      float fsin[10]={0.000000,1.000000,0.707107,0.3826834,0.1950903,0.09801713,0.04906767,0.02454123,0.01227154,0.00613588,};// 优化

      float fcos[10]={-1.000000,0.000000,0.7071068,0.9238796,0.9807853,0.99518472,0.99879545,0.9996988,0.9999247,0.9999812,};

      nn=1024;

      s=10;

      n1=nn/2;  n2=nn-1;

      j=1;

      for(i=1;i=nn;i++)

      {

        a[2*i]=ar[i-1];

        a[2*i+1]=ai[i-1];

      }

      for(l=1;ln2;l++)

      {

       if(lj)

       {

    t1=a[2*j];

    t2=a[2*j+1];

    a[2*j]=a[2*l];

    a[2*j+1]=a[2*l+1];

    a[2*l]=t1;

    a[2*l+1]=t2;

       }

       k=n1;

       while (kj)

       {

    j=j-k;

    k=k/2;

       }

       j=j+k;

     }

     for(i=1;i=s;i++)

     {

    u1=1;

    u2=0;

    m=(1i);

    k=m1;

    w1=fcos[i-1];

    w2=-fsin[i-1];

    for(j=1;j=k;j++)

    {

     for(l=j;lnn;l=l+m)

     {

        l1=l+k;

        t1=a[2*l1]*u1-a[2*l1+1]*u2;

        t2=a[2*l1]*u2+a[2*l1+1]*u1;

        a[2*l1]=a[2*l]-t1;

        a[2*l1+1]=a[2*l+1]-t2;

        a[2*l]=a[2*l]+t1;

        a[2*l+1]=a[2*l+1]+t2;

     }

     z=u1*w1-u2*w2;

     u2=u1*w2+u2*w1;

     u1=z;

    }

     }

     for(i=1;i=nn/2;i++)

     {

    ar[i]=a[2*i+2]/nn;

    ai[i]=-a[2*i+3]/nn;

    a[i]=4*sqrt(ar[i]*ar[i]+ai[i]*ai[i]);  // 幅值

     }

}

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