fft是改网上c代码的,fft2是自己写的,谁能不能给一个求任意长度(即不需要将输入补成2的指数倍)的fft程序
public class FftList{ //提供fft的输出
double[] fr;
double[] fi;
}
public class Fft2List{ //提供fft2的输出
double[][] outr;
double[][] outi;
}
//inr和ini为输入图像的实部和虚部,没有虚部只要令ini=null就可,w和h为输
//入图像的宽和高,ncols和nrows为输出图像的宽和高,注意ncols和nrows一定要是2的
//指数,ifft=true为逆fft2运算
//此函数相当于matlab下的fft2(X,nrows,ncols)
public Fft2List fft2(double[][] inr,double[][] ini,int w,int h,int n
cols,int nrows,boolean ifft)
{
int i,j,k;
double[] pr=new double[ncols]; //暂存每行实部
double[] pi=new double[ncols]; //暂存每行虚部
k=(int)(Math.log(ncols)/Math.log(2.0)); //k=log2(ncols)
Fft2List fft2list = new Fft2List();
fft2list.outr = new double[ncols][nrows];
fft2list.outi = new double[ncols][nrows];
////////////先对每行求fft/////////////
for(i=0;i
for(j=0;j
//inr和ini为输入图像的实部和虚部,没有虚部只要令ini=null就可,w和h为输
//入图像的宽和高,ncols和nrows为输出图像的宽和高,注意ncols和nrows一定要是2的
//指数,ifft=true为逆fft2运算
//此函数相当于matlab下的fft2(X,nrows,ncols)
public Fft2List fft2(double[][] inr,double[][] ini,int w,int h,int n
cols,int nrows,boolean ifft)
{
int i,j,k;
double[] pr=new double[ncols]; //暂存每行实部
double[] pi=new double[ncols]; //暂存每行虚部
k=(int)(Math.log(ncols)/Math.log(2.0)); //k=log2(ncols)
Fft2List fft2list = new Fft2List();
fft2list.outr = new double[ncols][nrows];
fft2list.outi = new double[ncols][nrows];
////////////先对每行求fft/////////////
for(i=0;i
for(j=0;j
pr[j]=0;
pi[j]=0;
}
for(j=0;j
pr[j]=inr[j][i];
if(ini != null) //输入可能没有虚部
pi[j]=ini[j][i];
}
FftList fftlist = this.fft(pr,pi,ncols,k,ifft);
for(j=0;j
fft2list.outr[j][i]=fftlist.fr[j];
fft2list.outi[j][i]=fftlist.fi[j];
}
}
/////////////////////////////////////////
pr=new double[nrows];
pi=new double[nr
pr[j]=0;
pi[j]=0;
}
for(j=0;j
pr[j]=inr[j][i];
if(ini != null) //输入可能没有虚部
pi[j]=ini[j][i];
}
FftList fftlist = this.fft(pr,pi,ncols,k,ifft);
for(j=0;j
fft2list.outr[j][i]=fftlist.fr[j];
fft2list.outi[j][i]=fftlist.fi[j];
}
}
/////////////////////////////////////////
pr=new double[nrows];
pi=new double[nrows];
k=(int)(Math.log(nrows)/Math.log(2.0)); //k=log2(nrows)
/////////////再对每列求fft/////////////////
for(i=0;i
for(j=0;j
pr[j]=0;
pi[j]=0;
}
for(j=0;j
pr[j]=fft2list.outr[i][j]; //将上一部分结果的每一
// 列赋给pr和pi
pi[j]=fft2list.outi[i][j];
}
FftList fftlist = this.fft(pr,pi,nrows,k,ifft);
for(j=0;j
fft2list.outr[i][j]=fftlist.fr[j];
fft2list.outi[i][j]=fftlist.fi[j];
}nt)(Math.log(nrows)/Math.log(2.0)); //k=log2(nrows)
/////////////再对每列求fft/////////////////
for(i=0;i
for(j=0;j
pr[j]=0;
pi[j]=0;
}
for(j=0;j
pr[j]=fft2list.outr[i][j]; //将上一部分结果的每一
// 列赋给pr和pi
pi[j]=fft2list.outi[i][j];
}
FftList fftlist = this.fft(pr,pi,nrows,k,ifft);
for(j=0;j
fft2list.outr[i][j]=fftlist.fr[j];
fft2list.outi[i][j]=fftlist.fi[j];
}
}
return fft2list;
}
//n=2^k,其他的计算会出错
public FftList fft(double[] pr,double[] pi,int n,int k,boolean ifft)
{
int it,m,is,i,j,nv,l0;
double p,q,s,vr,vi,poddr,poddi;
FftList cl = new FftList();
cl.fr = new double[n];
cl.fi = new double[n];
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;}
cl.fr[it]=pr[is]; cl.fi[it]=pi[is];
}
pr[0]=1.0; pi[0
return fft2list;
}
//n=2^k,其他的计算会出错
public FftList fft(double[] pr,double[] pi,int n,int k,boolean ifft)
{
int it,m,is,i,j,nv,l0;
double p,q,s,vr,vi,poddr,poddi;
FftList cl = new FftList();
cl.fr = new double[n];
cl.fi = new double[n];
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;}
cl.fr[it]=pr[is]; cl.fi[it]=pi[is];
}
pr[0]=1.0; pi[0]=0.0;
p=2*Math.PI/(1.0*n);
pr[1]=Math.cos(p); pi[1]=-Math.sin(p);
if (ifft) pi[1]=-pi[1]; //求逆fft
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=cl.fr[it]; vi=cl.fi[it];
cl.fr[it]=vr+cl.fr[it+1]; cl.fi[it]=vi+cl.fi[it+1];
cl.fr[it+1]=vr-cl.fr[it+1]; cl.fi[it+1]=vi-cl.fi[it+1];
}
m=n/2; nv=2;
for (l0=k-2; l0>=0; l0--)
{
m=m/2; nv=2*nv;
pr[1]=Math.cos(p); pi[1]=-Math.sin(p);
if (ifft) pi[1]=-pi[1]; //求逆fft
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=cl.fr[it]; vi=cl.fi[it];
cl.fr[it]=vr+cl.fr[it+1]; cl.fi[it]=vi+cl.fi[it+1];
cl.fr[it+1]=vr-cl.fr[it+1]; cl.fi[it+1]=vi-cl.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]*cl.fr[it+j+nv/2];
q=pi[m*j]*cl.fi[it+j+nv/2];
s=pr[m*j]+pi[m*j];
s=s*(cl.fr[it+j+nv/2]+cl.fi[it+j+nv/2]);
poddr=p-q; poddi=s-p-q;
cl.fr[it+j+nv/2]=cl.fr[it+j]-poddr;
cl.fi[it+j+nv/2]=cl.fi[it+j]-poddi;
cl.fr[it+j]=cl.fr[it+j]+poddr;
cl.fi[it+j]=cl.fi[it+j]+poddi;
}
}
if (ifft)
for (i=0;
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]*cl.fr[it+j+nv/2];
q=pi[m*j]*cl.fi[it+j+nv/2];
s=pr[m*j]+pi[m*j];
s=s*(cl.fr[it+j+nv/2]+cl.fi[it+j+nv/2]);
poddr=p-q; poddi=s-p-q;
cl.fr[it+j+nv/2]=cl.fr[it+j]-poddr;
cl.fi[it+j+nv/2]=cl.fi[it+j]-poddi;
cl.fr[it+j]=cl.fr[it+j]+poddr;
cl.fi[it+j]=cl.fi[it+j]+poddi;
}
}
if (ifft)
for (i=0; i<=n-1; i++)
{
cl.fr[i]=cl.fr[i]/(1.0*n);
cl.fi[i]=cl.fi[i]/(1.0*n);
}
return cl;
}