C 中的 2D 傅立叶变换 [英] 2D Fourier Transformation in C

问题描述

我使用此站点的公式实现了 2D DFT 和 IDFThttp://homepages.inf.ed.ac.uk/rbf/HIPR2/fourier.htm我认为这些是正确的并且很好地解释了.实现看起来像这样:

I implemented 2D DFT and IDFT using equation from this site http://homepages.inf.ed.ac.uk/rbf/HIPR2/fourier.htm I think these are correct and nicely explained. Implementation looks like that:

    for(int i=0;i<inImage.width;i++)
    {
     for(int j=0;j<inImage.height;j++)
     {
      float ak=0; 
      float bk=0;
          for(int ii=0;ii<inImage.width;ii++)
           {
             for(int jj=0;jj<inImage.height;jj++)
              {

               float x=-2.0*PI*i*ii/(float)inImage.width;
               float y=-2.0*PI*j*jj/(float)inImage.height;
            // ak+=inImage.pixels[i][j]*(cos(x)*cos(y)-sin(x)*sin(y));
           //  bk+=inImage.pixels[i][j]*(sin(x)*cos(y)+sin(y)*cos(x));
               ak+=inImage.pixels[i][j]*cos(x+y);
               bk+=inImage.pixels[i][j]*1.0*sin(x+y);
             }
           }
     DFTImageRE.pixels[i][j]=ak;
     DFTImageIM.pixels[i][j]=bk;
       }
     }

频域 (sqrt(ak * ak+bk * bk)) 看起来不像它应该的那样，并且图像重建(忽略虚部)没有使原始图像附近有任何东西.[0][0] 处的更多像素具有极高的值，并且没有像素范围从 0 到 255 作为原始像素.我做错了什么?

The frequency domain (sqrt(ak * ak+bk * bk)) doesnt look as it should, and the image reconstruction (ignoring the imaginary parts) doesnt make anything near the original image. What is more pixel at [0][0] have extremely high value and no pixels range from 0 to 255 as the original one. What am i doing wrong?

额外信息:

• inImage 和 DFTImages 只是构造原始 *.pgm 图像的结构，保存和加载图像有效，
• 我不能使用任何类(比如虚数)，因为这个实现将在 GPU 端，

谢谢

推荐答案

我找到了解决问题的方法.这只是索引问题.将 ii 和 jj 相加求出傅立叶变换

I found a solution for my problem. It was just indexing problem. Use ii and jj in sum to find a Fourier Transform

   for(int i=0;i<inImage.width;i++)
{
 for(int j=0;j<inImage.height;j++)
 {
  float ak=0; 
  float bk=0;
      for(int ii=0;ii<inImage.width;ii++)
       {
         for(int jj=0;jj<inImage.height;jj++)
          {

           float x=-2.0*PI*i*ii/(float)inImage.width;
           float y=-2.0*PI*j*jj/(float)inImage.height;
           ak+=inImage.pixels[ii][jj]*cos(x+y);
           bk+=inImage.pixels[ii][jj]*1.0*sin(x+y);
         }
       }
 DFTImageRE.pixels[i][j]=ak;
 DFTImageIM.pixels[i][j]=bk;
   }
 }

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