不同于cuda FFT和iFFT后的图像

问题描述

我想用CUDA对2D图像进行FFT - >斜坡滤波 - > iFFT。首先，作为一个测试，我试图做FFT和iFFt没有任何过滤器。在FFT和iFFT之后，图像看起来是相同的，但是在操作之前，图像像素值在0-255之间，并且在FFT和iFFT之后，图像包含〜10 ^ 7个值。

I'm trying to preform an FFT -> ramp filtering -> iFFT on a 2D image with CUDA. First, as a test I tried to do FFT and iFFt without any filters. After the FFT andthe iFFT the image seems the same, but before the operation the image pixel values were between 0-255 and after the FFT and iFFT the image contains ~10^7 values.

测试图像包含浮点数，尺寸为512 x 360.我使用cuffSinogram函数创建fft，使用cuInversefftSinogram函数创建iFFT。这是我写的两个函数：

The test image contains float numbers, and the dimensions are 512 x 360. I preform the fft with my "cuffSinogram" function, and the iFFT with the "cuInversefftSinogram" function. These are the two function what I wrote:

#define NX 512
#define NY 360

void cufftSinogram(cufftComplex* d_complex_Sinogram, float* d_real_sinogram){

cufftHandle plan;


/* Create a 2D FFT plan. */
if (cufftPlan2d(&plan, NX, NY, CUFFT_R2C) != CUFFT_SUCCESS){
    fprintf(stderr, "CUFFT Error: Unable to create plan\n"); return;
}

if (cufftExecR2C(plan, (cufftReal*)d_real_sinogram, d_complex_Sinogram) != CUFFT_SUCCESS){
    fprintf(stderr, "CUFFT Error: Unable to execute plan\n"); return;
}

if (cudaDeviceSynchronize() != cudaSuccess){
    fprintf(stderr, "Cuda error: Failed to synchronize\n"); return;
}

cufftDestroy(plan);}




void cuInversefftSinogram(float* d_real_sinogram, cufftComplex* d_complex_Sinogram){


cufftHandle plan;


/* Create a 2D FFT plan. */
if (cufftPlan2d(&plan, NX, NY, CUFFT_C2R)  != CUFFT_SUCCESS){
    fprintf(stderr, "CUFFT Error: Unable to create plan\n"); return;
}

if (cufftExecC2R(plan, d_complex_Sinogram, d_real_sinogram) != CUFFT_SUCCESS){
    fprintf(stderr, "CUFFT Error: Unable to execute plan\n"); return;
}

if (cudaDeviceSynchronize() != cudaSuccess){
    fprintf(stderr, "Cuda error: Failed to synchronize\n"); return;
}

cufftDestroy(plan);}

和修改的tiff图片可以在找到（我建议，打开与image）

One original and modified tiff image can be found here (I suggest, to open with imageJ)

推荐答案

CUDA FFT-> IFFT序列要求将结果值除以变换中的元素数，if您想要返回原始数据。

CUDA FFT->IFFT sequences require that you divide the resultant values by the number of elements in the transform, if you want to return back to the original data.

从：

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