C++ 关于cufft R2C和C2R

我曾经用过cufft做过我的研究，但是在使用它的过程中出现了一些问题。我的步骤如下：

__global__ void MultiplyKernel(cufftComplex *data, float *data1,cufftComplex *data2, unsigned vectorSize) {
    unsigned idx = blockIdx.x*blockDim.x+threadIdx.x;
    if (idx < vectorSize){
        data[idx].x = data2[idx].x*data1[idx];
        data[idx].y = data2[idx].y*data1[idx];
    }
}

__global__ void Scale(cufftReal *data, unsigned vectorSize) {
    unsigned idx = blockIdx.x*blockDim.x+threadIdx.x;
    if (idx < vectorSize){
        data[idx] = data[idx]/vectorSize;
    }
}

void ApplyKernel1(cufftReal *data2, float *ImageBuffer, float *KernelBuffer, unsigned int NX, unsigned int NY,unsigned int NZ)
{
      float *Akernel;
      cufftComplex *data_dev1, *data_dev2;
      cufftReal *data_dev3, *data_dev;
      cudaMalloc((void **)&Akernel, NX * NY * NZ * sizeof(float));
      cudaMalloc((void **)&data_dev3, NX * NY * NZ * sizeof(cufftReal));
      cudaMalloc((void **)&data_dev, NX * NY * NZ * sizeof(cufftComplex));
      cudaMalloc((void **)&data_dev1, NX * NY * NZ * sizeof(cufftComplex));
      cudaMalloc((void **)&data_dev2, NX * NY * NZ * sizeof(cufftComplex));
      cudaMemset(data_dev, 0, NX * NY * NZ * sizeof(cufftReal));
      cudaMemset(data_dev1, 0, NX * NY * NZ * sizeof(cufftComplex));
      cudaMemset(data_dev2, 0, NX * NY * NZ * sizeof(cufftComplex));
      //cufftComplex *resultFFT = (cufftComplex*)malloc(NX * NY * NZ * sizeof(cufftComplex));
      //cufftReal *resultIFFT = (cufftReal*)malloc(NX * NY * NZ * sizeof(cufftReal));

      cudaMemcpy(data_dev, ImageBuffer, NX * NY * NZ * sizeof(cufftReal), cudaMemcpyHostToDevice);

      cufftHandle plan;
      cufftPlan3d(&plan, NZ, NY, NX, CUFFT_R2C);
      cufftExecR2C(plan, data_dev, data_dev1);

      //Multiply kernel
      cudaMemcpy(Akernel, KernelBuffer, NX * NY * NZ * sizeof(float), cudaMemcpyHostToDevice);
      static const int BLOCK_SIZE = 1000;
      const int blockCount = (NX*NY*NZ+BLOCK_SIZE-1)/BLOCK_SIZE;
      MultiplyKernel <<<blockCount, BLOCK_SIZE>>> (data_dev2, Akernel, data_dev1, NX*NY*NZ);


      cufftDestroy(plan);
      //cufftPlan3d(&plan, NZ, NY, NX, CUFFT_C2R);
      cufftPlan3d(&plan, NZ,NY,NX, CUFFT_C2R);
      cufftExecC2R(plan, data_dev2, data_dev3 );
      Scale <<<blockCount, BLOCK_SIZE>>> (data_dev3, NX*NY*NZ);
      cudaMemcpy(data2, data_dev3, NZ * NY * NX * sizeof(cufftReal), cudaMemcpyDeviceToHost);

      cufftDestroy(plan);
      cudaFree(data_dev);
      cudaFree(data_dev1);
      cudaFree(data_dev2);
      cudaFree(data_dev3);
      cudaFree(Akernel);


}

使用R2C对图像进行前向FFT

将核系数与复数结果相乘

用C2R对乘法结果进行逆FFT

但是，当我使用复数结果对核进行乘法时，出现了一个严重的问题，即cuft复数结果与fftw的结果不相等，并且结果中存在大量的零。我知道R2C的结果大小是N1（N2/2+1），但我想得到完整的复杂结果。如何解决这个问题？i、 e.如何恢复R2C结果？如何将乘法结果放入C2R并得到正确答案

我的机具程序代码如下所示：

__global__ void MultiplyKernel(cufftComplex *data, float *data1,cufftComplex *data2, unsigned vectorSize) {
    unsigned idx = blockIdx.x*blockDim.x+threadIdx.x;
    if (idx < vectorSize){
        data[idx].x = data2[idx].x*data1[idx];
        data[idx].y = data2[idx].y*data1[idx];
    }
}

__global__ void Scale(cufftReal *data, unsigned vectorSize) {
    unsigned idx = blockIdx.x*blockDim.x+threadIdx.x;
    if (idx < vectorSize){
        data[idx] = data[idx]/vectorSize;
    }
}

void ApplyKernel1(cufftReal *data2, float *ImageBuffer, float *KernelBuffer, unsigned int NX, unsigned int NY,unsigned int NZ)
{
      float *Akernel;
      cufftComplex *data_dev1, *data_dev2;
      cufftReal *data_dev3, *data_dev;
      cudaMalloc((void **)&Akernel, NX * NY * NZ * sizeof(float));
      cudaMalloc((void **)&data_dev3, NX * NY * NZ * sizeof(cufftReal));
      cudaMalloc((void **)&data_dev, NX * NY * NZ * sizeof(cufftComplex));
      cudaMalloc((void **)&data_dev1, NX * NY * NZ * sizeof(cufftComplex));
      cudaMalloc((void **)&data_dev2, NX * NY * NZ * sizeof(cufftComplex));
      cudaMemset(data_dev, 0, NX * NY * NZ * sizeof(cufftReal));
      cudaMemset(data_dev1, 0, NX * NY * NZ * sizeof(cufftComplex));
      cudaMemset(data_dev2, 0, NX * NY * NZ * sizeof(cufftComplex));
      //cufftComplex *resultFFT = (cufftComplex*)malloc(NX * NY * NZ * sizeof(cufftComplex));
      //cufftReal *resultIFFT = (cufftReal*)malloc(NX * NY * NZ * sizeof(cufftReal));

      cudaMemcpy(data_dev, ImageBuffer, NX * NY * NZ * sizeof(cufftReal), cudaMemcpyHostToDevice);

      cufftHandle plan;
      cufftPlan3d(&plan, NZ, NY, NX, CUFFT_R2C);
      cufftExecR2C(plan, data_dev, data_dev1);

      //Multiply kernel
      cudaMemcpy(Akernel, KernelBuffer, NX * NY * NZ * sizeof(float), cudaMemcpyHostToDevice);
      static const int BLOCK_SIZE = 1000;
      const int blockCount = (NX*NY*NZ+BLOCK_SIZE-1)/BLOCK_SIZE;
      MultiplyKernel <<<blockCount, BLOCK_SIZE>>> (data_dev2, Akernel, data_dev1, NX*NY*NZ);


      cufftDestroy(plan);
      //cufftPlan3d(&plan, NZ, NY, NX, CUFFT_C2R);
      cufftPlan3d(&plan, NZ,NY,NX, CUFFT_C2R);
      cufftExecC2R(plan, data_dev2, data_dev3 );
      Scale <<<blockCount, BLOCK_SIZE>>> (data_dev3, NX*NY*NZ);
      cudaMemcpy(data2, data_dev3, NZ * NY * NX * sizeof(cufftReal), cudaMemcpyDeviceToHost);

      cufftDestroy(plan);
      cudaFree(data_dev);
      cudaFree(data_dev1);
      cudaFree(data_dev2);
      cudaFree(data_dev3);
      cudaFree(Akernel);


}

\uuuuu全局\uuuuuu无效多线程（cufftComplex*数据、浮点*data1、cufftComplex*data2、无符号向量大小）{
无符号idx=blockIdx.x*blockDim.x+threadIdx.x；
if（idx

将R2C fft的结果乘以复数时，结果不再对应于对称阵列

你不知道R2C和C2R变换是对称的，而CUFFT只计算了一半的解，因为这个？为什么不做C2C运算呢？是的，我用C2C做正向变换，将data.y设置为0，得到了正确的结果。但是，最后在乘以系数后，我想进行反向变换以获得真实结果，即C2R（反向），如何使用cufft库获得正确答案？（注意：我已经将一些系数乘以C2C结果。）感谢您的回答，我问如何从R2C结果中获得与C2C结果相等的完整结果。我想要整个结果，因为我想用它乘以一些系数，你能帮我吗？@KwuJohn复数的数目是实数的一半，因为如果输入数组是实数，频率n-k的响应是频率k响应的复共轭。如果您的过滤器不保留此属性，则过滤后的信号将不是纯真实的，使用C2R返回真实世界似乎有缺陷！