CUDA Inverse FFT Bug | 易学教程

问题

I have the following code that has bug when doing the inverse FFT. The forward FFT works as I printed the output and verified it. But the inverse does not seem to. Any ideas? Does it look like I have missed a concept?

Code - http://pastebin.com/iZYtdcqR

EDIT - I have essentially rewritten the code that comes with the CUDA toolkit samples. I am trying to perform a convolution using FFT but with a modified algorithm (DIF actually.)

EDIT2 - dding code to the question.

#include <stdio.h>
#include <stdlib.h>
#include <math.h>

#include <cuda_runtime.h>
#include <cufft.h>

typedef enum signaltype {REAL, COMPLEX} signal;

typedef float2 Complex;

void
printData(Complex *a, int size, char *msg) {

  if (msg == "") printf("\n");
  else printf("%s\n", msg);

  for (int i = 0; i < size; i++)
    printf("%f %f\n", a[i].x, a[i].y);
}

void
normData(Complex *a, int size, float norm) {

  for (int i = 0; i < size; i++) {
    a[i].x /= norm;
    a[i].y /= norm;
  }
}

void
randomFill(Complex *h_signal, int size, int flag) {

  // Real signal.
  if (flag == REAL) {
    for (int i = 0; i < size; i++) {
      h_signal[i].x = rand() / (float) RAND_MAX;
      h_signal[i].y = 0;
    }
  }
}

// FFT a signal that's on the _DEVICE_.
void
signalFFT(Complex *d_signal, int signal_size) {

  cufftHandle plan;
  if (cufftPlan1d(&plan, signal_size, CUFFT_C2C, 1) != CUFFT_SUCCESS) {
    printf("Failed to plan FFT\n");
    exit(0);
  }

  // Execute the plan.
  if (cufftExecC2C(plan, (cufftComplex *) d_signal, (cufftComplex *) d_signal, CUFFT_FORWARD) != CUFFT_SUCCESS) {
    printf ("Failed Executing FFT\n");
    exit(0);
  }

}

void
signalIFFT(Complex *d_signal, int signal_size) {

  cufftHandle plan;
  if (cufftPlan1d(&plan, signal_size, CUFFT_C2C, 1) != CUFFT_SUCCESS) {
    printf("Failed to plan IFFT\n");
    exit(0);
  }

  // Execute the plan.
  if (cufftExecC2C(plan, (cufftComplex *) d_signal, (cufftComplex *) d_signal, CUFFT_INVERSE) != CUFFT_SUCCESS) {
    printf ("Failed Executing IFFT\n");
    exit(0);
  }

}

int main()
{

  Complex *h_signal, *d_signal1;

  int alloc_size, i;

  alloc_size = 16;

  // Kernel Block and Grid Size.
  const dim3 blockSize(16, 16, 1);
  const dim3 gridSize(alloc_size / 16 + 1, alloc_size / 16 + 1, 1);

  h_signal = (Complex *) malloc(sizeof(Complex) * alloc_size);

  cudaMalloc(&d_signal1, sizeof(Complex) * alloc_size);
  if (cudaGetLastError() != cudaSuccess){
    printf("Cuda error: Failed to allocate\n");
    exit(0);
  }
  //cudaMalloc(&d_signal2, sizeof(Complex) * alloc_size);

  // Add random data to signal.
  randomFill(h_signal, alloc_size, REAL);
  printData(h_signal, alloc_size, "Random H1");

  cudaMemcpy(d_signal1, h_signal, sizeof(Complex) * alloc_size, cudaMemcpyHostToDevice);

  signalFFT(d_signal1, alloc_size);

  signalIFFT(d_signal1, alloc_size);

  cudaDeviceSynchronize();

  cudaMemcpy(h_signal, d_signal1, sizeof(Complex) * alloc_size, cudaMemcpyDeviceToHost);

  printData(h_signal, alloc_size, "IFFT");

  return 0;
}

回答1:

Suggestions for writing a good question:

Post your code in the question, not in an external link. Someday that link will become invalid, and so will your question for future readers.
Post your actual data (there isn't that much, in this case) in the question.
Post or identify what you would expect the data to be, and why.

Another note:

The cufft documentationn instructs you to use cufftComplex, not your own data type, although yours works ok. If the developers of cufft ever change their data layout for some odd reason, your code would break on recompile. If you use the recommended data type, it should not.

Now regarding your question, I ran your code and got results like this:

Random H1
0.840188 0.000000
0.394383 0.000000
0.783099 0.000000
0.798440 0.000000
0.911647 0.000000
0.197551 0.000000
0.335223 0.000000
0.768230 0.000000
0.277775 0.000000
0.553970 0.000000
0.477397 0.000000
0.628871 0.000000
0.364784 0.000000
0.513401 0.000000
0.952230 0.000000
0.916195 0.000000
IFFT
13.443005 0.000000
6.310127 -0.000000
12.529589 0.000000
12.775041 0.000000
14.586359 -0.000000
3.160823 0.000000
5.363565 0.000000
12.291674 -0.000000
4.444397 -0.000000
8.863521 0.000000
7.638353 0.000000
10.061934 -0.000000
5.836554 0.000000
8.214415 -0.000000
15.235678 -0.000000
14.659121 -0.000000

The only thing that seems to be missing from your code is that you didn't divide the results by the length of the transform (16 in this case) to get back the original data (as suggested in the sample code here). When I do that, I get what I believe is expected results:

Random H1
0.840188 0.000000
0.394383 0.000000
0.783099 0.000000
0.798440 0.000000
0.911647 0.000000
0.197551 0.000000
0.335223 0.000000
0.768230 0.000000
0.277775 0.000000
0.553970 0.000000
0.477397 0.000000
0.628871 0.000000
0.364784 0.000000
0.513401 0.000000
0.952230 0.000000
0.916195 0.000000
IFFT
0.840188 0.000000
0.394383 -0.000000
0.783099 0.000000
0.798440 0.000000
0.911647 -0.000000
0.197551 0.000000
0.335223 0.000000
0.768230 -0.000000
0.277775 -0.000000
0.553970 0.000000
0.477397 0.000000
0.628871 -0.000000
0.364785 0.000000
0.513401 -0.000000
0.952230 -0.000000
0.916195 -0.000000

By the way, kudos for providing a complete, compilable code sample.

来源：https://stackoverflow.com/questions/16781653/cuda-inverse-fft-bug

标签

cuda

fft

ifft