我以前幫助this答案實現就地轉換,它運作良好,但只有當我從真實的數據開始。如果我從複雜的數據開始,IFT + FFT之後的結果是錯誤的,而且這隻發生在原地版本中,我用這個變換的不合適版本得到了完美的結果。 這是代碼:cuFFT錯誤的結果只有當從複雜的開始
#include <stdio.h>
#include <stdlib.h>
#include <cuda_runtime.h>
#include <complex.h>
#include <cuComplex.h>
#include <cufft.h>
#include <cufftXt.h>
#define N 4
#define N_PAD (2*(N/2+1))
void print_3D_Real(double *array){
printf("\nPrinting 3D real matrix \n");
unsigned long int idx;
for (int z = 0; z < N; z++){
printf("---------------------------------------------------------------------------- plane %d below\n", z);
for (int x = 0; x < N; x++){
for (int y = 0; y < N; y++){
idx = z + N_PAD * (y + x * N);
printf("%.3f \t", array[idx]);
}
printf("\n");
}
}
}
void print_3D_Comp(cuDoubleComplex *array){
printf("\nPrinting 3D complex matrix \n");
unsigned long int idx;
for (int z = 0; z < (N/2+1); z++){
printf("---------------------------------------------------------------------------- plane %d below\n", z);
for (int x = 0; x < N; x++){
for (int y = 0; y < N; y++){
idx = z + (N/2+1) * (y + x * N);
printf("%+.3f%+.3fi \t", array[idx].x, array[idx].y);
}
printf("\n");
}
}
}
// Main function
int main(int argc, char **argv){
CU_ERR_CHECK(cudaSetDevice(0));
unsigned long int idx, in_mem_size, out_mem_size;
cuDoubleComplex *in = NULL, *d_in = NULL;
double *out = NULL, *d_out = NULL;
cufftHandle plan_r2c, plan_c2r;
in_mem_size = sizeof(cuDoubleComplex) * N*N*(N/2+1);
out_mem_size = in_mem_size;
in = (cuDoubleComplex *) malloc (in_mem_size);
out = (double *) in;
cudaMalloc((void **)&d_in, in_mem_size);
d_out = (double *) d_in;
cufftPlan3d(&plan_c2r, N, N, N, CUFFT_Z2D);
cufftPlan3d(&plan_r2c, N, N, N, CUFFT_D2Z);
memset(in, 0, in_mem_size);
memset(out, 0, out_mem_size);
// Initial complex data
for (int x = 0; x < N; x++){
for (int y = 0; y < N; y++){
for (int z = 0; z < (N/2+1); z++){
idx = z + (N/2+1) * (y + x * N);
in[idx].x = idx;
}
}
}
print_3D_Comp(in);
cudaMemcpy(d_in, in, in_mem_size, cudaMemcpyHostToDevice);
cufftExecZ2D(plan_c2r, (cufftDoubleComplex *)d_in, (cufftDoubleReal *)d_out);
cudaMemcpy(out, d_out, out_mem_size, cudaMemcpyDeviceToHost);
// Normalisation
for (int i = 0; i < N*N*N_PAD; i++)
out[i] /= (N*N*N);
print_3D_Real(out);
cudaMemcpy(d_out, out, out_mem_size, cudaMemcpyHostToDevice);
cufftExecD2Z(plan_r2c, (cufftDoubleReal *)d_out, (cufftDoubleComplex *)d_in);
cudaMemcpy(in, d_in, in_mem_size, cudaMemcpyDeviceToHost));
print_3D_Comp(in);
cudaDeviceReset();
return 0;
}
我的程序的輸出是這個pastebin。
有人可以指導我在正確的道路上?非常感謝你提前。
感謝您的回答羅伯特。你的確是正確的,我的不合適的版本不會產生與該輸入信號正確的結果,但它產生了與您發佈的輸入信號相同的結果,並且我以前使用它來測試它。傅立葉變換我不是很好,並不知道並非所有的複雜信號都可以用純粹的真實信號表示,因此我認爲我的代碼是錯誤的。我現在知道它對其他信號是正確的,如果它沒有編譯它一定是因爲我在這裏發佈時刪除了太多東西。再次感謝您的好評! – JohnWil