Question

我对这个程序有疑问：

#include <stdlib.h>
#include <stdio.h>
#include <string.h>
#include <math.h>
#include <cufft.h>
#include <cuComplex.h>

#define SIGNAL_SIZE        1024

int main(int argc, char **argv) {
   cudaEvent_t start, stop;
   cudaEventCreate(&start);
   cudaEventCreate(&stop);

   // Allocate host memory for the signal
   cuDoubleComplex *h_signal = (cuDoubleComplex *) malloc(sizeof(cuDoubleComplex) * SIGNAL_SIZE);

   // Initalize the memory for the signal
   for (unsigned int i = 0; i < SIGNAL_SIZE; ++i) {
      if((double)i/SIGNAL_SIZE>=0 && (double)i/SIGNAL_SIZE<0.5)  h_signal[i].x = (double)i/SIGNAL_SIZE;
      else if((double)i/SIGNAL_SIZE>=0.5 && (double)i/SIGNAL_SIZE<1)  h_signal[i].x = (double)i/SIGNAL_SIZE-1;
      h_signal[i].y = 0;
   }

// Allocate device memory for signal
   cuDoubleComplex *d_signal;

   cudaMalloc((void **) &d_signal, SIGNAL_SIZE*sizeof(cuDoubleComplex));
   // Copy host memory to device
   cudaMemcpy(d_signal, h_signal, SIGNAL_SIZE*sizeof(cuDoubleComplex), cudaMemcpyHostToDevice);


cudaEventRecord(start, 0);
   cufftHandle plan;
   cufftPlan1d(&plan, SIGNAL_SIZE , CUFFT_C2C, 1);

   // FFT computation
   cufftExecC2C(plan, (cufftComplex *) d_signal, (cufftComplex *) d_signal,
         CUFFT_FORWARD);

    cufftExecC2C(plan, (cufftComplex *) d_signal, (cufftComplex *) d_signal, CUFFT_INVERSE);

   cuDoubleComplex *h_signal_inv =(cuDoubleComplex *) malloc(sizeof(cuDoubleComplex) * SIGNAL_SIZE);
   cudaMemcpy(h_signal_inv, d_signal, sizeof(cuDoubleComplex) * SIGNAL_SIZE, cudaMemcpyDeviceToHost);


   cudaEventRecord(stop, 0);
   cudaEventSynchronize(stop);

   float elapsedTime;
   cudaEventElapsedTime(&elapsedTime, start, stop);
   printf("Elapsed Time:  %3.1f ms\n", elapsedTime);


    for(int i=0;i<SIGNAL_SIZE;i++) printf("\n%f %f", h_signal[i].x, h_signal_inv[i].x);

    cufftDestroy(plan);

   free(h_signal);
   free(h_signal_inv);

   cudaFree(d_signal);

   cudaDeviceReset();
   return 0;
}

我想转换一个信号，然后反过来回来，但是上半年的输出是错误的。

你能帮我找到错误吗？

非常感谢！

Answer 1

您的数据类型很混乱。

cufftDoubleComplex与cufftComplex不同。使用cufftDoubleComplex，your transform type should be Z2Z, not C2C。

时

另外，为了在进行正向变换后使用CUFFT进行逆变换时看到数据奇偶校验，它是necessary to divide the result by the signal size：

cuFFT执行非标准化FFT;也就是说，对输入数据集执行前向FFT，然后对得到的集合进行逆FFT，产生等于输入的数据，按元素数量进行缩放。通过数据集大小的倒数来缩放变换，留给用户按照适合的方式执行。

以下代码解决了上述问题，应该会给您带来更好的结果：

#include <stdlib.h>
#include <stdio.h>
#include <string.h>
#include <math.h>
#include <cufft.h>
#include <cuComplex.h>

#define SIGNAL_SIZE        1024

int main(int argc, char **argv) {
   cudaEvent_t start, stop;
   cudaEventCreate(&start);
   cudaEventCreate(&stop);

   // Allocate host memory for the signal
   cuDoubleComplex *h_signal = (cuDoubleComplex *) malloc(sizeof(cuDoubleComplex) * SIGNAL_SIZE);

   // Initalize the memory for the signal
   for (unsigned int i = 0; i < SIGNAL_SIZE; ++i) {
      if((double)i/SIGNAL_SIZE>=0 && (double)i/SIGNAL_SIZE<0.5)  h_signal[i].x = (double)i/SIGNAL_SIZE;
      else if((double)i/SIGNAL_SIZE>=0.5 && (double)i/SIGNAL_SIZE<1)  h_signal[i].x = (double)i/SIGNAL_SIZE-1;
      h_signal[i].y = 0;
   }

// Allocate device memory for signal
   cuDoubleComplex *d_signal;

   cudaMalloc((void **) &d_signal, SIGNAL_SIZE*sizeof(cuDoubleComplex));
   // Copy host memory to device
   cudaMemcpy(d_signal, h_signal, SIGNAL_SIZE*sizeof(cuDoubleComplex), cudaMemcpyHostToDevice);


cudaEventRecord(start, 0);
   cufftHandle plan;
   cufftPlan1d(&plan, SIGNAL_SIZE , CUFFT_Z2Z, 1);

   // FFT computation
   cufftExecZ2Z(plan, d_signal, d_signal, CUFFT_FORWARD);

    cufftExecZ2Z(plan, d_signal, d_signal, CUFFT_INVERSE);

   cuDoubleComplex *h_signal_inv =(cuDoubleComplex *) malloc(sizeof(cuDoubleComplex) * SIGNAL_SIZE);
   cudaMemcpy(h_signal_inv, d_signal, sizeof(cuDoubleComplex) * SIGNAL_SIZE, cudaMemcpyDeviceToHost);


   cudaEventRecord(stop, 0);
   cudaEventSynchronize(stop);

   float elapsedTime;
   cudaEventElapsedTime(&elapsedTime, start, stop);
   printf("Elapsed Time:  %3.1f ms\n", elapsedTime);


    for(int i=0;i<SIGNAL_SIZE;i++) printf("\n%f %f", h_signal[i].x, h_signal_inv[i].x/SIGNAL_SIZE);

    cufftDestroy(plan);

   free(h_signal);
   free(h_signal_inv);

   cudaFree(d_signal);

   cudaDeviceReset();
   return 0;
}

cuFFT输出不正确

1 个答案: