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c++ - 使用 cuda cuFFT 从复数转换为实数时输出不正确

转载 作者:行者123 更新时间:2023-11-30 01:09:32 27 4
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我正在使用 cuda 版本 7.5 cufft 来执行一些 FFT 和逆 FFT。使用 cufftExecC2R(.,.) 函数执行逆 FFT 时遇到问题。

实际上,当我在 cufftPlan1d(,) 中使用 batch_size = 1 时,我得到了正确的结果。但是,当我增加批量大小时,结果不正确。

我正在粘贴一个示例最小代码来说明这一点。请忽略代码的肮脏,因为我刚刚快速创建了它。

  #include <cufft.h>
#include <stdlib.h>
#include <stdio.h>
#include <string.h>
#include <math.h>
#include <ctime>
#include <iostream>

typedef float2 Complex;

void iTest(int argc, char** argv);

#define SIGNAL_SIZE 9
#define BATCH_SIZE 2

int main(int argc, char** argv) {

iTest(argc, argv);
return 0;

}

void iProcess(Complex *x, double *y, size_t n) {

cufftComplex *deviceData;
cudaMalloc(reinterpret_cast<void**>(&deviceData),
SIGNAL_SIZE * BATCH_SIZE * sizeof(cufftComplex));
cudaMemcpy(deviceData, x, SIGNAL_SIZE * sizeof(cufftComplex) * BATCH_SIZE,
cudaMemcpyHostToDevice);

cufftResult cufftStatus;
cufftHandle handle;
cufftStatus = cufftPlan1d(&handle, SIGNAL_SIZE, CUFFT_C2C, BATCH_SIZE);
if (cufftStatus != cudaSuccess) {
printf("cufftPlan1d failed!");
}

cufftComplex *d_complex;
cudaMalloc(reinterpret_cast<void**>(&d_complex),
sizeof(cufftComplex) * SIGNAL_SIZE * BATCH_SIZE);

cufftStatus = cufftExecC2C(handle, deviceData, d_complex, CUFFT_FORWARD);
if (cufftStatus != cudaSuccess) {
printf("cufftExecR2C failed!");
}

cufftComplex *hostOutputData = (cufftComplex*)malloc(
(SIGNAL_SIZE) * BATCH_SIZE * sizeof(cufftComplex));

cudaMemcpy(hostOutputData, d_complex,
SIGNAL_SIZE * sizeof(cufftComplex) * BATCH_SIZE,
cudaMemcpyDeviceToHost);

std::cout << "\nPrinting COMPLEX" << "\n";
for (int j = 0; j < (SIGNAL_SIZE) * BATCH_SIZE; j++)
printf("%i \t %f \t %f\n", j, hostOutputData[j].x, hostOutputData[j].y);


//! convert complex to real

cufftHandle c2r_handle;
cufftStatus = cufftPlan1d(&c2r_handle, SIGNAL_SIZE, CUFFT_C2R, BATCH_SIZE);
if (cufftStatus != cudaSuccess) {
printf("cufftPlan1d failed!");
}

cufftReal *d_odata;
cudaMalloc(reinterpret_cast<void**>(&d_odata),
sizeof(cufftReal) * SIGNAL_SIZE * BATCH_SIZE);
cufftStatus = cufftExecC2R(c2r_handle, d_complex, d_odata);

cufftReal odata[SIGNAL_SIZE * BATCH_SIZE];
cudaMemcpy(odata, d_odata, sizeof(cufftReal) * SIGNAL_SIZE * BATCH_SIZE,
cudaMemcpyDeviceToHost);

std::cout << "\nPrinting REAL" << "\n";
for (int i = 0; i < SIGNAL_SIZE * BATCH_SIZE; i++) {
std::cout << i << " \t" << odata[i]/(SIGNAL_SIZE) << "\n";
}


cufftDestroy(handle);
cudaFree(deviceData);
}

void iTest(int argc, char** argv) {

Complex* h_signal = reinterpret_cast<Complex*>(
malloc(sizeof(Complex) * SIGNAL_SIZE * BATCH_SIZE));

std::cout << "\nPrinting INPUT" << "\n";
for (unsigned int i = 0; i < SIGNAL_SIZE * BATCH_SIZE; ++i) {
h_signal[i].x = rand() / static_cast<float>(RAND_MAX);
h_signal[i].y = 0;

std::cout << i << "\t" << h_signal[i].x << "\n";
}
std::cout << "\n";

double y[SIGNAL_SIZE * BATCH_SIZE];
iProcess(h_signal, y, 1);

}

我无法找出代码中的错误在哪里以及缺少哪些信息。

使用 BATCH_SIZE = 1 时的示例输出

Image 1

使用 BATCH_SIZE = 2 时的示例输出 image 2

最佳答案

您缺少的信息是您不了解 C2C 转换与 C2R(或 R2C)预期的输入数据存在数据格式差异。

您应该先阅读 this sectionthis section CUFFT 文档。

注意它说:

Each of those functions demands different input data layout

但是您将对 C2C 转换正确的输入数据直接传递给 C2R 转换。那是行不通的。

IMO 最直接的解决方案是将您的所有工作转换为 C2C 转换类型。 C2C 转换可以支持正向(例如“实到复杂”)和反向(例如“复杂到实”)。您使用的 C2R 转换类型也可以支持“从复杂到真实”,但是您将用于 C2R 的数据排列不同与您将用于 C2C 的数据排列具有指定的反向路径,对于其他方面相同的转换。你没有考虑到这一点。

这是一个工作示例,显示了代码的修改版本,它对正向和反向路径使用 C2C,并正确地再现了批量大小为 2 的输入:

$ cat t19.cu
#include <cufft.h>
#include <stdlib.h>
#include <stdio.h>
#include <string.h>
#include <math.h>
#include <ctime>
#include <iostream>

typedef float2 Complex;

void iTest(int argc, char** argv);

#define SIGNAL_SIZE 9
#define BATCH_SIZE 2

int main(int argc, char** argv) {

iTest(argc, argv);
return 0;

}

void iProcess(Complex *x, double *y, size_t n) {

cufftComplex *deviceData;
cudaMalloc(reinterpret_cast<void**>(&deviceData),
SIGNAL_SIZE * BATCH_SIZE * sizeof(cufftComplex));
cudaMemcpy(deviceData, x, SIGNAL_SIZE * sizeof(cufftComplex) * BATCH_SIZE,
cudaMemcpyHostToDevice);

cufftResult cufftStatus;
cufftHandle handle;
cufftStatus = cufftPlan1d(&handle, SIGNAL_SIZE, CUFFT_C2C, BATCH_SIZE);
if (cufftStatus != cudaSuccess) {
printf("cufftPlan1d failed!");
}

cufftComplex *d_complex;
cudaMalloc(reinterpret_cast<void**>(&d_complex),
sizeof(cufftComplex) * SIGNAL_SIZE * BATCH_SIZE);

cufftStatus = cufftExecC2C(handle, deviceData, d_complex, CUFFT_FORWARD);
if (cufftStatus != cudaSuccess) {
printf("cufftExecR2C failed!");
}

cufftComplex *hostOutputData = (cufftComplex*)malloc(
(SIGNAL_SIZE) * BATCH_SIZE * sizeof(cufftComplex));

cudaMemcpy(hostOutputData, d_complex,
SIGNAL_SIZE * sizeof(cufftComplex) * BATCH_SIZE,
cudaMemcpyDeviceToHost);

std::cout << "\nPrinting COMPLEX" << "\n";
for (int j = 0; j < (SIGNAL_SIZE) * BATCH_SIZE; j++)
printf("%i \t %f \t %f\n", j, hostOutputData[j].x, hostOutputData[j].y);


//! convert complex to real

/* cufftHandle c2r_handle;
cufftStatus = cufftPlan1d(&c2r_handle, SIGNAL_SIZE, CUFFT_C2R, BATCH_SIZE);
if (cufftStatus != cudaSuccess) {
printf("cufftPlan1d failed!");
}
*/
cufftComplex *d_odata;
cudaMalloc(reinterpret_cast<void**>(&d_odata),
sizeof(cufftComplex) * SIGNAL_SIZE * BATCH_SIZE);
cufftStatus = cufftExecC2C(handle, d_complex, d_odata, CUFFT_INVERSE);

cufftComplex odata[SIGNAL_SIZE * BATCH_SIZE];
cudaMemcpy(odata, d_odata, sizeof(cufftComplex) * SIGNAL_SIZE * BATCH_SIZE,
cudaMemcpyDeviceToHost);

std::cout << "\nPrinting REAL" << "\n";
for (int i = 0; i < SIGNAL_SIZE * BATCH_SIZE; i++) {
std::cout << i << " \t" << odata[i].x/(SIGNAL_SIZE) << "\n";
}


cufftDestroy(handle);
cudaFree(deviceData);
}

void iTest(int argc, char** argv) {

Complex* h_signal = reinterpret_cast<Complex*>(
malloc(sizeof(Complex) * SIGNAL_SIZE * BATCH_SIZE));

std::cout << "\nPrinting INPUT" << "\n";
for (unsigned int i = 0; i < SIGNAL_SIZE * BATCH_SIZE; ++i) {
h_signal[i].x = rand() / static_cast<float>(RAND_MAX);
h_signal[i].y = 0;

std::cout << i << "\t" << h_signal[i].x << "\n";
}
std::cout << "\n";

double y[SIGNAL_SIZE * BATCH_SIZE];
iProcess(h_signal, y, 1);

}
$ nvcc -arch=sm_61 -o t19 t19.cu -lcufft
t19.cu: In function ‘void iProcess(Complex*, double*, size_t)’:
t19.cu:34:32: warning: comparison between ‘cufftResult {aka enum cufftResult_t}’ and ‘enum cudaError’ [-Wenum-compare]
if (cufftStatus != cudaSuccess) {
^
t19.cu:43:32: warning: comparison between ‘cufftResult {aka enum cufftResult_t}’ and ‘enum cudaError’ [-Wenum-compare]
if (cufftStatus != cudaSuccess) {
^
$ cuda-memcheck ./t19
========= CUDA-MEMCHECK

Printing INPUT
0 0.840188
1 0.394383
2 0.783099
3 0.79844
4 0.911647
5 0.197551
6 0.335223
7 0.76823
8 0.277775
9 0.55397
10 0.477397
11 0.628871
12 0.364784
13 0.513401
14 0.95223
15 0.916195
16 0.635712
17 0.717297


Printing COMPLEX
0 5.306536 0.000000
1 0.015338 -0.734991
2 -0.218001 0.740248
3 0.307508 -0.706533
4 1.022732 0.271765
5 1.022732 -0.271765
6 0.307508 0.706533
7 -0.218001 -0.740248
8 0.015338 0.734991
9 5.759857 0.000000
10 -0.328981 0.788566
11 0.055356 -0.521014
12 -0.127504 0.581872
13 0.014066 0.123027
14 0.014066 -0.123027
15 -0.127504 -0.581872
16 0.055356 0.521014
17 -0.328981 -0.788566

Printing REAL
0 0.840188
1 0.394383
2 0.783099
3 0.79844
4 0.911647
5 0.197551
6 0.335223
7 0.76823
8 0.277775
9 0.55397
10 0.477397
11 0.628871
12 0.364784
13 0.513401
14 0.95223
15 0.916195
16 0.635712
17 0.717297
========= ERROR SUMMARY: 0 errors
$

关于c++ - 使用 cuda cuFFT 从复数转换为实数时输出不正确,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/40196288/

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