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我正在尝试使用 cufft 库为 fft 1d 转换编写一个简单的代码。

我以 cufft 库教程（链接）上的代码为例，但转换前和逆变换后的数据不一样。为什么 ？

这是输出：

[1DCUFFT] starting...

[1DCUFFT] is starting...
Transforming signal cufftExecC2C
Transforming signal back cufftExecC2C
first : 0.840188 0.000000  after 2.020520 -2.465333
first : 0.394383 0.000000  after 2.690347 -2.105700
first : 0.783099 0.000000  after 3.155561 -1.491952
first : 0.798440 0.000000  after 3.309971 -0.761629
first : 0.911647 0.000000  after 3.139909 -0.092953
first : 0.197551 0.000000  after 2.734147 0.355526
first : 0.335223 0.000000  after 2.256154 0.501509
first : 0.768230 0.000000  after 1.887422 0.369626
first : 0.277775 0.000000  after 1.762471 0.083391
first : 0.553970 0.000000  after 1.920150 -0.179526
first : 0.477397 0.000000  after 2.289955 -0.252255
first : 0.628871 0.000000  after 2.718623 -0.039409
first : 0.364784 0.000000  after 3.026649 0.449287
first : 0.513401 0.000000  after 3.072947 1.104457
first : 0.952230 0.000000  after 2.803322 1.758350
first : 0.916195 0.000000  after 2.265463 2.245056

这是我的源代码，教程示例建议将数据集大小的输出划分为具有相同的值。

// includes, system
#include <stdlib.h>
#include <stdio.h>
#include <string.h>
#include <math.h>

// includes, project
#include <cufft.h>
#include <cutil_inline.h>
#include <shrQATest.h>
void runTest(int argc, char** argv);

// The filter size is assumed to be a number smaller than the signal size
#define SIGNAL_SIZE        16

////////////////////////////////////////////////////////////////////////////////
// Program main
////////////////////////////////////////////////////////////////////////////////
int main(int argc, char** argv) 
{
    runTest(argc, argv);
}

////////////////////////////////////////////////////////////////////////////////
//! Run a simple test for CUDA
////////////////////////////////////////////////////////////////////////////////
void runTest(int argc, char** argv) 
{

    printf("[1DCUFFT] is starting...\n");

    if( cutCheckCmdLineFlag(argc, (const char**)argv, "device") )
        cutilDeviceInit(argc, argv);
    else
        cudaSetDevice( cutGetMaxGflopsDeviceId() );
    cufftComplex* h_signal=(cufftComplex*)malloc(sizeof(cufftComplex) * SIGNAL_SIZE);
    // Allocate host memory for the signal
    //Complex* h_signal = (Complex*)malloc(sizeof(Complex) * SIGNAL_SIZE);
    // Initalize the memory for the signal
    for (unsigned int i = 0; i < SIGNAL_SIZE; ++i) {
        h_signal[i].x = rand() / (float)RAND_MAX;
        h_signal[i].y = 0;
    }




    int mem_size = sizeof(cufftComplex) * SIGNAL_SIZE;

    // Allocate device memory for signal
    cufftComplex* d_signal;
    cutilSafeCall(cudaMalloc((void**)&d_signal, mem_size));

    // Copy host memory to device
    cutilSafeCall(cudaMemcpy(d_signal, h_signal, mem_size,
                              cudaMemcpyHostToDevice));



    // CUFFT plan
    cufftHandle plan;
    cufftSafeCall(cufftPlan1d(&plan, mem_size, CUFFT_C2C, 1));

    // Transform signal 
    printf("Transforming signal cufftExecC2C\n");
    cufftSafeCall(cufftExecC2C(plan, (cufftComplex *)d_signal, (cufftComplex *)d_signal, CUFFT_FORWARD));


    // Transform signal back
    printf("Transforming signal back cufftExecC2C\n");
    cufftSafeCall(cufftExecC2C(plan, (cufftComplex *)d_signal, (cufftComplex *)d_signal, CUFFT_INVERSE));

    // Copy device memory to host
    cufftComplex* h_inverse_signal = (cufftComplex*)malloc(sizeof(cufftComplex) * SIGNAL_SIZE);;
    cutilSafeCall(cudaMemcpy(h_inverse_signal, d_signal, mem_size,
                              cudaMemcpyDeviceToHost));

    for(int i=0;i< SIGNAL_SIZE;i++){
        h_inverse_signal[i].x= h_inverse_signal[i].x/(float)SIGNAL_SIZE;
        h_inverse_signal[i].y= h_inverse_signal[i].y/(float)SIGNAL_SIZE;

        printf("first : %f %f  after %f %f \n",h_signal[i].x,h_signal[i].y,h_inverse_signal[i].x,h_inverse_signal[i].y);
    }  



    //Destroy CUFFT context
    cufftSafeCall(cufftDestroy(plan));

    // cleanup memory
    free(h_signal);

    free(h_inverse_signal);
    cutilSafeCall(cudaFree(d_signal));
    cutilDeviceReset();
}
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1 回答 1

4

你需要改变：

cufftSafeCall(cufftPlan1d(&plan, mem_size, CUFFT_C2C, 1));

到：

cufftSafeCall(cufftPlan1d(&plan, SIGNAL_SIZE, CUFFT_C2C, 1));

（cufftPlan1d需要数据点的数量，而不是内存大小（以字节为单位！）

于 2012-07-06T15:31:08.513 回答