c++ - 与fftw3相比，错误的2D CuFFT逆变换

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我正在尝试产生一些FFT数学，特别是它执行两个2D正向变换，将它们相乘，然后进行逆变换。在逆变换之前，一切都很好。我已经用fftw3做到了，但是在CuFFT中出了点问题。大多数值相似，但有些错误，这对将来的数学意义重大。这段代码有什么问题？

std::vector<complex> conv2dCUDA(complex *ui_anomaly, double *ds2,
complex *u0, int anx, int any, double factor) {
    cufftComplex *b1, *b2;
    int size = 2 * anx * 2 * any;
    int memsize = size *  sizeof(cufftComplex);
    b1 = (cufftComplex *)calloc(size, sizeof(cufftComplex));
    b2 = (cufftComplex *)calloc(size, sizeof(cufftComplex));

    // filling the matrixes

    cufftHandle plan;
    cufftComplex *ui, *g;
    checkCudaErrors(cudaMalloc((void**)&ui, memsize));
    checkCudaErrors(cudaMalloc((void**)&g,  memsize));
    checkCudaErrors(cufftPlan2d(&plan, 2 * anx, 2 * any, CUFFT_C2C));
    checkCudaErrors(cudaMemcpy(ui, (cufftComplex *)&b1[0], memsize, cudaMemcpyHostToDevice));
    checkCudaErrors(cudaMemcpy(g, (cufftComplex *)&b2[0], memsize, cudaMemcpyHostToDevice));
    checkCudaErrors(cufftExecC2C(plan, ui, ui, CUFFT_FORWARD));
    checkCudaErrors(cufftExecC2C(plan, g, g, CUFFT_FORWARD));
    int blockSize = 16;
    dim3 dimGrid(int(2 * any / blockSize) + 1, int(2 * anx / blockSize) + 1);
    dim3 dimBlock(blockSize, blockSize);

    ComplexMulAndScale<<<dimGrid, dimBlock>>>(ui, g, size, 1.0f);
    getLastCudaError("Kernel execution went wrong");

    checkCudaErrors(cudaMemcpy(b1, ui, memsize, cudaMemcpyDeviceToHost));
    std::cout << "After mult Cuda" << std::endl;
    for (auto i = 0; i < 2 * any; i++) {
       for (auto j = 0; j < 2 * anx; j++) {
            std::cout << b1[i * 2 * anx + j].x << " ";
       }
       std::cout << std::endl;
    }

    checkCudaErrors(cufftExecC2C(plan, ui, ui, CUFFT_INVERSE));
    cuComplex *inversed;
    inversed = (cuComplex*)malloc(memsize);
    checkCudaErrors(cudaMemcpy(inversed, ui, memsize, cudaMemcpyDeviceToHost));
    std::vector<complex> res(anx * any);
    for (auto i = 0; i < any; i++) {
        for (auto j = 0; j < anx; j++) {
            res[i * anx + j] = complex(inversed[i * anx * 2 + j].x * factor, inversed[i * anx * 2 + j].y * factor);
        }
    }
    std::cout << "CUDA"  << std::endl;
    for (auto i = 0; i < 2 * any; i++) {
        for (auto j = 0; j < 2 * anx; j++) {
            std::cout << inversed[i * 2 * anx + j].x << " ";
        }
        std::cout << std::endl;
    }
    checkCudaErrors(cudaFree(ui));
    checkCudaErrors(cudaFree(g));
    checkCudaErrors(cufftDestroy(plan));
    free(b1);
    free(b2);
    free(inversed);
    return res;
}

std::vector<complex> conv2d(complex *ui_anomaly, double *ds2, complex *u0, int anx, int any, double factor) {
    std::vector<complex> b1(anx * 2 * 2 * any, complex(0., 0.)), b2(anx * 2 * 2 * any, complex(0., 0.));

    // filling matrixes

    // forward fft 1 in-place
    fftw_plan p;
    p = fftw_plan_dft_2d(2 * any, 2 * anx, (fftw_complex *) (&b1[0]), (fftw_complex *) (&b1[0]),
                     FFTW_FORWARD, FFTW_ESTIMATE);
    fftw_execute(p);
    fftw_destroy_plan(p);
    // forward fft 2 in-place
    p = fftw_plan_dft_2d(2 * any, 2 * anx, (fftw_complex *) (&b2[0]), (fftw_complex *) (&b2[0]),
                     FFTW_FORWARD, FFTW_ESTIMATE);
    fftw_execute(p);
    fftw_destroy_plan(p);
    std::vector<complex> out(2 * anx * 2 * any, complex(0., 0.));

    for (auto i = 0; i < 2 * any * 2 * anx; i++) {
        out[i] = b1[i] * b2[i];
    }
    std::cout << "After mult fftw" << std::endl;
    for (auto i = 0; i < 2 * any; i++) {
        for (auto j = 0; j < 2 * anx; j++) {
            std::cout << out[i * 2 * anx + j].real() << " ";
       }
       std::cout << std::endl;
    }
    // inverse fft in-place
    p = fftw_plan_dft_2d(2 * (int) any, 2 * (int) anx, (fftw_complex *) (&out[0]), (fftw_complex *) (&out[0]),FFTW_BACKWARD, FFTW_ESTIMATE);
    fftw_execute(p);
    fftw_destroy_plan(p);

    std::vector<complex> res(anx * any);
    for (auto i = 0; i < any; i++) {
        for (auto j = 0; j < anx; j++) {
            res[i * anx + j] = out[i * anx * 2 + j] * factor;
       }
    }
    std::cout << "FFTW" << std::endl;
    for (auto i = 0; i < 2 * any; i++) {
        for (auto j = 0; j < 2 * anx; j++) {
            std::cout << out[i * 2 * anx + j].real() << " ";
        }
        std::cout << std::endl;
    }
    return res;
}

所以，这是我的代码。输出应同时在两个函数中

After mult fftw
8.34304e-08 -5.99259e-07 -4.7876e-07 5.30254e-07 9.55877e-07 4.28985e-07
-1.56375e-07 1.19699e-07 2.39276e-07 -1.68662e-08 -7.56988e-08 -3.69897e-07
-2.66505e-07 -2.33361e-07 -5.21763e-07 -5.29126e-07 1.8915e-07 1.68158e-07
-9.01859e-07 -2.37453e-07 -3.50661e-08 -4.11154e-07 4.14802e-07 -7.9879e-07
2.09404e-07 6.52034e-08 1.8915e-07 4.97805e-07 3.32612e-07 -2.33361e-07
-1.95738e-07 -3.69897e-07 -1.63577e-07 1.07737e-07 2.39276e-07 2.50198e-07
FFTW
-1.57349e-06 -7.5964e-06 -1.57349e-06 1.68876e-06 5.82335e-22 1.68876e-06
2.37158e-06 6.35275e-22 2.37158e-06 -1.18579e-06 1.05879e-22 -1.18579e-06
-1.57349e-06 -7.5964e-06 -1.57349e-06 1.68876e-06 1.97573e-22 1.68876e-06
3.14928e-06 2.37158e-06 3.14928e-06 -4.94164e-07 5.82335e-22 -4.94164e-07
2.11758e-22 -8.47033e-22 -1.05879e-22 5.29396e-22 1.41851e-23 1.05879e-22
3.14928e-06 2.37158e-06 3.14928e-06 -4.94164e-07 1.05879e-22 -4.94164e-07

After mult Cuda
8.34303e-08 -5.99259e-07 -4.78761e-07 5.30254e-07 9.55877e-07 4.28985e-07
-1.56375e-07 1.19699e-07 2.39276e-07 -1.68662e-08 -7.56988e-08 -3.69897e-07
-2.66505e-07 -2.33361e-07 -5.21763e-07 -5.29126e-07 1.8915e-07 1.68158e-07
-9.01859e-07 -2.37453e-07 -3.50661e-08 -4.11154e-07 4.14802e-07 -7.9879e-07
2.09404e-07 6.52034e-08 1.8915e-07 4.97805e-07 3.32612e-07 -2.33361e-07
-1.95738e-07 -3.69897e-07 -1.63577e-07 1.07737e-07 2.39276e-07 2.50198e-07
CUDA
-1.57349e-06 -7.5964e-06 -1.57349e-06 1.68876e-06 3.33981e-13 1.68876e-06
2.37158e-06 2.84217e-13 2.37158e-06 -1.18579e-06 1.10294e-13 -1.18579e-06
-1.57349e-06 -7.5964e-06 -1.57349e-06 1.68876e-06 -9.03043e-14 1.68876e-06
3.14928e-06 2.37158e-06 3.14928e-06 -4.94164e-07 4.62975e-13 -4.94164e-07
-3.2685e-13 -1.03562e-13 -3.59548e-13 -2.13163e-13 4.27658e-15 -2.43358e-14
3.14928e-06 2.37158e-06 3.14928e-06 -4.94164e-07 3.49288e-13 -4.94164e-07

可以看出，正向fft和乘法都正确，但是在cuda smth反向fft的情况下出错。

附言对不起，代码风格差

最佳答案

由于使用了FFTW，卷积信号的特征是数量众多，约为1e-6，少数数量约为1e-22。可能应该是零，但不是这样，因为这些零是使用双精度浮点数计算的。双精度数字的精度约为15位数字，因此可能会发生1e（-6-15）= 1e-21附近的错误。

使用cufft时，这些应该为零的数字约为1e-13，就好像使用单精度浮点数执行了计算一样。
确实是这样：类型cuComplex和cufftComplex是单精度复数，而fftw_complex是双精度复数。在complex likely defaults to double precision时，可以明确地将其指定为double complex。

要获得1e-22附近的数字，请尝试输入cufftDoubleComplex和cuDoubleComplex类型。引入的一次执行许多乘法的blockSize可能需要减少到8。但是，虽然很可能获得幅度为1e-22的数字，但这些数字仍可能与FFTW的数字有所不同。确实，由于算法可能不同，因此可能执行了不同的操作，并且精确度使得结果中大约1e-22的值与0的差别不大。

但是，更改为双精度数字可能会增加计算时间并明显增加内存占用量。如果卷积结果的六位数精度足以满足您的应用程序的要求，那么坚持使用单精度复杂DFT可能是正确的方法。

关于c++ - 与fftw3相比，错误的2D CuFFT逆变换，我们在Stack Overflow上找到一个类似的问题：https://stackoverflow.com/questions/57690934/