使用FFT(FFTW)计算两个函数的卷积

Question

我正在尝试使用FFT加速神经模拟器的计算.

等式是:

(1)\ sum(j = 1到N)(w(i-j)*s_NMDA [j])

其中s_NMDA是长度为N的向量,w由以下定义:

(2)W(j)的双曲正切= [1 /(2*西格玛*P)]*EXP(-abs(J)/(西格玛*P)]

sigma和p是常量.

(有没有更好的方法在stackoverflow上呈现方程？)

必须对N个神经元进行计算.由于(1)仅取决于绝对距离abs(i-j),因此应该可以使用FFT(卷积定理)来计算它.

我试图使用FFTW实现这一点,但结果与预期结果不符.我之前从未使用过FFTW,现在我不确定如果我对卷积定理的假设是假的,那么我的实现是不正确的.

void f_I_NMDA_FFT(
    const double     **states, // states[i][6] == s_NMDA[i]
    const unsigned int numNeurons)
{
    fftw_complex *distances, *sNMDAs, *convolution;
    fftw_complex *distances_f, *sNMDAs_f, *convolution_f;
    fftw_plan     p, pinv;
    const double scale = 1./numNeurons;

        distances = (fftw_complex *)fftw_malloc(sizeof(fftw_complex) * numNeurons);
        sNMDAs    = (fftw_complex *)fftw_malloc(sizeof(fftw_complex) * numNeurons);
        convolution = (fftw_complex *)fftw_malloc(sizeof(fftw_complex) * numNeurons);
        distances_f = (fftw_complex *)fftw_malloc(sizeof(fftw_complex) * numNeurons);
        sNMDAs_f    = (fftw_complex *)fftw_malloc(sizeof(fftw_complex) * numNeurons);
        convolution_f    = (fftw_complex *)fftw_malloc(sizeof(fftw_complex) * numNeurons);

        // fill input array for distances  
    for (unsigned int i = 0; i < numNeurons; ++i)
    {
        distances[i][0] = w(i);
        distances[i][1] = 0;
    }

        // fill input array for sNMDAs
    for (unsigned int i = 0; i < numNeurons; ++i)
    {
        sNMDAs[i][0] = states[i][6];
        sNMDAs[i][1] = 0;
    }

    p = fftw_plan_dft_1d(numNeurons,
                         distances,
                         distances_f,
                         FFTW_FORWARD,
                         FFTW_ESTIMATE);
    fftw_execute(p);

    p = fftw_plan_dft_1d(numNeurons,
                         sNMDAs,
                         sNMDAs_f,
                         FFTW_FORWARD,
                         FFTW_ESTIMATE);
    fftw_execute(p);

    // convolution in frequency domain
    for(unsigned int i = 0; i < numNeurons; ++i)
    {
        convolution_f[i][0] = (distances_f[i][0] * sNMDAs_f[i][0]
            - distances_f[i][1] * sNMDAs_f[i][1]) * scale;
        convolution_f[i][1] = (distances_f[i][0] * sNMDAs_f[i][1]
            - distances_f[i][1] * sNMDAs_f[i][0]) * scale;
    }

    pinv = fftw_plan_dft_1d(numNeurons,
                            convolution_f,
                            convolution,
                            FFTW_FORWARD,
                            FFTW_ESTIMATE);
    fftw_execute(pinv);

    // compute and compare with expected result
    for (unsigned int i = 0; i < numNeurons; ++i)
    {
            double expected = 0;

            for (int j = 0; j < numNeurons; ++j)
            {
                expected += w(i - j) * states[j][6];
            }
            printf("i=%d, FFT: r%f, i%f : Expected: %f\n", i, convolution[i][0], convolution[i][1], expected);
    }

    fftw_destroy_plan(p);
    fftw_destroy_plan(pinv);

    fftw_free(distances), fftw_free(sNMDAs), fftw_free(convolution);
    fftw_free(distances_f), fftw_free(sNMDAs_f), fftw_free(convolution_f);

这是20个神经元的示例输出:

i=0, FFT: r0.042309, i0.000000 : Expected: 0.041504
i=1, FFT: r0.042389, i0.000000 : Expected: 0.042639
i=2, FFT: r0.042466, i0.000000 : Expected: 0.043633
i=3, FFT: r0.042543, i0.000000 : Expected: 0.044487
i=4, FFT: r0.041940, i0.000000 : Expected: 0.045203
i=5, FFT: r0.041334, i0.000000 : Expected: 0.045963
i=6, FFT: r0.041405, i0.000000 : Expected: 0.046585
i=7, FFT: r0.041472, i0.000000 : Expected: 0.047070
i=8, FFT: r0.041537, i0.000000 : Expected: 0.047419
i=9, FFT: r0.041600, i0.000000 : Expected: 0.047631
i=10, FFT: r0.041660, i0.000000 : Expected: 0.047708
i=11, FFT: r0.041717, i0.000000 : Expected: 0.047649
i=12, FFT: r0.041773, i0.000000 : Expected: 0.047454
i=13, FFT: r0.041826, i0.000000 : Expected: 0.047123
i=14, FFT: r0.041877, i0.000000 : Expected: 0.046656
i=15, FFT: r0.041926, i0.000000 : Expected: 0.046052
i=16, FFT: r0.041294, i0.000000 : Expected: 0.045310
i=17, FFT: r0.042059, i0.000000 : Expected: 0.044430
i=18, FFT: r0.042144, i0.000000 : Expected: 0.043412
i=19, FFT: r0.042228, i0.000000 : Expected: 0.042253

结果似乎几乎是正确的,但误差随着神经元的数量而增加.而且,对于非常低或非常高的位置(i),结果似乎更准确.这里发生了什么？

更新:正如Oli Charlesworth所建议的那样,我在八度音程中实现了算法,看它是否是一个实现或数学问题:

input = [0.186775; 0.186775; 0.186775; 0.186775; 0.186775; 0; 0.186775; 0.186775; 0.186775; 0.186775];

function ret = _w(i)
  ret = tanh(1 / (2* 1 * 32)) * exp(-abs(i) / (1 * 32));
end

for i = linspace(1, 10, 10)
  expected = 0;
  for j = linspace(1, 10, 10)
    expected += _w(i-j) * input(j);
  end
  expected
end

distances = _w(transpose(linspace(0, 9, 10)));

input_f = fft(input);
distances_f = fft(distances);

convolution_f = input_f .* distances_f;

convolution = ifft(convolution_f)

结果:

expected =  0.022959
expected =  0.023506
expected =  0.023893
expected =  0.024121
expected =  0.024190
expected =  0.024100
expected =  0.024034
expected =  0.023808
expected =  0.023424
expected =  0.022880
convolution =

   0.022959
   0.023036
   0.023111
   0.023183
   0.023253
   0.022537
   0.022627
   0.022714
   0.022798
   0.022880

结果非常相似.因此,我对卷积定理/ FFT的理解肯定有问题.

Answer 1

要通过FFT卷积2个信号,通常需要这样做:

1. 根据需要为每个信号添加尽可能多的零,使其长度成为原始信号的累积长度 - 1(即卷积结果的长度).
2. 如果您的FFT库要求输入长度为2的幂,则为每个信号添加尽可能多的零以满足该要求.
3. 计算信号1的DFT(通过FFT).
4. 计算信号2的DFT(通过FFT).
5. 将两个DFT按元素相乘.它应该是一个复杂的乘法,顺便说一句.
6. 计算乘法DFT的逆DFT(通过FFT).这将是你的卷积结果.

在我的代码中,我看到FFTW_FORWARD了所有3个FFT.我猜这不是问题,它是它的一部分.最后一个FFT应该是"向后",而不是"向前".

另外,我认为你在第二个表达式中需要"+",而不是" - ":

convolution_f[i][0] = (distances_f[i][0] * sNMDAs_f[i][0]
            - distances_f[i][1] * sNMDAs_f[i][1]) * scale;

convolution_f[i][1] = (distances_f[i][0] * sNMDAs_f[i][1]
            - distances_f[i][1] * sNMDAs_f[i][0]) * scale;