abs() 为不同的 FFT 输入返回相同的输出

Question

我有 1024 个样本，为了对其执行 FFT，我将其分成 32 个块，以下是 FFT 的输出：

(3.13704,2.94588) (12.9193,14.7706) (-4.4401,-6.21331) (-1.60103,-2.78147) (-0.84114,-1.86292) (-0.483564,-1.43068) (-0.272469,-1.17551) (-0.130891,-1.00437) (-0.0276415,-0.879568) (0.0523422,-0.782884) (0.117249,-0.704425) (0.171934,-0.638322) (0.219483,-0.580845) (0.261974,-0.529482) (0.300883,-0.48245) (0.337316,-0.438409) (0.372151,-0.396301) (0.40613,-0.355227) (0.439926,-0.314376) (0.474196,-0.27295) (0.509637,-0.23011) (0.54704,-0.184897) (0.587371,-0.136145) (0.631877,-0.0823468) (0.682262,-0.021441) (0.740984,0.0495408) (0.811778,0.135117) (0.900701,0.242606) (1.01833,0.384795) (1.18506,0.586337) (1.44608,0.901859) (1.92578,1.48171)

(-3.48153,2.52948) (-16.9298,9.92273) (6.93524,-3.19719) (3.0322,-1.05148) (1.98753,-0.477165) (1.49595,-0.206915) (1.20575,-0.047374) (1.01111,0.0596283) (0.869167,0.137663) (0.759209,0.198113) (0.669978,0.247168) (0.594799,0.288498) (0.52943,0.324435) (0.471015,0.356549) (0.417524,0.385956) (0.367437,0.413491) (0.319547,0.439819) (0.272834,0.4655) (0.226373,0.491042) (0.17926,0.516942) (0.130538,0.543728) (0.0791167,0.571997) (0.0236714,0.602478) (-0.0375137,0.636115) (-0.106782,0.674195) (-0.18751,0.718576) (-0.284836,0.772081) (-0.407084,0.839288) (-0.568795,0.928189) (-0.798009,1.0542) (-1.15685,1.25148) (-1.81632,1.61402)

(-1.8323,-3.89383) (-6.57464,-18.4893) (1.84103,7.4115) (0.464674,3.17552) (0.0962861,2.04174) (-0.0770633,1.50823) (-0.1794,1.19327) (-0.248036,0.982028) (-0.29809,0.827977) (-0.336865,0.708638) (-0.368331,0.611796) (-0.394842,0.530204) (-0.417894,0.459259) (-0.438493,0.395861) (-0.457355,0.337808) (-0.475018,0.283448) (-0.491906,0.231473) (-0.508378,0.180775) (-0.524762,0.130352) (-0.541376,0.0792195) (-0.558557,0.0263409) (-0.57669,-0.0294661) (-0.596242,-0.089641) (-0.617818,-0.156045) (-0.642245,-0.231222) (-0.670712,-0.318836) (-0.705033,-0.424464) (-0.748142,-0.55714) (-0.805167,-0.732645) (-0.885996,-0.981412) (-1.01254,-1.37087) (-1.24509,-2.08658)

为了证明它们是不同的值，我只包含了 32 个块中的 3 个块。

在获取此输出并将其提供给 abs() 函数以计算幅度后，我注意到每个块都得到相同的输出！ （下例）

4.3034 19.6234 7.63673 3.20934 2.04401 1.51019 1.20668 1.01287 0.880002 0.784632 0.714117 0.661072 0.62093 0.590747 0.568584 0.553159 0.543646 0.539563 0.54071 0.547141 0.559178 0.577442 0.602943 0.63722 0.682599 0.742638 0.822946 0.932803 1.08861 1.32218 1.70426 2.42983

4.3034 19.6234 7.63673 3.20934 2.04401 1.51019 1.20668 1.01287 0.880002 0.784632 0.714117 0.661072 0.62093 0.590747 0.568584 0.553159 0.543646 0.539563 0.54071 0.547141 0.559178 0.577442 0.602943 0.63722 0.682599 0.742638 0.822946 0.932803 1.08861 1.32218 1.70426 2.42983

4.3034 19.6234 7.63673 3.20934 2.04401 1.51019 1.20668 1.01287 0.880002 0.784632 0.714117 0.661072 0.62093 0.590747 0.568584 0.553159 0.543646 0.539563 0.54071 0.547141 0.559178 0.577442 0.602943 0.63722 0.682599 0.742638 0.822946 0.932803 1.08861 1.32218 1.70426 2.42983

为什么我从不同的输入中得到完全相同的输出？这正常吗？

这是我正在执行所有这些计算的代码的一部分：

int main(int argc, char** argv)
{
    int i;
    double y;
    const double Fs = 100;//How many time points are needed i,e., Sampling Frequency
    const double  T = 1 / Fs;//# At what intervals time points are sampled
    const double f = 4;//frequency
    int chuck_size = 32; // chunk size (N / 32=32 chunks)
    Complex chuck[32];
    int j = 0;
    int counter = 0;
    for (int i = 0; i < N; i++)
    {
        t[i] = i * T;
        in[i] = { (0.7 * cos(2 * M_PI * f * t[i])), (0.7 * sin(2 * M_PI * f * t[i])) };// generate (complex) sine waveform
        chuck[j] = in[i];
        //compute FFT for each chunk
        if (i + 1 == chuck_size) // for each set of 32 chunks, apply FFT and save it all in a 1d array (magnitude)
        {
            chuck_size += 32;
            CArray data(chuck, 32);
            fft(data);
            j = 0;
            for (int h = 0; h < 32; h++)
            {
                magnitude[counter] = abs(data[h]);
                std::cout << abs(data[h]) << " ";
                counter++;
            }
            printf("\n\n");

        }
        else
            j++;
    }
}

频谱图（归一化）：

Answer 1

您的信号是正弦波。你把它切碎。每个段将具有相同的频率分量，只是不同的相位（偏移）。 FFT 为您提供每个频率分量的幅度和相位，但在abs 之后，仅保留幅度。对于您的所有块，这些量级必须相同。