abs() returns 不同 FFT 输入的相同输出
abs() returns the same output for different FFT inputs
我有一个 1024 个样本，我将它分成 32 个块以便对其执行 FFT，下面是 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)
我只包括了 3 个 32 的块，以证明它们每个都是不同的值。
在获取此输出并将其提供给 abs() 函数来计算震级后，我注意到我为每个块获得了相同的输出！ （示例如下）
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）：
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 之后只剩下幅度。这些量级对于所有块都必须相同。
我有一个 1024 个样本，我将它分成 32 个块以便对其执行 FFT，下面是 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)
我只包括了 3 个 32 的块，以证明它们每个都是不同的值。
在获取此输出并将其提供给 abs() 函数来计算震级后，我注意到我为每个块获得了相同的输出！ （示例如下）
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）：
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 之后只剩下幅度。这些量级对于所有块都必须相同。