问题描述
我有一个 FFT 结果.它们存储在两个 double 数组中:实部数组和虚部数组.如何确定与这些数组中的每个元素对应的频率?
I have an FFT result. These are stored in two double arrays: a real part array and an imaginary part array. How do I determine the frequencies that correspond to each element in these arrays?
换句话说，我想创建一个数组来存储我的 FFT 的每个实部和虚部的频率.
In other words, I would like have create an array that stores the frequencies for each real and imaginary component of my FFT.
推荐答案
FFT 中的第一个 bin 是 DC (0 Hz)，第二个 bin 是 Fs/N，其中 Fs 是采样率，N 是 FFT 的大小.下一个 bin 是 2 * Fs/N.概括地说，nth 个 bin 是 n * Fs/N.
The first bin in the FFT is DC (0 Hz), the second bin is Fs / N, where Fs is the sample rate and N is the size of the FFT. The next bin is 2 * Fs / N. To express this in general terms, the nth bin is n * Fs / N.
因此，如果您的采样率 Fs 为 44.1 kHz，而您的 FFT 大小 N 为 1024，则 FFT 输出箱位于:
So if your sample rate, Fs is say 44.1 kHz and your FFT size, N is 1024, then the FFT output bins are at:
0: 0 * 44100 / 1024 = 0.0 Hz
1: 1 * 44100 / 1024 = 43.1 Hz
2: 2 * 44100 / 1024 = 86.1 Hz
3: 3 * 44100 / 1024 = 129.2 Hz
4: ...
5: ...
...
511: 511 * 44100 / 1024 = 22006.9 Hz
请注意，对于实数输入信号(虚部全为零)，FFT 的后半部分(从 N/2 + 1 到 N - 1 的区间)包含没有有用的附加信息(它们与前 N/2 - 1 个 bin 具有复共轭对称性).最后一个有用的 bin(用于实际应用)位于 N/2 - 1，对应于上述示例中的 22006.9 Hz.N/2 处的 bin 表示奈奎斯特频率下的能量，即 Fs/2(在本例中为 = 22050 Hz)，但这通常没有任何实际用途，因为抗混叠滤波器通常会衰减 Fs/2 及以上的任何信号.
Note that for a real input signal (imaginary parts all zero) the second half of the FFT (bins from N / 2 + 1 to N - 1) contain no useful additional information (they have complex conjugate symmetry with the first N / 2 - 1 bins). The last useful bin (for practical aplications) is at N / 2 - 1, which corresponds to 22006.9 Hz in the above example. The bin at N / 2 represents energy at the Nyquist frequency, i.e. Fs / 2 ( = 22050 Hz in this example), but this is in general not of any practical use, since anti-aliasing filters will typically attenuate any signals at and above Fs / 2.
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