快速傅立叶变换结果：频率轴刻度？ [英] Fast Fourier Transform results: frequency axis scale?

问题描述

我使用Apple的Accelerate Framework（在iOS设备上执行）成功地实现了获取数组数据并对其进行快速傅里叶变换的代码。

I successfully implemented code that takes array data and runs a fast fourier transform on it, using Apple's Accelerate Framework (performed on iOS device).

我现在的问题是频率轴的刻度是多少？结果在某些频率范围内达到了预期的峰值，但是我不确定该频率是多少。 Accelerate Framework的FFT函数接收一个数组，并使用相同（或更多）数量的数据点吐出一个数组。是否假设所有这些点在时间上均等？它不会将采样频率或时间变量作为输入。频率轴的标度（即每个点的频率增量）是否只是采样周期除以2 * Pi（或类似的值？），我在文档中找不到很多信息。我一直在网上寻找类似的问题，但没有发现任何东西。

My question now is what is the scale of the frequency axis? The results have peaks as expected in certain frequency ranges, but I'm not sure what the frequency should be. The Accelerate Framework's FFT functions take in an array and spit out an array with the same (or more) number of data points. Does it assume that all those points are equally spaced in time? It doesn't take the sampling frequency or time variable as input. Is the scale of the frequency axis (i.e. frequency increment on each point) just the sampling period divided by 2*Pi (or something similar to that?) I couldn't find a lot of information in the documentation on this. I've been looking for similar questions online and haven't found anything.

从某种程度上讲，这是一个数学问题，尽管它很大程度上取决于Accelerate Framework的实现。

This is in some ways a math question, although it depends heavily on the Accelerate Framework implementation.

谢谢

EDIT
我问了一个后续问题此处，但尚未有人回答。请看看！

EDIT I asked a follow-up question here but no one has answered it yet. Please take a look!

推荐答案

FFT可以为您提供线性间隔的频点，直至采样频率。这意味着两个条带之间的间隔是（采样频率）/（条带数量）。

The FFT gives you linearly spaced frequency bins up to the sampling frequency. This means that the spacing between the bins is (sample frequency) / (number of bins).

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