了解 FFT 输出 [英] Understanding FFT output
问题描述
我需要一些帮助来理解 DFT/FFT 计算的输出.
I need some help understanding the output of the DFT/FFT computation.
我是一名经验丰富的软件工程师，需要解读一些智能手机加速度计读数，例如找出主要频率.不幸的是，十五年前，我在大学 EE 课程中的大部分时间都睡了，但最近几天我一直在阅读 DFT 和 FFT(显然，收效甚微).
I'm an experienced software engineer and need to interpret some smartphone accelerometer readings, such as finding the principal frequencies. Unfortunately, I slept through most of my college EE classes fifteen years ago, but I've been reading up on DFT and FFT for the last several days (to little avail, apparently).
请不要回复去参加 EE 课程".如果我的雇主付钱给我，我实际上打算这样做.:)
Please, no responses of "go take an EE class". I'm actually planning to do that if my employer will pay me. :)
所以这是我的问题:
我捕获了一个 32 Hz 的信号.这是我在 Excel 中绘制的 32 个点的 1 秒示例.
I've captured a signal at 32 Hz. Here is a 1 second sample of 32 points, which I've charted in Excel.
然后我得到了一些 FFT 代码 用哥伦比亚大学的 Java 编写(遵循Reliable和 Java 中的快速 FFT").
I then got some FFT code written in Java from Columbia University (after following the suggestions in a post on "Reliable and fast FFT in Java").
这个程序的输出如下.我相信它正在运行就地 FFT，因此它为输入和输出重新使用相同的缓冲区.
The output of this program is as follows. I believe it is running an in-place FFT, so it re-uses the same buffer for both input and output.
Before:
Re: [0.887 1.645 2.005 1.069 1.069 0.69 1.046 1.847 0.808 0.617 0.792 1.384 1.782 0.925 0.751 0.858 0.915 1.006 0.985 0.97 1.075 1.183 1.408 1.575 1.556 1.282 1.06 1.061 1.283 1.701 1.101 0.702 ]
Im: [0.0 0.0 0.0 0.0 0.0 0.0 0.0 0.0 0.0 0.0 0.0 0.0 0.0 0.0 0.0 0.0 0.0 0.0 0.0 0.0 0.0 0.0 0.0 0.0 0.0 0.0 0.0 0.0 0.0 0.0 0.0 0.0 ]
After:
Re: [37.054 1.774 -1.075 1.451 -0.653 -0.253 -1.686 -3.602 0.226 0.374 -0.194 -0.312 -1.432 0.429 0.709 -0.085 0.0090 -0.085 0.709 0.429 -1.432 -0.312 -0.194 0.374 0.226 -3.602 -1.686 -0.253 -0.653 1.451 -1.075 1.774 ]
Im: [0.0 1.474 -0.238 -2.026 -0.22 -0.24 -5.009 -1.398 0.416 -1.251 -0.708 -0.713 0.851 1.882 0.379 0.021 0.0 -0.021 -0.379 -1.882 -0.851 0.713 0.708 1.251 -0.416 1.398 5.009 0.24 0.22 2.026 0.238 -1.474 ]
因此，在这一点上，我无法确定输出的正面或反面.我理解 DFT 概念，例如实部是分量余弦波的幅度，虚部是分量正弦波的幅度.我还可以从伟大的书中数字信号处理的科学家和工程师指南"中的这张图表进行操作:
So, at this point, I can't make heads or tails of the output. I understand the DFT concepts, such as the real portion being the amplitudes of the component cosine waves and the imaginary portion being the amplitudes of the component sine waves. I can also follow this diagram from the great book "The Scientist and Engineer's Guide to Digital Signal Processing":
所以我的具体问题是:
从 FFT 的输出中，我如何找到最常出现的频率"?这是我对加速度计数据的分析的一部分.我应该读取实数(余弦)还是虚数(正弦)数组?
From the output of the FFT, how do I find the "most occurring frequencies"? This is part of my analysis of my accelerometer data. Should I read the real (cosine) or imaginary (sine) arrays?
我在时域中有一个 32 点输入.FFT 的输出不应该是实数的 16 元素数组和虚数的 16 元素数组吗?为什么程序给了我大小为 32 的实数和虚数数组输出?
I have a 32-point input in the time domain. Shouldn't the output of the FFT be a 16-element array for reals and a 16-element array for imaginary? Why does the program give me real and imaginary array outputs both of size 32?
关于上一个问题，如何解析输出数组中的索引?鉴于我以 32 Hz 采样的 32 个样本的输入，我的理解是，16 元素阵列输出的索引应该均匀分布到采样率(32 Hz)的 1/2，所以我理解每个元素是否正确数组的 表示 (32 Hz * 1/2)/16 = 1 Hz?
Related to the previous question, how do I parse the indexes in the output arrays? Given my input of 32 samples sampled at 32 Hz, my understanding is that a 16-element array output should have its index uniformly spread up to 1/2 the sampling rate (of 32 Hz), so am I correct in understanding that each element of the array represents (32 Hz * 1/2) / 16 = 1 Hz?
为什么 FFT 输出有负值?我认为这些值代表正弦曲线的幅度.例如，对于频率为 3 的余弦波，Real[3] = -1.075 的输出应该意味着幅度为 -1.075.对吗?振幅怎么可能是负的?
Why does the FFT output have negative values? I thought the values represent amplitudes of a sinusoid. For example, the output of Real[ 3 ] = -1.075 should mean an amplitude of -1.075 for a cosine wave of frequency 3. Is that right? How can an amplitude be negative?
推荐答案
您不应该寻找复数的实部或虚部(即您的实数和虚数数组是什么).相反，您想要查找定义为 sqrt (real * real + imag * imag) 的频率幅度.这个数字永远是正数.现在您只需搜索最大值(忽略数组中的第一个条目.这是您的 DC 偏移，不包含频率相关信息).
You should neither look for the real or imaginative part of a complex number (that what's your real and imaginary array is). Instead you want to look for the magnitude of the frequency which is defined as sqrt (real * real + imag * imag). This number will always be positive. Now all you have to search is for the maximum value (ignore the first entry in your array. That is your DC offset and carries no frequency dependent information).
您得到 32 个实数和 32 个虚数输出，因为您使用的是复数到复数的 FFT.请记住，您已通过用零虚部对其进行扩展将 32 个样本转换为 64 个值(或 32 个复数值).这导致对称 FFT 输出，其中频率结果出现两次.一旦准备好在输出 0 到 N/2 中使用，并在输出 N/2 到 N 中镜像.在您的情况下，最简单的方法是忽略输出 N/2 到 N.您不需要它们，它们是只是一个关于如何计算 FFT 的神器.
You get 32 real and 32 imaginary outputs because you are using a complex to complex FFT. Remember that you've converted your 32 samples into 64 values (or 32 complex values) by extending it with zero imaginary parts. This results in a symetric FFT output where the frequency result occurs twice. Once ready to use in the outputs 0 to N/2, and once mirrored in the outputs N/2 to N. In your case it's easiest to simply ignore the outputs N/2 to N. You don't need them, they are just an artifact on how you calculate your FFT.
fft-bin 方程的频率是 (bin_id * freq/2)/(N/2) 其中 freq 是您的采样频率(又名 32 Hz，N 是您的 FFT 的大小).在您的情况下，这简化为每个 bin 1 Hz.箱 N/2 到 N 代表负频率(奇怪的概念，我知道).对于您的情况，它们不包含任何重要信息，因为它们只是前 N/2 个频率的镜像.
The frequency to fft-bin equation is (bin_id * freq/2) / (N/2) where freq is your sample-frequency (aka 32 Hz, and N is the size of your FFT). In your case this simplifies to 1 Hz per bin. The bins N/2 to N represent negative frequencies (strange concept, I know). For your case they don't contain any significant information because they are just a mirror of the first N/2 frequencies.
每个 bin 的实部和虚部构成一个复数.如果实部和虚部为负，而频率本身的幅度为正，则没关系(请参阅我对问题 1 的回答).我建议你阅读复数.解释它们是如何工作的(以及它们为什么有用)超出了在单个 stackoverflow 问题中所能解释的范围.
Your real and imaginary parts of each bin form a complex number. It is okay if real and imaginary parts are negative while the magnitude of the frequency itself is positive (see my answer to question 1). I suggest that you read up on complex numbers. Explaining how they work (and why they are useful) exceeds what is possible to explain in a single stackoverflow-question.
注意:您可能还想了解自相关是什么，以及如何使用它来找到信号的基频.我有一种感觉，这就是你真正想要的.
Note: You may also want to read up what autocorrelation is, and how it is used to find the fundamental frequency of a signal. I have a feeling that this is what you really want.
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