使用八度 fft 进行外推
使用 GNU 倍频程，我对一段信号计算 fft，然后消除一些频率，最后重建信号。这给了我一个很好的信号近似值；但它没有给我一种推断数据的方法。
基本上假设我绘制了三个半周期
f: x -> sin(x) + 0.5*sin(3*x) + 1.2*sin(5*x)
，然后添加了一段低幅度、零中心随机噪声。使用 fft/ifft，我可以轻松消除大部分噪声；但是我如何推断出另外 3 个周期的信号数据呢？ （当然还有复制信号）。
数学方法很简单：将函数分解为正弦/余弦的无限和，只需提取部分和并将其应用到任何地方。但我不太明白编程方式......
谢谢！
Using GNU octave, I'm computing a fft over a piece of signal, then eliminating some frequencies, and finally reconstructing the signal. This give me a nice approximation of the signal ; but it doesn't give me a way to extrapolate the data.
Suppose basically that I have plotted three periods and a half of
f: x -> sin(x) + 0.5*sin(3*x) + 1.2*sin(5*x)
and then added a piece of low amplitude, zero-centered random noise. With fft/ifft, I can easily remove most of the noise ; but then how do I extrapolate 3 more periods of my signal data? (other of course that duplicating the signal).
The math way is easy : you have a decomposition of your function as an infinite sum of sines/cosines, and you just need to extract a partial sum and apply it anywhere. But I don't quite get the programmatic way...
Thanks!
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离散傅里叶变换依赖于时域数据是周期性的假设，因此您可以重复时域数据
令人恶心- 不需要显式外推。当然，如果您的各个组件周期不是 DFT 输入窗口持续时间的精确约数，这可能不会给您带来预期的结果。这就是为什么我们通常应用窗口函数的原因之一，例如转换之前的汉宁窗。The Discrete Fourier Transform relies on the assumption that your time domain data is periodic, so you can just repeat your time domain data
ad nauseam- no explicit extrapolation is necessary. Of course this may not give you what you expect if your individual component periods are not exact sub-multiples of the DFT input window duration. This is one reason why we typically apply window functions such as the Hanning Window prior to the transform.