Matlab的快速傅立叶变换/ FFT时间和速度 [英] Matlab Fast Fourier Transform / fft for time and speed

问题描述

我有数据的子集的时间和速度，像这样一个2列向量：

I have a 2 column vector with times and speeds of a subset of data, like so:

5 40
10 37
15 34
20 39

等。我想要得到的傅里叶变换速度得到的频率。我怎么会去用快速傅立叶这样变换（FFT）？

And so on. I want to get the fourier transform of speeds to get a frequency. How would I go about doing this with a fast fourier transform (fft)?

如果我的矢量的名字是sampleData在，我已经试过

If my vector name is sampleData, I have tried

fft(sampleData);

但是，让我实数和虚数的矢量。为了能够得到合理的数据绘制的，我怎么会去这样做？

but that gives me a vector of real and imaginary numbers. To be able to get sensible data to plot, how would I go about doing this?

推荐答案

傅立叶变换将产生一个复杂的矢量，当你FFT你得到频率的矢量，每个人都有一个光谱相位。这些阶段可以是非常重要的！ （它们包含大部分的时域信号的信息，你不会看到没有他们等干扰效应...）。如果你要绘制的功率谱，可以

Fourier Transform will yield a complex vector, when you fft you get a vector of frequencies, each has a spectral phase. These phases can be extremely important! (they contain most of the information of the time-domain signal, you won't see interference effects without them etc...). If you want to plot the power spectrum, you can

plot(abs(fft(sampleData)));

要完成这个故事，你可能需要fftshift，并且还产生一个频率向量。下面是一个更复杂的code：

To complete the story, you'll probably need to fftshift, and also produce a frequency vector. Here's a more elaborate code:

% Assuming 'time' is the 1st col, and 'sampleData' is the 2nd col: 
N=length(sampleData);  
f=window(@hamming,N)';
dt=mean(diff(time)); 
df=1/(N*dt); % the frequency resolution (df=1/max_T)
if mod(N,2)==0
    f_vec= df*((1:N)-1-N/2); % frequency vector for EVEN length vector
    else
    f_vec= df*((1:N)-0.5-N/2); 
end

fft_data= fftshift(fft(fftshift(sampleData.*f))) ;

plot(f_vec,abs(fft_data))

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