Matlab，DFT，FFT，频率范围 [英] Matlab, DFT, FFT, frequency range;

问题描述

我真的很难理解MATLAB中的fft函数如何工作. 根据:

http://www.mathworks.de /help/matlab/math/fast-fourier-transform-fft.html

fs/n是频谱中采样点之间的距离，其中fs是采样频率，n是信号的长度.他们提供的代码确实提取了频率(尽管我不知道为什么)，但是根据fft的实现，他们提供了:

http://www.mathworks.de/help/matlab/ref /fft.html

光谱中各点之间的距离应为:1/fs.因为不是插入索引j，而是插入j * T，其中T = 1/fs是采样时间，然后您可以计算点之间的距离，该距离不应为fs/n.

如果有人能向我解释频域中点之间的距离是什么，为什么会这样，我将非常感激:)

这不是特定于Matlab的问题.关于傅里叶变换和离散傅里叶变换与频率轴的标度/单位之间的关系的问题更多.可以在以下 PDF文档中找到很好的解释.第3页.

[/EDIT]

解决方案

在两种情况下，频谱上频率点之间的距离均为fs/n，这是正确的值.如果我们将代码放在 http://www.mathworks.co.uk中/help/matlab/ref/fft.html ，我们有:

>> Fs = 1000;                    % Sampling frequency
>> T = 1/Fs;                     % Sample time
>> L = 1000;                     % Length of signal
>> t = (0:L-1)*T;                % Time vector
>> NFFT = 2^nextpow2(L); % Next power of 2 from length of y
>> f = Fs/2*linspace(0,1,NFFT/2+1);
>> f(2)-f(1)
ans =  0.97656
>> Fs/NFFT
ans =  0.97656

您可以仔细检查所有其他f(n+1)-f(n)，它们都是一样的.

I am really having a hard time to understand how the fft-function in MATLAB works. According to:

http://www.mathworks.de/help/matlab/math/fast-fourier-transform-fft.html

fs/n is the distance between the sampled points in the Spectrum, where fs is the sampling frequency and n is the length of the signal. The code that they present, does extract the frequencies (although I do not know why), but according to the implementation of fft, that they present:

http://www.mathworks.de/help/matlab/ref/fft.html

the distance between the points in the spectrum should rather be: 1/fs. Because instead of the index j one inserts j*T, where T = 1/fs is the sampling time and then you can calculate the distance between the points, which should not be fs/n.

I would really be grateful if someone could explain me what the distance between the points in the frequency domain is and why this is so:)

[EDIT]
This is not a Matlab-specific problem. It is more a problem about the relationship between the Fourier-Transformation and the Discrete-Fourier-Transformation and the scaling/units of the frequency-axis. A pretty good explanation can be found in this PDF-Document at page 3.

[/EDIT]

解决方案

The distance between the frequency points on the spectrum is fs/n in both cases, and that is the correct value. If we take the code in http://www.mathworks.co.uk/help/matlab/ref/fft.html, we have:

>> Fs = 1000;                    % Sampling frequency
>> T = 1/Fs;                     % Sample time
>> L = 1000;                     % Length of signal
>> t = (0:L-1)*T;                % Time vector
>> NFFT = 2^nextpow2(L); % Next power of 2 from length of y
>> f = Fs/2*linspace(0,1,NFFT/2+1);
>> f(2)-f(1)
ans =  0.97656
>> Fs/NFFT
ans =  0.97656

You can double-check all the other f(n+1)-f(n), they're all the same.

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