FFT function in matlab

Question

siyu lin 2018-3-12

0
链接

此问题的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/387676-fft-function-in-matlab

编辑： David Goodmanson 2018-3-12

long_6washer.mat

if true
  % close all;clear all;clc
load('short_6washer')
t = data(:,1);
Amp = data(:,2);
Fs = 1/(t(2) - t(1)); % Sampling frequency
L = length(data(:,1)); % Length of signal
f = Fs*(0:(L/2))/L;

figure subplot(2,1,1) plot(t,Amp) xlabel('t (seconds)') ylabel('X(t) [V]') Y = abs(fft(Amp)); % f = (0:length(Y)-1)*10000/length(Y); P = 2*abs(Y(1:length(f)))/L; subplot(2,1,2) plot(f,P) axis([-5 15 0 6]) xlabel('f (Hz)') ylabel('amplitude [V]')% end

I am doing the FFT of the attached file in matlab. The image is what I got but it seems wrong to me. If it is wrong, can someone debug the code for me? The sample frequency is 10000Hz and the data was collected in 10 seconds.

2 个评论
显示无隐藏无

Joost Meulenbeld 2018-3-12

编辑：Joost Meulenbeld 2018-3-12

With no attached code, it's hard to debug your code; what exactly seems wrong to you? How did you compute the real one-sided fourier spectrum?

siyu lin 2018-3-12

I have attached the code. Since the spike is not clear and hard to determine.

请先登录，再进行评论。

请先登录，再回答此问题。

Answer 1

David Goodmanson 2018-3-12

0
链接

此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/387676-fft-function-in-matlab#answer_309553

编辑：David Goodmanson 2018-3-12

在 MATLAB Online 中打开

Hi siyu,

Things are actually working correctly, and the only real problem is that your wave is oscillating about the value 2.5. That means that there is a large DC component of that size. (It also means that if you are going to be doubling the size of Y array elements, you should not double the size of the first array element. That element corresponds to frequency = 0, i.e. DC. That peak was the correct size, 2.5, before it got doubled).

Assuming you are not particularly interested in the DC component, then you will get better results by taking it out, with

Y = abs(fft(Amp-mean(Amp)));