带FFT的加速度计-奇怪的输出 [英] Accelerometer with FFT - strange output

问题描述

在阅读了许多研究和针对该主题的著作之后，在将FFT应用于加速度计数据时，我仍然遇到问题.我的大部分代码来自官方的MATLAB示例: FFT和加速度计数据:为什么我会得到此输出?哪里有使用开窗的建议.因此，在进行更多阅读之后，我在代码中添加了汉明窗.

After reading a lot of research and works on subject I still have got problems applying FFT to my accelerometer data. Most of my code is taken from official MATLAB example: FFT for one dimension. After more reading I've found this question: FFT and accelerometer data: why am I getting this output? where there was suggestion to use windowing. So after some more reading I've added hamming window to my code.

我的数据如下图所示:

My data looks like that on plot:

这是我用于FFT的代码:

And this is the code that I am using for FFT:

fs = 1/0.02; %0.02 comes from picking sample each 20ms
m = size(data,1);
w = hanning(m);
yw = w.*data;
n = pow2(nextpow2(yw));
y = fft(yw,size(n,1));
f = (0:size(n,1)-1)*(fs/size(n,1));
power = y.*conj(y)/size(n,1);
figure
plot(f,power)

问题是我从这段代码中得出的情节看起来像这样:

The problem is that my plot from this code looks like that:

有人可以告诉我我的代码有什么问题吗?老实说，我希望它看起来会更好(例如: http://imgur.com/wGs43)所以这就是为什么我问这个问题.

Can someone tell me what is wrong with my code? To be honest I'd excepted it would look better (something like this:http://imgur.com/wGs43) so that's why I am asking this question.

我的数据可以在这里找到: https://dl.dropboxusercontent.com/u/58774274 /exp.txt

My data can be found here: https://dl.dropboxusercontent.com/u/58774274/exp.txt

推荐答案

您的Fs是50，因此数据中的最高频率可以是Fs/2 = 25Hz.

Your Fs is 50 so the highest frequency in your data can be Fs/2 = 25Hz.

看看此代码是否有帮助.

See if this code helps.

fid = fopen('1.txt','r');
C = textscan(fid, '%f');
fclose(fid);
data = C{1};
fs = 50;
m = length(data);
nfft = 2^nextpow2(m);
y = fft(data,nfft)/m;
f = fs/2 * linspace(0,1,nfft/2+1);
power = abs(y);
subplot(211)
plot(f,power(1:nfft/2+1)) 
t = (0 : m-1)/fs;
s0 = .8*fs : 3.2*fs;  % .8 sec to 3.2 sec
p(s0) = .5*cos(2*pi*3.3*t(s0)+.25*pi); 
p = p + mean(data); 
subplot(212)
plot(t,data);hold on
plot(t,p,'r')

这是您在频域中的数据.

This is your data in frequency domain.

在3.3 Hz处有一个峰.

作为证明，我绘制了频率为3.3 Hz的正弦曲线以及您的数据，

As a proof I plotted a sinusoidal with frequency of 3.3 Hz along with your data,

如您所见，它与您的数据完全匹配.

As you can see, it totally matches your data.

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