Question

我对振动和使用matalb fft相当新。我得到一组长度为15000的数据（1D数组）（不确定这是否相关）我试图弄清楚是否有任何波浪埋藏在根据这些数据。我被指示可能使用matlab fft。这是正确的方法吗？我期待看到什么？我真的不确定如何解释我会得到的任何结果。请让我知道你们的想法。谢谢，如果需要更多细节，我会提供。例如：

% Df=[array is given to me and it is of size 15000];
% t=[time used for the above array, and if it is of the same size, also provided to me]


N_0= length(t);

fs_0=length(Dfxz);

Y_0=fft(Dfxz,N_0);

k_0=-N_0/2:N_0/2-1;

%Find the phase angle
p_0 = (angle(Y_0));
R_0 = norm(Y_0);

ff_0 = (0:length(Y_0)-1)'/length(Y_0)*100;    % Frequency vector
FT_power1_0 = abs(Y_0);

plot(k_0*fs_0/N_0,fftshift(abs(Y_0)))

我只看到1次频率= 0，但我确信有非零频率，我做错了什么？谢谢！ PS：我不知道如何选择采样频率？任何提示请（请记住，我不知道原始频率）

Answer 1

试试我的版本。在我看来，您拥有在数据中找到频率峰值所需的所有信息（如果存在）。如果所有你能看到的是零频率的大峰值，你可能会有一个巨大的DC偏移，它会淹没你所有的其他数据。我在我的代码中提出了一个补救措施。。

x = randn(15000,1); %//This is the data you were given - I'll use noise for demonstration - replace x with your Df data

%//If youre getting a massive peak at zero that dwarfs everything else, you
%//probably have a large DC offset. Easily removed in the time domain using
%//the following ..
x = x-mean(x);

tAxis = linspace(3/15000,3,15000); %//You said you have this too - I'll make up something for demonstration - make sure you replace this with your t data
dt = diff(tAxis(1:2)); %//sample period from time axis
fs = 1/dt;%//sample rate from sample period

NFFT = numel(x); %//number of fft bins - change if you like
Y = abs(fft(x, NFFT)).^2; %power spectrum

%//Calculate frequency axis
df = fs/NFFT;
fAxis = 0:df:(fs-df);

%//Plot it all
figure; plot(fAxis(1:NFFT/2), Y(1:NFFT/2))
xlabel('Frequency in Hz')
ylabel('Power')

如果确定无误，您可以通过查看another FFT answer on stackoverflow来深入了解。

给定一系列数据，用fft提取可能的频率，如何？

1 个答案: