Frequency based on acceleration data

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Hi,
I want to find the frequency during testing based on the acceleration data collected using accelerometer. I already try several method using MATLAB but I'm not sure if mine is correct. Different method give me different frequency value. For the info, the machine frequency is 6.67 Hz. I attach the code and file for reference.
clc;
clear all;
close all;
Time = xlsread('Book5','A1:A585');
Acceleration = xlsread('Book5','B1:B585');
n = length(Acceleration);
Ts = Time(2)-Time(1);
Fs = 1/Ts;
NFFT = 2^nextpow2(n);
Y = fft(Acceleration,NFFT)/n;
f = Fs/2*linspace(0,1,NFFT/2+1);
Iv = 1:length(f);
figure(1)
plot(f,abs(Y(Iv)))
xlabel('Frequency (Hz)')
ylabel('Amplitude (m)')
num=csvread("Book5.csv");
Fs = 1/mean(diff(num(:,1)));
Fn = Fs/2;
Tr = linspace(num(1,1), num(1,end), size(num,1));
Dr = resample(num(:,4), Tr);
Dr_mc = Dr - mean(Dr,1);
FDr_mc = fft(Dr_mc, [], 1);
Fv = linspace(0, 1, fix(size(FDr_mc,1)/2)+1)*Fn;
Iv = 1:numel(Fv);
figure(13)
plot(Fv, abs(FDr_mc(Iv,:))*2)
grid
xlabel('Frequency (Hz)')
ylabel('Amplitude')
Thank you.

回答(1 个)

Mathieu NOE
Mathieu NOE 2023-4-14
hello
I tried to do my best finding the fft length and overlap that gives me the best info - as your signal is quite unstationnary
this is what I could tell from your data
Spectrogram : look at the red area as this is the dominant frequencies (depend of time)
the amplitude (color) is expressed in dB (log scale of amplitude)
Averaged FFT spectrum , basicaly the time average of the spectrogram , here you lose the information of time dependance but you can probably find your frequencies (no 6.67 Hz found ! )
Code
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
% load signal
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
filename = ('Book5.xlsx');
data = readmatrix(filename);
time = data(:,1);
signal = data(:,2);
dt = mean(diff(time));
Fs = 1/dt; % sampling rate
[samples,channels] = size(signal);
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
% FFT parameters
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
NFFT = 200; %
OVERLAP = 0.95; % must be between 0 and 0.95
% spectrogram dB scale
spectrogram_dB_scale = 80; % dB range scale (means , the lowest displayed level is XX dB below the max level)
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
% options
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
% if you are dealing with acoustics, you may wish to have A weighted
% spectrums
% option_w = 0 : linear spectrum (no weighting dB (L) )
% option_w = 1 : A weighted spectrum (dB (A) )
option_w = 0;
%% decimate (if needed)
% NB : decim = 1 will do nothing (output = input)
decim = 1;
if decim>1
for ck = 1:channels
newsignal(:,ck) = decimate(signal(:,ck),decim);
Fs = Fs/decim;
end
signal = newsignal;
end
samples = length(signal);
time = (0:samples-1)'*1/Fs;
%%%%%% legend structure %%%%%%%%
for ck = 1:channels
leg_str{ck} = ['Channel ' num2str(ck) ];
end
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
% display 1 : time domain plot
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
figure(1),plot(time,signal);grid on
title(['Time plot / Fs = ' num2str(Fs) ' Hz ']);
xlabel('Time (s)');ylabel('Amplitude');
legend(leg_str);
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
% display 2 : averaged FFT spectrum
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
[freq, sensor_spectrum] = myfft_peak(signal,Fs,NFFT,OVERLAP);
% convert to dB scale (ref = 1)
sensor_spectrum_dB = 20*log10(sensor_spectrum);
% apply A weigthing if needed
if option_w == 1
pondA_dB = pondA_function(freq);
sensor_spectrum_dB = sensor_spectrum_dB+pondA_dB;
my_ylabel = ('Amplitude (dB (A))');
else
my_ylabel = ('Amplitude (dB (L))');
end
figure(2),plot(freq,sensor_spectrum_dB);grid on
df = freq(2)-freq(1); % frequency resolution
title(['Averaged FFT Spectrum / Fs = ' num2str(Fs) ' Hz / Delta f = ' num2str(df,3) ' Hz ']);
xlabel('Frequency (Hz)');ylabel(my_ylabel);
legend(leg_str);
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
% display 3 : time / frequency analysis : spectrogram demo
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
for ck = 1:channels
[sg,fsg,tsg] = specgram(signal(:,ck),NFFT,Fs,hanning(NFFT),floor(NFFT*OVERLAP));
% FFT normalisation and conversion amplitude from linear to dB (peak)
sg_dBpeak = 20*log10(abs(sg))+20*log10(2/length(fsg)); % NB : X=fft(x.*hanning(N))*4/N; % hanning only
% apply A weigthing if needed
if option_w == 1
pondA_dB = pondA_function(fsg);
sg_dBpeak = sg_dBpeak+(pondA_dB*ones(1,size(sg_dBpeak,2)));
my_title = ('Spectrogram (dB (A))');
else
my_title = ('Spectrogram (dB (L))');
end
% saturation of the dB range :
% saturation_dB = 60; % dB range scale (means , the lowest displayed level is XX dB below the max level)
min_disp_dB = round(max(max(sg_dBpeak))) - spectrogram_dB_scale;
sg_dBpeak(sg_dBpeak<min_disp_dB) = min_disp_dB;
% plots spectrogram
figure(2+ck);
imagesc(tsg,fsg,sg_dBpeak);colormap('jet');
axis('xy');colorbar('vert');grid on
df = fsg(2)-fsg(1); % freq resolution
title([my_title ' / Fs = ' num2str(Fs) ' Hz / Delta f = ' num2str(df,3) ' Hz / Channel : ' num2str(ck)]);
xlabel('Time (s)');ylabel('Frequency (Hz)');
end
function pondA_dB = pondA_function(f)
% dB (A) weighting curve
n = ((12200^2*f.^4)./((f.^2+20.6^2).*(f.^2+12200^2).*sqrt(f.^2+107.7^2).*sqrt(f.^2+737.9^2)));
r = ((12200^2*1000.^4)./((1000.^2+20.6^2).*(1000.^2+12200^2).*sqrt(1000.^2+107.7^2).*sqrt(1000.^2+737.9^2))) * ones(size(f));
pondA = n./r;
pondA_dB = 20*log10(pondA(:));
end
function [freq_vector,fft_spectrum] = myfft_peak(signal, Fs, nfft, Overlap)
% FFT peak spectrum of signal (example sinus amplitude 1 = 0 dB after fft).
% Linear averaging
% signal - input signal,
% Fs - Sampling frequency (Hz).
% nfft - FFT window size
% Overlap - buffer percentage of overlap % (between 0 and 0.95)
[samples,channels] = size(signal);
% fill signal with zeros if its length is lower than nfft
if samples<nfft
s_tmp = zeros(nfft,channels);
s_tmp((1:samples),:) = signal;
signal = s_tmp;
samples = nfft;
end
% window : hanning
window = hanning(nfft);
window = window(:);
% compute fft with overlap
offset = fix((1-Overlap)*nfft);
spectnum = 1+ fix((samples-nfft)/offset); % Number of windows
% % for info is equivalent to :
% noverlap = Overlap*nfft;
% spectnum = fix((samples-noverlap)/(nfft-noverlap)); % Number of windows
% main loop
fft_spectrum = 0;
for i=1:spectnum
start = (i-1)*offset;
sw = signal((1+start):(start+nfft),:).*(window*ones(1,channels));
fft_spectrum = fft_spectrum + (abs(fft(sw))*4/nfft); % X=fft(x.*hanning(N))*4/N; % hanning only
end
fft_spectrum = fft_spectrum/spectnum; % to do linear averaging scaling
% one sidded fft spectrum % Select first half
if rem(nfft,2) % nfft odd
select = (1:(nfft+1)/2)';
else
select = (1:nfft/2+1)';
end
fft_spectrum = fft_spectrum(select,:);
freq_vector = (select - 1)*Fs/nfft;
end
  3 个评论
Mathieu NOE
Mathieu NOE 2023-4-17
in terms of signal processing, you can decide wheter you make
  • one fft computation with the fft length equal to the signal length (and in this case there is of course no averaging
  • or you split your signal into smaller chunks (buffers) and you do the fft on each buffer and then you average the spectra ; this reduce the impact of spurious / non correlated noises on your resulting spectrum. You can increase the number of averages by introducing some overlap between buffers. For spectrograms, having / increasing the overlap makes the rendering more progressive (smooth) at the expense of more fft computations
  • also the fft spectrum frequency resolution df depends of the fft length nfft and sampling rate (Fs) :
  • df = Fs / nfft, so for a given Fs, if you want more averages you will have to reduce nfft (and maybe use a higher overlap), but globally you cannot have a the same time a very small frequency resolution and lot of averages. It's all about finding the best compromise depending of your signal characteristics (stationnary or not)
Mathieu NOE
Mathieu NOE 2023-5-12
hello again
did my submission fullfil your request ?

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