How to make an FRF with my accelerometer Data and FFT?

조회 수: 25 (최근 30일)
Jo98 2023년 12월 4일
댓글: Mathieu NOE 2023년 12월 4일
Dear, Matlab Users.
I have some inquiries about my accelerometer Data, that was processed by a Dewesoft Software.
So, I am gonna share my code and data, and I would like to know if there any suggestion to this resolution.
Sensitivy: 102 mV/g, or 10.40 mV/m/s^2
clear all
whos
close all
%Plot the array according to the data
%Data of Channel 1
g1= transpose(Data1_352C33OLD)
t1= transpose(Data1_time_352C33OLD); % time
dt=1/(Sample_rate) % period
L=length(g1)
%Iterate
% Figure of Channel 1
figure(1)
plot(t1,g1,'b.-')
set(gca,'Fontsize',14)
xlabel('time(s)')
ylabel('g')
xlim([min(t1) max(t1)])
%ylim([-6 6])
grid on
% Calculate Fourier transform of force
points=round((t1(L))/dt) %Calculate the points
A_F= g1; % This is the Force for setting in the fft.
freq= (0:1/(dt*points):(1-1/(2*points))/dt)'; %Hz
freq=freq(1:round(points/2+1),:); %Hz
length(freq)
length(fft(A_F))
%In this part I need help
A_fft=fft(g1); %without the conversion by 9.8 m/s^2
abs_A=abs(A_fft)
figure(4)
plot(freq,abs_A(1:length(freq)))
xlim([0 6]) % Because 5 Hz is 300 RPM
%ylim([-400 400])
xlabel('Frequency (Hz)')
ylabel('g')
%According to the practice in the laboratory the result is the following
%image.
This was my process to obtain the FFT, I would like to know if this process is correct?. Because when I plot this data is not the same values in y-axis that I had in the Dewesoft software.
Dewesoft fft(g) (attached) vs matlab fft(g) (attached)
Question #2
According to a Mechanical vibration book, I have the next equation to find the FRF. I applied this but my FRF y axis value is too high so, I don't know If I didn't applied correctly. That information is in my code and I use the following equation.
w= natural frequency = 31.42 rad/s^2.
I assumpted that A/F is g because according to the book, it says the next information: "Because the acceloremeter produces a voltage that is proportional to acceleration, we obtain the accelerance, A/F(w), from the dynamic signal analyzer".
My graph result:
This graph it doesn't have sense, because I need values aroun of 0.15-0.17 in the first vibration mode. So, I would like to know if there any suggestion how to resolve this.
Best Regards,
Jo98.

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채택된 답변

Mathieu NOE 2023년 12월 4일
hello
I got the correct result if I assume that your data is in g and not volts (so your analyser has already used the sensor sensivity when storing the data)
there can be some minor difference in terms of frequency and amplitude as I am not aware of what type of window / nfft is used on your side
%Plot the array according to the data
%Data of Channel 1
signal = double(Data1_352C33OLD);
time= Data1_time_352C33OLD; % time
[samples,channels] = size(signal);
dt = mean(diff(time));
Fs = 1/dt;
%% decimate (if needed)
% NB : decim = 1 will do nothing (output = input)
decim = 10;
if decim>1
Fs = Fs/decim;
for ck = 1:channels
newsignal(:,ck) = decimate(signal(:,ck),decim);
end
signal = newsignal;
end
samples = length(signal);
time = (0:samples-1)*1/Fs;
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
% FFT parameters
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
NFFT = 1000; % for maximal resolution take NFFT = signal length (but then no averaging possible)
OVERLAP = 0.75; % overlap will be used only if NFFT is shorter than signal length
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
% display 1 : time domain plot
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
figure(1),
plot(time,signal,'b');grid on
title(['Time plot / Fs = ' num2str(Fs) ' Hz ']);
xlabel('Time (s)');ylabel('Amplitude');
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
% display 2 : averaged FFT spectrum
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
[freq, spectrum_raw] = myfft_peak(signal,Fs,NFFT,OVERLAP);
figure(2),plot(freq,spectrum_raw,'b');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('Amplitude');
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
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Jo98 2023년 12월 4일
Yes, I'm gonna contact you by DM, asking you the email.
Regard with your FRF, I replace this vector with the formula that I post in the original post, and it works. So A_F is (spectrum_raw*9.8 m/s^2). This because the conversion between g and m/s^2.
I assure that A_F is g because the theory of the book that I learnt about FRF says:" Because the acceloremeter produces a voltage that is proportional to acceleration, we obtain the accelerance, A/F(w), from the dynamic signal analyzer".
Best Regards.
Mathieu NOE 2023년 12월 4일
NB that what we display above is only the autospectrum of the acceleration signal
for the accelerance FRF we need a force signal , as already mentionned above, it's not in your data file.
Also why do you want an accelerance FRF at only one frequency ?

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