![](https://www.mathworks.com/matlabcentral/answers/uploaded_files/1559694/image.png)
How to make an FRF with my accelerometer Data and FFT?
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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
%Load data
clear all
load("C:\Users\joant\Desktop\Field_Research\Matlab_Code\data_test.mat")
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)
![](https://www.mathworks.com/matlabcentral/answers/uploaded_files/1559599/image.jpeg)
![](https://www.mathworks.com/matlabcentral/answers/uploaded_files/1559604/image.jpeg)
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.
![](https://www.mathworks.com/matlabcentral/answers/uploaded_files/1559609/image.png)
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:
![](https://www.mathworks.com/matlabcentral/answers/uploaded_files/1559614/image.jpeg)
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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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
![](https://www.mathworks.com/matlabcentral/answers/uploaded_files/1559694/image.png)
load("data_test.mat")
%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
댓글 수: 6
Mathieu NOE
2023년 12월 4일
tx for accepting my answer
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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