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fft finding oscillation frequency and strouhal number for lift

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Megi Clen
Megi Clen on 25 Nov 2020
Commented: Mathieu NOE on 29 Nov 2020
Hello all,
I have a data which comes from an oscillating plate in a laminar flow. (please see the attachment). my sampling frequency is 2000 Hz. I am trying find out frequency magnitude and strouhal number graphs with fft. For the frequency plot, I would like to neglect all frequencies above the Nyquist frequency. I have looked at the examples on the page, followed them and created my scripts as below but at some points I became confused and not sure about my script is right or not. So, I am wondering does any of you help me to solve this problem ?
Thanks,
Megi

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Accepted Answer

Mathieu NOE
Mathieu NOE on 25 Nov 2020
hello
it's difficult to test your code if you share it only as a picture
it looks ok but if you want to double check this is a code I regularly share for fft spectral analysis
looks like the oscillation frequency is around 8 Hz
nb : your time vector is not equally spaced in your file , maybe due to rounding somewhere
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
% FFT parameters
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
NFFT = 1024; %
OVERLAP = 0.85;
% 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;
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
% load signal
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%% test data
A = readmatrix('lift.xlsx');
time = A(:,1); % NB : time stamps are not equally spaced
signal = A(:,2);
dt = mean(diff(time));
Fs = 1/dt;
Fs = 2000; % had to force the value because time stamps are not equally spaced
% % decimate
% decim = 12;
% signal = decimate(signal,decim);
% Fs = Fs/decim;
% %[data,Fs]=wavread('Approach_Gear_Drop_Aft Ctr.wav '); %(older matlab)
% % or
% [data,Fs]=audioread('myWAVaudiofile.wav'); %(newer matlab)
% channel = 1;
% signal = data(:,channel);
% samples = length(signal);
dt = 1/Fs;
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
% display 1 : 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(1),plot(freq,sensor_spectrum_dB,'b');grid
title(['Averaged FFT Spectrum / Fs = ' num2str(Fs) ' Hz / Delta f = ' num2str(freq(2)-freq(1)) ' Hz ']);
xlabel('Frequency (Hz)');ylabel(my_ylabel);
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
% display 2 : time / frequency analysis : spectrogram demo
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
[sg,fsg,tsg] = specgram(signal,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 :
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);
imagesc(tsg,fsg,sg_dBpeak);colormap('jet');
axis('xy');colorbar('vert');grid
title([my_title ' / Fs = ' num2str(Fs) ' Hz / Delta f = ' num2str(fsg(2)-fsg(1)) ' Hz ']);
xlabel('Time (s)');ylabel('Frequency (Hz)');
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%ù
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 [f,fft_spectrum] = myfft_peak(s, fs, nfft, Overlap)
%FFT peak spectrum of signal (example sinus amplitude 1 = 0 dB after fft).
% s - input signal,
% fs - Sampling frequency (Hz).
% spectsize - FFT window size
% spectnum - number of windows to analyze
f = [0:fs/nfft:fs/2];
f = f(:);
dt = 1/fs;
samples = length(s);
% fill s with zeros if length is lower than nfft
if samples<nfft
s_tmp = zeros(nfft,1);
s_tmp((1:samples)) = s;
s = s_tmp;
end
% window : hanning
window = hanning(nfft);
window = window(:);
% compute fft with overlap
offset = floor((1-Overlap)*nfft);
spectnum = 1+ floor((length(s)-nfft)/offset); % nb of averages
fft_spectrum = 0;
% main loop
for i=1:spectnum
start = (i-1)*offset;
sw = s((1+start):(start+nfft)).*window;
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 (index= (1:L/2+1))
fft_spectrum = fft_spectrum(1:nfft/2+1);
end
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

  14 Comments

Mathieu NOE
Mathieu NOE on 29 Nov 2020
hello Megi
below I have a bit modified your code to improve the fft computation
and also compute the frequency of the first peak in the fft (plotted as a red star)
the frequency of the peak is around 7 Hz, not 1.9531 Hz
the value you mentionned (1.9531 Hz) is my fft frequency resolution, which is something different;
the fft spectrum is computed on a frequency vector and its resolution is related to fs and nfft by df = fs / nfft
that means the frequency increment is df = fs / nfft
in your case the nfft is equal to the length of the record, so we cannot reduce (refine) the frequency resolution;
you should (if needed) increase the time duration of your record
you could also reduce the sampling frequency because 2000 Hz is very fast for analysing signals below 10 Hz
you could easily reduce fs to 100 Hz, that would reduce the record file size and post processing time;
a=xlsread('lift.xlsx');
x=a(:,1);
y=a(:,2);
N=length(a); % sample number
fs=2000; % [Hz] sampling frequency
nfft = N; % no averaging : the fft length = the samples qty
% X=2*fft(y)/N;
window = hanning(nfft);
% fft_spectrum=abs(fft(y.*window))*4/nfft; % the fft amplitude is correct in this fomula if you stick to the hanning window
% 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;
% let's find frequency of first peak
[peaks,loc] = findpeaks(fft_spectrum);
freq_of_first_peak = freq_vector(loc(1));
semilogx(freq_vector,fft_spectrum,'b',freq_of_first_peak,peaks(1),'*r');grid
xlabel('frequency');
ylabel('fft magnitude');
title('lift');
% strouhal number
St = freq_of_first_peak*D/U;
Megi Clen
Megi Clen on 29 Nov 2020
Hi Mathieu. I got the new plot as shown in the picture. Also, I must tell you that the new script needs a little modification, when % symbol in the line 7 and 9 is removed, it is working fine.
Many thanks for your help and information.
Have a nice day, Megi.

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