FFT from CSV data file

Hello guys,

I have a .csv data from a distance laser. The laser measured an amplitude of a swinging modul for drilling and now I need to find out what the main frequency of the swinging is. The tool rotated with n=500 1/min and the sampling frequency was 20000 Hz. Could anyone help me how to find out what the frequency is?

Thank you for any solution.

Best regards

Mathias

[William Rose]

@Mathias Braun, what have you tried so far?

[edit: correct typo in my comments]

First, you need to get your data into the Matlab workspace:

data=importdata('test_amplitude_50.2 (1).csv');
y=data.data;

How long is the vector of measurements?

N=length(y); disp(N)

      120349

Now you know the length of the data.

You know the sampling rate. Let's call it fs:

fs=20000;

What will be the frequency vector associated with the fft?

Answer 1

[Star Strider]

This corroborates previous analyses.

What other information do you want from the data?

% S = fileread('test_amplitude_50.2 (1).csv')

T1 = readtable('test_amplitude_50.2 (1).csv', 'HeaderLines',2, 'VariableNamingRule','preserve');
VN = T1.Properties.VariableNames;
T1.('Protocol TimeStamp').Format = 'yyyy-MM-dd HH:mm:ss.SSSSSS'

T1 = 120349×2 table
        Protocol TimeStamp        Sensor1 Distance1 (scaled)
    __________________________    __________________________

    2023-11-30 15:53:30.146061              7.0187          
    2023-11-30 15:53:30.146111              7.0188          
    2023-11-30 15:53:30.146161              7.0188          
    2023-11-30 15:53:30.146211              7.0187          
    2023-11-30 15:53:30.146261               7.019          
    2023-11-30 15:53:30.146311               7.019          
    2023-11-30 15:53:30.146361               7.019          
    2023-11-30 15:53:30.146411               7.019          
    2023-11-30 15:53:30.146461              7.0191          
    2023-11-30 15:53:30.146511              7.0193          
    2023-11-30 15:53:30.146561              7.0194          
    2023-11-30 15:53:30.146611              7.0196          
    2023-11-30 15:53:30.146661              7.0194          
    2023-11-30 15:53:30.146711              7.0194          
    2023-11-30 15:53:30.146761              7.0194          
    2023-11-30 15:53:30.146811              7.0193          

t = T1{:,1};
s = T1{:,2};

figure
plot(t, s)
grid

Fs = 1/mean(seconds(diff(T1{:,1})));
Fn = Fs/2;
L = numel(t);

L = 120349

NFFT = 2^nextpow2(L);
FTs = fft((s-mean(s)).*hann(L), NFFT)/L;
Fv = linspace(0, 1, NFFT/2+1)*Fn;
Iv =  1:numel(Fv);

[pks,locs] = findpeaks(abs(FTs(Iv))*2, 'MinPeakProminence',0.025);

figure
plot(Fv, abs(FTs(Iv))*2)
grid
xlim([0 100])
text(Fv(locs), pks, sprintf('\\leftarrow Magnitude = %.4f Scaled Distance Units\n    Frequency = %.4f Hz\n    Period = %.4f s', pks,Fv(locs),1/Fv(locs)), 'Vert','top')

.

[Mathias Braun]

Thank you for this great answer :)

[Star Strider]

My pleasure!

Answer 2

[William Rose]

@Mathias Braun,

data=importdata('test_amplitude_50.2 (1).csv');
y=data.data;        % signal
N=length(y);        % length of signal
fs=20000;           % sampling rate (Hz)
f=fs*(0:N-1)/N;     % vector of frequencies
Y=fft(y-mean(y));   % Y=Fourier transform of y
[~,ind]=max(abs(Y(1:round(N/2))));     % index of max value of Y in the lower half of the 2-sided spectrum
fmaxpow=f(ind);
fprintf('Frequency of max. power=%.1f Hz.\n',fmaxpow)

Frequency of max. power=13.0 Hz.

Plot the FFT and highlight the peak. I will plot the full FFT and I will zoom in on the frequencies of interest.

figure; subplot(211); semilogy(f,abs(Y),'-r',f(ind),abs(Y(ind)),'b*'); 
grid on; xlabel('Frequency (Hz)');
subplot(212); plot(f,abs(Y),'-r',f(ind),abs(Y(ind)),'b*'); 
grid on; xlabel('Frequency (Hz)');  xlim([0 50])

This is not the frequency I expected, based on your description ("500 cycles/minute", which equals 8.3 Hz). Therefore let us plot the original signal, to see the oscillation in the time domain. Plot one-second-long segments of y(t), at the start and the middle. Plot final 2 seconds.

t=(1:N)/fs;
figure; subplot(311), plot(t,y,'-r'); grid on; xlim([0 1]); xlabel('Time (s)');
subplot(312), plot(t,y,'-r'); grid on; xlim([(N/2)/fs, (N/2)/fs+1]); xlabel('Time (s)');
subplot(313), plot(t,y,'-r'); grid on; xlim([4 6.1]); xlabel('Time (s)');

The time domain plots confirm that the oscillation frequency is 13 Hz.

Good luck.

[Mathias Braun]

Thank you too for your answer. This is all right because i wanted a Frequenz of 13 Hz. The Tool I'm using does 1,5 oscillation per rotation of 360°.

[William Rose]

@Mathias Braun, you are welcome.

Answer 3

[Harald]

Hi Mathias,

after you have imported the signal using readmatrix, readtable, or readtimetable, you can calculate the fft of the amplitude. The documentation of fft has nice examples. You can then, in the easiest case, just look at the frequency with maximum power.

You may find some of the Onramps helpful: https://matlabacademy.mathworks.com/#getting-started, especially the Signal Processing Onramp.

Best wishes,

Harald