Obtaining result in Frequency Domain using FFT in Matlab

Attached here are the two vectors: Time and Displacement

Vector displ.mat is the displacement vector obtained after running a dynamic response analysis under a load.

I want to obtain the fundamental frequency which can be easily picked up by plotting the two vectors using `plot(Time,displ)`. In this plot, one can easily see a sine wave with a frequency of roughly 0.5 Hz. So, I expect to see this peak of 0.5 Hz using `fft` function of Matlab.

I am using the following code and but not getting the expected peak at 0.5 Hz:

    load Time.mat
    load displ.mat

    t = Time;
    Fs = 149;
    x = displ;
    x = detrend(x,0);
    xdft = fft(x);
    freq = 0:Fs/length(x):Fs/2;
    xdft = xdft(1:length(x)/2+1);

    plot(freq,abs(xdft));
    [~,I] = max(abs(xdft));
    fprintf('Maximum occurs at %d Hz.\n',freq(I));
    plot(Time,displ)

Answer 1

[Star Strider]

Here you go:

T = load('Rehan Rehan Time.mat');
D = load('Rehan Rehan displ.mat');

t = T.Time;
d = D.displ;

L = length(t);                                              % Signal Length (samples)
Ts = mean(diff(t));                                         % Sampling Interval (seconde)
Fs = 1/Ts;                                                  % Sampling Frequency (Hz)
Fn = Fs/2;                                                  % Nyquist Frequency (Hz)

dm = d-mean(d);
ft_d = fft(dm)/L;                                           % Normalised Fourier Transform
Fv = linspace(0, 1, fix(L/2)+1)*Fn;                         % Frequency Vector
Iv = 1:length(Fv);                                          % Index Vector

[pks,idx] = findpeaks(abs(ft_d(Iv))*2);
lbl = sprintf('Frequency = %.3f\nAmplitude = %.3f', Fv(idx), pks);

figure(1)
plot(Fv, abs(ft_d(Iv))*2)
grid
xlabel('Frequency (Hz)')
ylabel('Amplitude (m)')
text(Fv(idx)+0.1, pks, lbl, 'HorizontalAlignment','left', 'VerticalAlignment','top')

[Rehan Rehan]

Thanks a lot for a perfect answer!

In my case, I am now getting an error at "findpeaks" ...what do i do?

Thanks

[Star Strider]

My pleasure!

The findpeaks function is in the Signal Processing Toolbox. Since you have only one peak, you can get the same result with the max function (I verified that the results are the same):

[pks,idx] = max(abs(ft_d(Iv))*2);