Help interpreting Fast Fourier Transform.

I followed this video : https://www.youtube.com/watch?v=cw9eoSzln_k&ab_channel=Study%26Tutor

to perform a FFT on data I collected that reperesented voltage vs time. The data can be found here https://mega.nz/file/JX81BI5L#ivW7WgkKpxjhqMBXC2nUYuc8ykC_sXahZuAu1UaMc6w

The code I used is as follows :

t = readmatrix('fftpoop.xlsx');

>> dt = 16.788

dt = 16.7880

>> fs = 1/dt;

>> plot(t(1:16789,2))

>> %time

>> t = t(1:16789,2);

>> t_f = fft(t);

>> m = length(t_f);

>> freq = (-m/2:(m/2-1))*fs/(m-1);

>> plot(freq,fftshift(abs(t_f)))

And i recieved a plot that looks like this :

I manually counted the average time between peaks from the original time data to be .37s. I used this time to calculate the a distance, D = 8.3*10^-7 m/s * .37 s = 307 nm* (2) = 614 nm. This is very close to the value I was expecting (633 nm).

I believe that If I find the frequency from this fft that has the highest peak, then I can use that to use the equation, time = 2pi/freq and verify my time is correct / make my time variable more accurate, and get a better value for D calculated above

I tried using the x value centered at 0 and the next highest peak from the 0 peak in the above graph as my frequency, but they all produce numbers wayyyy bigger from the expected distance.

Am I using this data correctly?

Answer 1

[Star Strider]

My analysis —

M1 = readmatrix('fftpoop')

M1 = 16789×2

6.7970 -0.3610 6.7980 -0.3620 6.7990 -0.3650 6.8000 -0.3680 6.8010 -0.3680 6.8020 -0.3650 6.8030 -0.3600 6.8040 -0.3590 6.8050 -0.3610 6.8060 -0.3660

t = M1(:,1);
s = M1(:,2);

L = size(M1,1)

L = 16789

Ts = mean(diff(t));
Fs = 1/Ts;
Fn = Fs/2;

figure
plot(t, s)
grid
xlabel('Time (s)')
ylabel('Amplitude (V)')

NFFT = 2^nextpow2(L)

NFFT = 32768

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',0.015);

time = 1/Fv(locs)

time = 0.4201

figure
plot(Fv, abs(FTs(Iv))*2)
grid
xlabel('Frequency (Hz)')
ylabel('Magnitude (V)')
xlim([0 20])
ylim([min(ylim) 0.022])
text(Fv(locs), pks, sprintf('\\leftarrow %.3f Hz\n     %.3f V', Fv(locs), pks), 'Horiz','left', 'Vert','top')

I assume your ‘time’ is based on radian frequencies, so I calculated it as the period of the peak frequency in Hz (since my plot is in Hz), giving the period in seconds. I am not certain what the data are or what your analysis requires, so I will defer to you for those.

.