how can find frequency from an fft function?

hi, I'm doing a coupled oscillators experiment involving two measured angles changing over time. i imported the data -csv - with one column (867x1) for each variable; time, angle 1, angle 2.

when i enter: plot(abs(fftshift(fft(Y1)))); (where Y1 = the column vector for angle 1.) i am unsure of the meaning of the figure that is produced, specifically the numbers on the x axis.

i have been told that the y axis is the amplitude and that the x axis is the index, but i do not understand how i am meant to understand the frequency of the oscillation from these two things.

additionally every post i have found connected to this problem refers to the following equation: Frequency = i * (Fs/N)

am i correct in assuming that "i" is the x coordinate? that N is 867? and Fs i am entirely confused as to how to calculate.

i have attached the figure, please advise.

Answer 1

[Rick Rosson]

   Fs = 40;                   % samples per second
   N = length(Y1);            % samples
   dF = Fs/N;                 % hertz per sample

   f = -Fs/2:dF:Fs/2-dF + (dF/2)*mod(N,2);      % hertz

   Y1_fft = fftshift(fft(Y1))/N;

   figure;
   plot(f,abs(Y1_fft));

[Ullah Nadeem]

Hello Rick!

Hope you're fine, I guess dF = Fs/N is hertz {(samples per sec)/(samples) = 1/sec = hertz} not hertz per sample, hertz per sample would be dF/N. Please correct me if I'm wrong...

Thank you!

Answer 2

[Guillaume]

It sounds like you're lacking the basics of fourier transform analysis. You would probably better off learning about it before trying to apply it.

Yes, i is your sample number, your x, axis. N is the number of sample (usually you try to make it a power of 2). Fs is your sampling frequency. You must have acquired your measurement at a fixed frequency.

[mrkovi]

i understand that i'm lacking the basics Guillaume, i'm a first year physics student thrown into an advanced physics lab. the data was acquired by an electronic sensor that measured the movement of the oscillation... if i look at the csv, it appears for every 1.00 seconds that passes, 40 measurements were taken i.e. every 0.025 seconds... pardon my ignorance but how do i get the sampling frequency from that?

[Guillaume]

Your sampling frequency is thus 40 Hz (40 samples per second).

Answer 3

[Star Strider]

The current R2015b documentation is confusing (at least in my opinion). See the R2015a documentation for fft for a more appropriate implementation. The only correction that needs to be made to the code between the first two plot figures is to multiply the result of the fft by 2 with a one-sided fft.

This will give the correct amplitudes:

Y = fft(y,NFFT)*2/L;