How to determine and adjust the x axis after taking the 1D FFT?

조회 수: 44 (최근 30일)
L'O.G. 2022년 12월 19일
답변: Paul 2022년 12월 19일
I am trying to take an FFT, but have trouble figuring out the x axis. The time vector is of length N = 100000. The vector has uniformly spaced time points, i.e. dt = t(2)-t(1). How do I find out what the values of f should be? And how can I change the spacing df?

채택된 답변

Star Strider
Star Strider 2022년 12월 19일
I usually do something like this —
t = linspace(0, 100, 10000); % Time Vector
s = sum(sin([1 5 10 15 20 25 30 35 40 45].'*2*pi*t)); % Signal (Here A Vector)
plot(t, s)
L = numel(t); % Length Of The Signal Vector
dt = t(2)-t(1); % Sampling Time Increment
Fs = 1/dt; % Sampling Frequency
Fn = Fs/2; % Nyquist Frequency
NFFT = 2^nextpow2(L); % For Efficiency
FTs = fft(s(:).*hamming(L),NFFT)/L; % Fourier Transform (Windowing Not Required, Shown For Information)
Fv = linspace(0, 1, NFFT/2+1)*Fn; % Frequency Vector For One-Sided Fourier Transform Plot
Iv = 1:numel(Fv); % Index Vector (For Convenience)
plot(Fv, abs(FTs(Iv))*2) % Multiply By 2 To Correct For Energy Division Between Positive And Negative Frequencies
xlabel('Frequency (Units)')
ylabel('Magnitude (Units)')
I am not certain what you are doing, however something like this should work.
  댓글 수: 4
L'O.G. 2022년 12월 19일
Thank you so much!
Star Strider
Star Strider 2022년 12월 19일
As always, my pleasure!

댓글을 달려면 로그인하십시오.

추가 답변 (1개)

Paul 2022년 12월 19일
Hi L'O.G.
Let Y be the output of fft (with or without zero padding)
Let NFFT = numel(Y)
Let Ts = dt = t(2) - t(1) and Fs = 1/Ts.
Then the frequency vector that corresponds to the values in Y is given by
f = (0: (NFFT-1))/NFFT*C
where the proportionality constant C is often one of:
C = 1 % cycle/sample
C = 2*pi % rad/sample
C = Fs % cycle/sec, i.e. Hz
C = 2*pi/Ts % = 2*pi*Fs rad/sec.
If Y is the output of fftshift, then f is
f = (-(NFFT-1)/2 : (NFFT-1)/2)/NFFT*C % N odd
f = ((-NFFT/2) : (NFFT/2-1))/NFFT*C % N even


Help CenterFile Exchange에서 Spectral Measurements에 대해 자세히 알아보기





Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by