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

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L'O.G.
L'O.G. am 19 Dez. 2022
Beantwortet: Paul am 19 Dez. 2022
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?

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Star Strider
Star Strider am 19 Dez. 2022
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)
figure
plot(t, s)
grid
xlabel('Time')
ylabel('Amplitude')
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)
figure
plot(Fv, abs(FTs(Iv))*2) % Multiply By 2 To Correct For Energy Division Between Positive And Negative Frequencies
grid
xlabel('Frequency (Units)')
ylabel('Magnitude (Units)')
I am not certain what you are doing, however something like this should work.
.
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Paul
Paul am 19 Dez. 2022
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

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R2021b

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