Confusion with FFT for a specific part of the whole signal
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Ancalagon8
am 4 Feb. 2017
Kommentiert: Star Strider
am 27 Jan. 2021
As a newbie i am working on a signal from an accelerometer and i am trying to produce an fft for a specific part of the whole signal. I am a little bit confused about which code for the fft is the appropriate to use for my case.
here is my code:
clear all
close all
load signal
tvec=noact(:,1);
acc=noact(:,2);
%% ALL SIGNAL
L=length(signal); % Length of signal
t_max=max(signal(:,1)); % Tmax
Fs=L/t_max; % Sampling Rate
T = 1/Fs; % Sample time
%% PART OF SIGNAL
trace1=signal(Fs*25:Fs*32,:); % select part of signal
matrice1=trace1(:,2); % accelerations column
L1=length(matrice1); % Length of trace (number of samples)
t1 = (0:L1-1)*T; % Time vector
% Frequency vector
f1=(0:Fs/length(matrice1):(length(matrice1)-1)*Fs/length(matrice1));
FrecSign=abs(fft(matrice1))*T;
figure
plot(FrecSign);
xlabel Frequency(Hz)
ylabel Amplitude
title 'FFT'
or
NFFT = 2^nextpow2(L); % Next power of 2 from length of matrice1
Y = fft(matrice1,NFFT)/L;
f = Fs/2*linspace(0,1,NFFT/2+1); % create frequency vector
figure % Plot single-sided amplitude spectrum.
plot(f,2*abs(Y(1:NFFT/2+1)))
title('Single-Sided Amplitude Spectrum of y(t)')
xlabel('Frequency (Hz)')
ylabel('|Y(f)|')
Thanks in advance
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Akzeptierte Antwort
Star Strider
am 4 Feb. 2017
I cannot follow our code.
However, you cannot ‘produce an fft for a specific part of the whole signal’, since you have to take the fft (link) of the entire signal if you want valid data.
If you want to view only part of the signal, use the axis function to limit the frequency (here x) axis to the range you want.
If you want to eliminate specific frequencies from your accelerometer signals (for example a constant offset, low-frequency baseline variations, high-frequency noise), use a discrete filter. There are several ways to to this, with the designfilt function being the easiest to use (in my opinion).
Weitere Antworten (2)
Ancalagon8
am 6 Feb. 2017
1 Kommentar
Star Strider
am 6 Feb. 2017
I’m not sufficiently certain that I understand your data well enough to address that. I would analyse one variable at a time in each call to periodogram, so most likely (ii).
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