After using "FFT" getting NAN. Why?

58 vues (au cours des 30 derniers jours)
Danish Ahmad
Danish Ahmad le 11 Avr 2019
Commenté : Star Strider le 2 Sep 2020
clc
close all
clear all
%%
data = xlsread('freq15.xlsx','Group Name #2');
v= data(:,1);
plot (v);
title('Voltage signal with noise')
xlabel('Samples')
ylabel('Amplitudes')
%%
%plot magnitude specturm of a signal
clc
format long
X_mag=abs((fft(v)))
figure(1)
plot(X_mag)
xlabel('DFT Bins')
ylabel('Magnitude')
%%
Above is a code using FFT in MATLAB. The outcome of the code is NAN+NANi. How to get rid of NAN or what is the reason of this outcome?
  2 commentaires
Jan
Jan le 11 Avr 2019
Please use the buttons to format the code. It is a good idea to post a question in the best readable format.
Omit the brute clearing header "clc; close all; clear all". This is a waste of time and cargo-cult-programming. If you want to keep your workspace clean, use functions.
Without knowing, what the contents of v is, there is no chance to understand, why abs((fft(v))) is NaN. So please show us, what your input data are.
Danish Ahmad
Danish Ahmad le 14 Avr 2019
Thank you for your response.
Regards

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Star Strider
Star Strider le 11 Avr 2019
... what is the reason of this outcome?
You probably either have a NaN in your data, or an empty cell. Both will appear as NaN in the xlsread output.
If that is the situation, your best option is to interpolate using the fillmissing (link) function (R2016b and later releases). The reason is that Fourier transforms require regularly-sampled time-domain signals for them to provide reliable output. Omitting the NaN value disrupts this regularity and results in an inaccurate fft result. Interpolating the NaN values will not provide as accurate a result has having the original value, however it is the only reliable alternative.
  12 commentaires
Diego Soto Chávez
Diego Soto Chávez le 2 Sep 2020
Hi Star Strider
know you another option like "fillmissing" but for previosly versiones of matlab? Thanks!!
Star Strider
Star Strider le 2 Sep 2020
Something like this woulld likely work:
t = 1:8
y = rand(size(t))
y([3 5]) = NaN
ti = t;
ti(isnan(y)) = []
y(isnan(y)) = []
ynew = interp1(ti, y, t)
.

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