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time integration via frequency domain

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Hekseli on 25 Jun 2014
Commented: Antonio Leanza on 13 Mar 2023
I have been trying to get this integration by fft and dividing by 2*pi*f*i working but the amplitude is wrong all the time. Basically my code is the same as in this example:
function int_time_data = integrate(data,dt)
N1 = length(data);
N = 2^nextpow2(N1);
if N > N1
data(N1+1:N) = 0; % pad array with 0's
df = 1 / (N*dt); % frequency increment
Nyq = 1 / (2*dt); % Nyquist frequency
freq_data = (fftshift(fft(data)));
int_freq_data = zeros(size(freq_data));
f = -Nyq:df:Nyq-df;
for i = 1 : N
if f(i) ~= 0
int_freq_data(i) = freq_data(i)/(2*pi*f(i)*sqrt(-1));
int_freq_data(i) = 0;
int_time_data = ifft(ifftshift((int_freq_data)));
int_time_data = int_time_data(1:N1);
dt = 0.01; % seconds per sample
N = 512; % number of samples
t = 0 : dt : (N-1)*dt; % in seconds
wave_freq = 17.1; % in Hertz
data = sin(2*pi*wave_freq*t);
integrated_time_data = integrate(data,dt);
integrated_time_data_analytical = -1/(2*pi*wave_freq)*cos(2*pi*wave_freq*t);
hold on;
But this produces wrong results. What is wrong?
  1 Comment
Antonio Leanza
Antonio Leanza on 13 Mar 2023
In this code the detrendization via high pass filtering is missing.

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Answers (3)

Star Strider
Star Strider on 25 Jun 2014
You did not say what was wrong about the amplitude. I did not run that code, but see if changing the freq_data line to:
freq_data = (fftshift(fft(data)/length(data)));
solves the problem.
Star Strider
Star Strider on 25 Jun 2014
That may be required, since according to the ifft documentation, it also normalizes by dividing the computed ifft by the length of the argument.
Antonio Leanza
Antonio Leanza on 13 Mar 2023
Edited: Antonio Leanza on 13 Mar 2023
The original code works well regarding that line.

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Hekseli on 25 Jun 2014
Thanks for the hint. It didn't solve the problem totally. When dividing by the length in fft I assume I should multiply by the length after the ifft? Otherwise the amplitude is far too small.
Now after the multiplication I get the following result which is attached

Michael on 27 Aug 2014
You sometimes need to perform a high pass filter on your data when you convert it from acceleration to position. That should get rid of any low frequency drift you see in your results.


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