# inverse Fourier transform without error problem

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Moses Eshovo OJO on 6 Sep 2021
I have a MATLAB code that i need to perform inverse fourier transform, but it does not produce any output. It appears that some data points in the frequency domain were skipped, maybe this could be the reason why MATLAB could not perform the inverse fourier transform to time domain.
clear all;close all;clc
%OBJECTIVE OF THIS CODE; Single shot characterisation of a given THZ pulse i.e. to characterize a THz pulse by using chirped pulse
N=1024*2^14;
dt=1e-14;
t=(0:N/2-1)*dt;
df=1/(N*dt);
f=(0:N/2-1)*df;w=2*pi*f;
d = 500e-6;
m=(289.27-(6*(f*1e-12).^2))./(29.16-((f*1e-12).^2));
s=3.2394;c=3e8;
T=0.15e-12;
first_term=exp(-(w).^2*T.^2/4).*(sqrt(2)*pi*T./(q.^2 - s^2));
second_term=0.5.*(1+ (s./q)).*(exp(i.*w.*q*d./c));
third_term=(0.5*(1- (s./q)).*exp(-i.*w.*q.*d./c));
fourth_term=exp(i.*w.*s.*d./c);
result= first_term.*(second_term + third_term - fourth_term);
RESULT=abs(result).^2;
figure(1);plot(f,RESULT);title('frequency plot')
figure(2);plot(t, ifft(RESULT));title('time plot')

David Goodmanson on 6 Sep 2021
Hi MO,
It turns out that the variable m is positive in most of its range, but is negative for the array elements
>> [min(find(m<0)) max(find(m<0))]
ans = 905971 1164920
because the denominator of the expression for m is negative and the numerator is positive. Since you define q as
q=real(sqrt(m)),
q = 0 in that region. The 2nd and 3rd terms contain a factor of s/q, so the 2nd and 3rd terms are infinite in that region, and subtracting the 2nd and 3rd terms means that the result has nans in that region. You can plot result or RESULT since plotting ignores nans. But the ifft doesn't work.
So you will need to decide what q should be doing when m is negative.
Moses Eshovo OJO on 6 Sep 2021
Thank you David.