# Approximating the FT with the FFT

Example Consider the function $f(t)=e^{-|t|}$. Its Fourier transform is $\hat f(\omega) = \sqrt{2/\pi} (1+\omega^2)^{-1}$. Numerically approximate $\hat f$ using the FFT with interval $[-2\pi, 2\pi]$ and $n$ = 2048. Graph both $\hat f$ and its approximation $\hat f_{ap}$ over values of $\omega$ in an interval about $\omega=0$. We will use the matlab m-file function given below.

function [yhat,omega]=ft_approx(y,a,b)
% FT_APPROX computes an approximate Fourier transform from samples
% contained in the vector y. The arguments a and b are the intial
% and final times for samples to be taken. The outputs are the
% approximate Fourier transform, contained in yhat, and the
% corresponding frequencies, contained in omega. The frequencies
% are in radians per second.
%
% F. J. Narcowich, Mar. 20, 2012.
%
s=size(y);
y=y(:).';
N=floor(length(y)/2);
if N < length(y)/2,
M=N;
else
M=N-1;
end
dt=(b-a)/(2*N);
del_omega=2*pi/(b-a);
omega=((-N):(M))*del_omega;
phase_factor=exp(-i*a*omega);
yhat=phase_factor.*fftshift(fft(y))*(dt/sqrt(2*pi));
yhat=reshape(yhat,s);
omega=reshape(omega,s);


Solution First define f and its Fourier transform in matlab using the anonymous'' function method:

f = @(x) exp(-abs(x))
fhat = @(x) sqrt(2/pi)*(1+x.^2).^(-1)


Next, define arrays for dealing with n = 2L time samples.

n=2048;
t=(-2*pi):4*pi/n:(2*pi - 4*pi/n);
y=f(t);


Finally, find the approximate FT, yhat, and the values of the frequency on which it is calculated, omega. Do the same for the actual FT, and then plot the results. (The variable ind restricts the points plotted to a small frequency band about ω =0.)

[yhat,omega]=ft_approx(y,-2*pi,2*pi);
yhat_actual = fhat(omega);
ind=1000:1048;
subplot(2,1,1), plot(omega(ind),yhat_actual(ind))
subplot(2,1,2), plot(omega(ind),real(yhat(ind)))



Updated 3/21/2014