fft-based differentiation on a non-periodic time span

Hi everybody

I'm puzzled with a basic property of the Fourier transform. We all know that if the fourier transform of f(t) is F(w), then the fourier transform of df/dt is simply jwF(w). Well, to show this numerically using fft you need your f(t) to be periodic on your time span. Suppose that you define f(x) = sin(x) for -pi<x<pi. I want to show that fft(cos(x)) = jw*fft(sin(x)). Consider the following code:

clc

clear

close all

N = pow2(6);

a = pi;

dx = 2*a/N;

x = -a+dx*(0:N-1);

m = 2*pi/(2*a)*[0:N/2-1, 0, -N/2+1:-1];

%%

f = sin(x);

fx = cos(x); % df/dx

Ff = fft(f);

Ffx = 1i*m.*Ff;

iFfx = real(ifft(Ffx(:),'symmetric'));

plot(x,iFfx,'b')

hold on

plot(x,fx,'ro')

which is perfect. But the problem arises when the function is not periodic over the specified domain. For example, if I choose a = 1 in the example above, I wouldn't be satisfied with the results anymore. Are there any tricks so that I can solve this problem?

Any suggestion is appreciated.

[zhoumu wu]

sunaina2018-Calculating Numerical Derivatives using Fourier Transform some pitfalls and how to avoid them