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Suppose you have two real functions $g(x)$ and $f(x)$ which are non-zero only for $x \geq 0$. Now you want to compute the convolution of the two functions and you can approximate it by integrating from zero to some finite value $T$ provided $T$ is sufficiently large. This is also equivalent to multiplying either $g(x)$ or $f(x)$ by a one-sided top hat function which is equal to one for $0 \leq x \leq T$ and zero otherwise.

Now what would the Fourier transform of $g \ast f$ be? If $T$ is sufficiently large then the truncated convolution integral should be a good approximation of $g \ast f$, but then the FT of the one-sided top hat function is given approximately by $Te^{i\omega T/2}$ which means that the larger $T$ is the more this linear term in $\omega$ dominates in the Fourier transform. By the convolution theorem then, there is an additional phase shift that is linear in $\omega$ added onto FT($g \ast f$).

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As the functions $g$ and $f$ are nonzero only for $x\geq0$, the convolution integral has in fact a lower and an upper boundary. It is given by $$(g \star f)(x) = \int_0^x\!dy \,g(y) f(x-y).$$

The Fourier Transform is not the proper framework to treat such convolutions. Rather you should take a look at the Laplace transform. We have that $$G(s) = \mathcal{L} g(s) = \int_0^\infty\!dt\,g(t) e^{-st} .$$ With that you find that $$\mathcal{L}(g\star f)(s) =G(s) F(s).$$

If you really insist on Fourier transforms, you can observe that the Laplace-transform is given by the Fourier transform evaluated at $s= -i \omega$ (depending a bit on your definition of the Fourier transform). So you have that $$\widehat{g\star f}(-i \omega) = \hat{G}(-i\omega) \,\hat{F}(-i\omega)$$ with $\hat\cdot$ denoting the Fourier transform.

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  • $\begingroup$ I see, thanks. But is the problem I observed "real"? As in, a truncated convolution integral would not give me the expected frequency dependency for a non-truncated $g \ast f$? $\endgroup$ – David Young Mar 27 '17 at 9:58
  • $\begingroup$ Also I think the same problem also exists for functions that are not one-sided? Since the FT of a top hat function for $-T \leq x \leq T$ is a sinc function. $\endgroup$ – David Young Mar 27 '17 at 11:22
  • $\begingroup$ @DavidYoung: could you please explain again which problem you are referring to (if the problem is more generic then it would be useful to state it as simple as possible maybe even involving no convolution???) $\endgroup$ – Fabian Mar 27 '17 at 11:24
  • $\begingroup$ The problem is that, I can compute $g \ast f$ approximately by integrating over the domain $-T \leq x \leq T$ provided that $T$ is sufficiently large. But then this is also equivalent to multiplying $g$ or $f$ by a top hat function which equals to 1 over $-T \leq x \leq T$. However the FT of the approximated convolution obtained $g \ast f$ would not be the same as the actual FT of $g \ast f$ as there would be an extra factor involving $sinc(\omega T)$. I am unable to see how they converge to the same thing as $T \rightarrow \infty$. $\endgroup$ – David Young Mar 27 '17 at 11:32
  • $\begingroup$ These limits are tricky. The correct statement is that there is a convolution with a sinc-function. In the limit $T\to\infty$ this function approaches a delta-distribution. And it is a feature of this distribution that $\delta \star f = f$. $\endgroup$ – Fabian Mar 27 '17 at 11:37

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