Truncated Convolution Integral

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($g \ast f$).

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.
• 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 $g \ast f$? – David Young Mar 27 '17 at 9:58
• 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. – David Young Mar 27 '17 at 11:22
• 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 $T \rightarrow \infty$. – David Young Mar 27 '17 at 11:32
• 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 $\delta \star f = f$. – Fabian Mar 27 '17 at 11:37