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I am struggling to fully get how to choose proper limits of integration when calculating convolutions. Right now I am stuck on a problem where I have to show that when taking the Fourier transform of the convolution of two rectangular functions, the result equals the square sinc function.

So, I have defined:

$f(t) = 1$ for $-1 < t < 1$

$f(t) = 0$ otherwise

By taking the convolution of this function with itself, I have set up:

$$f \ast f (t)= \int_{- \infty}^{\infty} f(\tau) f(t - \tau) d \tau$$

And this is where I struggle to find the correct limits of integration. I know that when I solve this, it is an easy matter to set up a Fourier transform, and since this transform is integrated with respect to $t$, it is easy to see that the limits of integration then will be $-1$ and $1$. But how do I determine the limits here when we integrate with respect to $\tau$? Any help will be greatly appreciated!

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You will never find sinc here. – 1015 Feb 8 '13 at 12:52
I know that I will not find sinc^2 by just performing the convolution. I have to also take the FT. Or is it some other reason for that? Anyways, my main question here is really about how to choose proper limits of integration. – Kristian Feb 8 '13 at 13:06

1 Answer 1

up vote 1 down vote accepted

First, since $f(\tau)=0$ outside of $[-1,1]$, we have $$ f\ast f(t)=\int_{-1}^1f(t-\tau)d\tau. $$

Now change the variable by $u=t-\tau$: $$ f\ast f(t)=\int_{t-1}^{t+1}f(u)du. $$

If $t+1\leq -1$ (ie $t \leq -2$), or if $t-1\geq 1$ (ie $t\geq 2$), then $(-1,1)\cap (t-1,t+1)=\emptyset$ so $f\ast f(t)=0$.

Let us consider the case $-2<t<2$ now.

First, if $-2<t\leq 0$, then $1+t\leq -1$ and $[t-1,t+1]\cap [-1,1]=[-1,t+1]$ so $$ f\ast f(t)=\int_{-1}^{t+1}1du=2+t. $$

Second, if $0\leq t<2$ then $1-t\geq 1$ and $[t-1,t+1]\cap [-1,1]=[t-1,1]$ so $$ f\ast f(t)=\int_{t-1}^{1}1du=2-t. $$

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Great! Thanks a lot! – Kristian Feb 8 '13 at 13:51
Hey, you're welcome! – 1015 Feb 8 '13 at 13:55
Hi, if I have $-2 \le t \le 0$, I add +1 to both sides to get $-1 \le t+1 \le 1$. Why do you say then that $1+t \le -1$? – badmax Jun 13 '14 at 4:03

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