Convolution of f with itself

Question

I was trying the following exercise:

(a) Find the fourier transform of the unit rectangular distribution of

\begin{equation} f(t) = \begin{cases} 1, \ |t|<1 \\ 0 & \text{otherwise} \end{cases} \end{equation}

which I found to be $\frac{2sin(w)}{w\sqrt{2\pi}}$

(b) Determine the convolution of f(t) with itself:

I have learned convolutions recently and I'm confuse on how to calculate:

$(f*f)(t)=\frac{1}{\sqrt2\pi}\int^{\infty}_{-\infty}f(u)f(t-u)du$

can someone explain me with detail the range changes of the integral?

Thanks!

Answer 1

$(f*f)(t)=\int_{(-1,1)\cap (t-1,t+1)} du= m((-1,1)\cap (t-1,t+1))$ where $m$ is Lebesgue measure. Consider then case $t \leq -2, -2<t\leq 0, 0 <t\leq 2$ and $t >2$ to evaluate this.

Answer 2

$$ \int_{-\infty}^\infty f(u)f(t-u)du$$

the only way for the integrand not to vanish is $|u|<1$ and $|t-u|<1$, so you need to have $$u\in (-1,1)\cap (-1+t,1+t)$$

Check for which $t$ the intersection is not empty, and compute the integral inside the intersection (since $f=1$ where it doesn't vanish, it's just the length of the intersection).

Answer 3

One can apply the convolution theorem: $\mathcal{F}\big\{(f*f)(t)\big\} = (\hat{F}(w))^2$ and then transform it back.