# Calculate Convolution of a funcion

Given $$f(t)=\mathbb{1}_{[-\frac a 2,\frac a 2]}(t)=\cases{1\qquad t\in[-\frac a 2,\frac a 2]\\0\qquad\text{otherwise}}\quad(0<a<\pi)$$, Calculate $f\ast f$, the convolution on $\mathbb{T}$

Namely, it's $$h(t)=\int_{\mathbb{T}}f(t-s)g(s)ds=\int_{[-\frac a 2,\frac a 2]}f(t-s)ds=\int_{[-\frac a 2,\frac a 2]}f(s-t)ds$$ but here I'm stuck. That's clear that $(f\ast f)(0)=a$ but I'm not sure about other values of the convolution here. How can I finish the calculation?

• Make sure you got the right answer! – science Feb 7 '15 at 23:24

$\int_{- \frac{a}{2}}^{\frac{a}{2}} f(t-s)ds$
can be solved by the Substitution: $u = t-s$ (treat $t$ like a constant; don't Forget adjusting the Integration interval!). Other hint:
$\int f(u)du =$ (0 for $u < - \frac{a}{2}$)($u$ for $u > - \frac{a}{2}$ and $u < \frac{a}{2}$)(constant elsewhere).