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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?

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  • $\begingroup$ Make sure you got the right answer! $\endgroup$ – science Feb 7 '15 at 23:24
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You are on te right way. The integral

$\int_{- \frac{a}{2}}^{\frac{a}{2}} f(t-s)ds$

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).

Finally evaluate at the endpoints of the integral and you are done.

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  • $\begingroup$ Now I really feel stupid. thanks. $\endgroup$ – user65985 Feb 7 '15 at 16:18

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