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I have to find the convolution of two identical square functions.

[tex]

F(t)=\left\{\begin{array}{cc}1,&\mbox{ if } 0\leq t < 1\\0, & \mbox otherwise\end{array}\right.

[/tex]

I need to calculate F * F

[tex](F * F)(t) = \int_{\infty}^{-\infty} F(u)F(t-u) du[/tex]

The first thing I did was change integration limit to [0,1] and F(u) to 1. My main problem is F(t-u).

F(t-u) = 1 when

0 < t-u < 1

-1 < u-t < 0

t-1 < u < t

I think thus far I haven't made any mistakes. Now what do I do to substitue F(t-u)? I think my options are to either redefine the integration limit (but it should be [0,1]) or somehow write F(t-u) as a algebraic function. I don't see how I can do the latter. If someone can point me in the right direction I would appreciate it.

oh and the resulting convolution is a hat function.

[tex]

F(t)=\left\{\begin{array}{cc}1,&\mbox{ if } 0\leq t < 1\\0, & \mbox otherwise\end{array}\right.

[/tex]

I need to calculate F * F

[tex](F * F)(t) = \int_{\infty}^{-\infty} F(u)F(t-u) du[/tex]

The first thing I did was change integration limit to [0,1] and F(u) to 1. My main problem is F(t-u).

F(t-u) = 1 when

0 < t-u < 1

-1 < u-t < 0

t-1 < u < t

I think thus far I haven't made any mistakes. Now what do I do to substitue F(t-u)? I think my options are to either redefine the integration limit (but it should be [0,1]) or somehow write F(t-u) as a algebraic function. I don't see how I can do the latter. If someone can point me in the right direction I would appreciate it.

oh and the resulting convolution is a hat function.

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