- #1
CrusaderSean
- 44
- 0
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.
Last edited: