Any convolution is defined as
\((f*g)(t)=\displaystyle\int^{\infty}_{-\infty}f(x)g(t-x)dx=\int^{\infty}_{-\infty}g(x)f(t-x)dx\)
Trying to do it... Never done this type of question before.
\((f*h)(3)=\displaystyle\int^{\infty}_{-\infty}f(x)h(3-x)dx=\int^{\infty}_{\infty}h(x)f(3-x)dx\)
But f(x) is only defined in {-1<=x<=1|x\(\in\mathbb{Z}\)}
h(x) is only defined in {1<=x<=5|x\(\in\mathbb{Z}\)}
So Idk what does it mean by integrating f(x)h(3-x) from infinity to negative infinity... How on Mars could I do that?
