In my ODE class, I was taught that the convolution of two functions is calculated by: $$\left(f*g\right)\left(t\right)=\int_0^t{f\left(u\right)g\left(t-u\right)du}$$ According to Wikipedia, the convolution is defined as: $$\left(f*g\right)\left(t\right)=\int_{-\infty}^\infty{f\left(u\right)g\left(t-u\right)du}$$ and can be simplified to the definition I was taught if the functions $f\left(t\right)$ and $g\left(t\right)$ are only defined for $t\in\left[0, \infty\right)$. $$$$ It makes sense why the lower bound would become $0$ when the functions are only defined for $t\in\left[0, \infty\right)$, but I am confused why the upper bound becomes $t$ instead of remaining at $\infty$.
- Wouldn't this change the value of the convolution?
- If so, do both definitions have the property that $\mathcal{L}^{-1}\left\{F\left(s\right)G\left(s\right)\right\} = \left(f*g\right)\left(t\right)$?