L { Integral (t- T)^2 Cos(2T) dT }
T = tau
The integral goes from 0 to t
I am unsure how to get to the answer, which is 2/ (s^2)(s^2+4)
Given $\displaystyle f(t)$ and $\displaystyle g(t)$ two functions and $\displaystyle F(s)$ and $\displaystyle G(s)$ their L-trasforms. The convolution is defined as...
$\displaystyle f*g = \int_{0}^{t} f(t-\tau)\cdot g(\tau)\cdot d\tau$ (1)
Is...
$\displaystyle \mathcal{L} \{f*g\} = F(s)\cdot G(s)= \mathcal{L}\{t^{2}\}\cdot \mathcal{L} \{\cos 2t\} = \frac{2}{s^{3}} \cdot \frac{s}{s^{2} + 4} = \frac{2}{s^{2}\cdot (s^{2} +4)}$ (2)
Kind regards
$\displaystyle \chi$ $\displaystyle \sigma$