Laplace Transform : Convolution

**naikn** · April 4th 2010, 10:59 AM

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)

**chisigma** · April 4th 2010, 12:00 PM

Given $f(t)$ and $g(t)$ two functions and $F(s)$ and $G(s)$ their L-trasforms. The convolution is defined as...

$f*g = \int_{0}^{t} f(t-\tau)\cdot g(\tau)\cdot d\tau$ (1)

Is...

$\mathcal{L} \{f*g\} = F(s)\cdot G(s)= \mathcal{L}\{t^{2}\}\cdot \mathcal{L} \{\cos 2t\} = \frac{2}{s^{3}} \cdot \frac{s}{s^{2} + 4} = \frac{2}{s^{2}\cdot (s^{2} +4)}$ (2)

Kind regards

$\chi$ $\sigma$

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