Laplace transform of the integral of a difference equation

[Roberto]

Hi, Please I need some help, how can I get the Laplace transform of the integration of a difference equation??

$\int _{ 0 }^{ \infty }{ { e }^{ -st } } \int _{ -\tau }^{ 0 }{ G(\theta )x(t+\theta )d\theta } dt$

Many thanks in advanced.

 

[chisigma]

Hi, Please I need some help, how can I get the Laplace transform of the integration of a difference equation??

$\int _{ 0 }^{ \infty }{ { e }^{ -st } } \int _{ -\tau }^{ 0 }{ G(\theta )x(t+\theta )d\theta } dt$

Many thanks in advanced.

Wellcome on MHB Roberto!...

... defining the convolution of g(t) and x(t) as...

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

... for the 'convolution theorem' is...

$\displaystyle \mathcal{L} \{ f * g\ \} = \int_{0}^{\infty} (f * g)\ e^{- s\ t}\ d t = G(s)\ X(s)\ (2)$

 

[I like Serena]

Hi, Please I need some help, how can I get the Laplace transform of the integration of a difference equation??

$\int _{ 0 }^{ \infty }{ { e }^{ -st } } \int _{ -\tau }^{ 0 }{ G(\theta )x(t+\theta )d\theta } dt$

Many thanks in advanced.

Hi Roberto! Welcome to MHB! :)

Swap the order of integration:
$$\int _{ -\tau }^{ 0 }{ G(\theta ) } \int _{ 0 }^{ \infty }{ { e }^{ -st } x(t+\theta )\,dt\,d\theta }$$
Now the inner integral is a time shifted Laplace transform of x.