Laplace transform with Convolution

  • #1
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Homework Statement



Use convolution theorem to solve:

[tex]\mathfrak{L} \left \{t\int_{0}^{t} \sin \tau d\tau \right \}[/tex]

Do not solve the integral.

Homework Equations



"Convolution Theorem" in textbook is stated as:

[tex]\mathfrak{L}\left \{ f*g \right \}=F(s)G(s)[/tex]


[tex]f*g=\int_{0}^{t} \ f(\tau )g(t-\tau ) d\tau[/tex]



The Attempt at a Solution



Not quite sure how to approach this one with convolution and not solving the integral.

I need a t-τ instead of a t, but I can't have τ's in my Laplace because they won't go to S-space with any meaning.

Answer in book is :

[tex]\frac{3s^2+1}{s^2(s^2+1)^2}[/tex]


working backwards only leads me to a partial fraction type inverse transform.

Thanks in advance for help!
 
Last edited:

Answers and Replies

  • #2
Perhaps the idea is to use the property
$$\mathfrak{L}\{t f(t)\} = -F'(s) $$ where
$$f(t) = \int_0^t \sin \tau \, d\tau, $$ and to find F(s) using the convolution theorem.
 
Last edited:
  • #3
Thanks Vela, that's exactly it
 

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