# Laplace transform with Convolution

1. Feb 7, 2012

### NJJ289

1. The problem statement, all variables and given/known data

Use convolution theorem to solve:

$$\mathfrak{L} \left \{t\int_{0}^{t} \sin \tau d\tau \right \}$$

Do not solve the integral.

2. Relevant equations

"Convolution Theorem" in textbook is stated as:

$$\mathfrak{L}\left \{ f*g \right \}=F(s)G(s)$$

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

3. 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.

$$\frac{3s^2+1}{s^2(s^2+1)^2}$$

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

Last edited: Feb 7, 2012
2. Feb 8, 2012

### vela

Staff Emeritus
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: Feb 8, 2012
3. Feb 8, 2012

### NJJ289

Thanks Vela, that's exactly it