Laplace Transforms: Convolution Theorem

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Convolution Theorem is very powerful for finding inverse Laplace Transforms. This theorem states that: If $f(t)$ and $g(t)$ are piecewise continuous on $[0, \infty)$ and of exponential order, then

$\mathcal{L}\{f*g\} = \mathcal{L}\{f(t) \}\mathcal{L}\{g(t) \} = F(s)G(s)$ .

Let's apply Convolution Theorem to find

$\mathcal{L}^{-1}\begin{Bmatrix} \dfrac{k}{\left(s^{2} + k^{2} \right)^{2}}\end{Bmatrix}$ .

Let $F(s) = \dfrac{1}{s^2 + k^2}, \ G(s) = \dfrac{k}{s^2 + k^2}$ ,

so that $f(t) = \frac{1}{k}\mathcal{L}^{-1}\begin{Bmatrix} \dfrac{k}{s^2 + k^2} \end{Bmatrix} = \frac{1}{k}\text{sin}kt$ ,

$g(t) = \mathcal{L}^{-1}\begin{Bmatrix} \dfrac{k}{s^2 + k^2} \end{Bmatrix} = \text{sin}kt$ .

Thus $\mathcal{L}^{-1}\begin{Bmatrix} \dfrac{k}{\left(s^{2} + k^{2} \right)^{2}}\end{Bmatrix} = \frac{1}{k}\int_0^t\text{sin}k\tau\text{sin}(t-\tau)d\tau$

Recall from trigonometry:

$\text{cos}(A+B) = \text{cos}A\text{cos}B - \text{sin}A\text{sin}B$ , and

$\text{cos}(A-B) = \text{cos}A\text{cos}B + \text{sin}A\text{sin}B$ ,

so $\text{sin}A\text{sin}B = \frac{1}{2}\begin{bmatrix}\text{cos}(A-B) - \text{sin}(A+B) \end{bmatrix}$

If we set $A = k\tau, \ B = k(t-\tau)$ ;

$\mathcal{L}^{-1}\begin{Bmatrix} \dfrac{k}{\left(s^{2} + k^{2} \right)^{2}}\end{Bmatrix} = \frac{1}{2k}\int_0^t\begin{bmatrix}\text{cos}k(2\tau - t) - \text{cos}kt\end{bmatrix}d\tau$

= $\frac{1}{2k}\begin{bmatrix}\frac{1}{2k}\text{sin}k(2\tau - t) - \tau\text{cos}kt\end{bmatrix}_0^t$

= $\frac{\text{sin}kt - kt\text{cos}kt}{2k^2}$

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