Inverse Laplace transform involving an Laplace transform and an absolute sine function

Question

I want to compute this inverse Laplace transform that involve in itself a Laplace transform, is there a general approach for this? And how can I find what the inverse looks like?

$$\mathcal{L}_\text{s}^{-1}\left[\frac{1}{1+\text{L}\cdot\text{C}_2\cdot\text{s}^2}\cdot\mathcal{L}_t\left[\left|\sin\left(\omega t+\theta\right)\right|\right]_{\left(\text{s}\right)}\right]_{\left(t\right)}=$$ $$\mathcal{L}_\text{s}^{-1}\left[\frac{1}{1+\text{L}\cdot\text{C}_2\cdot\text{s}^2}\cdot\left\{\int_0^\infty e^{-\text{s}t}\cdot\left|\sin\left(\omega t+\theta\right)\right|\space\text{d}t\right\}\right]_{\left(t\right)}=$$ $$\mathcal{L}_\text{s}^{-1}\left[\frac{1}{1+\text{L}\cdot\text{C}_2\cdot\text{s}^2}\cdot\left\{\int_0^\infty e^{-\text{s}t}\cdot\left(\frac{2}{\pi}-\frac{4}{\pi}\sum_{\text{n}\ge1}\frac{\cos\left(2\text{n}\left(\omega t+\theta\right)\right)}{4\text{n}^2-1}\right)\space\text{d}t\right\}\right]_{\left(t\right)}$$

Thanks in advance

Answer 1

According to the convolution theorem:

$$\mathcal{L}[f_1(t)*f_2(t)]=F_1(s)F_2(s)$$

where $*$ is convolution.

So here

$$\begin{align} \mathcal{L}^{-1}\left[\frac{1}{1+{L}C_2 {s}^2}\cdot\mathcal{L}\left[\left|\sin\left(\omega t+\theta\right)\right|\right]\right]&=\mathcal{L}^{-1}\left[\frac{1/{L}{C}_2}{1/{L}{C}_2+{s}^2}\right]*|\sin\left(\omega t+\theta\right)|\\[10pt] &=\frac{1}{\sqrt{LC_2}}\sin(\frac{1}{\sqrt{LC_2}}t)*|\sin\left(\omega t+\theta\right)| \end{align}$$