How to find the inverse Laplace transform of $F(s) = {se^{-\frac{s}{2}} + \pi e^{-s} \over (s^2 + \pi^2)}$?

Question

How to find the inverse Laplace transform of $$ F(s) = {se^{-\frac{s}{2}} + \pi e^{-s} \over (s^2 + \pi^2)} $$

Can't use partial fractions, it isn't in any of the standard forms either. I have no clue how to approach this question. Any hints please?

Answer 1

Given that : $$L^{-1}\left\{\frac{e^{-\frac{s}{2}}s+\pi e^{-s}}{s^2+\pi ^2}\right\}$$

Expand : $$=L^{-1}\left\{\frac{e^{-\frac{s}{2}}s}{s^2+\pi ^2}+\frac{\pi e^{-s}}{s^2+\pi ^2}\right\}$$

$$\mathrm{Use\:the\:linearity\:property\:of\:Inverse\:Laplace\:Transform:}$$ $$\mathrm{For\:functions\:}f\left(s\right),\:g\left(s\right)\mathrm{\:and\:constants\:}a,\:b:$$ $$L^{-1}\left\{a\cdot f\left(s\right)+b\cdot g\left(s\right)\right\}=a\cdot L^{-1}\left\{f\left(s\right)\right\}+b\cdot L^{-1}\left\{g\left(s\right)\right\}$$

We will get : $$=L^{-1}\left\{\frac{se^{-\frac{1}{2}s}}{s^2+\pi ^2}\right\}+\pi L^{-1}\left\{\frac{e^{-s}}{s^2+\pi ^2}\right\}$$

Solve first part : $$L^{-1}\left\{\frac{se^{-\frac{1}{2}s}}{s^2+\pi ^2}\right\}$$

Notice that : $$\mathrm{Apply\:inverse\:transform\:rule:\quad if\:}L^{-1}\left\{F\left(s\right)\right\}=f\left(t\right)$$ $$\mathrm{\:then}\:L^{-1}\left\{e^{-as}F\left(s\right)\right\}=H\left(t-a\right)f\left(t-a\right))$$ $$\mathrm{Where\:}\text{H}\left(t\right)\mathrm{\:is\:Heaviside\:step\:function}$$ $$\mathrm{For\:}\frac{se^{-\frac{1}{2}s}}{s^2+\pi ^2}:\quad F\left(s\right)=\frac{s}{s^2+\pi ^2},\:\quad a=\frac{1}{2}$$

We will get : $$=\text{H}\left(t-\frac{1}{2}\right)L^{-1}\left\{\frac{s}{s^2+\pi ^2}\right\}\left(t-\frac{1}{2}\right)$$ $$=\text{H}\left(t-\frac{1}{2}\right)\cos \left(\sqrt{\pi ^2}\left(t-\frac{1}{2}\right)\right))$$

Solve the second part : $$L^{-1}\left\{\frac{e^{-s}}{s^2+\pi ^2}\right\}$$

Notice that : $$\mathrm{Apply\:inverse\:transform\:rule:\quad if\:}L^{-1}\left\{F\left(s\right)\right\}=f\left(t\right)$$ $$\mathrm{\:then}\:L^{-1}\left\{e^{-as}F\left(s\right)\right\}=H\left(t-a\right)f\left(t-a\right))$$ $$\mathrm{Where\:}\text{H}\left(t\right)\mathrm{\:is\:Heaviside\:step\:function}$$ $$\mathrm{For\:}\frac{e^{-s}}{s^2+\pi ^2}:\quad F\left(s\right)=\frac{1}{s^2+\pi ^2},\:\quad a=1$$

We will get : $$=\text{H}\left(t-1\right)L^{-1}\left\{\frac{1}{s^2+\pi ^2}\right\}\left(t-1\right)$$ $$\text{H}\left(t-1\right)\frac{1}{\sqrt{\pi ^2}}\sin \left(\sqrt{\pi ^2}\left(t-1\right)\right)$$

Combine first and second part : $$=\text{H}\left(t-\frac{1}{2}\right)\cos \left(\sqrt{\pi ^2}\left(t-\frac{1}{2}\right)\right)+\pi \text{H}\left(t-1\right)\frac{1}{\sqrt{\pi ^2}}\sin \left(\sqrt{\pi ^2}\left(t-1\right)\right)$$

Refine and we will get : $$=\text{H}\left(t-\frac{1}{2}\right)\cos \left(\sqrt{\pi ^2}\left(t-\frac{1}{2}\right)\right)+\text{H}\left(t-1\right)\sin \left(\sqrt{\pi ^2}\left(t-1\right)\right)$$

Answer 2

From the property of Laplace transform: time shifting, $$ \boxed{ \mathcal{L}\{g(t-a)\theta(t-a)\}(s) = e^{-as}G(s)} $$ where $a>0$, $\theta(t)$ is the Heaviside step function. Let $$ F(s) = e^{-\frac{s}{2}}\underbrace{\frac{s}{s^2+\pi^2}}_{G_1(s)} + e^{-s}\underbrace{\frac{\pi}{s^2+\pi^2}}_{G_2} $$ Then $$ \mathcal{L}\left\{g_1\left(t-\frac{1}{2}\right) \theta\left(t-\frac{1}{2}\right) + g_2(t-1)\theta(t-1)\right\}(s) =e^{-\frac{s}{2}} G_1(s) + e^{-s}G_2(s) $$ Note that $$ \mathcal{L}\{\cos \omega t\}(s) = \frac{p}{p^2+\omega^2},\quad \mathcal{L}\{\sin \omega t\}(s) = \frac{\omega}{p^2+\omega^2} $$ So you can obtain $g_1(t)$ and $g_2(t)$: $$ g_1(t) = \mathcal{L}^{-1}\left\{\frac{s}{s^2+\pi^2}\right\}=\cos \pi t,\quad g_2(t) = \mathcal{L}^{-1}\left\{\frac{\pi}{s^2+\pi^2}\right\}=\sin \pi t $$ Thus, the inverse Laplace transform of $F(s)$ is: $$ \boxed{ \begin{align} \mathcal{L}^{-1}\{F(s)\}&= \mathcal{L}^{-1}\left\{\frac{s e^{-\frac{s}{2}}+\pi e^{-s}}{s^2+\pi^2}\right\}\\[.2cm] &=g_1\left(t-\frac{1}{2}\right) \theta\left(t-\frac{1}{2}\right) + g_2(t-1)\theta(t-1)\\[.2cm] &=\boxed{\left[\theta\left(t-\frac{1}{2} \right)- \theta(t-1)\right] \sin \pi t} \end{align} } $$