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I am trying to find the inverse Laplace transform of:

$F(S) = \frac{e^{-\pi s}+ 2 + s}{s^2 +2s + 2}$.

I really have no idea how to approach this, since I cannot use partial fractions decomposition of the denominator.

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    $\begingroup$ Complete the square. $\endgroup$ – David Jul 29 '19 at 0:38
  • $\begingroup$ This might by a silly question, but what do you mean? $\endgroup$ – jakvah Jul 29 '19 at 0:40
  • $\begingroup$ Put the denominator in the form $(s+a)^2+b$ where $a,b$ are real numbers. Searching up 'complete the square' will give plenty of results. $\endgroup$ – Toby Mak Jul 29 '19 at 0:44
  • $\begingroup$ By doing this, I can see how to find the inverse of $\frac{s+2}{(s+ 2)^2 +2}$, but not how to find the inverse of $\frac{e^{-\pi s}}{(s+2)^2 +2}$ $\endgroup$ – jakvah Jul 29 '19 at 0:48
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    $\begingroup$ Mathematica gives: $\frac{1}{2} e^{(-1-i) t} \left(i e^{\pi } \left(-1+e^{2 i t}\right) \theta (t-\pi )+(1-i) e^{2 i t}+(1+i)\right)$ where $\theta$ is the Heaviside step function. $\endgroup$ – David G. Stork Jul 29 '19 at 1:10
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$$f(t)=\mathcal{L}^{-1}\{F(s)\}=\mathcal{L}^{-1}\left(\frac{e^{-\pi s}+ 2 + s}{s^2 +2s + 2}\right)$$ $$=\mathcal{L}^{-1}\left(\frac{e^{-\pi s}}{s^2 +2s + 2}\right)+\mathcal{L}^{-1}\left(\frac{s+2}{s^2 +2s + 2}\right)\tag1$$ Now $$\mathcal{L}^{-1}\left(\frac{s+2}{s^2 +2s + 2}\right)=\mathcal{L}^{-1}\left(\frac{s+1}{(s+1)^2+ 1}\right)+\mathcal{L}^{-1}\left(\frac{1}{(s+1)^2+ 1}\right)=e^{-t}~\cos t~+~e^{-t}~\sin t$$and $$\mathcal{L}^{-1}\left(\frac{e^{-\pi s}}{s^2 +2s + 2}\right)=\text{H}\left(t-\pi \right)e^{-\left(t-\pi \right)}\sin \left(t-\pi \right)\qquad \text{(by second shifting property )}$$

where $~H(x)=\begin{cases}1 & ,\text{if} ~~~~~x\ge 0\\0 &,\text{if} ~~~~~x\lt 0\end{cases}~$ be the Heaviside function.

Now from $(1)$ , $$f(t)=\mathcal{L}^{-1}\{F(s)\}=e^{-t}~\cos t~+~e^{-t}~\sin t~+~\text{H}\left(t-\pi \right)e^{-\left(t-\pi \right)}\sin \left(t-\pi \right)$$


Remarks:

  • $$\mathcal{L}=\{e^{at}\}=\frac{1}{p-a}\implies \mathcal{L}^{-1}\left\{\frac{1}{p-a}\right\}=e^{at}$$
  • $$\mathcal{L}=\{\sin(at)\}=\frac{a}{p^2+a^2}\implies \mathcal{L}^{-1}\left\{\frac{a}{p^2+a^2}\right\}=\sin(at)$$
  • $$\mathcal{L}=\{\cos(at)\}=\frac{p}{p^2+a^2}\implies \mathcal{L}^{-1}\left\{\frac{p}{p^2+a^2}\right\}=\cos(at)$$

Second Shifting Property: If $~\mathcal{L}^{-1}\left\{F(p)\right\} =f(t)~$, then $~\mathcal{L}^{-1}\left\{F(p)~e^{-ap}\right\} =g(t)~$ where $$~g(t)=\begin{cases} f(t-a) \quad \text{if} ~~~t~\gt~a\\ 0 \quad ~~~~~~~~~~~~\text{if} ~~~t~\lt ~a\end{cases}~$$

which can also be written as $$\mathcal{L}^{-1}\{e^{-ap}~F(p)\}=f(t-a)~H(t-a)$$ where $~H(x)=\begin{cases}1 & ,\text{if} ~~~~~x\ge 0\\0 &,\text{if} ~~~~~x\lt 0\end{cases}~$ be the Heaviside function.

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Break the numerator into three terms and use linearity:

$-\frac{1}{2} i e^{(-1-i) (t-\pi )} \left(-1+e^{2 i (t-\pi )}\right) \theta (t-\pi ), $

$-i e^{(-1-i) t} \left(-1+e^{2 i t}\right),$

and

$\left(\frac{1}{2}+\frac{i}{2}\right) e^{(-1-i) t} \left(e^{2 i t}-i\right).$

Add them.

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