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I want to calculate the inverse laplace of $$F(s)=e^{-3s}\frac{1+s}{s^3+2s^2+2s}$$ And i'm wondering if applying the derivative theorem is correct.

To keep it simple it split them up: $$F(s)=F_1(s).F_2(s)$$ with $$F_1(s)=e^{-3s}$$ $$F_2(s)=\frac{1+s}{s^3+2s^2+2s}$$

$F_2(s)$

Factorizing $F_2(s)$: $$\frac{1+s}{s(s^2+2s+2)}=\frac{A}{s}+\frac{Bs+C}{s^2+2s+2}$$ $$1+s=As^2+2As+2A+Bs^2+Cs$$ $$1+s=s^2(A+B)+s(2A+C)+2A$$

  • $A=\frac{1}{2}$
  • $B=-\frac{1}{2}$
  • $C=0$

$$F_2(s)=\frac{1}{2s}-\frac{1}{2}(\frac{s}{s^2+2s+2})$$ Rewritten as: $$F_2(s)=\frac{1}{2s}-\frac{1}{2}(\frac{s}{(s+1)^2+1})$$

So here my question arises, considering the second part of $F_2(s)$:

Is it okay, to apply the derivative-theorem which states that: $s.F(s)-f(0)=\frac{df(t)}{dt}$

$$F_2(s)=\frac{1}{2s}-\frac{1}{2}(\frac{s}{(s+1)^2+1})$$ $$=\frac{1}{2s}-\frac{s}{2}(\frac{1}{(s+1)^2+1})$$ Using the table with Laplace-transforms $$f_2(t)=\frac{1}{2}(\mathcal{H}(t)-\frac{d}{dt}e^t.sin(t))$$ $$=\frac{1}{2}(\mathcal{H}(t)-\frac{d}{dt}e^t.sin(t))$$ $$=\frac{1}{2}(\mathcal{H}(t)-(e^t.sin(t)+e^t.cos(t))$$

Putting them all together: $$f(t)=\frac{1}{2}\delta(t).(\mathcal{H}(t)-(e^t.sin(t)+e^t.cos(t))$$

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The inverse Laplace transform considered is $$F(s)=e^{-3s}\frac{1+s}{s^3+2s^2+2s}$$ and can be reduced as follows. First notice that \begin{align} F(s) = \frac{e^{-3 s}}{s} \cdot \frac{s+1}{(s+1)^{2} + 1} = f_{1}(s) \cdot f_{2}(s) \end{align} where \begin{align} f_{1}(s) = \frac{e^{-3 s}}{s} \hspace{10mm} f_{2}(s) = \frac{s+1}{(s+1)^{2} + 1}. \end{align} The inverses of $f_{1,2}(s)$ are \begin{align} f_{1}(t) = H(t-3) \hspace{10mm} f_{2}(t) = e^{-t} \, \cos t. \end{align} Now by the convolution theorem the inverse desired is \begin{align} F(t) &= \int_{0}^{t} H(u-3) \, e^{-(u-3)} \, \cos(u-3) \, du \\ &= \int_{3}^{t} e^{-(u-3)} \, \cos(u-3) \, du \\ &= \int_{0}^{t-3} e^{-x} \, \cos x \, dx \\ &= \frac{1}{2} \, \left[ e^{-x} \, (\sin x - \cos x) \right]_{0}^{t-3} \\ &= \frac{1}{2} \left[ 1 + e^{-(t-3)} \, (\sin(t-3) - \cos(t-3)) \right] \end{align}

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