S2.

Angolstruktura » Introduction to the mathematics of medical imaging » Mathematical Methods of Linear Model Based Image Processing Procedures » Appendix. » Solution of problems » S2.

Solution:
Let’s attempt to derive the Laplace transformation of $$l(x)=e^{-\alpha x}e^{j\omega x}=e^{-(\alpha-j\omega)x}$ function instead of $f(x) .

$$ \mathcal{L}\left\{l(x)\right\} = \mathcal{L}\left\{e^{-(\alpha-j\omega )x}\right\} = \frac{1}{s+(\alpha-j\omega)}$ ; as it is known $$\mathcal{L}\left\{Ae^{-\alpha x}\right\} = \frac{A}{s+\alpha}$

Then, let’s make the following operations!

$$ \left.\begin{matrix} \mathcal{L}\left\{l(x)\right\} = \frac{1}{s+(\alpha-j\omega)} = \frac{1}{(s+\alpha)-j\omega} = \frac{(s+\alpha)+j\omega}{(s+\alpha)^{2}+\omega^{2}} = \frac{s+\alpha}{(s+\alpha)^{2}+\omega^{2}}+j\frac{\omega}{(s+\alpha)^{2}+\omega^{2}}\\ \\ \mathcal{L}\left\{l(x)\right\} = \mathcal{L}\left\{e^{-\alpha x}\text{cos}x+je^{-\alpha x}\text{sin}x\right\} = \mathcal{L}\left\{e^{-\alpha x}\text{cos}x\right\} + j\mathcal{L}\left\{e^{-\alpha x}\text{sin}x\right\} \end{matrix}\right\}$

Consequently the obtained results:
$$ \mathcal{L}\left\{e^{-\alpha x}\text{cos}x\right\} + j\mathcal{L}\left\{e^{-\alpha x}\text{sin}x \right\} = \frac{s+\alpha}{(s+\alpha)^{2}+\omega^{2}}+j\frac{\omega}{(s+\alpha)^{2}+\omega^{2}}$

$$ \mathcal{L}\left\{e^{-\alpha x}\text{cos}x\right\} = \frac{s+\alpha}{(s+\alpha)^{2}+\omega^{2}}; \quad \mathcal{L}\left\{e^{-\alpha x}\text{sin}x \right\} = \frac{\omega}{(s+\alpha)^{2}+\omega^{2}}$

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S2.

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