sin凾数のラプラス変換

ラプラス変換

$$\begin{eqnarray} \mathfrak{L}\left[ {f\left( t \right)} \right] &=&\int_0^\infty {f\left( t \right){e^{–st}}}\mathrm{d}t \end{eqnarray}$$

\( \sin{\left(\omega t\right)} \)のラプラス変換

$$\begin{eqnarray} f\left( t \right)&=&\sin{\left(\omega t\right)} \\\mathfrak{L}\left[ {f\left( t \right)} \right] &=& \int_0^\infty {\sin{\left(\omega t\right)}{e^{ –st}}}\mathrm{d}t \\&=& \left[ \sin{\left(\omega t\right)} \cdot {\frac{–1}{s}e^{-st}} \right]_0^{\infty} -\int_0^\infty {\omega \cos{\left(\omega t\right)} \cdot {\frac{–1}{s}e^{–st}}}\mathrm{d}t \\&&\;\cdots\;\int_a^b{f'\left( t \right) g\left( t \right) }\mathrm{d}t=\left[f\left( t \right) g\left( t \right)\right]_a^b-\int_a^b{f\left( t \right) g'\left( t \right) }\mathrm{d}t\;\;(f':fの微分,\;g':gの部分) \\&&\;\cdots\;\int e^{at} \mathrm{d}t =\frac{1}{a}e^{at}+C, \frac{\mathrm{d}}{\mathrm{d}t}\sin{\left(\omega t\right)}=\omega \cos{\left(\omega t\right)} \\&=& \left[ \sin{\left(\omega \infty\right)} \cdot {\frac{–1}{s}e^{-s\infty}} - \sin{\left(\omega 0\right)} \cdot {\frac{–1}{s}e^{-s0}}\right] -\left(\frac{-\omega}{s}\right)\int_0^\infty {\cos{\left(\omega t\right)} {e^{–st}}}\mathrm{d}t \\&=& \left[ 0 - 0\right] +\frac{\omega}{s}\int_0^\infty {\cos{\left(\omega t\right)} {e^{–st}}}\mathrm{d}t \\&&\;\cdots\;e^{-\infty}=0,\;\sin{\left( 0\right)}=0 \\&=& \frac{\omega}{s}\int_0^\infty {\cos{\left(\omega t\right)} {e^{–st}}}\mathrm{d}t \\&=& \frac{\omega}{s}\left[ \left[ \cos{\left(\omega t\right)} \cdot {\frac{–1}{s}e^{-st}} \right]_0^{\infty} -\int_0^\infty {-\omega \sin{\left(\omega t\right)} \cdot {\frac{–1}{s}e^{–st}}}\mathrm{d}t \right] \\&&\;\cdots\;\int_a^b{f'\left( t \right) g\left( t \right) }\mathrm{d}t=\left[f\left( t \right) g\left( t \right)\right]_a^b-\int_a^b{f\left( t \right) g'\left( t \right) }\mathrm{d}t\;\;(f':fの微分,\;g':gの微分) \\&&\;\cdots\;\int e^{at} \mathrm{d}t =\frac{1}{a}e^{at}+C, \frac{\mathrm{d}}{\mathrm{d}t}\cos{\left(\omega t\right)}=-\omega \sin{\left(\omega t\right)} \\&=& \frac{\omega}{s}\left[ \left[ \cos{\left(\omega \infty\right)} \cdot {\frac{–1}{s}e^{-s\infty}} - \cos{\left(\omega 0\right)} \cdot {\frac{–1}{s}e^{-s0}}\right] -\frac{\omega}{s}\int_0^\infty {\sin{\left(\omega t\right)} {e^{–st}}}\mathrm{d}t \right] \\&=& \frac{\omega}{s}\left[ \left[0 - 1 \cdot \frac{–1}{s}\cdot 1\right] -\frac{\omega}{s}\int_0^\infty {\sin{\left(\omega t\right)} {e^{–st}}}\mathrm{d}t \right] \\&&\;\cdots\;e^{-\infty}=0,\; \cos{\left( 0\right)}=1,\; e^{0}=1 \\&=& \frac{\omega}{s}\left[ \frac{1}{s} -\frac{\omega}{s}\int_0^\infty {\sin{\left(\omega t\right)} {e^{–st}}}\mathrm{d}t \right] \\&=& \frac{\omega}{s^2}-\frac{\omega^2}{s^2}\int_0^\infty {\sin{\left(\omega t\right)} {e^{–st}}}\mathrm{d}t \\&=& \frac{\omega}{s^2}-\frac{\omega^2}{s^2}\mathfrak{L}\left[ {f\left( t \right)} \right] \\\mathfrak{L}\left[ {f\left( t \right)} \right]+\frac{\omega^2}{s^2}\mathfrak{L}\left[ {f\left( t \right)} \right] &=& \frac{\omega}{s^2} \\\mathfrak{L}\left[ {f\left( t \right)} \right]\left(1+\frac{\omega^2}{s^2}\right) &=& \frac{\omega}{s^2} \\\mathfrak{L}\left[ {f\left( t \right)} \right]&=& \frac{\omega}{s^2}\frac{1}{1+\frac{\omega^2}{s^2}} \\&=&\frac{\omega}{s^2\left(1+\frac{\omega^2}{s^2}\right)} \\&=&\frac{\omega}{s^2+\omega^2} \end{eqnarray}$$