삼각함수의 라플라스 변환
Laplace Transform of Trigonometric Function
공식1
사인과 코사인의 라플라스 변환은 다음과 같다.
$$ \mathcal{L} \left\{ \sin (at) \right\} = \dfrac{a}{s^2+a^2},\quad s>0 $$
$$ \mathcal{L} \left\{ \cos (at) \right\} = \dfrac{s}{s^2+a^2},\quad s>0 $$
유도
$\sin (at)$
$$ \begin{align*} \mathcal{L} \left\{ \sin (at) \right\}& =\displaystyle \int_0^\infty e^{-st}\sin(at)dt \\ &= \lim \limits_{A \to \infty} \left[-\dfrac{1}{a}e^{-st}\cos (at) \right]_0^A+ \lim \limits_{A \to \infty} \int _0^\infty -\dfrac{s}{a}e^{-st} \cos (at)dt \\ &= \dfrac{1}{a} - \lim \limits_{A \to \infty} \dfrac{s}{a} \left[ \dfrac{1}{a} \left[ e^{-st}\sin (at) \right]_0^A + \dfrac{s}{a}\int _0^A e^{-st} \sin (at) dt \right] \\ &=\dfrac{1}{a} - \dfrac{s^2}{a^2} \int_0^\infty e^{-st} \sin (at) dt \end{align*} $$
여기서 $\mathcal{L} \left\{ \sin (at) \right\} = \displaystyle \int_0^\infty e^{-st} \sin (at) dt$가 성립하므로,
$$ \begin{align*} \implies& &\dfrac{a^2+s^2}{a^2} \int _0^\infty e^{-st} \sin (at) dt &= \dfrac{1}{a} \\ \implies& &\int_0^\infty e^{-st} \sin (at)dt &=\dfrac{a}{s^2+a^2} \end{align*} $$
단, $\lim \limits_{A \to \infty} e^{-sA}\sin (aA)=0$을 만족해야하므로 $s>0$
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$\cos (at)$
$\sin$의 결과를 이용하면 $\cos$의 라플라스 변환은 훨씬 쉽고 짧게 구할 수 있다.
$$ \begin{align*} \mathcal{ L } \left\{ \cos (at) \right\} &=\int _0^\infty e^{-st} \cos (at) dt \\ &= \lim \limits_{A \to \infty} \dfrac{1}{a} \left[ e^{-st} \sin (at) \right]_0^A + \dfrac{s}{a} \int_0^\infty e^{-st} \sin (at) dt \\ &= \dfrac{s}{a} \dfrac{a}{s^2+a^2} \\ &=\dfrac{s}{s^2+a^2} \end{align*} $$
단, $s>0$
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같이보기
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William E. Boyce, Boyce’s Elementary Differential Equations and Boundary Value Problems (11th Edition, 2017), p246 ↩︎