(동영상 - 역라플라스 변환, 도함수의 라플라스 변환)
$\displaystyle \left{f(t)\right} \overset{\longrightarrow^{\mathcal{L}}}{\longleftarrow_{\mathcal{L}^{-1}}} \left{F(s)\right}$
$\displaystyle \mathcal{L}^{-1}\left{\frac1s\right}=1?$
$\displaystyle \mathcal{L}^{-1}\left{\frac1{s^2}\right}=t?$
$\displaystyle \mathcal{L}^{-1}\{\cdots\}$ 이 유일한가? 즉 $\displaystyle \mathcal{L}$ 이 단사인가? 답은 yes & no. 연속함수 중에서 고르면 유일하다.
$\displaystyle \mathcal{L}^{-1}\left{\frac1s\right}=1$
$\displaystyle \mathcal{L}^{-1}\left{\frac1{s^{n+1}}\right}=\frac1{n!}t^n \;\;\;n=1,2,\cdots$
$\displaystyle \mathcal{L}^{-1}\left{\frac1{s-a}\right}=e^{at}$
$\displaystyle \mathcal{L}^{-1}\left{\frac{k}{s^2+k^2}\right}=\sin kt$
$\displaystyle \mathcal{L}^{-1}\left{\frac{s}{s^2+k^2}\right}=\cos kt$
$\displaystyle \mathcal{L}^{-1}\left{\frac{k}{s^2-k^2}\right}=\sinh kt$
$\displaystyle \mathcal{L}^{-1}\left{\frac{s}{s^2-k^2}\right}=\cosh kt$
sk 아니고 ks순이다..
예
$\displaystyle \mathcal{L}^{-1}\left{\frac{1}{(s-1)(s+2)(s+4)}\right}$ ?
$\displaystyle \frac{1}{(s-1)(s+2)(s+4)}=\frac{A}{s-1}+\frac{B}{s+2}+\frac{C}{s+4}$
미정계수법을 쓰면
$\displaystyle A=\frac1{15},B=-\frac16,C=\frac1{10}$
$\displaystyle =\mathcal{L}^{-1}\left{\frac{1}{15}\cdot\frac1{s-1}-\frac16\cdot\frac1{s+2}+\frac1{10}\cdot\frac1{s+4}\right}$
선형성을 쓰면
$\displaystyle =\frac1{15}\mathcal{L}^{-1}\left{\frac{1}{s-1}\right}-\frac16\mathcal{L}^{-1}\left{\frac1{s+2}\right}+\frac1{10}\mathcal{L}^{-1}\left{\frac1{s+4}\right}$
$\displaystyle =\frac1{15}e^{t}-\frac16e^{-2t}+\frac1{10}e^{-4t}$