Difference between r1.19 and the current
@@ -73,7 +73,6 @@
$\mathcal{L}\left\lbrace t^2 \right\rbrace=\frac2{s^3}=\frac{2!}{s^3}$
$\mathcal{L}\left\lbrace t^3 \right\rbrace=\frac6{s^4}=\frac{3!}{s^4}$
$\vdots$
$\mathcal{L}\left\lbrace t^n \right\rbrace=\frac{n!}{s^{n+1}}$
$\mathcal{L}\left\lbrace \sin kt \right\rbrace=\frac{k}{s^2+k^2}$
$\mathcal{L}\left\lbrace t^3 \right\rbrace=\frac6{s^4}=\frac{3!}{s^4}$
$\vdots$
$\mathcal{L}\left\lbrace \sin kt \right\rbrace=\frac{k}{s^2+k^2}$
cursive L: 𝓛
좋은 점
불연속함수를 자연스럽게 다룰 수 있다
1. 정의 ¶
$\displaystyle t\ge0$ 에서 졍의된 함수 $\displaystyle f(t)$ 에 대해 이상적분
$\displaystyle \int_0^{\infty}e^{-st}f(t)dt=\lim_{b\to\infty}\int_0^be^{-st}f(t)dt$
가 수렴할 때 그 극한을 $\displaystyle f(t)$ 의 라플라스 변환이라 하고$\displaystyle F(s)=\mathcal{L}\left\lbrace f(t)\right\rbrace=\int_0^{\infty}e^{-st}f(t)dt$
라 쓴다함수 $\displaystyle f(t)$ 의 정의역이 $\displaystyle [0,\infty)$ 이고 여기서 piecewise continuous 라고 하자.
(연속인데 완벽한 연속일 필요는 없고, 몇 군데에서 끊어져도 됨)
이것은 $\displaystyle s$ 에 대한 함수이다. (존재할 경우)
(연속인데 완벽한 연속일 필요는 없고, 몇 군데에서 끊어져도 됨)
$\displaystyle \mathcal{L}\{f(t)\}(s):=\int_{0}^{\infty} e^{-st}f(t)dt$을 $\displaystyle f(t)$ 의 라플라스 변환이라고 한다. (Laplace transform of $\displaystyle f(t)$ )
이것은 $\displaystyle s$ 에 대한 함수이다. (존재할 경우)
예)
$\displaystyle \mathcal{L}\{1\}(s)=\int_0^{\infty}e^{-st}\cdot1dt$ ( $\displaystyle s\le0$ 일 때는 정의 안 됨 )
$\displaystyle \mathcal{L}\{t\}(s)=\int_0^{\infty}e^{-st}tdt$
$\displaystyle \mathcal{L}\{1\}(s)=\int_0^{\infty}e^{-st}\cdot1dt$ ( $\displaystyle s\le0$ 일 때는 정의 안 됨 )
$\displaystyle =\left[-\frac1s e^{-st}\right]_0^{\infty}$
$\displaystyle =\lim_{a\to\infty}\left[-\frac1s e^{-st}\right]_0^a$
$\displaystyle =\lim_{a\to\infty}\left(-\frac1s e^{-sa}+\frac1s\right)$
$\displaystyle =\frac1s$
예)$\displaystyle =\lim_{a\to\infty}\left[-\frac1s e^{-st}\right]_0^a$
$\displaystyle =\lim_{a\to\infty}\left(-\frac1s e^{-sa}+\frac1s\right)$
$\displaystyle =\frac1s$
$\displaystyle \mathcal{L}\{t\}(s)=\int_0^{\infty}e^{-st}tdt$
이것을 부분적분하면,
$\displaystyle =\left[-\frac1se^{-st}\cdot t\right]_0^{\infty}-\int\nolimits_0^{\infty}\left(-\frac1se^{-st}\right)\cdot1dt$
$\displaystyle s>0$ 이므로,
$\displaystyle =(0-0)+\frac1s\int\nolimits_0^{\infty}e^{-st}dt$
$\displaystyle =\frac1{s^2}$
........이하 $\displaystyle \int_0^{\infty}$ 의 아래끝 위끝 생략.........$\displaystyle =\left[-\frac1se^{-st}\cdot t\right]_0^{\infty}-\int\nolimits_0^{\infty}\left(-\frac1se^{-st}\right)\cdot1dt$
$\displaystyle s>0$ 이므로,
$\displaystyle =(0-0)+\frac1s\int\nolimits_0^{\infty}e^{-st}dt$
$\displaystyle =\frac1{s^2}$
예)
$\displaystyle \mathcal{L}\{e^{-3t}\}(s)=\int e^{-st}e^{-3t}dt$
$\displaystyle \mathcal{L}\{e^{-at}\}=\frac1{s+a}\quad(s>-a)$
$\displaystyle \mathcal{L}\{e^{-3t}\}(s)=\int e^{-st}e^{-3t}dt$
$\displaystyle =\int e^{-(s+3)t}dt$ s>-3이므로
$\displaystyle =\frac1{s+3}$
따라서 지수함수를 넣으면 유리함수가 나온다.$\displaystyle =\frac1{s+3}$
$\displaystyle \mathcal{L}\{e^{-at}\}=\frac1{s+a}\quad(s>-a)$
예)
$\displaystyle \mathcal{L}\{\sin2t\}=\int e^{-st}\sin2tdt$
부분적분하면
그러므로 정리하면
$\displaystyle \mathcal{L}\lbrace\sin 2t\rbrace=\frac2{s^2}-\frac4{s^2}\cdot\mathcal{L}\left\lbrace\sin 2t\right\rbrace$
$\displaystyle (1+\frac4{s^2})\mathcal{L}\{\sin2t\}=\frac2{s^2}$
$\displaystyle \mathcal{L}\{\sin2t\}=\frac{\frac{2}{s^2}}{1+\frac{4}{s^2}}=\frac2{s^2+4}\quad (s>0)$
$\displaystyle \mathcal{L}\{\cos2t\}=\frac{s}{2}\cdot\mathcal{L}\{\sin2t\}=\frac{s}{s^2+4}\quad (s>0)$
$\displaystyle \mathcal{L}\{\sin2t\}=\int e^{-st}\sin2tdt$
부분적분하면
$\displaystyle =\left[-\frac1se^{-st}\sin2t\right]_0^{\infty}-\int\left(-\frac1se^{-st}\cdot2\cos2t\right)dt$
$\displaystyle =0+\frac2s\int e^{-st}\cos 2t dt$
$\displaystyle =\frac2s\cdot\mathcal{L}\{\cos 2t\}$
$\displaystyle =\frac2s\left[\left[-\frac1se^{-st}\cos2t\right]_0^{\infty}-\int(-\frac1s e^{-st})(-2\cdot\sin 2t)dt\right]$
$\displaystyle =\frac2s\left[\left(0-(-\frac1s)\right)-\frac2s\int_0^{\infty}e^{-st}\sin 2tdt\right]$
이렇게 원래 구하고자 하던 $\displaystyle \mathcal{L}\{\sin2t\}$ 꼴이 나옴$\displaystyle =0+\frac2s\int e^{-st}\cos 2t dt$
$\displaystyle =\frac2s\cdot\mathcal{L}\{\cos 2t\}$
$\displaystyle =\frac2s\left[\left[-\frac1se^{-st}\cos2t\right]_0^{\infty}-\int(-\frac1s e^{-st})(-2\cdot\sin 2t)dt\right]$
$\displaystyle =\frac2s\left[\left(0-(-\frac1s)\right)-\frac2s\int_0^{\infty}e^{-st}\sin 2tdt\right]$
그러므로 정리하면
$\displaystyle \mathcal{L}\lbrace\sin 2t\rbrace=\frac2{s^2}-\frac4{s^2}\cdot\mathcal{L}\left\lbrace\sin 2t\right\rbrace$
$\displaystyle (1+\frac4{s^2})\mathcal{L}\{\sin2t\}=\frac2{s^2}$
$\displaystyle \mathcal{L}\{\sin2t\}=\frac{\frac{2}{s^2}}{1+\frac{4}{s^2}}=\frac2{s^2+4}\quad (s>0)$
$\displaystyle \mathcal{L}\{\cos2t\}=\frac{s}{2}\cdot\mathcal{L}\{\sin2t\}=\frac{s}{s^2+4}\quad (s>0)$
2. 선형성 ¶
$\displaystyle \mathcal{L}\{3t-5\sin2t\}$
$\displaystyle =3\mathcal{L}\{t\}-5\mathcal{L}\{\sin 2t\}$
$\displaystyle =3\cdot\frac1{s^2}-5\cdot\frac2{s^2+4}$
$\displaystyle =3\cdot\frac1{s^2}-5\cdot\frac2{s^2+4}$
3. 기본적인 것 몇가지... ¶
$\displaystyle \mathcal{L}\left\lbrace 1 \right\rbrace=\frac1s$
$\displaystyle \mathcal{L}\left\lbrace t \right\rbrace=\frac1{s^2}=\frac{1!}{s^2}$
$\displaystyle \mathcal{L}\left\lbrace t^2 \right\rbrace=\frac2{s^3}=\frac{2!}{s^3}$
$\displaystyle \mathcal{L}\left\lbrace t^3 \right\rbrace=\frac6{s^4}=\frac{3!}{s^4}$
$\displaystyle \mathcal{L}\left\lbrace t \right\rbrace=\frac1{s^2}=\frac{1!}{s^2}$
$\displaystyle \mathcal{L}\left\lbrace t^2 \right\rbrace=\frac2{s^3}=\frac{2!}{s^3}$
$\displaystyle \mathcal{L}\left\lbrace t^3 \right\rbrace=\frac6{s^4}=\frac{3!}{s^4}$
$\displaystyle \vdots$
$\displaystyle \mathcal{L}\left\lbrace t^n \right\rbrace=\frac{n!}{s^{n+1}}$$\displaystyle \mathcal{L}\left\lbrace \sin kt \right\rbrace=\frac{k}{s^2+k^2}$
$\displaystyle \mathcal{L}\left\lbrace \cos kt \right\rbrace=\frac{s}{s^2+k^2}$
$\displaystyle \mathcal{L}\left\lbrace \sinh kt \right\rbrace=\frac{k}{s^2-k^2}$
$\displaystyle \mathcal{L}\left\lbrace \cosh kt \right\rbrace=\frac{s}{s^2-k^2}$
4. 표: Transform of Some Basic Functions ¶
| x | 𝓛{x} | 정의되는 경우 |
| $\displaystyle 1$ | $\displaystyle \frac1s$ | (s>0) |
| $\displaystyle t^n$ | $\displaystyle \frac{n!}{s^{n+1}}\;\;n=1,2,3,\cdots$ | (s>0) |
| $\displaystyle e^{at}$ | $\displaystyle \frac1{s-a}$ | (s>a) |
| $\displaystyle \sin kt$ | $\displaystyle \frac{k}{s^2+k^2}$ | (s>0) |
| $\displaystyle \cos kt$ | $\displaystyle \frac{s}{s^2+k^2}$ | (s>0) |
| $\displaystyle \sinh kt$ | $\displaystyle \frac{k}{s^2-k^2}$ | (s>0) |
| $\displaystyle \cosh kt$ | $\displaystyle \frac{s}{s^2-k^2}$ | (s>0) |
참고로 𝓛{1/t}는 존재하지 않는다. (s≤0일 때도, s>0일 때도 ∞로 발산)
라플라스변환의 약점(?).
5. 역라플라스변환 Inverse Laplace Transform ¶
(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{\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..
$\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 \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}$
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}$
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}$
6. Laplace Transform of Derivatives ¶
$\displaystyle f(t) {\mathcal{L}\atop\longrightarrow} F(s)$
↓미분
$\displaystyle f'(t) {\mathcal{L}\atop\longrightarrow} s\cdot F(s)-f(0)$$\displaystyle \mathcal{L}\{f'(t)\}=\int_0^{\infty}e^{-st}f'(t)dt$
$\displaystyle =\left[e^{-st}f(t)\right]_0^{\infty}-\int_0^{\infty}(-s)e^{-st}\cdot f(t)dt$
$\displaystyle =(0-f(0))+s\int_0^{\infty}e^{-st}f(t)dt$
$\displaystyle =-f(0)+s\cdot F(s)$
i.e.$\displaystyle =(0-f(0))+s\int_0^{\infty}e^{-st}f(t)dt$
$\displaystyle =-f(0)+s\cdot F(s)$
$\displaystyle F(s)=\mathcal{L}\{f(t)\}$ 일 때,
$\displaystyle \mathcal{L}\{f'(t)\}(s)=s\cdot F(s)-f(0)$
여러번 미분한 것은?
$\displaystyle \mathcal{L}\{f''(t)\}=s\cdot\mathcal{L}\{f'(t)\}-f'(0)$
$\displaystyle F(s)=\mathcal{L}\{f(t)\}$ 일 때,
$\displaystyle =s\cdot(s\cdot F(s)-f(0))-f'(0)$
$\displaystyle =s^2\cdot F(s)-f(0)\cdot s-f'(0)$
정리$\displaystyle =s^2\cdot F(s)-f(0)\cdot s-f'(0)$
$\displaystyle F(s)=\mathcal{L}\{f(t)\}$ 일 때,
$\displaystyle \mathcal{L}\{f^{(n)}(t)\}=s^n\cdot F(s)-f(0)\cdot s^{n-1}-f'(0)\cdot s^{n-2}-(\cdots)-f^{(n-1)}(0)$가정: $\displaystyle s>0,\;f,f',f'',\cdots,f^{(n-1)}:$ subexponential
예)
$\displaystyle \mathcal{L}\{\cos kt\}=s\cdot\mathcal{L}\{\frac1{k}\sin kt\}-0$
$\displaystyle \mathcal{L}\{\cos kt\}=s\cdot\mathcal{L}\{\frac1{k}\sin kt\}-0$
$\displaystyle =\frac{s}{k}\cdot\mathcal{L}\{\sin kt\}$
$\displaystyle =\frac{s}{k}\cdot\frac{k}{s^2+k^2}$
$\displaystyle =\frac{s}{s^2+k^2}$
$\displaystyle =\frac{s}{k}\cdot\frac{k}{s^2+k^2}$
$\displaystyle =\frac{s}{s^2+k^2}$
7. First Translation Theorem ¶
(from http://www.kocw.net/home/search/kemView.do?kemId=1189624 제1, 2 평행이동 정리)
$\displaystyle a\in\mathbb{R},$
$\displaystyle \mathcal{L}\left{f(t)\right}=F(s)$
$\displaystyle \Rightarrow\;\mathcal{L}\left{e^{at}f(t)\right}=F(s-a)$
$\displaystyle \mathcal{L}\left{f(t)\right}=F(s)$
$\displaystyle \Rightarrow\;\mathcal{L}\left{e^{at}f(t)\right}=F(s-a)$
8. Inverse Form of the First Translation Theorem ¶
$\displaystyle \mathcal{L}^{-1}\left\lbrace F(s)\right\rbrace=f(t)$
$\displaystyle \Rightarrow\;\mathcal{L}^{-1}\left\lbrace F(s-a)\right\rbrace=e^{at}f(t)$
$\displaystyle \Rightarrow\;\mathcal{L}^{-1}\left\lbrace F(s-a)\right\rbrace=e^{at}f(t)$
10. References ¶
Notes from 덕성여대 http://www.kocw.net/home/search/kemView.do?kemId=1189624 라플라스 변환의 정의와 예
정의 from https://www.youtube.com/watch?v=j6zT-mDWyM4
정의 from https://www.youtube.com/watch?v=j6zT-mDWyM4
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