강의노트 라플라스 역변환

식(1)을 이용하여 라플라스 역변환을 구할 수 있다.

$$\tag{1} f(t)=\mathcal{L}^{-1}[F(s)]=\dfrac{1}{2\pi j}\int_{\sigma -j\infty}^{\sigma +j\infty}f(s)e^{st}ds$$

어려운 적분을 피해 $F(s)$를 부분 분수로 분해(partial fraction expansion)하고 라플라스 변환 테이블 이용하여 $f(t)$를 구한다.

부분 분수에 의한 라플라스 역변환

실수 단근인 경우

$$G(s)=\dfrac{N(s)}{(s +p_{1})(s + p_{2})\cdots(s + p_{n})}=\dfrac{K_{1}}{s + p_{1}}+\dfrac{K_{2}}{s + p_{2}}+\cdots +\dfrac{K_{n}}{s + p_{n}}$$

$$\left . K_{i}=\lim_{s\to -p_{i}}(s+p_{i})G(s)\right.$$

$$\begin{align*} \mathcal{L}^{-1}[G(s)] & =\mathcal{L}^{-1}[\dfrac{K_{1}}{s + p_{1}}]+\mathcal{L}^{-1}[\dfrac{K_{2}}{s + p_{2}}]+\cdots +\mathcal{L}^{-1}[\dfrac{K_{n}}{s + p_{n}}]\\&= K_{1}e^{-p_{1}t}+ K_{2}e^{-p_{2}t}+\cdots + K_{n}e^{-p_{n}t} \end{align*}$$

[예] $F(s)=\dfrac{10}{s^{3}+ 8 s^{2}+ 15s}$

[풀이]

$$F(s)=\dfrac{10}{s^{3}+ 8 s^{2}+ 15s} = \dfrac{10}{s(s+3)(s+5)} = \dfrac{k_1}{s}+\dfrac{k_2}{s+3}+\dfrac{k_3}{s+5}$$

$$k_1 = \lim_{s \to 0} s \cdot \dfrac{10}{s(s+3)(s+5)} = \dfrac{10}{15} = \dfrac{2}{3}$$

$$k_2 = \lim_{s \to -3} (s+3) \cdot \dfrac{10}{s(s+3)(s+5)} = \dfrac{10}{(-3)(-3+5)} = -\dfrac{5}{3}$$

$$k_3 = \lim_{s \to -5} (s+5) \cdot \dfrac{10}{s(s+3)(s+5)} = \dfrac{10}{(-5)(-5+3)} = 1$$

$$F(s) = \dfrac{2}{3} \dfrac{1}{s} -\dfrac{5}{3}\dfrac{1}{s+3} +1\dfrac{1}{s+5}$$

$$\therefore \quad f(t) = \dfrac{2}{3} u(t) -\dfrac{5}{3}e^{-3t} +1 e^{-5t}$$

공액 복소근을 포함한 경우

$$\begin{align*} G(s) &=\dfrac{N(s)}{[(s +\alpha)^{2}+\omega^{2}](s + p_{3})\cdots(s + p_{n})}\\\\&=\dfrac{s A + B}{(s +\alpha)^{2}+\omega^{2}}+\dfrac{K_{3}}{s + p_{3}}+\cdots +\dfrac{K_{n}}{s + p_{n}} \end{align*}$$

• $K_i$는 위에서 구한 방법으로 구하고
• A,B는 다음과 같이 구한다.

$$\lim_{s\to-\alpha +j\omega}s A + B =\lim_{s\rightarrow -\alpha + j\omega}[(s+\alpha)^{2}+\omega^{2}]G(s)$$

[예] $F(s)=\dfrac{3}{s^{3}+ 2 s^{2}+ 2s}$

[풀이]

$$F(s)=\dfrac{3}{s^{3}+ 2 s^{2}+ 2s} = \dfrac{3}{s((s+1)^2+1^2)}= \dfrac{k_1}{s} + \dfrac{sA+B}{s((s+1)^2+1^2)}$$

$$k_1 = \lim_{s \to 0} s \cdot \dfrac{3}{s((s+1)^2+1^2)} = \dfrac{3}{(0+1)^2+1} = \dfrac{3}{2}$$

$$\begin{aligned} \lim_{s\to -1+j1}s A + B &=\lim_{s\rightarrow -1+ j1}[(s+1)^{2}+1^{2}] \dfrac{3}{s((s+1)^2+1^2)} \\& = \dfrac{3}{-1+ j1} \cdot \dfrac{-1-j1}{-1-j1} = -1.5 - 1.5j\end{aligned}$$

$$(-1+j1)A+B = -1.5 - 1.5j$$

$$\begin{cases} -A +B &= -1.5 \\ A &= -1.5 \end{cases}$$

$$\therefore \quad A = -1.5, \quad B = -3$$

$$\begin{aligned} F(s)&= \dfrac{1.5}{s} -1.5 \cdot \dfrac{1s +2}{(s+1)^2+1^2} \\ &= \dfrac{1.5}{s} -1.5 \cdot \left[ \dfrac{s +1}{(s+1)^2+1^2} + \dfrac{1}{(s+1)^2+1^2} \right] \end{aligned}$$

$$\therefore \quad f(t) = 1.5u(t) -1.5 \left[ \cos( t) e^{-t} + \sin(t) e^{-t}\right]$$

다중근인 경우

$$\begin{align*} G(s) &=\dfrac{N(s)}{(s +p_{1})^{m}(s + p_{2})\cdots(s + p_{n})}\\\\&=\dfrac{K_{11}}{(s + p_{1})^{m}}+\dfrac{K_{12}}{(s + p_{1})^{m-1}}+\cdots +\dfrac{K_{1m}}{(s + p_{1})}+\dfrac{K_{2}}{s + p_{2}}+\cdots +\dfrac{K_{n}}{s + p_{n}} \end{align*}$$

$$K_{11}=\lim_{s\to -p_{i}}(s+p_{i})^{m}G(s)$$

$$K_{12}=\lim_{s\to -p_{i}}\dfrac{d}{ds}\left[(s+p_{i})^{m}G(s)\right]$$

$$\vdots$$

$$K_{1m}=\lim_{s\to -p_{i}}\dfrac{1}{(m-1)!}\dfrac{d^{m-1}}{ds^{m-1}}\left[(s+p_{i})^{m}G(s)\right]$$

[예] $F(s)=\dfrac{s+2}{s^{3}(s+1)^{2}}$

[풀이]

$$F(s)=\dfrac{s+2}{s^{3}(s+1)^{2}} = \dfrac{k_{11}}{s^3} + \dfrac{k_{12}}{s^2}+\dfrac{k_{13}}{s} + \dfrac{k_{21}}{(s+1)^2}+\dfrac{k_{22}}{(s+1)}$$

$$k_{11} = \lim_{s\to 0} s^{3} \cdot \dfrac{s+2}{s^{3}(s+1)^{2}} = 2$$

$$\begin{aligned}k_{12}&=\lim_{s\to 0} \dfrac{d}{ds}\left[\dfrac{s+2}{(s+1)^{2}} \right] = \lim_{s\to 0} \left[\dfrac{1 \cdot (s^2+2s+1)-(s+2) \cdot (2s+2)}{(s+1)^{4}} \right] \\ \\ &= \lim_{s\to 0} \left[\dfrac{s^2+2s+1-2s^2 -6s -4}{(s+1)^{4}} \right] = -3 \end{aligned}$$

$$k_{13}= \lim_{s\to 0} \dfrac{d}{ds}\left[\dfrac{-(s+3)}{(s+1)^{3}} \right] = \lim_{s\to 0} \left[\dfrac{-1 \cdot (s+1)^3 +(s+3)3(s^2+2s+1) }{(s+1)^{6}} \right] = 8$$

$$k_{21} = \lim_{s\to -1} (s+1)^{2} \cdot \dfrac{s+2}{s^{3}(s+1)^{2}} = -1$$

$$\begin{aligned}k_{22}&=\lim_{s\to -1} \dfrac{d}{ds}\left[\dfrac{s+2}{s^{3}} \right] = \lim_{s\to -1} \left[\dfrac{1 \cdot (s^3)-(s+2) \cdot (3s^2)}{s^{6}} \right] \\ \\ &= \lim_{s\to -1} \left[\dfrac{-2s^3-6s^2}{(s)^{6}} \right] = -4 \end{aligned}$$

$$\therefore F(s)=\dfrac{s+2}{s^{3}(s+1)^{2}} = \dfrac{2}{s^3} + \dfrac{-3}{s^2}+\dfrac{8}{s} + \dfrac{-1}{(s+1)^2}+\dfrac{-4}{(s+1)}$$

$$f(t) = \mathcal{L}^{-1}F(s) = t^2-3t+8u(t)-te^{-t}-4e^{-t}$$

라플라스 변환에 의한 미분 방정식의 해법

$$\dfrac{d^{2}y}{dt^{2}}+ 5\dfrac{dy}{dt}+ 6y = u_{s(t)} \quad ; \quad y(0)=1 , \quad y^{(1)}(0)= 0$$

$$\dfrac{d^{2}c(t)}{dt^{2}}+ 5\dfrac{d c(t)}{dt}+ 6c(t)= 10u(t) \quad ; \quad c(0)=0$$

$$\dfrac{d^{2}y}{dt^{2}}+ 3\dfrac{dy}{dt}+ 2y = 5 \quad ; \quad y(0)=-1, \quad y^{(1)}(0)= 2$$