Lecture 라플라스 변환의 정의

Views 1362 • Comments 0 • Last Updated at 2 months ago Full screen  
  • 라플라스 변환
  • 기본수학
  • 라플라스 변환

라플라스 변환

f(t)f(t)의 라플라스 변환 L[f(t)]\mathcal{L}[f(t)]는 다음과 같이 정의한다.

F(s)=L[f(t)]=0f(t)estdtF(s) = \mathcal{L}[f(t)] = \int_{0}^{\infty} f(t)e^{-s t} dt

라플라스 변환 예제

단위 계단함수

u(t)={1if t00if t<0(1)\tag{1} u (t) = \begin{cases} 1 &\text{if } t \geqslant 0 \\ 0 &\text{if } t<0 \end{cases}

L[u(t)]=0estdt=1sest0=1s\mathcal{L}[u(t)]=\int_{0}^{\infty}e^{-st}dt=-\dfrac{1}{s}e^{-st}\mid_{0}^{\infty}=\dfrac{1}{s}

지수 감쇠함수

f(t)=eatf(t)= e^{at}

L[f(t)]=0eatestdt=0e(as)tdt=1sae(sa)t0=1sa\mathcal{L}[f(t)] = \int_{0}^{\infty}e^{at}e^{-st}dt = \int_{0}^{\infty}e^{(a-s)t}dt \\ = -\dfrac{1}{s-a}e^{-(s-a)t}\mid_{0}^{\infty} = \dfrac{1}{s-a}

단위 램프 함수

f(t)=tf(t)= t

L[f(t)]=0testdt=0(test)dt01estsdt=1s2\mathcal{L}[f(t)]=\int_{0}^{\infty}te^{-st}dt=\int_{0}^{\infty}\left(te^{-st}\right)^{'}dt - \int_{0}^{\infty}1 \dfrac{e^{-st}}{-s}dt = \dfrac{1}{s^2}

정현파 함수

f(t)=sinw0tf(t)=\sin w_{0}t

L[f(t)]=0sin(ωot)estdt=0(sin(ωot)ests)dt0ωocos(ωot)estsdt=sin(ωot)ests0+ωos0cos(ωot)estdt=ωos[0(cos(ωot)ests)dt0ωosin(ωot)estsdt]=ωos[cos(ωot)ests0ωos0sin(ωot)estdt]=ωos[1sωos0sin(ωot)estdt]\begin{matrix} \mathcal{L}[f(t)]&=&\int_{0}^{\infty}\sin (\omega_o t) e^{-st}dt \\ &=&\int_{0}^{\infty}\left(\sin (\omega_o t) \dfrac{e^{-st}}{-s}\right)^{'}dt - \int_{0}^{\infty}\omega_o \cos (\omega_o t) \dfrac{e^{-st}}{-s}dt \\ &=& \dfrac{\sin (\omega_o t) e^{-st}}{-s}\mid_0^{\infty}+ \dfrac{\omega_o}{s}\int_{0}^{\infty}\cos (\omega_o t)e^{-st}dt\\ &=& \dfrac{\omega_o}{s}\left[ \int_{0}^{\infty}\left(\cos (\omega_o t) \dfrac{e^{-st}}{-s}\right)^{'}dt - \int_{0}^{\infty}\omega_o \sin (\omega_o t) \dfrac{e^{-st}}{-s}dt \right] \\ &=& \dfrac{\omega_o}{s}\left[ \dfrac{\cos (\omega_o t) e^{-st}}{-s}\mid_0^{\infty}- \dfrac{\omega_o}{s} \int_{0}^{\infty}\sin (\omega_o t) e^{-st}dt \right] \\ &=& \dfrac{\omega_o}{s}\left[ \dfrac{1}{s}- \dfrac{\omega_o}{s} \int_{0}^{\infty}\sin (\omega_o t) e^{-st}dt \right] \end{matrix}

그러므로

0sin(ωot)estdt=ωos2+ωo2\int_{0}^{\infty}\sin (\omega_o t) e^{-st}dt = \dfrac{\omega_o}{s^2+\omega_o^2}

단위 임펄스 함수

f(t)=δ(t)f(t)=\delta(t)

중요한 라플라스 변환

f(t) F(s) f(t) F(s)
δ(t)\delta(t) 11 u(t)u(t) 1s \dfrac{1}{s}
tt 1s2 \dfrac{1}{s^2} t2t^2 2s3 \dfrac{2}{s^3}
tnt^n n!sn+1 \dfrac{n!}{s^{n+1}} eate^{-at} 1s+a \dfrac{1}{s+a}
teatt e^{-at} 1(s+a)2 \dfrac{1}{(s+a)^2} tneatt^n e^{-at} n!(s+a)n+1 \dfrac{n!}{(s+a)^{n+1}}
1eat1-e^{-at} as(s+a) \dfrac{a}{s(s+a)} (1at)eat(1-at)e^{-at} s(s+a)2 \dfrac{s}{(s+a)^2}
sinωt \sin \omega t ωs2+ω2\dfrac{\omega}{s^2+\omega^2} cosωt \cos \omega t ss2+ω2\dfrac{s}{s^2+\omega^2}
eatsinωte^{-at} \sin \omega t ω(s+a)2+ω2\dfrac{\omega}{(s+a)^2+\omega^2} eatcosωte^{-at} \cos \omega t (s+a)(s+a)2+ω2\dfrac{(s+a)}{(s+a)^2+\omega^2}
first article
next article
Comments
Feel free to ask a question, answer or comment, etc.