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raccourcis clavier
graph LR
    Diff{{differential equations}} -- "Laplace transform" --> Algebraic{{algebraic equations}} -- "inverse Laplace transform" --> End{{time domain solution}}
L{f(t)}=0f(t)stdt=F(s)\mathcal{L} \{f(t)\} = \int_0^{\infty}f(t)^{-st}dt = F(s) Item no.f(t)F(s)1.δ(t)12.u(t)1s3.tu(t)1s24.tnu(t)n!sn+15.eatu(t)1s+a6.sin(ωt)u(t)ωs2+ω27.cos(ωt)u(t)ss2+ω2\begin{array}{c c c} \hline \text{Item no.} & f(t) & F(s) \\ \hline 1. & \delta(t) & 1 \\ 2. & u(t) & \frac{1}{s} \\ 3. & tu(t) & \frac{1}{s^2} \\ 4. & t^n u(t) & \frac{n!}{s^{n+1}} \\ 5. & e^{-at}u(t) & \frac{1}{s + a} \\ 6. & \sin(\omega t)u(t) & \frac{\omega}{s^2 + \omega^2} \\ 7. & \cos(\omega t)u(t) & \frac{s}{s^2 + \omega^2} \\ \hline \end{array} δ(t)=0,t0,0δ(t)dt=1\delta{(t)} = 0, \quad t \neq 0,\quad \int_0^{\infty}{\delta{(t)}}dt=1

example: Given a unit step function u(t)={0t<01t0u(t) = \begin{cases} 0 & t < 0 \\ 1 & t \ge 0 \end{cases}

U(s)=L{u(t)}=0u(t)estdt=1sestdtU(s)=1s(01)=1s\begin{aligned} U(s) = \mathcal{L} \{u(t)\} &= \int_{0}^{\infty} u(t) e^{-st} dt = -\frac{1}{s} e^{-st} dt \\ U(s) &= -\frac{1}{s}(0-1) = \frac{1}{s} \end{aligned}

inverse form

L1{F(s)}=12πjlimωσjωσ+jωF(s)estds\mathcal{L}^{-1} \{ F(s) \} = \frac{1}{2\pi j} \lim_{\omega \to \infty} \int_{\sigma-j\omega}^{\sigma+j\omega} F(s) e^{st} \, ds