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IT SEEMS LIKE THE VARIABLES s AND t WERE INTERCHANGED BELOW.

Laplace Transform Pairs and Properties
Definition
Laplace Transform $ F(s)=\int_{-\infty}^\infty f(t) e^{-st}dt, \ s\in {\mathbb C} \ $
Inverse Laplace Transform add formula here
Properties of the Laplace Transform
$ f(s) $ $ F(t) $
$ af_1(s)+bf_2(s) $ $ aF_1(t)+bF_2(t) $
$ f(s/a) $ $ aF(at) $
$ f(s-a) $ $ e^{at}F(t) $
$ e^{-as}f(s) $ $ u(t-a) = \begin{cases} F(t-a) & t>a \\ 0 & t<a \end{cases} $
$ sf(s)-F(0) $ $ F'(t) $
$ s^2f(s)-sF(0)-F'(0) $ $ F''(t) $
$ s^{n}f(s)-\sum_{k=1}^ns^{n-k}F^{(k)}(0) $ $ F^{(n)}(t) $
$ f'(s) $ $ -tF(t) $
$ f''(s) $ $ t^2F(t) $
$ f^{(n)}(s) $ $ (-1)^{(ntn)}F(t) $
$ \frac{f(s)}s $ $ \int_{0}^{t} F(u) du $
$ \frac{f(s)}{s^n} $ $ \int_{0}^{t}...\int_{0}^{t}F(u)du^n = \int_{0}^{t}\frac{{(t-u)}^{n-1}}{(n-1)!} F(u)du $
$ f(s)g(s) $ $ \int_{0}^{t}F(u)G(t-u)du $
$ \int_{s}^{\infty}f(u)du $ $ \frac{F(t)}t $
$ \frac1{1-e^{-sT}}\int_{0}^{T}e^{-su}F(u)du $ $ F(t)=F(t+T) $
$ \frac{f(\sqrt{s})}s $ $ \frac{1}{\sqrt{{\pi}t}}\int_{0}^{\infty}e^{-\frac{u^2}4t}F(u)du $
$ \frac1sf(\frac1s) $ $ \int_{0}^{\infty}J_0(2\sqrt{ut})F(u)du $
$ \frac1{g^{n+1}}f(\frac1s) $ $ t^{\frac{n}2}\int_{0}^{\infty}u^{-\frac{n}2}J_n(2\sqrt{ut})F(u)du $
$ \frac{s+\frac1s}{s^2+1} $ $ \int_{0}^{t}J_0(2\sqrt{u(t-u)})F(u)du $
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Laplace Transform Pairs
notes Signal Laplace Transform ROC
unit impulse/Dirac delta $ \,\!\delta(t) $ 1 $ \text{All}\, s \in {\mathbb C} $
unit step function $ \,\! u(t) $ $ \frac{1}{s} $ $ \mathcal{R} \mathfrak{e} \lbrace s \rbrace > 0 $
$ \,\! -u(-t) $ $ \frac{1}{s} $ $ \mathcal{R} \mathfrak{e} \lbrace s \rbrace < 0 $
$ \frac{t^{n-1}}{(n-1)!}u(t) $ $ \frac{1}{s^{n}} $ $ \mathcal{R} \mathfrak{e} \lbrace s \rbrace > 0 $
$ -\frac{t^{n-1}}{(n-1)!}u(-t) $ $ \frac{1}{s^{n}} $ $ \mathcal{R} \mathfrak{e} \lbrace s \rbrace < 0 $
$ \,\!e^{-\alpha t}u(t) $ $ \frac{1}{s+\alpha} $ $ \mathcal{R} \mathfrak{e} \lbrace s \rbrace > -\alpha $
$ \,\! -e^{-\alpha t}u(-t) $ $ \frac{1}{s+\alpha} $ $ \mathcal{R} \mathfrak{e} \lbrace s \rbrace < -\alpha $
$ \frac{t^{n-1}}{(n-1)!}e^{-\alpha t}u(t) $ $ \frac{1}{(s+\alpha )^{n}} $ $ \mathcal{R} \mathfrak{e} \lbrace s \rbrace > -\alpha $
$ -\frac{t^{n-1}}{(n-1)!}e^{-\alpha t}u(-t) $ $ \frac{1}{(s+\alpha )^{n}} $ $ \mathcal{R} \mathfrak{e} \lbrace s \rbrace < -\alpha $
$ \,\!\delta (t - T) $ $ \,\! e^{-sT} $ $ \text{All}\,\, s\in {\mathbb C} $
$ \,\cos( \omega_0 t)u(t) $ $ \frac{s}{s^2+\omega_0^{2}} $ $ \mathcal{R} \mathfrak{e} \lbrace s \rbrace > 0 $
$ \, \sin( \omega_0 t)u(t) $ $ \frac{\omega_0}{s^2+\omega_0^{2}} $ $ \mathcal{R} \mathfrak{e} \lbrace s \rbrace > 0 $
$ \,e^{-\alpha t}\cos( \omega_0 t) u(t) $ $ \frac{s+\alpha}{(s+\alpha)^{2}+\omega_0^{2}} $ $ \mathcal{R} \mathfrak{e} \lbrace s \rbrace > -\alpha $
$ \, e^{-\alpha t}\sin( \omega_0 t)u(t) $ $ \frac{\omega_0}{(s+\alpha)^{2}+\omega_0^{2}} $ $ \mathcal{R} \mathfrak{e} \lbrace s \rbrace > -\alpha $
$ u_n(t) = \frac{d^{n}\delta (t)}{dt^{n}} $ $ \,\!s^{n} $ $ All\,\, s $
$ u_{-n}(t) = \underbrace{u(t) *\dots * u(t)}_{n\,\,times} $ $ \frac{1}{s^{n}} $ $ \mathcal{R} \mathfrak{e} \lbrace s \rbrace > 0 $

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