Laplace Transforms

F (s)

Converting functions into Laplace Transforms makes them easier to manipulate, which can then be untransformed

\begin{aligned} L {c} & = \frac{c}{s} (constant) \\ L {e^{a t}} & = \frac{1}{s - a} \\ L {\cos (w t)} & = \frac{s}{s^{2} + w^{2}} \\ L {\sin (w t)} & = \frac{w}{s^{2} + w^{2}} \\ L {\frac{e^{x} - e^{- x}}{2}} = L {h \sin (w t)} & = \frac{w}{s^{2} - w^{2}} \end{aligned}

Transforms of derivatives:

\begin{aligned} L {f^{'} (t)} & = s L {f (t)} - f (0) \\ L {f^{(n)} (t)} & = s^{n} L {f (t)} - s^{n - 1} f (0) - s^{n - 2} f^{'} (0) - \dots s f^{(n - 2)} (0) - f^{(n - 1)} (0) \end{aligned}

Transforms of powers:

L {t^{n}} = \frac{n!}{s^{n + 1}}, s > 0, n \in N

Doesn't work for negative powers, as the integrals needed for the definition of the laplace transform doesn't converge
for non-integers, $γ$ function is used to calculate the exponential of fractions

Transform of $e^{a t} f (t)$

L {e^{a t} f (t)} = F (s - a)

Solution obtained is only valid for $t \geq 0$

Inverse Laplace Transforms

L^{- 1} {F (s)} = f (t)

The inverse Laplace Transform is also a linear operator

L^{- 1} {\frac{1}{(s - 1)^{2}} + \frac{10}{s - 1}} = t e^{t} + 10 e^{t}

This can be used with factoring, Partial Fractions and other techniques to simplify the inversing

u (t) = {\begin{cases} 0, t < 0 \\ 1, t > 0 \end{cases}

L {u (t - a) f (t - a)} = e^{- a s} F (s)

L^{- 1} {e^{- a s} F (s)} = u (t - a) f (t - a)

An impulse function representing instantaneous behaviour

δ_{ϵ} (t) = {\begin{cases} \frac{1}{2 ϵ} if - ϵ < t < ϵ \\ 0 if | t | \geq ϵ \end{cases}

$\int δ (t) d t = 1$