Laplace Transform

$F(s) = \int_0^{\infty} f(t)\, e^{-st}\, dt$

Derivatives become multiplication by $s$ (with initial conditions), so circuits can be analysed with impedances $R$, $sL$ and $1/(sC)$.

$f(t)$	$F(s)$
$\delta(t)$	1
$u(t)$	$1/s$
$e^{-at}$	$1/(s+a)$
$\sin\omega t$	$\omega/(s^2+\omega^2)$
$\cos\omega t$	$s/(s^2+\omega^2)$

A transfer function $H(s) = Y(s)/X(s)$ has poles (roots of the denominator) and zeros. A system is stable if all poles lie in the left half of the s-plane. Setting $s = j\omega$ gives the frequency response.