Going from digital to analog filters
The main trick is knowing how typical lumped circuit element impedances are represented in the $s$-domain. Recall the current-voltage relationship for a capacitor:

$$
i(t) = C \frac{dv(t)}{dt}
$$

Transforming this to the Laplace domain yields:

$$
I(s) = CsV(s)
$$

Note that voltage and current still have a proportional relationship in this domain, just like the time domain. This allows us to write a (frequency-dependent) $s$-domain impedance for the capacitor. This impedance can be used with Ohm's law and techniques like nodal analysis to calculate what a circuit's Laplace-domain output would be for a given Laplace-domain input signal (e.g. an impulse, whose Laplace transform is unity).

For a capacitor, this impedance is:

$$
Z(s) = \frac{V(s)}{I(s)} = \frac{1}{Cs}
$$

This gives capacitors their known highpass characteristic; for $s=j\omega$, as $\omega$ gets larger, the impedance gets smaller. For the limiting case of $\omega=0$, $Z(0) = \infty$; a capacitor passes no DC current.



We can follow the same procedure for an inductor, starting with its current-voltage relationship:

$$
v(t) = L\frac{di(t)}{dt}
$$

Apply the Laplace transform:

$$
V(s) = LsI(s)
$$

and again, we can calculate the inductor's $s$-domain impedance:

$$
Z(s) = \frac{V(s)}{I(s)} = Ls
$$

As the dual to capacitors, inductors have a lowpass characteristic. For small frequencies ($s \to 0$), $Z(s) \to 0$ as well.



Given these impedances, it becomes very straightforward to apply basic circuit analysis techniques to determine the $s$-domain transfer function of a network containing resistors, capacitors, and inductors.