On top are low-pass filters, and on bottom are high-pass filters. The resistance of the resistor in ohms should be the square root of the inductor in Henries divided by the capacitor in Farads. The equation would look like √(L/C)*. The filters on right are sharper than the filters on left, but the ones on right have a small spot on the frequency spectrum where it actually amplifies the input because of resonance.
For purely resistive filters (as described in the next paragraph) you can put copies of the filter (and other purely resistive filters) in series to change the behavior. For example, putting two identical low-pass filters in series will attenuate the high frequency twice as much (e.g. -20dB turns to -40dB). If a high-pass and low-pass are put in series it’ll create a band-pass. One thing to note is that putting filters in series can create certain spots in the frequency spectrum where it actually amplifies slightly due to resonance, but this effect is usually small.
The filters on the right are purely resistive from the point of view of the voltage source (the current is in phase with the voltage). The filters on the left however do not have this property, but are partially reactive as well. This can be fixed with a capacitor or an inductor in parallel with the voltage source. For a low-pass filter use a capacitor. For a high-pass filter use an inductor.
Here are the equations for the inductors and capacitors given the frequency and output resistor:
L = R/(τf)
C = 1/(τfR)
Where
τ = 2π
f = frequency (Hertz)
R = resistance (Ohms)
L = inductance (Henries)
C = capacitance (Farads)
All inductors and capacitors within a single filter will have the values given by the above equations. However, when you chain filters some inductors or capacitors will have half or double the value due to two ending up in series or parallel and thus being able to be combined into one with a different value.
*Credit to @2ctiby for the sqrt symbol.
