A simple low-pass filter you can build yourself:


              R
   Vin o-----MMM----+---------o  Vout
                    |   
                    |
   GND o-+         === C    +-o  GND
         |          |       |
         +----------+-------+
Where:
R
Resistance
C
Capacitance
Vin
Input Voltage
Vout
Output Voltage
GND
Ground
===
Capacitor
MMM
Resistor

The attenuation (A) of this circuit is (with w = 2*PI*f, where f is the frequency):
                    1
  A = -------------------------------
      SQRT(1 + (R^2)(C^2)(w^2))
For this quick circuit analysis, we postulate that there is no current out Vout (i.e., it has infinite resistance). We know that the resistor's resistance (for this discussion, impedance) is constant with respect to frequency, but the capacitor's is not. Indeed, if we let Z be the impedance of the capacitor, Z is proportional to 1/(C*w). Thus, at low frequency, w goes to 0 and 1/w goes to infinity. Thus, the voltage at Vout is equal to Vin (minus the voltage drop across the resistor). At high frequency, w goes to infinity and 1/w goes to 0. Thus, Vout goes to ground, since the voltage drop across C goes to zero (no impedance).

NOTE, please, that this is only a very, very superficial treatment of this circuit. The phase might be altered by this circuit, and funky things happen (plot the attenuation! It's not very neat!). It's not anywhere near perfect, and there are other, much better filters out there. Basically, don't blame me if you toast stuff. That said, this little puppy can do some spiff stuff. If you liked this writeup, learn circuit analysis. Maybe I'll be in your Circuits I class. Grin