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