The low pass filter attenuates high frequencies and lets low frequencies through. The basic circuit using a resistor and a capacitor is as follows:

Let X_{C} be the capacitive reactance
and ω be the angular frequency.
To calculate the cut-off frequency we proceed as follows:

V_{in}= IZ = I√(R^{2}+X_{C}^{2}) V_{out}= IX_{C}V_{out}/V_{in}= 1/(ωC√(R^{2}+(1/(ω^{2}C^{2})))) [multiply top and bottom by ωC] = 1/√(1+R^{2}ω^{2}C^{2})

Now, at the cut-off frequency,

1/√2 = 1/√(1+R^{2}ω^{2}C^{2}) 1/2 = 1/(1+R^{2}ω^{2}C^{2}) 1 = R^{2}ω^{2}C^{2}= RωC

Taking a 1kΩ resistor and 100nF capacitor we can calculate ω at the cut-off frequency which will give us the cut-off frequency, f, when we divide that by 2π. So,

C = 100nF = 1×10^{-7}R = 1kΩ = 1×10^{3}RC = 1×10^{-4}ω = 1/RC = 1×10^{4}f = ω/2π = 1.590kHz

The circuit was built and both V_{in} and V_{out} were measured with an oscilloscope at various
frequencies.
V_{in} was supplied by a signal generator.
The first measurement was taken at roughly the cut-off frequency of 1.59kHz.
In the image below, channel 1 (blue trace) is V_{in} and the channel 2 (yellow trace) is V_{out}.
As expected, V_{out}/V_{in} is roughly 0.7.

The next measurement was taken at a frequency of roughly 100Hz which is much lower than the cut-off frequency of 1.59kHz.
As expected, V_{out}/V_{in} is roughly 1.
This low frequency signal is passing through the filter.

The last measurement was taken at a frequency of roughly 15.9kHz which is much higher than the cut-off frequency of 1.59kHz.
As expected, V_{out}/V_{in} shows us that the signal has been attenuated.

Fischer-Cripps. A.C., *The Electronics Companion.* Institute of Physics, 2005.

Copyright © 2014 Barry Watson. All rights reserved.