The high pass filter attenuates low frequencies and lets high frequencies through. We can build such a filter with a RC circuit or a RL circuit.

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 V_{out}= IR V_{out}/V_{in}= R/√(R^{2}+X_{C}^{2}) = R/√(R^{2}+1/(ω^{2}C^{2}) [Multiply top and bottom by RωC] = RωC/√(1+R^{2}ω^{2}C^{2})

Now, at the cut-off frequency,

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

Taking a 1kΩ resistor and a 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 157Hz which is much lower than the cut-off frequency of 1.59kHz. This low frequency signal is attenuated.

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} is roughly 1.
This signal is passing through the filter.

The basic circuit using a resistor and an inductor is as follows:

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

V_{in}= IZ = I√(R^{2}+ω^{2}L^{2}) V_{out}= IX_{L}= ωL V_{out}/V_{in}= ωL/√(R^{2}+ω^{2}L^{2}) = 1/√(R^{2}/ω^{2}L^{2}+1)

Now, at the cut-off frequency,

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

Taking a 1kΩ resistor and a 85mH inductor 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,

L = 0.085H R = 1kΩ ω = R/L = 11764.7 f = ω/2π = 1.873kHz

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 1.87kHz.
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 187Hz which is much lower than the cut-off frequency of 1.87kHz. This low frequency signal is attenuated.

The last measurement was taken at a frequency of roughly 18.7kHz which is much higher than the cut-off frequency of 1.87kHz.
As expected, V_{out}/V_{in} is roughly 1.
This signal is passing through the filter.

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

Copyright © 2014 Barry Watson. All rights reserved.