The tutor gives background along with a basic explanation of how a high-pass filter works.
A high-pass filter will send along high frequency signals but block low frequencies. It can do so because the impedance, Xc, of a capacitor of capacitance C, at frequency f, is
At very high frequency, therefore, Xc≈0
The total impedance, Z, of a resistor R and capacitor C in series is given by
Z=(R2 + Xc2)0.5
Therefore, with input voltage V, and instantaneous current i, Ohm’s Law gives
V=i(R2 + Xc2)0.5
The voltage across the resitor is
Therefore, the ratio of the voltage across the resistor to the source is
VR/V = iR/i(R2 + Xc2)0.5 = R/(R2 + Xc2)0.5
As mentioned earlier, for high frequency, Xc≈0, giving
VR/V ≈ R/(R2)0.5 = 1
With a high-pass filter, the maximum output voltage for a high frequency signal equals the input voltage.
Convention says that the critical frequency, fc, is that at which VR/V = 1/(2)0.5 = 0.707, which occurs when Xc = R:
VR/V = R/(R2 + Xc2)0.5 = R/(R2 + R2)0.5 = R/(2R2)0.5
which leads to
VR/V = 1/20.5 = 0.707
Therefore, a high-pass filter will pass any frequency higher than the critical frequency fc, where fc is calculated from
Xc = R
1/(2πfcC) = R
1 = 2πfcCR
1/(2πCR) = fc
By that reasoning, a 10kΩ resistor, in series with a 12pF (picoFarad) capacitor, placed in series, should produce a high-pass filter with critical frequency fc = 1.3MHz. The output would be read across the resistor R.
Serway, Raymond A. Physics for Scientists and Engineers with modern physics. Toronto: Saunders College Publishing, 1986.
Jack of Oracle Tutoring by Jack and Diane, Campbell River, BC.