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, X_c, of a capacitor of capacitance C, at frequency f, is

X_c=1/(2πfC)

At very high frequency, therefore, X_c≈0

The total impedance, Z, of a resistor R and capacitor C in series is given by

Z=(R² + X_c²)^0.5

Therefore, with input voltage V, and instantaneous current i, Ohm’s Law gives

V=i(R² + X_c²)^0.5

The voltage across the resitor is

V_R=iR

Therefore, the ratio of the voltage across the resistor to the source is

V_R/V = iR/i(R² + X_c²)^0.5 = R/(R² + X_c²)^0.5

As mentioned earlier, for high frequency, X_c≈0, giving

V_R/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, f_c, is that at which V_R/V = 1/(2)^0.5 = 0.707, which occurs when X_c = R:

V_R/V = R/(R² + X_c²)^0.5 = R/(R² + R²)^0.5 = R/(2R²)^0.5

which leads to

V_R/V = 1/2^0.5 = 0.707

Therefore, a high-pass filter will pass any frequency higher than the critical frequency f_c, where f_c is calculated from

X_c = R

1/(2πf_cC) = R

1 = 2πf_cCR

1/(2πCR) = f_c

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 f_c = 1.3MHz. The output would be read across the resistor R.

Source:

www.electronics-tutorials.ws

ecee.colorado.edu

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.