The tutor briefly explains the low-pass filter.

This explanation draws on ideas from that of a high-pass filter (see my article here).

A low-pass filter sends along low frequencies but blocks higher ones. The one we’re looking at today has a resistor and a capacitor in series. As detailed in my article on the series high-pass filter, we have total impedance Z=(R² + X_c²)^0.5, where X_c = 1/(2πfC).

As the frequency decreases, the impedance of the capacitor increases, so its share of the voltage output rises. A low-pass filter will read the voltage across the capacitor. Relative to the input voltage for the circuit, V, its output will be

V_out/V = X_c/(R² + X_c²)^0.5

At very low frequency, the impedance of the capacitor X_c = 1/(2πfC)>>R, so

V_out/V ≈ X_c/(X_c²)^0.5 = X_c/X_c = 1

The critical frequency, f_c, is when V_out/V = 0.707. f_c happens when R=X_c:

V_out/V = X_c/(X_c² + X_c²)^0.5 = X_c/(2X_c²)^0.5 = 1/2^0.5 = 0.707.

To find f_c we set R=X_c=1/(2πfC), then arrive at f = 1/(2πRC). A series low-pass filter with capacitor 4700pF and resistor 10kΩ will have critical frequency f_c = 1/(2π*1×10⁴*4700*10^-12) = 3390Hz.

Source:

ecee.colorado.edu

www.electronics-tutorials.ws

Jack of Oracle Tutoring by Jack and Diane, Campbell River, BC.