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 + Xc2)0.5, where Xc = 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
Vout/V = Xc/(R2 + Xc2)0.5
At very low frequency, the impedance of the capacitor Xc = 1/(2πfC)>>R, so
Vout/V ≈ Xc/(Xc2)0.5 = Xc/Xc = 1
The critical frequency, fc, is when Vout/V = 0.707. fc happens when R=Xc:
Vout/V = Xc/(Xc2 + Xc2)0.5 = Xc/(2Xc2)0.5 = 1/20.5 = 0.707.
To find fc we set R=Xc=1/(2πfC), then arrive at f = 1/(2πRC). A series low-pass filter with capacitor 4700pF and resistor 10kΩ will have critical frequency fc = 1/(2π*1×104*4700*10-12) = 3390Hz.
Jack of Oracle Tutoring by Jack and Diane, Campbell River, BC.