# 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=(R^{2} + X_{c}^{2})^{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}/(R^{2} + X_{c}^{2})^{0.5}

At very low frequency, the impedance of the capacitor X_{c} = 1/(2πfC)>>R, so

V_{out}/V ≈ X_{c}/(X_{c}^{2})^{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_{c}^{2} + X_{c}^{2})^{0.5} = X_{c}/(2X_{c}^{2})^{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*1×10^{4}*4700*10^{-12}) = 3390Hz.

Source:

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