This explanation draws on ideas from that of a high-pass filter (see my article here<\/a>).<\/p>\n 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,<\/a> we have total impedance Z=(R2<\/sup> + Xc<\/sub>2<\/sup>)0.5<\/sup>, where Xc<\/sub> = 1\/(2\u03c0fC).<\/p>\n 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.<\/strong> Relative to the input voltage for the circuit, V, its output will be<\/p>\n Vout<\/sub>\/V = Xc<\/sub>\/(R2<\/sup> + Xc<\/sub>2<\/sup>)0.5<\/sup><\/p>\n At very low frequency, the impedance of the capacitor Xc<\/sub> = 1\/(2\u03c0fC)>>R, so<\/p>\n Vout<\/sub>\/V \u2248 Xc<\/sub>\/(Xc<\/sub>2<\/sup>)0.5<\/sup> = Xc<\/sub>\/Xc<\/sub> = 1<\/p>\n The critical frequency, fc<\/sub>, is when Vout<\/sub>\/V = 0.707. fc<\/sub> happens when R=Xc<\/sub>:<\/p>\n Vout<\/sub>\/V = Xc<\/sub>\/(Xc<\/sub>2<\/sup> + Xc<\/sub>2<\/sup>)0.5<\/sup> = Xc<\/sub>\/(2Xc<\/sub>2<\/sup>)0.5<\/sup> = 1\/20.5<\/sup> = 0.707.<\/p>\n To find fc<\/sub> we set R=Xc<\/sub>=1\/(2\u03c0fC), then arrive at f = 1\/(2\u03c0RC). A series low-pass filter with capacitor 4700pF and resistor 10k\u2126 will have critical frequency fc<\/sub> = 1\/(2\u03c0*1×104<\/sup>*4700*10-12<\/sup>) = 3390Hz.<\/p>\n Source:<\/p>\n ecee.colorado.edu<\/a><\/p>\n www.electronics-tutorials.ws<\/a><\/p>\n