How this instrument works
A resistor-capacitor low-pass filter works because a capacitor's opposition to alternating current, its reactance, shrinks as frequency climbs: X_C = 1 ⁄ (2πfC). Tap the output across the capacitor and, at low frequencies, that reactance is large compared with the resistor, so nearly all the input voltage appears at the output. Push the frequency up and the reactance falls, the capacitor increasingly shorts the signal to ground, and the output shrinks with it. Cutoff frequency is simply the crossover point: the one frequency at which the capacitor's reactance equals the resistor's value, X_C = R, splitting the incoming power evenly between the two.
That equal split is why the cutoff is also called the −3 dB or half-power point. At f_c the output voltage has dropped to 1 ⁄ √2 of the input, about 70.7 percent, which corresponds to half the power reaching the output stage. Below f_c the filter barely touches the signal; well above it, output amplitude keeps falling at a steady −20 dB per decade, or −6 dB every time frequency doubles, because this is a single-pole filter with exactly one reactive element doing the work. A sharper cutoff needs more stages, or an inductor added to make it resonate instead of just roll off.
The formula quietly assumes a source with zero output impedance feeding a load that draws no current at all. Neither exists in practice. A real signal source has some series resistance, which stacks with R and lowers the true f_c a little; a real load draws current through whatever sits after the capacitor, which pulls it the other way. Op-amp buffers and high-impedance ADC inputs get close enough that these effects stay under a percent or two, which is why the plain two-component formula remains the everyday design tool, but a designer chasing a precise corner in a finished circuit still measures it on the bench rather than trusting the datasheet arithmetic alone.
- Enter the resistor's value in Resistance — the field defaults to kΩ; switch to Ω for small values like 100 or 220.
- Enter the capacitor's value in Capacitance, matching whatever nF or µF rating is printed on the part's body.
- Read the result in Cutoff frequency, in Hz or kHz — this is the −3 dB point where output power has halved.
- To hit a target frequency instead of just checking one, pick a standard capacitor value first, then adjust Resistance until Cutoff frequency lands where the design calls for.
Worked example — a 10 kΩ resistor with a 100 nF capacitor
Take a 10 kΩ resistor paired with a 100 nF capacitor, two values that sit in nearly every hobbyist's parts drawer. Converted to base units, R = 10,000 Ω and C = 1 × 10⁻⁷ F. Multiply them with 2π: 2π × 10,000 × 0.0000001 = 0.00628318530718. Take the reciprocal and the cutoff frequency comes out to 159.154943092 Hz, which the results panel rounds and displays as 159.155 Hz.
That exact pairing turns up when someone wires a microcontroller's PWM pin through a resistor and capacitor to fake an analog control voltage. A typical 490 Hz Arduino PWM rate sits well above the 159 Hz cutoff, so most of the switching ripple gets knocked down while the RC time constant of 1 millisecond still lets the output track duty-cycle changes in a few milliseconds. Swap in a 200 nF capacitor instead and the cutoff falls to 79.577 Hz — heavier ripple rejection, at the cost of a sluggish response.
Questions
What does the −3 dB cutoff point actually mean?
It marks where the output voltage has fallen to 1 ⁄ √2, about 70.7 percent, of the input, which works out to half the input power reaching the far side of the filter. It is the conventional boundary between a filter's passband, where the signal passes through nearly untouched, and its stopband, where attenuation grows steadily with frequency.
Why does a high-pass filter built from the same two parts share this exact cutoff?
Because f_c = 1 ⁄ (2πRC) only cares about the values of R and C, not which component the output is measured across. Tap the output across the capacitor and low frequencies pass freely — a low-pass. Tap it across the resistor instead, using the same 10 kΩ and 100 nF, and the identical 159 Hz point becomes the frequency above which the signal gets through — a high-pass.
What happens to the cutoff if I double the capacitor?
It is cut in half. Cutoff frequency is inversely proportional to C, so raising this circuit's 100 nF to 200 nF while holding R at 10 kΩ drops f_c from 159.155 Hz to 79.577 Hz. Halving R instead — to 5 kΩ — has the identical effect in reverse, doubling the cutoff to 318.310 Hz, since R and C enter the formula the same way.
How steep is the roll-off once frequency climbs past cutoff?
About −20 dB per decade, or −6 dB every time frequency doubles. That gentle, steady slope is the signature of a single-pole filter with one resistor and one capacitor. Steeper rejection needs cascading more RC stages or building an active filter with an op-amp, since a passive single-stage RC pair cannot roll off any faster.
Does the load connected after the filter change the real cutoff frequency?
Yes, if it draws meaningful current. The formula assumes the output feeds an infinite-impedance load and is driven by a zero-impedance source; a real load resistance effectively parallels the capacitor and nudges the true corner upward, while a source with its own series resistance adds to R and pulls it down. A high-impedance buffer keeps the built calculation close to what a bench measurement shows.
Can this circuit smooth PWM output into a usable analog voltage?
Yes, provided the PWM frequency sits well above the filter's cutoff. A 10 kΩ and 100 nF pair, cutoff near 159 Hz, will heavily attenuate the switching ripple of a several-hundred-hertz PWM signal while still following slower changes in duty cycle, which is exactly why this pairing shows up so often in microcontroller projects that need a crude digital-to-analog output.