Audited 31 Jul 2026·Last updated 15 Sept 2026·3 citations·Tier 2·0 uses

Cutoff Frequency Calculator

RC cutoff frequency calculator: f_c = 1/(2πRC). Find the −3 dB corner of a low-pass or high-pass RC filter and the rolloff beyond it.

Cutoff Frequency Calculator

Ω
F
RC cutoff frequency
159.1549
Result of f_c = 1 / (2πRC) using the entered coherent-SI magnitudes.
Model scope
Ideal first-order RC -3 dB corner only; component tolerance, source/load impedance, parasitics, op-amp limits, filter order, topology, and frequency-dependent components are outside it.

Background.

One resistor and one capacitor make the simplest filter electronics has, and a single number characterises it: the cutoff frequency f_c = 1/(2πRC). Below the cutoff a low-pass RC passes signals essentially untouched; above it, attenuation deepens at 6 dB per octave. Swap the two components' positions and the same corner frequency describes a high-pass instead.

The cutoff is not a wall. At f_c itself the output has fallen to 1/√2 of the input — the −3 dB point, where half the signal power is lost — and the transition is gentle on both sides. One octave above the corner a low-pass still passes about 45% of the amplitude; a decade above, 10%. Treating f_c as a brick-wall boundary is the most common misreading of a first-order filter, and the reason audio crossovers and anti-aliasing filters cascade several stages.

The product RC has units of seconds — it is the filter's time constant τ, the same τ that governs how the capacitor charges through the resistor. The frequency-domain corner and the time-domain step response are two views of one number: f_c = 1/(2πτ). A 1 kΩ resistor with a 1 µF capacitor has τ = 1 ms and therefore corners near 159 Hz, whether you look at it with a signal generator or an oscilloscope step.

This page computes the ideal first-order corner. Component tolerance (electrolytics commonly ±20%), source and load impedance loading the divider, and parasitic elements all move a built circuit's corner; the scope note beside the result keeps that boundary honest.

What is cutoff frequency calculator?

The cutoff (or corner, or −3 dB) frequency of a first-order RC filter is the frequency at which the output amplitude falls to 1/√2 ≈ 70.7% of the input — equivalently, where output power is halved. It is set entirely by the resistor-capacitor product: f_c = 1/(2πRC). At this frequency the capacitor's reactance equals the resistance, the phase shift is 45°, and the same value marks the corner whether the RC pair is arranged as a low-pass or a high-pass.

How to use this calculator.

  1. Enter the resistance in ohms — the value in the signal path for a low-pass, the shunt value for a high-pass.
  2. Enter the capacitance in farads: 1 µF is 1e-6, 100 nF is 1e-7, 22 pF is 2.2e-11 — unit slips of a thousand are the classic error here.
  3. Read f_c and mark it mentally as the −3 dB point, not a cliff: signals an octave inside the passband are already down about 1 dB.
  4. To hit a target corner instead, pick a convenient capacitor and rearrange to R = 1/(2πf_cC) — resistors come in finer value steps than capacitors.
  5. Derate for reality: ±20% capacitor tolerance moves the corner ±20%, and a load impedance comparable to R drags the corner and passband gain with it.

The formula.

f_c = 1 / (2πRC)

An RC low-pass is a voltage divider whose lower leg is the capacitor's reactance X_C = 1/(2πfC). At low frequency X_C is enormous and the divider passes nearly everything; at high frequency X_C collapses and the output follows it down. The crossover point where X_C = R defines the corner: solving 1/(2πf_cC) = R gives f_c = 1/(2πRC). At that frequency the divider's magnitude is |1/(1+j)| = 1/√2, hence −3 dB and a 45° phase lag. Above the corner each doubling of frequency halves the output — the 6 dB/octave (20 dB/decade) rolloff characteristic of any single pole. The same corner written as ω_c = 1/RC connects to the step response e^(−t/RC): fast filters settle fast, and a corner chosen for smoothing sets the settling time you must accept. The engine computes the reciprocal with Decimal arithmetic, rounding once to twelve significant digits.

A worked example.

Example

An RC low-pass filter is built from a 1 kΩ resistor and a 1 µF capacitor. Where is its corner? The -3 dB cutoff is f_c = 1/(2πRC). The RC product is 1,000 × 1×10⁻⁶ = 10⁻³ seconds, so f_c = 1/(2π × 10⁻³) ≈ 159.15 Hz. At that frequency the output has fallen to 70.7% of the input voltage (half power), with roughly a 6 dB-per-octave slide beyond it — 320 Hz emerges near −7 dB, 1.6 kHz near −20 dB. Because R and C enter as a product, the same corner is available from many pairs: 10 kΩ with 100 nF, or 100 kΩ with 10 nF, all give 159 Hz — the choice among them is set by source and load impedance, not by the corner. This page computes the ideal first-order corner only; component tolerance alone (a ±10% capacitor) moves the real one by ±10%.

capacitance F0
resistance Ohm1,000

Frequently asked questions.

What does −3 dB actually mean at the cutoff?
Amplitude is down to 1/√2 ≈ 70.7% of the input, which is exactly half the power — 10·log₁₀(½) ≈ −3.01 dB. It is a conventional reference point on a smooth curve, not where filtering ‘begins’: at half the cutoff frequency a low-pass is already down about 1 dB, and at twice the cutoff it is down about 7 dB.
Is the cutoff different for a low-pass and a high-pass RC filter?
No — the same R and C give the same f_c = 1/(2πRC) in both arrangements; what changes is which side of the corner is passed. Taking the output across the capacitor makes a low-pass; across the resistor, a high-pass. At the corner both versions sit at −3 dB, one rolling off upward and the other downward.
How fast does the signal die off past the cutoff?
At 6 dB per octave — every doubling of frequency halves the amplitude — equivalently 20 dB per decade. A single RC stage therefore attenuates only 10× at ten times the corner, which is often not enough: anti-aliasing and audio applications cascade stages or use higher-order topologies to steepen the slope, at 6 dB/octave per added pole.
How is the cutoff related to the RC time constant?
They are the same physics in two domains: τ = RC seconds, and f_c = 1/(2πτ). The worked example's 1 ms time constant corners at 159 Hz. This duality is a design constraint — a smoothing filter with a 1 Hz corner necessarily has τ ≈ 0.16 s and takes several tenths of a second to settle after any step, which no amount of cleverness with a single pole avoids.
Why does my built filter corner at a different frequency than calculated?
Three usual suspects, in order of size: capacitor tolerance (electrolytics are commonly −20/+20% or worse, shifting f_c proportionally); loading — a load impedance comparable to R forms a new divider and moves both corner and passband level; and source impedance adding to R. Parasitic capacitance matters at radio frequencies. Measuring the actual −3 dB point with a sweep beats trusting nominal component values.

How this page was produced

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Quanta Calculator
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3 cited below
Method
f_c = 1 / (2πRC)
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