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Electrical

Cutoff frequency of a first-order RC filter

Frequency at which a first-order RC filter passes about 71 % (−3 dB) of the input amplitude.

Ref
ELE-39
Rev
v1
Field
Electrical
Result
fc [Hz]

A first-order RC filter has a cutoff (corner) frequency fc = 1/(2πRC), where the output amplitude has fallen to 1/√2 (0.707, or −3 dB) of its passband value. Below fc a low-pass filter passes the signal; above it, the output falls at 20 dB per decade. Here R is entered in kΩ and C in nF, and the result is in Hz.

Frequency at which a first-order RC filter passes about 71 % (−3 dB) of the input amplitude. It is the same for the low-pass and high-pass arrangements.

Calculator

Solve for
Hz
kΩ
0.1100
nF
110,000
  • Time constant, RC100 µs
  • Angular cutoff frequency, 1/(RC)10,000 rad/s
  • Gain at ten times the cutoff frequency0.0995

What drives the result

Each variable is moved across its typical range while the others stay at the values above. The longest bar is the key driver.

Cutoff frequency (Hz) current value 1,592 Hz
CCapacitanceKey driver
15.9159,000
RResistance
15.915,900
  • A 1 kΩ resistor with a 100 nF capacitor has a corner at 1,592 Hz and a time constant of 100 µs.
  • To move the corner by a decade, change R or C by a factor of ten; two cascaded stages with buffering give a roll-off of 40 dB per decade.
Chart data as tables

Ranked variables

VariableTypical rangeLowest resultHighest resultSwing (Hz)Local stepChange in result
C Capacitance1 to 10,000 nF15.92159,200159,100+10%−9.09 %
R Resistance0.1 to 100 kΩ15.9215,92015,900+10%−9.09 %

Sweep of Capacitance

Capacitance (nF)Cutoff frequency (Hz)
1159,200
3.16250,330
1015,920
31.625,033
1001,592
316.2503.3
1,000159.2
3,16250.33
10,00015.92

Sweep one variable

Cutoff frequency (Hz)Hover, touch, or use the arrow keys on the chart
050k100k150k 110100100010k Capacitance (nF), log scale now 1,592

Variables and how each one contributes

fc

Cutoff frequency

Result · Hz

What the equation gives you. Everything else feeds it.

R

Resistance

kΩ · typical 0.1 to 100

The resistance, in kΩ. Lowers the cutoff frequency in inverse proportion: a larger resistance charges the capacitor more slowly. Raising it by 0.5 kΩ (to 1.5 kΩ) moves the result by −33.3 %, from 1,592 to 1,061 Hz.

C

Capacitance

nF · typical 1 to 10,000

The capacitance, in nF. Lowers the cutoff frequency in inverse proportion: a larger capacitor takes longer to charge. Raising it by 10 nF (to 110 nF) moves the result by −9.09 %, from 1,592 to 1,447 Hz.

Worked example with the values above

  1. Time constant: RC = 1 kΩ × 100 nF = 100 µs = 1×10−4 s.
  2. Denominator: 2π × 1×10−4 = 6.283×10−4 s.
  3. fc = 1 / 6.283×10−4 = 1,592 Hz.

Assumptions and limits

  • An ideal resistor and capacitor driven from a low-impedance source into a high-impedance load, so nothing else loads the filter.
  • Sinusoidal signals in the steady state.

Common mistakes

  • Using kΩ with µF or nF without converting to the consistent system.
  • Confusing the cutoff frequency in Hz with the angular frequency in rad/s (factor 2π).

Statistics and uncertainty

Capacitors of 10 % or 20 % tolerance and resistors of 1 % give a corner uncertain by 10 % or more, which is dominated by the capacitor. Use C0G or film capacitors if the corner frequency matters, since ceramic types change with voltage and temperature.

Where it is used

  • Anti-aliasing and noise filters in sensor and data-acquisition circuits.
  • Setting the bandwidth of a coupling or smoothing circuit.

Maths you need

  • Capacitor: stores energy in an electric field
  • Decibel: 20 log₁₀ of an amplitude ratio

Related formulas

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