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Low Pass Filter Calculator

Electronics & PhysicsCircuits and Components
Filter type

Shows how much a sine wave at this frequency is attenuated.

Cutoff frequency fc (Hz) —
Cutoff with SI prefix
—
Time constant τ
—
Angular cutoff ωc (rad/s)
—
Voltage gain Vout/Vin at signal
—
Gain at signal (dB)
—
Phase shift at signal (°)
—

A first-order RC low-pass filter has a cutoff frequency fc = 1 ÷ (2πRC); an RL version uses fc = R ÷ (2πL). At fc the output is 70.7 % of the input, or −3.01 dB. A 1 kΩ resistor with a 1 µF capacitor gives fc = 159.15 Hz and a 1 ms time constant.

About this tool

Anyone sizing an anti-aliasing stage, smoothing a PWM output or taming switch noise on a sensor line needs the corner frequency of a simple low-pass network. Choose RC or RL, enter the part values with their prefixes, and the page gives the −3 dB cutoff in hertz, the same figure with an SI prefix, the angular cutoff and the time constant. Set the signal frequency to see how strongly that tone is reduced, both as a voltage ratio and in decibels, and how far its phase lags. The model is an ideal single-pole network driven from a zero-impedance source into an open load; a real load resistance or source impedance shifts the corner, and component tolerances of 5 % to 20 % move it by the same proportion.

How to use it

  1. Pick RC or RL

    The capacitance or inductance field appears to match the network you are building.

  2. Enter component values

    Type each value and choose its prefix from the unit menu inside the field, such as kΩ, nF or mH.

  3. Set a test frequency

    Enter a signal frequency in hertz to see its gain, decibel attenuation and phase shift.

  4. Check the working

    The steps show the formula with your values substituted, so you can verify each figure.

Examples

RC: 1 kΩ with 1 µF

Filter type
RC (resistor, capacitor)
Resistance R
1
Resistance unit
kΩ
Capacitance C
1
Capacitance unit
µF
Signal frequency to test
1000

Result Cutoff frequency fc (Hz): 159.15
Cutoff with SI prefix: 159.2 Hz
Time constant τ: 1 ms
Angular cutoff ωc (rad/s): 1,000
Voltage gain Vout/Vin at signal: 0.1572
Gain at signal (dB): -16.07
Phase shift at signal (°): -80.96

  1. R = 1 kΩ
  2. C = 1 µF
  3. fc = 1 ÷ (2π × R × C) = 1 ÷ (2π × 1 kΩ × 1 µF) = 159.2 Hz
  4. Time constant τ = R × C = 1 ms
  5. At 1 kHz: f ÷ fc = 6.2832, |H| = 1 ÷ √(1 + 39.4784) = 0.1572 (-16.07 dB), phase = −atan(6.2832) = -80.96°

A common textbook pair: the time constant is 1 ms and the −3 dB point falls just under 160 Hz.

RC: 10 kΩ, 10 nF, signal at fc

Filter type
RC (resistor, capacitor)
Resistance R
10
Resistance unit
kΩ
Capacitance C
10
Capacitance unit
nF
Signal frequency to test
1591.55

Result Cutoff frequency fc (Hz): 1,591.55
Cutoff with SI prefix: 1.592 kHz
Time constant τ: 100 µs
Angular cutoff ωc (rad/s): 10,000
Voltage gain Vout/Vin at signal: 0.7071
Gain at signal (dB): -3.01
Phase shift at signal (°): -45

  1. R = 10 kΩ
  2. C = 10 nF
  3. fc = 1 ÷ (2π × R × C) = 1 ÷ (2π × 10 kΩ × 10 nF) = 1.592 kHz
  4. Time constant τ = R × C = 100 µs
  5. At 1.592 kHz: f ÷ fc = 1, |H| = 1 ÷ √(1 + 1) = 0.7071 (-3.01 dB), phase = −atan(1) = -45°

Feeding a sine exactly at the cutoff shows the defining point: about 70.7 % of the input voltage and a 45° lag.

RL: 1 kΩ with 10 mH

Filter type
RL (resistor, inductor)
Resistance R
1
Resistance unit
kΩ
Inductance L
10
Inductance unit
mH
Signal frequency to test
1000

Result Cutoff frequency fc (Hz): 15,915.49
Cutoff with SI prefix: 15.92 kHz
Time constant τ: 10 µs
Angular cutoff ωc (rad/s): 100,000
Voltage gain Vout/Vin at signal: 0.998
Gain at signal (dB): -0.02
Phase shift at signal (°): -3.6

  1. R = 1 kΩ
  2. L = 10 mH
  3. fc = R ÷ (2π × L) = 1 kΩ ÷ (2π × 10 mH) = 15.92 kHz
  4. Time constant τ = L ÷ R = 10 µs
  5. At 1 kHz: f ÷ fc = 0.0628, |H| = 1 ÷ √(1 + 0.0039) = 0.998 (-0.02 dB), phase = −atan(0.0628) = -3.6°

With a series inductor and a shunt resistor the cutoff is R over 2πL, here close to 15.9 kHz.

How it is calculated

RC: fc = 1 ÷ (2π·R·C); RL: fc = R ÷ (2π·L); |H(f)| = 1 ÷ √(1 + (f/fc)²); φ = −atan(f/fc)

R
resistance in ohms
C
capacitance in farads
L
inductance in henries
f
signal frequency in hertz
τ
time constant, R·C or L÷R, measured in s

Both networks are single-pole filters whose transfer function is 1 ÷ (1 + j·f/fc). The magnitude falls to 1/√2 at the cutoff and then by 20 dB per decade, or about 6 dB per octave. The cutoff equals 1 ÷ (2πτ), so the time constant and the corner frequency carry the same information. Gain in decibels is 20·log10 of the voltage ratio.

When not to use it

  • Second-order or active designs such as Sallen-Key or Butterworth stages need their own Q-dependent formulas.
  • When the load resistance is comparable to R, the divider changes both the corner and the passband gain.

Common mistakes

  • Leaving the prefix menu on µF when the part is marked in nF, which shifts the cutoff by a factor of 1000.
  • Confusing the angular cutoff in rad/s with the cutoff in hertz; they differ by 2π.

Frequently asked questions

Why is the cutoff called the −3 dB point?

At fc the output voltage is 1/√2, about 0.707, of the input. 20·log10(0.707) is −3.01 dB, which also means the output power is halved at that frequency.

How fast does a first-order low-pass roll off?

Well above the cutoff the gain falls by 20 dB for every tenfold rise in frequency, about 6 dB per octave. A tone at ten times fc comes out at roughly −20 dB, a tenth of its voltage.

How do I choose R and C for a target cutoff?

Rearrange the formula to C = 1 ÷ (2π·R·fc). Choose R first, often 1 kΩ to 100 kΩ so the source can drive it and the load does not swamp it, then pick the nearest standard capacitor and recheck the cutoff here.

What is the phase shift at the cutoff frequency?

The output lags the input by 45° at fc. The lag approaches 0° well below the cutoff and approaches 90° far above it, which matters in feedback loops and audio crossovers.

Does a heavy load change the result?

Yes. A load resistor across the capacitor forms a divider with R, lowering the passband gain and raising the cutoff, since the effective resistance becomes R in parallel with the load. Buffer the output or keep the load at least ten times R.

How is the time constant related to the cutoff?

τ equals R·C for the RC version and L÷R for the RL version, and fc = 1 ÷ (2πτ). After one time constant a step input has risen to about 63 % of its final value.

Cite this tool

Using this tool in an assignment, article or report? Copy a citation, or a link for your website.

APA

Tools Directory Editorial Team. (2026, October 17). Low Pass Filter Calculator. Tools Directory. Retrieved October 27, 2026, from https://tools.milunir.com/electronics-physics/circuits-and-components/low-pass-filter-calculator/

MLA

Tools Directory Editorial Team. “Low Pass Filter Calculator.” Tools Directory, 17 Oct. 2026, https://tools.milunir.com/electronics-physics/circuits-and-components/low-pass-filter-calculator/. Accessed 27 Oct. 2026.

Chicago

Tools Directory Editorial Team. “Low Pass Filter Calculator.” Tools Directory. Last modified October 17, 2026. Accessed October 27, 2026. https://tools.milunir.com/electronics-physics/circuits-and-components/low-pass-filter-calculator/.

Link<a href="https://tools.milunir.com/electronics-physics/circuits-and-components/low-pass-filter-calculator/">Low Pass Filter Calculator – Tools Directory</a>