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Low-pass filter Created 2026-10-14 Updated 2026-10-18

A low-pass filter preferentially transmits low frequencies and attenuates higher frequencies. It can smooth rapid noise in a slowly varying signal, but it also suppresses wanted variations outside the useful signal bandwidth. An RC low-pass filter gives a concrete continuous-time example; a moving average implements related smoothing on a discrete-time signal.

The useful question is how much change the measurement can tolerate. Preserving the low frequencies of a temperature record while reducing a fast unwanted oscillation can improve readability. Preserving the sharp onset of a pulse requires higher frequencies, so excessive smoothing changes its timing and shape. The RC low-pass frequency response and RC low-pass step response make this signal bandwidth versus settling time tradeoff quantitative.

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RC low-pass step response 2026-10-18

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The step response describes a sudden persistent input change. Suppose the input becomes a constant

V_{s}

t = 0

, and the initial capacitor voltage is

v_{0}

. Finite electric current leaves that voltage continuous. Subtract

V_{s}

from the RC low-pass filter equation: the remaining error satisfies

τ d e / d t + e = 0

. Separating variables gives

v_{out} (t) = V_{s} + (v_{0} - V_{s}) e^{- t / τ} (t \geq 0) .

(1)

For an initially uncharged capacitor and a

0

3 V

step in the constructed example,

v_{out} (t) = 3 (1 - e^{- t / (103.4 μ s)}) V .

(2)

At one time constant,

t = 103.4 μ s

, the voltage is

1.8964 V

, about

63.2%

of its final value. The voltage error then continues its exponential decay. This is different from a pure delay: the output starts changing immediately, yet takes time to approach the target.

Figure 1.
RC low-pass frequency response and RC low-pass step response from one time constant
. Original analytical curves for $R = 2.2 k Ω$ and $C = 47 nF$ ; no experimental points. Left: a Bode plot of voltage gain and output-minus-input phase difference, with cutoff frequency $1.5392 kHz$ . Right: the response to a $0$ to $3 V$ step, compared with doubling $C$ at fixed $R$ . Doubling the time constant halves the cutoff frequency and doubles the time to reach the same fraction of the final voltage. **Details**.

Define settling time relative to the total voltage change: require

∣ v_{out} - V_{s} ∣ \leq ϵ ∣ v_{0} - V_{s} ∣

thereafter, with

0 < ϵ < 1

. The exponential decay is monotone, so

t_{ϵ} = τ ln \frac{1}{ϵ}, f_{c} t_{ϵ} = \frac{l n ( 1/ ϵ )}{2 π} .

(3)

For

1%

settling,

t_{0.01} = 4.6052 τ \approx 476.17 μ s

; the

10%

90%

rise takes

τ ln 9 \approx 227.19 μ s

. Reducing the cutoff frequency by doubling

C

doubles both times. This reciprocal relation is the useful connection between signal bandwidth and response speed: stronger suppression of rapid noise comes with slower tracking of wanted changes.

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