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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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Past exam of the physics 2 paper of the University of Oxford / 2019 resit / 5 / Solution 2026-10-25

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Let

v_{C}

denote the capacitor voltage with the same polarity as the applied voltage. By Kirchhoff's voltage law, Ohm's law and

I = C d v_{C} / d t

,

RC \frac{d v _{C}}{d t} + v_{C} = V, v_{C} (0) = 0.

(1)

Solving this first-order differential equation gives the RC low-pass step response

v_{C} (t) = V (1 - e^{- t / τ}), τ = RC = 0.050 s .

(2)

The capacitor voltage is continuous at switching because an instantaneous finite jump would require an impulsive electric current.

For the square-wave response of an RC low-pass filter, take equal positive and negative half-periods. Let

A = 10 V

,

T = 1/ f

,

h = T /2

and

α = e^{- h / τ}

. After the transient, symmetry makes the capacitor voltage alternate between

- v_{p}

and

+ v_{p}

. Over the positive half-period,

v_{C} (t) = A - (A + v_{p}) e^{- t / τ}, 0 \leq t < h .

(3)

Requiring

v_{C} (h) = v_{p}

gives

v_{p} = A \frac{1 - α}{1 + α} = A tanh (\frac{T}{4 τ}) .

(4)

The next half-period is the sign-reversed copy. These exact expressions explain the requested sketches:

At $1 kHz$ , $h / τ = 0.01$ . The voltage barely changes in each half-period, so the output is nearly triangular about zero, with extrema about $\pm 0.0500 V$ . Its small-signal slope is approximately $\pm A / τ = \pm 200 V s^{- 1}$ .
At $10 Hz$ , $h = τ$ . Clearly curved exponential rises and falls reach only $\pm 4.621 V$ ; the output never settles to either input level.
At $0.1 Hz$ , $h = 100 τ$ . The output closely follows the input plateaux at $\pm 10 V$ , joined by continuous exponential transitions with time constant $0.050 s$ .

Figure 1.
Steady periodic capacitor voltages for three square-wave frequencies
. The blue curves use the exact periodic RC low-pass filter solution with $RC = 0.050 s$ . Dashed grey curves show the applied square wave in the lower panels; the first panel uses an expanded output scale. Time is measured in input periods. **Details**.

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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}

at

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

to

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%

to

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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