A low-pass filter has R = 4.7\ k and C = 100\ nF. Find the cut-off frequency. [2 marks]

Express a voltage gain of × 50 in decibels. [2 marks]

WJEC Eduqas Eduqas 2021-style practice: A low-pass RC filter uses a 10\ k resistor and a 22\ nF capacitor. Calculate the cut-off frequency, and state whether a 50\ Hz signal and a 50\ kHz signal are passed or attenuated.

Cut-off frequency (up to 3 marks): f_c = 12π R C = 12π × 10000 × 22 × 10^-9 = 11.38 × 10^-3 = 723\ Hz. Signals (up to 2 marks): a low-pass filter passes frequencies below f_c and attenuates those above. The 50\ Hz signal is well below 723\ Hz, so it is passed; the 50\ kHz signal is far above f_c, so it is heavily attenuated. Markers reward f_c ≈ 723\ Hz, identifying 50\ Hz as passed and 50\ kHz as attenuated.

WJEC Eduqas Eduqas 2019-style practice: An amplifier has an output voltage of 2.0\ V for an input of 25\ mV. Calculate the voltage gain, and express it in decibels.

Voltage gain (up to 2 marks): A_v = V_outV_in = 2.025 × 10^-3 = 80. Gain in decibels (up to 2 marks): G_dB = 20 _10(A_v) = 20 _10(80) = 20 × 1.903 = 38\ dB. Markers reward the linear gain 80 and the decibel value 38\ dB using G_dB = 20 _10(A_v).

EnglandElectronicsSyllabus dot point

How does an RC network pass some frequencies and block others, and what is the cut-off frequency?

Passive filters: RC low-pass and high-pass filters, the cut-off frequency, voltage gain in decibels, and reading a frequency-response (Bode) plot.

An Eduqas A-Level Electronics answer on passive filters: how RC low-pass and high-pass networks select frequencies, the cut-off frequency formula, voltage gain expressed in decibels, and how to read a frequency-response (Bode) plot including the half-power point.

Generated by Claude Opus 4.813 min answerUpdated 2026-06-02

Reviewed by: AI editorial process; not yet individually human-reviewed

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Quick answer

A passive filter is an RC network that passes some frequencies and blocks others, using the capacitor's frequency-dependent reactance. A low-pass filter (output across the capacitor) passes low frequencies and attenuates high ones; a high-pass filter (output across the resistor) does the reverse. The cut-off frequency is where the output power has fallen to half (output voltage to $\frac{1}{\sqrt 2} \approx 70.7\%$ ):

f_c = \frac{1}{2\pi RC}.

Voltage gain is often quoted in decibels:

G_\text{dB} = 20\log_{10}\left(\frac{V_\text{out}}{V_\text{in}}\right)

, so the cut-off (half-power) point is at

-3\ \text{dB}

. A frequency-response (Bode) plot shows gain in decibels against frequency on a logarithmic scale; a simple RC filter rolls off at

20\ \text{dB}

per decade beyond

f_c

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What this dot point is asking

Eduqas wants you to analyse passive RC low-pass and high-pass filters, calculate the cut-off frequency, express voltage gain in decibels, and read a frequency-response (Bode) plot. Filters shape every signal that passes through an electronic system, removing noise and selecting bands.

The answer

Low-pass and high-pass RC filters

The cut-off frequency

Voltage gain in decibels

Reading a frequency-response plot

Designing a high-pass filter for a chosen cut-off

Design a high-pass RC filter with a cut-off frequency of $1.0\ \text{kHz}$ using a $15\ \text{nF}$ capacitor.

step 1: Rearrange the cut-off formula for R

$f_c = \frac{1}{2\pi RC}$ , so $R = \frac{1}{2\pi f_c C}$ .

step 2: Substitute the values

$R = \frac{1}{2\pi \times 1000 \times 15 \times 10^{-9}} = \frac{1}{9.42 \times 10^{-5}} = 1.06 \times 10^{4}\ \Omega \approx 10.6\ \text{k}\Omega$ .

step 3: Choose a preferred value

The nearest E24 value is $11\ \text{k}\Omega$ , giving $f_c = \frac{1}{2\pi \times 11000 \times 15 \times 10^{-9}} = 965\ \text{Hz}$ , close to target. Taking the output across the resistor makes it high-pass.

Examples in context

Passive filters are the first line of signal conditioning: a low-pass filter removes high-frequency noise and is the anti-aliasing filter before an analogue-to-digital converter, a high-pass filter blocks DC and mains hum from an audio signal, and a band-pass combination selects a channel in a receiver. Decibel gain is the universal language of amplifiers and audio. The first-order RC roll-off here is the foundation for the steeper active filters built with op-amps.

Try this

Q1. A low-pass filter has $R = 4.7\ \text{k}\Omega$ and $C = 100\ \text{nF}$ . Find the cut-off frequency. [2 marks]

Cue. $f_c = \frac{1}{2\pi \times 4700 \times 100 \times 10^{-9}} = 339\ \text{Hz}$ .

Q2. Express a voltage gain of $\times 50$ in decibels. [2 marks]

Cue. $G_\text{dB} = 20\log_{10}(50) = 34\ \text{dB}$ .

Q3. State the output voltage fraction at the cut-off frequency of an RC filter. [1 mark]

Cue. $\frac{1}{\sqrt 2} \approx 70.7\%$ of the input (the half-power point).

Exam-style practice questions

Practice questions written in the style of WJEC Eduqas exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

Eduqas 20215 marksA low-pass RC filter uses a

10\ \text{k}\Omega

resistor and a

22\ \text{nF}

capacitor. Calculate the cut-off frequency, and state whether a

50\ \text{Hz}

signal and a

50\ \text{kHz}

signal are passed or attenuated.

Show worked answer →

Cut-off frequency (up to 3 marks): $f_c = \dfrac{1}{2\pi R C} = \dfrac{1}{2\pi \times 10000 \times 22 \times 10^{-9}} = \dfrac{1}{1.38 \times 10^{-3}} = 723\ \text{Hz}$ .

Signals (up to 2 marks): a low-pass filter passes frequencies below $f_c$ and attenuates those above. The $50\ \text{Hz}$ signal is well below $723\ \text{Hz}$ , so it is passed; the $50\ \text{kHz}$ signal is far above $f_c$ , so it is heavily attenuated.

Markers reward $f_c \approx 723\ \text{Hz}$ , identifying $50\ \text{Hz}$ as passed and $50\ \text{kHz}$ as attenuated.

Eduqas 20194 marksAn amplifier has an output voltage of

2.0\ \text{V}

for an input of

25\ \text{mV}

. Calculate the voltage gain, and express it in decibels.

Show worked answer →

Voltage gain (up to 2 marks): $A_v = \dfrac{V_\text{out}}{V_\text{in}} = \dfrac{2.0}{25 \times 10^{-3}} = 80$ .

Gain in decibels (up to 2 marks): $G_\text{dB} = 20 \log_{10}(A_v) = 20 \log_{10}(80) = 20 \times 1.903 = 38\ \text{dB}$ .

Markers reward the linear gain $80$ and the decibel value $38\ \text{dB}$ using $G_\text{dB} = 20 \log_{10}(A_v)$ .

Related dot points

Sources & how we know this

Eduqas GCE AS/A Level Electronics specification (A410QS) — WJEC Eduqas (2017)