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Learn7408Paper 3B · Option: ElectronicsAnalogue and digital signalsSignals, sampling and LC filters
Analogue and digital signals · Subtopic 1 of 1

Signals, Sampling and LC Filters

An analogue signal varies continuously; a digital one has discrete levels. Digitising samples the signal at least twice per cycle of its highest frequency, and an nn-bit sample has 2n2^n levels; bit rate == samples per second ×\times bits per sample. An LC circuit resonates at f0=12πLCf_0=\frac1{2\pi\sqrt{LC}}, picking out one band of frequencies; its sharpness is Q=f0fBQ=\frac{f_0}{f_B}.

AQA 7408 Paper 3B≈ 40 minRevision 1 · updated 2026-10-27
Section 1 of 14

Learning objectives

  • Convert between binary and decimal; find quantisation levels.
  • Use the sampling rule and calculate bit rates.
  • Use f0=12πLCf_0=\frac1{2\pi\sqrt{LC}} and Q=f0fBQ=\frac{f_0}{f_B}.
Pause and think

A radio aerial picks up hundreds of stations at once. How does the radio choose just one?

Section 2 of 14

Core idea

Start hereMusic as numbers

A song on your phone is just a long list of numbers. Every fraction of a millisecond the sound wave was measured and stored. Do it often enough, with enough levels, and the playback sounds identical to the original.

The big idea

Sample at least twice the highest frequency; nn bits give 2n2^n levels; bit rate =fs×n=f_s\times n. LC circuits pick out frequencies near f0=12πLCf_0=\frac1{2\pi\sqrt{LC}}.

1. Digitising

Worked exampleCD bit rate

44.144.1 kHz, 1616 bits, stereo.

Goal
Bit rate.
Strategy
Multiply.
  1. 44 100×16=705 60044\,100\times16=705\,600
    One channel.
  2. ×2=1.41×106 bit s−1\times2=1.41\times10^6\text{ bit s}^{-1}
    Two channels.
Answer1.411.41 Mbit s⁻¹.
✓ CheckAbout 1010 MB per minute of audio ✓.

2. LC filters

Resonant frequency
  1. XL=2πfL,XC=12πfCX_L=2\pi fL,\quad X_C=\frac1{2\pi fC}
    Reactances of inductor and capacitor.
  2. XL=XC at resonanceX_L=X_C\ \text{at resonance}
    Their effects cancel.
  3. 4π2f02LC=14\pi^2f_0^2LC=1
    Cross-multiply.
  4. f0=12πLCf_0=\frac1{2\pi\sqrt{LC}}
    Rearrange.
Worked exampleTuning a radio

L=0.25L=0.25 mH; tune to 1.01.0 MHz.

Goal
CC.
Strategy
C=14π2f02LC=\frac1{4\pi^2f_0^2L}.
  1. 4π2(106)2(2.5×10−4)=9.87×1094\pi^2(10^6)^2(2.5\times10^{-4})=9.87\times10^9
    Denominator.
  2. C=1.01×10−10 FC=1.01\times10^{-10}\text{ F}
    Invert.
AnswerAbout 100100 pF.
✓ CheckTypical variable capacitors are tens to hundreds of pF ✓.
Digital signal chain

Analogue to digital and back.

Key takeawayQ factor Q=f0fBQ=\frac{f_0}{f_B}; high QQ is selective (narrow band), useful for tuning.
Common misconceptionAliasing Sampling too slowly makes a high frequency masquerade as a low one.
Pause and think

Why does adding one bit per sample halve quantisation error?

Summary

  • 2n2^n levels.
  • fs≥2fmax⁡f_s\ge2f_{\max}.
  • f0=12πLCf_0=\frac1{2\pi\sqrt{LC}}; Q=f0fBQ=\frac{f_0}{f_B}.
Key words
Quantisation
Rounding a sample to a fixed level.
Sampling rate
Samples taken per second.
Q factor
Sharpness of resonance.
HistoryNyquist. Harry Nyquist showed in 1928 that sampling at twice the highest frequency is enough to capture a signal exactly.
Fun factWhy 44.1 kHz? CDs sample at 44.144.1 kHz, just over twice the 2020 kHz limit of human hearing.
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