Skip to main content
EnglandElectronicsSyllabus dot point

How is a continuous analogue signal turned into a digital code, and what sets its accuracy?

Analogue-to-digital conversion: sampling, quantisation and resolution, the sampling theorem and aliasing, quantisation error, and the trade-off between resolution and data rate.

An Eduqas A-Level Electronics answer on analogue-to-digital conversion: sampling a continuous signal, quantisation and resolution, the Nyquist sampling theorem and aliasing, quantisation error, and the trade-off between resolution, sampling rate and the data rate produced.

Generated by Claude Opus 4.813 min answer

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

Have a quick question? Jump to the Q&A page

Jump to a section
  1. What this dot point is asking
  2. The answer
  3. Examples in context
  4. Try this

What this dot point is asking

Eduqas wants you to describe analogue-to-digital conversion: sampling, quantisation and resolution, the sampling theorem and aliasing, quantisation error, and the resolution-versus-data-rate trade-off. This is how the real analogue world enters a digital system.

The answer

Sampling and quantisation

Resolution

The sampling theorem, aliasing and quantisation error

The resolution-versus-data-rate trade-off

Examples in context

Analogue-to-digital conversion is the front door of every digital device that senses the world: a microcontroller's ADC reads a sensor voltage, a sound card digitises audio, and a digital camera converts light to numbers. The sampling theorem sets the sampling rate (CD audio at 44.1 kHz44.1\ \text{kHz} for a 20 kHz20\ \text{kHz} audio band), the resolution sets the accuracy, and the anti-aliasing low-pass filter (from the filters topic) protects the conversion. The binary codes produced are exactly the numbers handled in the number-systems topic.

Try this

Q1. State how many levels a 10-bit ADC has. [1 mark]

  • Cue. 210=10242^{10} = 1024 levels.

Q2. An 8-bit ADC covers 00 to 4.0 V4.0\ \text{V}. Find its resolution. [2 marks]

  • Cue. 4.0256=1.56×102 V15.6 mV\frac{4.0}{256} = 1.56 \times 10^{-2}\ \text{V} \approx 15.6\ \text{mV} per step.

Q3. State the minimum sampling rate for a signal whose highest frequency is 15 kHz15\ \text{kHz}. [1 mark]

  • Cue. 2×15=30 kHz2 \times 15 = 30\ \text{kHz} (the Nyquist rate).

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 20216 marksAn 8-bit analogue-to-digital converter has an input range of 00 to 5.0 V5.0\ \text{V}. Calculate the number of quantisation levels, the resolution (the voltage represented by one step), and the digital output code for an input of 2.0 V2.0\ \text{V}.
Show worked answer →

Levels (up to 2 marks): an nn-bit converter has 2n2^n levels. For 8 bits, 28=2562^8 = 256 levels.

Resolution (up to 2 marks): the resolution is the full range divided by the number of steps: 5.0256=1.95×102 V19.5 mV\dfrac{5.0}{256} = 1.95 \times 10^{-2}\ \text{V} \approx 19.5\ \text{mV} per step (using 2n1=2552^n - 1 = 255 steps gives 19.6 mV19.6\ \text{mV}, either is accepted).

Output code (up to 2 marks): the code is the input divided by the resolution, rounded: 2.019.5×103=102.6\dfrac{2.0}{19.5 \times 10^{-3}} = 102.6, so code 103103 (binary 011001110110\,0111).

Markers reward 256256 levels, a resolution near 19.5 mV19.5\ \text{mV}, and an output code of about 102102 or 103103.

Eduqas 20195 marksState the sampling theorem, explain what aliasing is, and state how it is prevented.
Show worked answer →

Sampling theorem (up to 2 marks): to reconstruct a signal faithfully it must be sampled at a rate at least twice the highest frequency present in the signal (the Nyquist rate), fsample2fmaxf_\text{sample} \ge 2 f_\text{max}.

Aliasing (up to 2 marks): if the signal is sampled too slowly (below the Nyquist rate), high-frequency components are misread as lower frequencies that were not present, appearing as false (alias) signals that distort the reconstruction.

Prevention (up to 1 mark): pass the signal through a low-pass anti-aliasing filter before sampling to remove any frequencies above half the sampling rate, and/or sample faster.

Markers reward the at-least-twice sampling rate, aliasing as high frequencies appearing as false low frequencies, and the anti-aliasing low-pass filter (or faster sampling) as the cure.

Related dot points

Sources & how we know this