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How is an analogue signal turned into digital form, and what sets the quality of the conversion?

Analogue-to-digital conversion: sampling, the sampling rate and the Nyquist criterion, quantisation, resolution and the number of bits, and quantisation error.

A focused answer to WJEC A-Level Electronics analogue-to-digital conversion, covering sampling and the sampling rate, the Nyquist criterion, quantisation into levels, resolution and the number of bits, and quantisation error.

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  1. What this dot point is asking
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What this dot point is asking

Real-world signals are analogue, but processing and storage are digital, so conversion sits at the heart of modern electronics. WJEC expects you to explain sampling and the sampling rate, state the Nyquist criterion and the consequence of breaking it (aliasing), describe quantisation into levels, and calculate resolution from the number of bits. Resolution calculations and the Nyquist/aliasing argument are reliable, high-mark exam content.

The answer

Sampling and the Nyquist criterion

If the rate is too low, aliasing occurs: high frequencies masquerade as false lower frequencies that were never in the original. This is why audio is sampled at 44.1 kHz, comfortably above twice the 20 kHz limit of human hearing.

Quantisation and resolution

Quantisation error

Because each sample is rounded to the nearest level, there is always a small quantisation error, at most half a step. Increasing the number of bits shrinks the step and so shrinks this error, which is why high-quality audio uses 16 or 24 bits.

Examples in context

Example 1. Digitising audio for a music file
A microphone signal is sampled at 44.1 kHz and quantised to 16 bits, giving 216=65,5362^{16} = 65{,}536 levels. The high sample rate captures frequencies up to about 22 kHz (satisfying Nyquist for human hearing) and the many levels keep the quantisation error inaudible.
Example 2. A microcontroller reading a sensor
A temperature sensor's analogue voltage is read by the 10-bit ADC built into a PIC, giving 1024 levels across the reference voltage. The resolution sets the smallest temperature change the system can detect, which is why a higher-bit ADC is chosen when fine measurement matters.
Example 3. Anti-aliasing before sampling
Because frequencies above half the sample rate would alias, a low-pass anti-aliasing filter is placed before the ADC to remove them first. This filter is essential: without it, out-of-band noise would fold down into the signal band and could not be removed afterwards.

Try this

Q1. An ADC samples at 40 kHz40\,\text{kHz}. What is the highest signal frequency it can capture without aliasing? [1 mark]

  • Cue. Half the sample rate, so 20 kHz20\,\text{kHz} (the Nyquist frequency).

Q2. A 4-bit ADC has a full-scale of 8.0 V8.0\,\text{V}. Find the number of levels and the resolution. [2 marks]

  • Cue. Levels =24=16= 2^4 = 16; resolution =8.016=0.50 V= \frac{8.0}{16} = 0.50\,\text{V} per step.

Exam-style practice questions

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

WJEC Eduqas 20205 marksAn ADC has an 8-bit output and a reference (full-scale) voltage of 5.0 V5.0\,\text{V}. Calculate the number of quantisation levels, the resolution (the voltage of one step), and explain what happens to the resolution if the number of bits is increased to 10.
Show worked answer →

An n-bit ADC divides the input range into 2n2^n levels.

Levels for 8 bits: 28=2562^8 = 256.

Resolution (one step): Vfull2n=5.0256=0.0195 V≈20 mV\dfrac{V_{full}}{2^n} = \dfrac{5.0}{256} = 0.0195\,\text{V} \approx 20\,\text{mV} per step.

Increasing to 10 bits gives 210=10242^{10} = 1024 levels, so the resolution becomes 5.01024=4.9 mV\dfrac{5.0}{1024} = 4.9\,\text{mV} per step. More bits give smaller steps, so the digital value tracks the analogue signal more finely and the quantisation error falls.

Markers reward the 256 levels, the resolution of about 20 mV20\,\text{mV}, and the statement that more bits give finer resolution and less quantisation error.

WJEC Eduqas 20184 marksState the Nyquist criterion and explain what happens to a 5 kHz5\,\text{kHz} tone if it is sampled at 8 kHz8\,\text{kHz}.
Show worked answer →

The Nyquist criterion states that a signal must be sampled at a rate of at least twice its highest frequency component for it to be reconstructed faithfully.

For a 5 kHz5\,\text{kHz} tone, the minimum sampling rate is 2×5=10 kHz2 \times 5 = 10\,\text{kHz}.

Sampling at only 8 kHz8\,\text{kHz} is below this minimum, so the signal is under-sampled and aliasing occurs: the 5 kHz5\,\text{kHz} tone appears as a false lower-frequency signal (here 8−5=3 kHz8 - 5 = 3\,\text{kHz}) that was never present, corrupting the reconstruction.

Markers reward the at-least-twice statement, the 10 kHz10\,\text{kHz} minimum, and aliasing producing a false lower-frequency tone.

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