Why is "Confusing sampling rate with sample resolution — for example describing the sampling rate as '16 bits'." a common mistake in Representing sound?

Why it happens: Both are numeric settings that control recording quality and both increase file size, so under pressure the two names attach to the wrong quantities. How to avoid it: Use the **units** as the anchor: **sampling rate is in hertz** (samples per second) and **sample resolution is in bits** (per sample). Rate acts along the **time** axis — how often — while resolution acts along the **amplitude** axis — how precisely. Source: AQA 8525 specification 3.3.7 defines sampling rate in hertz and sample resolution as the number of bits per sample.

Why is "Substituting a duration in minutes directly into the formula instead of converting to seconds." a common mistake in Representing sound?

Why it happens: The duration is given in minutes and the formula asks for a number, so the value is used as it appears — the units mismatch is easy to overlook when working quickly. How to avoid it: The formula requires **seconds**, so multiply by 60 first: 3 minutes is **180 seconds**. Substituting 3 makes the answer **sixty times too small**, and the conversion usually carries a mark of its own. Source: AQA 8525 specification 3.3.7 defines the formula as rate × res × secs, with secs meaning the number of seconds.

Why is "Giving the answer in bits when the question asked for bytes, kB or MB." a common mistake in Representing sound?

Why it happens: The multiplication feels like the complete calculation, and the unit conversion reads as a formality rather than part of the answer. How to avoid it: The formula produces **bits**. Divide by **8** for bytes, then by **1,000** for kB and again for MB. Sense-check the result — a three-minute song coming out at 127 MB rather than 16 MB is exactly a factor of eight out. Source: The ÷8 is awarded as a separate method mark in sound file-size questions.

Why is "Ignoring the word 'stereo' and calculating for a single channel only." a common mistake in Representing sound?

Why it happens: The formula makes no mention of channels, so the detail in the question feels like background context rather than a required part of the calculation. How to avoid it: **Stereo means two channels**, so the result must be **multiplied by 2**. Mono is one channel and needs no adjustment. Underline the word 'stereo' when you see it — AQA's formula assumes a single channel unless the question states otherwise. Source: Questions specifying stereo award a mark for the doubling, which is lost if the detail is overlooked.

Why is "Working backwards from a file size given in bytes without converting it to bits first." a common mistake in Representing sound?

Why it happens: The size is stated in bytes so it seems natural to substitute it directly — the fact that the formula is defined in **bits** is easy to forget when rearranging. How to avoid it: The formula gives **bits**, so a size given in bytes must be **multiplied by 8** before rearranging. Note the direction reverses: forwards you divide by 8, backwards you multiply. Skipping this makes the answer eight times out. Source: Rearrangement questions asking for a missing duration, rate or resolution appear regularly and require this first step.

Why is "Defining a sample vaguely, such as 'a small piece of the sound' or 'part of the recording'." a common mistake in Representing sound?

Why it happens: The everyday meaning of 'sample' is a small portion of something, so the technical definition never displaces the intuitive one. How to avoid it: Use the specification's exact wording: a sample is **a measure of amplitude at a point in time**. Both halves matter — it is a **measurement of amplitude**, and it is taken at a **single instant** rather than over a period. Source: AQA 8525 specification 3.3.7 additional information: 'Students should understand that a sample is a measure of amplitude at a point in time.'

AQA GCSE Computer Science (8525)

Representing sound

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Detailed Study Notes

Detailed notes on Fundamentals of Data Representation for AQA GCSE Computer Science, covering key concepts, explanations, examples, and exam-focused revision points.

Representing Sound — AQA GCSE Computer Science 8525 (3.3.7)

The sister topic to images, with one formula given verbatim: File size (bits) = rate × res × secs. You must understand that sound is analogue and must be converted to digital for storage, that analogue signals are sampled, that a sample is a measure of amplitude at a point in time, and the two properties that define the recording — sampling rate (samples per second, in hertz) and sample resolution (bits per sample). Get the formula automatic and this subtopic is reliable marks.

At a glance

Sound is analogue — a continuously varying wave — and must be converted to digital for storage and processing.
Sampling means measuring the analogue signal at regular intervals.
A sample is a measure of amplitude at a point in time.
Sampling rate = the number of samples taken per second, measured in hertz (1 Hz = 1 sample per second).
Sample resolution = the number of bits per sample.
File size (bits) = rate × res × secs.
Divide by 8 for bytes, then by 1,000 for kB (3.3.3).
Higher sampling rate = more samples per second = better quality, larger file.
Higher sample resolution = more bits per sample = more accurate amplitude, larger file.
The digital version is always an approximation of the original analogue wave.

What you’ll learn

Mapped to the AQA GCSE 8525 syllabus (2027-2029).

Understand that sound is analogue and that it must be converted to a digital form for storage and processing.
Understand that analogue signals are sampled to create the digital version of sound, and that a sample is a measure of amplitude at a point in time.
Describe the digital representation of sound in terms of sampling rate and sample resolution, knowing that sampling rate is the number of samples taken in a second measured in hertz, and sample resolution is the number of bits per sample.
Calculate sound file sizes based on the sampling rate and the sample resolution, using File size (bits) = rate × res × secs.

Sound is analogue and must be converted

A continuous wave becomes a series of discrete measurements. That is sampling.

Why conversion is necessary

Sound is analogue. It is a continuously varying wave — air pressure rising and falling smoothly, with infinitely many possible values at infinitely many moments in time.

Computers can only store discrete binary values (3.3.1). A continuous wave cannot be stored directly, so it must be converted to a digital form for storage and processing.

Sampling — how the conversion works

Sampling means measuring the amplitude of the analogue signal at regular intervals.

A sample is a measure of amplitude at a point in time — that is the specification's own definition and is worth reproducing exactly.

The smooth blue wave is the original analogue signal; the red dots are the only values actually stored.

The digital version is an approximation

Between the samples, the original wave is simply not recorded. The stored version is therefore an approximation of the original — and how close an approximation depends on the two properties in the next section.

The two conversions

Direction	What it does	Where it happens
Analogue → digital	Samples the wave for storage	When recording, e.g. a microphone input
Digital → analogue	Reconstructs a wave from the samples	When playing back through speakers

AQA tip. In a "why must sound be converted?" question, give both halves: sound is analogue (a continuously varying wave), and computers can only store discrete binary values — so a conversion is required.

Sound is analogue — a continuously varying wave.
Computers store only discrete binary values, so conversion is required.
Sampling means measuring the signal at regular intervals.
A sample is a measure of amplitude at a point in time.
The digital version is an approximation, because the wave between samples is not recorded.

See the full worked example for representing sound →

Sampling rate and sample resolution

Rate = samples per second (hertz). Resolution = bits per sample. Do not confuse them.

These two properties define a digital sound recording, and confusing them is the most common error in this subtopic.

Sampling rate

The sampling rate is the number of samples taken in one second, measured in hertz (Hz), where 1 hertz = 1 sample per second.

Rate	Quality
8,000 Hz (8 kHz)	Telephone quality
44,100 Hz (44.1 kHz)	CD quality
48,000 Hz	Professional audio

Higher rate = samples taken more often = the shape of the wave is captured more accurately.

Sample resolution

The sample resolution is the number of bits used per sample.

Each sample stores an amplitude value, and the resolution determines how precisely that amplitude can be recorded:

Resolution	Amplitude levels	Quality
8 bits	2⁸ = 256	Basic
16 bits	2¹⁶ = 65,536	CD quality
24 bits	2²⁴ ≈ 16.7 million	Professional

Higher resolution = each amplitude is recorded more accurately.

The distinction that matters

Rate works along the time axis; resolution works along the amplitude axis. Both affect quality and both affect file size.

	Sampling rate	Sample resolution
Measures	How often samples are taken	How accurately each is stored
Units	Hertz (samples per second)	Bits per sample
Axis	Time (horizontal)	Amplitude (vertical)
Increasing it	Captures the wave's shape better	Records each amplitude more precisely

Both increase quality and both increase file size — but they do so in different ways, and questions frequently ask you to distinguish them.

AQA tip. State the units precisely. Sampling rate is in hertz; sample resolution is in bits. Writing "the sampling rate is 16 bits" confuses the two and loses the mark.

Sampling rate = number of samples per second, measured in hertz.
Sample resolution = number of bits per sample.
Rate acts along time; resolution acts along amplitude.
Resolution of n bits gives 2ⁿ possible amplitude levels.
Both increase quality and file size, but in different ways.

See the full worked example for representing sound →

Calculating sound file size

File size (bits) = rate × res × secs. Then ÷8 for bytes.

The specification gives the formula verbatim:

$\text{File size (bits)} = \text{rate} \times \text{res} \times \text{secs}$

where rate = sampling rate (Hz), res = sample resolution (bits), secs = number of seconds.

The method

Identify rate, resolution and duration.
Convert the duration to seconds if given in minutes.
Multiply all three → size in bits.
Divide by 8 → bytes, then by 1,000 for kB, again for MB.
Sense-check against real-world scale.

Worked example

A 30-second recording is sampled at 44,100 Hz with a resolution of 16 bits. Calculate the file size in MB.

Step 1 — bits: $44{,}100 \times 16 \times 30 = 21{,}168{,}000 \text{ bits}$

Step 2 — bytes: $21{,}168{,}000 \div 8 = 2{,}646{,}000 \text{ bytes}$

Step 3 — megabytes: $2{,}646{,}000 \div 1{,}000{,}000 = 2.646 \text{ MB}$

Sense-check: CD-quality audio runs at roughly 5 MB per minute, so 30 seconds at around 2.6 MB is exactly right ✓

Converting minutes to seconds

A very common step — and a very common slip:

$3 \text{ minutes} = 3 \times 60 = 180 \text{ seconds}$

Always convert before substituting. Putting "3" into the formula instead of 180 makes the answer sixty times too small.

Working backwards

Rearrange as needed:

$\text{secs} = \frac{\text{size in bits}}{\text{rate} \times \text{res}} \qquad \text{res} = \frac{\text{size in bits}}{\text{rate} \times \text{secs}}$

A file of 1,411,200 bits was recorded at 44,100 Hz with 16-bit resolution. How long is it?

$\frac{1{,}411{,}200}{44{,}100 \times 16} = \frac{1{,}411{,}200}{705{,}600} = 2 \text{ seconds}$

Stereo

If a question mentions stereo, the sound has two channels, so multiply by 2. Mono has one channel and needs no adjustment. AQA's formula assumes a single channel unless the question states otherwise.

AQA tip. Show every step and label the units. Mark schemes award the multiplication and the ÷8 as separate method marks, so working protects marks even if the arithmetic slips — and the minutes-to-seconds conversion is often a mark in its own right.

File size (bits) = rate × res × secs.
Convert minutes to seconds BEFORE substituting — multiply by 60.
Divide by 8 for bytes, then by 1,000 per prefix step.
Rearrange to find a missing duration, rate or resolution.
For stereo, multiply by 2 for the two channels.

See the full worked example for representing sound →

How rate and resolution affect quality and file size

Both improve accuracy and both enlarge the file, proportionally.

Effect of sampling rate

Rate	Effect on quality	Effect on size
Higher	Samples taken more often, so the shape of the wave is captured more accurately and the playback sounds closer to the original	Larger file, proportionally
Lower	Fewer samples, so detail between them is lost and the sound is less faithful	Smaller file

Doubling the sampling rate doubles the file size, because twice as many samples are stored.

Effect of sample resolution

Resolution	Effect on quality	Effect on size
Higher	More bits per sample, so each amplitude is recorded more precisely with less rounding	Larger file, proportionally
Lower	Amplitudes rounded to fewer levels, producing a less accurate, potentially distorted sound	Smaller file

Doubling the resolution doubles the file size, because each sample takes twice as many bits.

Worked comparison

A 10-second recording:

Rate	Resolution	Bits	Bytes
22,050 Hz	8 bits	1,764,000	220,500
44,100 Hz	8 bits	3,528,000	441,000
44,100 Hz	16 bits	7,056,000	882,000

Doubling the rate doubled the size; doubling the resolution doubled it again. Both are directly proportional.

Why the digital version is never perfect

However high the settings, the stored version remains an approximation:

Between samples, the wave is not recorded at all.
Each amplitude is rounded to the nearest value the resolution allows.

Raising the rate and resolution reduces both losses but never eliminates them.

Choosing appropriately

Use	Typical settings	Why
Telephone speech	8 kHz, 8 bits	Speech is intelligible at low quality; bandwidth matters more
Music (CD)	44.1 kHz, 16 bits	High enough that most listeners cannot detect the difference
Professional recording	48 kHz+, 24 bits	Quality must survive later editing

AQA tip. When describing effects, use numbers: "Doubling the sampling rate from 22,050 Hz to 44,100 Hz doubles the file size from 220 kB to 441 kB, but captures the wave more accurately." A worked comparison turns a 2-mark answer into a 4-mark one.

Higher sampling rate captures the wave's shape more accurately.
Higher sample resolution records each amplitude more precisely.
Both are directly proportional to file size — doubling either doubles the size.
The digital version is always an approximation: gaps between samples, and rounded amplitudes.
Support descriptions with worked numbers.

See the full worked example for representing sound →

Quick recap

Sound is analogue and must be converted to digital because computers store only discrete binary values.
Sampling means measuring the analogue signal at regular intervals.
A sample is a measure of amplitude at a point in time.
Sampling rate is the number of samples per second, measured in hertz.
Sample resolution is the number of bits per sample.
File size (bits) = rate × res × secs; divide by 8 for bytes.
Convert minutes to seconds before substituting into the formula.
Higher rate or resolution improves quality but increases file size proportionally.
The digital version is always an approximation of the original wave.

Memorise this

Verbatim phrases and definitions Cambridge mark schemes credit.

Sound is ANALOGUE and must be converted to digital for storage and processing.
A sample is a measure of AMPLITUDE at a POINT IN TIME.
Sampling rate = samples per second, measured in HERTZ.
Sample resolution = number of BITS PER SAMPLE.
File size (bits) = rate × res × secs.
Then ÷ 8 for bytes, ÷ 1,000 for kB.
Convert minutes to seconds first — × 60.
Doubling the rate OR the resolution doubles the file size.
Stereo = 2 channels, so × 2.

How it’s examined

3.3.7 is assessed on Paper 2 and is examined in close parallel with 3.3.6, often as the second half of a data-representation question. Four shapes recur. (1) Recall, 1–3 marks: define a sample, state what sampling rate and sample resolution mean, or explain why sound must be converted to digital (AO1) — the definition of a sample as a measure of amplitude at a point in time is the specification's own wording and is directly creditworthy. (2) Calculate the file size, 3–5 marks: apply rate × res × secs and convert to bytes or MB, with separate method marks for the multiplication, the minutes-to-seconds conversion and the ÷8 (AO2). (3) Work backwards, 3–4 marks: given the file size, find the duration, rate or resolution by rearranging (AO2). (4) Describe or justify the effect of changing settings, 3–6 marks: explain how rate and resolution each affect quality and size, ideally with worked numbers, or recommend settings for a scenario such as a podcast versus a music recording (AO2/AO3). The two errors that cost most are confusing rate with resolution — they have different units, hertz versus bits — and forgetting to convert minutes into seconds, which makes the answer sixty times too small.

Sources: AQA GCSE Computer Science 8525 specification (v1.3, 16 June 2025) — 3.3.7 Representing sound; AQA GCSE Computer Science Paper 2 past papers, mark schemes and examiner reports. Last reviewed 2026-07-29.

Take this whole topic with you

Step-by-step worked examples — Representing sound

Step-by-step solutions to past-paper-style questions on representing sound, written exactly the way a tutor would explain them at the board.

Question type:

1Explain why sound must be converted to digital (3 marks)
Getting startedIdentify & classify• Paper 2 style — 3 marks (AO1)• analogue, digital, sampling
Question
(a) Explain why sound must be converted before a computer can store it. (b) State what is meant by sampling. (c) State what a sample is. (3 marks)
Step-by-step solution
1. Step 1
  (a) Sound is analogue — a continuously varying wave with infinitely many possible values — whereas computers can store only discrete binary values.
2. Step 2
  (b) Sampling means measuring the analogue signal at regular intervals to produce a series of discrete values.
3. Step 3
  (c) Use the specification's wording: a sample is a measure of amplitude at a point in time.
Answer
(a) Sound is analogue, meaning it is a continuously varying wave, whereas a computer can only store discrete binary values — so the signal must be converted into a digital form for storage and processing (1).

(b) Sampling means measuring the amplitude of the analogue signal at regular intervals (1).

(c) A sample is a measure of amplitude at a point in time (1). (3)
Examiner tip
Part (c) uses the specification's exact wording, and both halves matter — amplitude and at a point in time. Vague answers such as 'a small piece of the sound' do not convey that a specific quantity is being measured at a specific moment.
2Distinguish sampling rate from sample resolution (4 marks)
Getting startedIdentify & classify• Paper 2 style — 4 marks (AO1)• sampling-rate, resolution, hertz, bits
Question
(a) State what is meant by sampling rate and give its units. (b) State what is meant by sample resolution and give its units. (c) A recording uses a sample resolution of 8 bits. State how many different amplitude levels can be recorded. (4 marks)
Step-by-step solution
1. Step 1
  (a) Sampling rate is the number of samples taken in one second, measured in hertz — 1 Hz means 1 sample per second.
2. Step 2
  (b) Sample resolution is the number of bits used per sample, measured in bits.
3. Step 3
  (c) With 8 bits per sample, the number of levels is 2⁸ = 256.
  $2^8 = 256$
Answer
(a) The sampling rate is the number of samples taken in one second (1), measured in hertz (Hz), where 1 hertz is 1 sample per second (1).

(b) The sample resolution is the number of bits used per sample (1), measured in bits.

(c) 2⁸ = 256 different amplitude levels (1). (4)
Examiner tip
The units are the fastest way to keep these apart: hertz for rate, bits for resolution. Writing 'the sampling rate is 16 bits' confuses the two and loses the mark. Note that resolution relates to amplitude levels by the same 2ⁿ rule used throughout this unit.
3Calculate a sound file size (4 marks)
Getting startedDirect calculation• Paper 2 style — 4 marks (AO2)• file-size, formula, calculate
Question
A sound is recorded for 20 seconds at a sampling rate of 8,000 Hz with a sample resolution of 8 bits.

Calculate the file size in bytes, showing all your working. (4 marks)
Step-by-step solution
1. Step 1
  Identify the values. rate = 8,000, res = 8, secs = 20.
2. Step 2
  Apply the formula: File size (bits) = rate × res × secs.
  $8{,}000 \times 8 \times 20 = 1{,}280{,}000 \text{ bits}$
3. Step 3
  Convert to bytes by dividing by 8.
  $1{,}280{,}000 \div 8 = 160{,}000 \text{ bytes}$
4. Step 4
  Sense-check. 160,000 bytes is 160 kB for 20 seconds of telephone-quality audio, which is entirely plausible.
Answer
Size in bits = rate × res × secs = 8,000 × 8 × 20 (1) = 1,280,000 bits (1)

Size in bytes = 1,280,000 ÷ 8 (1) = 160,000 bytes (1)

(Equivalently 160 kB.)
Examiner tip
The multiplication and the ÷8 are awarded as separate method marks, so both must appear — a bare final answer risks scoring nothing if the arithmetic slips. Note the coincidence that the resolution here is also 8, which makes it tempting to cancel; resist that and show each step.
4Handle a duration given in minutes (5 marks)
Building confidenceMulti-step problem• Paper 2 style — 5 marks (AO2)• file-size, minutes, conversion, megabytes
Question
A song lasts 3 minutes and is recorded at CD quality — a sampling rate of 44,100 Hz and a sample resolution of 16 bits.

Calculate the file size in megabytes, showing all your working. (5 marks)
Step-by-step solution
1. Step 1
  Convert the duration to seconds first. 3 × 60 = 180 seconds. Substituting 3 instead would make the answer sixty times too small.
  $3 \times 60 = 180 \text{ seconds}$
2. Step 2
  Apply the formula: 44,100 × 16 × 180.
  $44{,}100 \times 16 \times 180 = 127{,}008{,}000 \text{ bits}$
3. Step 3
  Convert to bytes: divide by 8.
  $127{,}008{,}000 \div 8 = 15{,}876{,}000 \text{ bytes}$
4. Step 4
  Convert to megabytes: divide by 1,000,000.
  $15{,}876{,}000 \div 1{,}000{,}000 = 15.876 \text{ MB}$
5. Step 5
  Sense-check. CD audio runs at roughly 5 MB per minute, so about 16 MB for three minutes is exactly right.
Answer
Convert to seconds: 3 × 60 = 180 seconds (1)

Size in bits = 44,100 × 16 × 180 = 127,008,000 bits (1)

Size in bytes = 127,008,000 ÷ 8 = 15,876,000 bytes (1)

Size in MB = 15,876,000 ÷ 1,000,000 = 15.876 MB (1)

Sense-check: CD-quality audio is roughly 5 MB per minute, so approximately 16 MB for a three-minute track is plausible (1).
Examiner tip
The minutes-to-seconds conversion carries its own mark and is the step most frequently omitted — substituting 3 rather than 180 produces an answer sixty times too small. Note that this is the uncompressed size; a real MP3 of the same track would be far smaller, which links forward to 3.3.8.
5Rearrange the formula to find a missing value (5 marks)
Building confidenceMulti-step problem• Paper 2 style — 5 marks (AO2/AO3)• rearrange, duration, resolution
Question
(a) A file of 1,411,200 bits was recorded at 44,100 Hz with a 16-bit resolution. Calculate its duration in seconds. (b) A 10-second recording at 22,050 Hz has a file size of 220,500 bytes. Calculate its sample resolution. (5 marks)
Step-by-step solution
1. Step 1
  (a) Rearrange: secs = size in bits ÷ (rate × res).
2. Step 2
  Calculate the denominator: 44,100 × 16 = 705,600 bits per second.
3. Step 3
  Divide: 1,411,200 ÷ 705,600 = 2 seconds.
  $\frac{1{,}411{,}200}{705{,}600} = 2$
4. Step 4
  (b) Convert bytes to bits first, because the formula works in bits: 220,500 × 8 = 1,764,000 bits.
  $220{,}500 \times 8 = 1{,}764{,}000 \text{ bits}$
5. Step 5
  Rearrange: res = size in bits ÷ (rate × secs) = 1,764,000 ÷ (22,050 × 10) = 1,764,000 ÷ 220,500 = 8 bits.
  $\frac{1{,}764{,}000}{220{,}500} = 8$
Answer
(a) Rearranging: secs = size in bits ÷ (rate × res) 44,100 × 16 = 705,600 bits per second (1) 1,411,200 ÷ 705,600 = 2 seconds (1)

(b) Convert the size to bits first: 220,500 × 8 = 1,764,000 bits (1) Rearranging: res = size in bits ÷ (rate × secs) 22,050 × 10 = 220,500 (1) 1,764,000 ÷ 220,500 = 8 bits (1)
Examiner tip
Part (b)'s essential first move is converting the given bytes into bits, since the formula is defined in bits — working in bytes throughout gives an answer eight times out. Note the direction reverses when working backwards: you multiply by 8 rather than dividing.
6Compare the effect of rate and resolution (5 marks)
Building confidenceMulti-step problem• Paper 2 style — 5 marks (AO2)• quality, file-size, compare
Question
A 10-second recording is made at 22,050 Hz with 8-bit resolution.

(a) Calculate the file size in bytes. (b) Calculate the new size if the sampling rate is doubled to 44,100 Hz. (c) Calculate the new size if instead the resolution is doubled to 16 bits. (d) Describe the different effect each change has on the recording's quality. (5 marks)
Step-by-step solution
1. Step 1
  (a) 22,050 × 8 × 10 = 1,764,000 bits ÷ 8 = 220,500 bytes.
  $22{,}050 \times 8 \times 10 \div 8 = 220{,}500$
2. Step 2
  (b) Doubling the rate doubles the size: 44,100 × 8 × 10 = 3,528,000 bits ÷ 8 = 441,000 bytes.
3. Step 3
  (c) Doubling the resolution also doubles the size: 22,050 × 16 × 10 = 3,528,000 bits ÷ 8 = 441,000 bytes — identical to (b).
4. Step 4
  (d) Higher rate means samples are taken more often, so the shape of the wave over time is captured more accurately.
5. Step 5
  Higher resolution means each sample stores the amplitude more precisely, with less rounding to the nearest available level.
Answer
(a) 22,050 × 8 × 10 = 1,764,000 bits ÷ 8 = 220,500 bytes (1)

(b) 44,100 × 8 × 10 = 3,528,000 bits ÷ 8 = 441,000 bytes (1) — exactly double

(c) 22,050 × 16 × 10 = 3,528,000 bits ÷ 8 = 441,000 bytes (1) — also exactly double, and identical to (b)

(d) Although both changes have the same effect on file size, they improve quality differently. Increasing the sampling rate means samples are taken more often, so the shape of the wave over time is captured more accurately and rapid changes in the sound are not missed (1). Increasing the sample resolution means each sample stores its amplitude more precisely, with less rounding to the nearest available level, so individual measurements are more faithful (1).
Examiner tip
The striking result is that both changes produce the same file size but improve different aspects of the recording — rate acts along the time axis, resolution along the amplitude axis. That distinction is exactly what higher-tariff questions probe.
7Justify settings for two different recordings (6 marks)
StretchWord problem• Paper 2 style — 6 marks (AO3)• justify, quality, scenario
Question
A company records two things: a spoken podcast and a live orchestral concert. Storage space is limited.

(a) Calculate the size of a 60-second recording at 8,000 Hz with 8-bit resolution. (b) Calculate the size of a 60-second recording at 44,100 Hz with 16-bit resolution. (c) Justify which settings should be used for each recording. (6 marks)
Step-by-step solution
1. Step 1
  (a) 8,000 × 8 × 60 = 3,840,000 bits ÷ 8 = 480,000 bytes = 480 kB.
  $8{,}000 \times 8 \times 60 \div 8 = 480{,}000$
2. Step 2
  (b) 44,100 × 16 × 60 = 42,336,000 bits ÷ 8 = 5,292,000 bytes ≈ 5.29 MB — about eleven times larger.
  $44{,}100 \times 16 \times 60 \div 8 = 5{,}292{,}000$
3. Step 3
  (c) The podcast carries only speech, which occupies a narrow frequency range and needs to be intelligible rather than perfectly faithful.
4. Step 4
  So the lower settings suffice, saving a great deal of storage — significant when podcasts are long and numerous.
5. Step 5
  The orchestral concert contains a wide range of frequencies and subtle dynamics, so a low rate would lose high frequencies and a low resolution would flatten quiet passages.
6. Step 6
  Conclude by weighing the storage cost against the purpose of each recording.
Answer
(a) 8,000 × 8 × 60 = 3,840,000 bits ÷ 8 = 480,000 bytes (480 kB) (1)

(b) 44,100 × 16 × 60 = 42,336,000 bits ÷ 8 = 5,292,000 bytes (≈ 5.29 MB) (1) — roughly eleven times larger

(c) The podcast should use the lower settings. Speech occupies a relatively narrow frequency range and only needs to be clearly intelligible, not perfectly faithful, so the lower sampling rate and resolution are sufficient (1). Given that podcasts are typically long and produced regularly, using 480 kB per minute rather than 5.29 MB saves a very large amount of storage over time (1).

The concert should use the higher settings. Orchestral music contains a wide range of frequencies and subtle changes in volume, so a low sampling rate would fail to capture the higher frequencies and a low resolution would round away the quiet passages, noticeably degrading the recording (1). Since the purpose of a concert recording is faithful reproduction, the larger file is justified — and storage could be reduced afterwards through compression rather than by recording at lower quality in the first place (1).
Examiner tip
A 'justify' question requires the decision to be tied to the nature of the content, not to a general preference for quality. The closing observation — that compression can reduce the stored size without sacrificing the recording quality — is a genuinely strong AO3 point linking to 3.3.8.
8Calculate a stereo recording's file size (6 marks)
StretchMulti-step problem• Paper 2 style — 6 marks (AO2/AO3)• stereo, channels, file-size
Question
A 4-minute track is recorded in stereo (two channels) at 44,100 Hz with 16-bit resolution.

(a) Calculate the file size in MB. (b) Calculate how many such tracks fit on a 1 GB device. (c) Explain why the actual number would be higher in practice. (6 marks)
Step-by-step solution
1. Step 1
  Convert to seconds: 4 × 60 = 240 seconds.
2. Step 2
  Calculate for one channel: 44,100 × 16 × 240 = 169,344,000 bits.
  $44{,}100 \times 16 \times 240 = 169{,}344{,}000$
3. Step 3
  Double for stereo, since two channels are recorded: 169,344,000 × 2 = 338,688,000 bits.
4. Step 4
  Convert: ÷8 gives 42,336,000 bytes, then ÷1,000,000 gives 42.336 MB.
  $338{,}688{,}000 \div 8 = 42{,}336{,}000 \text{ bytes}$
5. Step 5
  (b) 1 GB = 1,000 MB, so 1,000 ÷ 42.336 = 23.6, meaning 23 complete tracks.
6. Step 6
  (c) Real music files are compressed, so each occupies far less than the uncompressed size calculated here.
Answer
(a) Convert to seconds: 4 × 60 = 240 seconds (1) One channel: 44,100 × 16 × 240 = 169,344,000 bits (1) Stereo has two channels, so × 2 = 338,688,000 bits (1) In bytes: 338,688,000 ÷ 8 = 42,336,000 bytes In MB: 42.336 MB (1)

(b) 1 GB = 1,000 MB, so 1,000 ÷ 42.336 = 23.6 → 23 complete tracks (1)

(c) Music files are almost always compressed — an MP3 might be around a tenth of the uncompressed size — so far more than 23 tracks would fit in practice. The formula gives the size if every sample were stored individually at full resolution, which is the uncompressed figure (1).
Examiner tip
Stereo requires the result to be doubled for the two channels; AQA's formula assumes a single channel unless the question states otherwise. Part (b) must round down, since a partial track cannot be stored, and part (c) connects the uncompressed calculation to compression in 3.3.8.
9Explain why the digital version is an approximation (6 marks)
StretchWord problem• Paper 2 style — 6 marks (AO3)• approximation, sampling, quality, explain
Question
Explain why a digitally recorded sound can never be a perfect reproduction of the original, and explain how increasing the sampling rate and the sample resolution each reduce — but do not eliminate — the loss. (6 marks)
Step-by-step solution
1. Step 1
  Loss 1 — gaps between samples. Sampling records the amplitude only at discrete moments; what happens between those moments is not recorded at all.
2. Step 2
  Consequence: any rapid change occurring between two samples is lost entirely.
3. Step 3
  Loss 2 — rounded amplitudes. Each sample is stored using a fixed number of bits, so the measured amplitude must be rounded to the nearest value that resolution allows.
4. Step 4
  Increasing the sampling rate narrows the gaps between samples, so less is missed — but there are still gaps, however small.
5. Step 5
  Increasing the resolution provides more amplitude levels, so the rounding is finer — but rounding still occurs.
6. Step 6
  Conclude: the original is continuous with infinitely many values, whereas any digital version is discrete and finite, so some loss is unavoidable in principle.
Answer
Loss 1 — the gaps between samples. Sampling records the amplitude only at discrete points in time, so whatever the wave does between two samples is not recorded at all (1). Any rapid change occurring in that interval is lost completely.

Loss 2 — rounding of amplitudes. Each sample is stored using a fixed number of bits, so the measured amplitude must be rounded to the nearest value the resolution allows — an 8-bit resolution provides only 256 levels, so any true amplitude falling between two of them is stored inaccurately (1).

Increasing the sampling rate takes samples more often, narrowing the gaps so that less of the wave's behaviour is missed (1) — but however high the rate, there are still gaps between samples, so some information is always lost (1).

Increasing the sample resolution provides more amplitude levels, so each value is rounded far less severely (1) — but rounding still occurs, because a finite number of bits can never represent every possible amplitude.

Fundamentally, the original analogue signal is continuous, with infinitely many possible values at infinitely many instants, whereas any digital version is discrete and finite — so while higher settings reduce the difference to the point where it is imperceptible, some loss is unavoidable in principle (1).
Examiner tip
This is the most conceptually demanding question in the subtopic and rewards identifying two distinct and independent sources of loss — one along the time axis, one along the amplitude axis — and explaining that each improvement reduces but cannot eliminate its own loss. The closing continuous-versus-discrete point is what secures the top band.

Model Answers — Representing sound

High-scoring sample answers for representing sound on the AQA GCSE 8525 paper, with examiner-style notes mapping each response to the mark scheme and assessment objectives.

Question 1
Paper 2 style — AO11 mark
State what is meant by a sample in the context of digital sound. (1 mark)
Model answer
A sample is a measure of amplitude at a point in time (1).
Why this scores
This is the specification's own wording and both halves are essential — amplitude and at a point in time. Answers such as 'a small part of the sound' do not convey that a specific quantity is measured at a specific instant.
Question 2
Paper 2 style — AO12 marks
State what the sampling rate measures and the unit in which it is expressed. (2 marks)
Model answer
The sampling rate measures the number of samples taken in one second (1), and is expressed in hertz (Hz), where 1 hertz means 1 sample per second (1).
Why this scores
The unit is what keeps sampling rate distinct from sample resolution, which is measured in bits. Confusing the two units is the commonest error in this subtopic.
Question 3
Paper 2 style — AO12 marks
Explain why sound must be converted from analogue to digital before a computer can store it. (2 marks)
Model answer
Sound is analogue, meaning it is a continuously varying wave that can take infinitely many values (1). A computer can only store discrete binary values, so the continuous signal must be sampled and converted into a digital form before it can be stored or processed (1).
Why this scores
Both halves are needed: the analogue nature of sound and the digital limitation of computers. This connects directly to 3.3.1, where the specification states that computers use binary to represent all data.
Question 4
Paper 2 style — AO24 marks
A recording lasts 45 seconds, uses a sampling rate of 16,000 Hz and a sample resolution of 8 bits. Calculate the file size in kilobytes, showing all working. (4 marks)
Model answer
Size in bits = rate × res × secs = 16,000 × 8 × 45 (1) = 5,760,000 bits (1)

Size in bytes = 5,760,000 ÷ 8 = 720,000 bytes (1)

Size in kB = 720,000 ÷ 1,000 = 720 kB (1)
Why this scores
Three conversion factors appear: the multiplication for bits, ÷8 for bytes, and ÷1,000 for kilobytes using the decimal prefix from 3.3.3. Label the units at each stage — losing track of which unit you are in is the commonest cause of error.
Question 5
Paper 2 style — AO1/AO24 marks
Explain the difference between sampling rate and sample resolution, and state how each affects the quality of a recording. (4 marks)
Model answer
Sampling rate is the number of samples taken per second, measured in hertz (1). Increasing it means samples are taken more often, so the shape of the wave over time is captured more accurately and rapid changes in the sound are less likely to be missed (1).

Sample resolution is the number of bits used per sample, measured in bits (1). Increasing it provides more possible amplitude levels, so each individual measurement is stored more precisely with less rounding — 16 bits gives 65,536 levels compared with only 256 at 8 bits (1).
Why this scores
The clean way to hold these apart is that rate acts along the time axis and resolution along the amplitude axis. Both increase quality and both increase file size proportionally, but they improve genuinely different aspects of the recording.
Question 6
Paper 2 style — AO23 marks
A sound file is 3,528,000 bits in size. It was recorded at 44,100 Hz with a sample resolution of 8 bits. Calculate its duration in seconds. (3 marks)
Model answer
Rearranging the formula: secs = size in bits ÷ (rate × res) (1)

rate × res = 44,100 × 8 = 352,800 bits per second (1)

secs = 3,528,000 ÷ 352,800 = 10 seconds (1)
Why this scores
Calculating rate × res first gives the bits needed per second, which makes the final division straightforward. Note that the size is already in bits here — had it been given in bytes, it would need multiplying by 8 before rearranging.
Question 7
Paper 2 style — AO2/AO38 marks
A podcast episode lasts 25 minutes, recorded in mono at 22,050 Hz with a sample resolution of 16 bits.

(a) Calculate the file size in MB. (b) The producer wants to halve the file size. Suggest two ways of doing this and state the effect of each on quality. (c) Explain why compression might be preferable to either option. (8 marks)
Model answer
(a) Convert to seconds: 25 × 60 = 1,500 seconds (1) Size in bits = 22,050 × 16 × 1,500 = 529,200,000 bits (1) Size in bytes = 529,200,000 ÷ 8 = 66,150,000 bytes (1) Size in MB = 66.15 MB (1)

(b) Option 1 — halve the sampling rate to 11,025 Hz (1). This halves the file size, but samples are taken half as often, so the shape of the wave is captured less accurately and higher frequencies may be lost, making the speech sound duller (1).

Option 2 — halve the sample resolution to 8 bits (1). This also halves the file size, but each amplitude is stored with only 256 levels instead of 65,536, so quiet passages in particular are rounded more severely and may sound distorted (1).

(c) Compression reduces the file size without changing the recording settings at all, so the sampling rate and resolution — and therefore the underlying quality of the capture — are preserved. Lossless compression would allow the original data to be restored exactly, whereas reducing the rate or resolution discards information permanently and cannot be undone.
Why this scores
Part (c) is the AO3 discriminator and links directly to 3.3.8: the key insight is that lowering the rate or resolution destroys information irreversibly, whereas compression can reduce storage while preserving what was captured. Note that speech tolerates reduced settings better than music, which is a further creditworthy observation.
Question 8
Paper 2 style — AO36 marks
Compare the formula for calculating a bitmap image file size with the formula for a sound file size, explaining what the two have in common. (6 marks)
Model answer
The two formulae. For a bitmap image, Size (bits) = W × H × D, where W × H gives the number of pixels and D is the colour depth in bits per pixel (1). For sound, File size (bits) = rate × res × secs, where rate × secs gives the number of samples and res is the resolution in bits per sample (1).

What they have in common. Both work by multiplying how many measurements are stored by how many bits each measurement requires (1). In an image the measurements are pixels and in sound they are samples, but the structure is identical: number of items × bits per item (1).

Both also require the same follow-up. Each formula produces a result in bits, so both must be divided by 8 to give bytes and then by 1,000 per prefix step for kB or MB (1).

And both give the uncompressed size. Neither formula accounts for compression, so a real file of either type is usually considerably smaller than the calculated figure — which is precisely the baseline that compression questions ask you to compare against (1).
Why this scores
Recognising that both formulae share the structure number of measurements × bits per measurement makes them far easier to hold and far harder to confuse. The closing point that both give the uncompressed size is the natural bridge into 3.3.8.
Question 9
Paper 2 style — AO36 marks
Explain why a digital recording is only an approximation of the original sound, and explain what happens to the sound if the sampling rate is set too low. (6 marks)
Model answer
Why it is an approximation. Sampling measures the amplitude only at discrete points in time, so whatever the wave does between those points is never recorded (1). In addition, each sample is stored using a fixed number of bits, so the amplitude must be rounded to the nearest level the resolution allows rather than stored exactly (1). The original analogue signal is continuous, with infinitely many possible values, whereas the stored version is discrete and finite — so some difference is unavoidable in principle (1).

If the sampling rate is too low. Samples are taken too infrequently to follow the wave accurately, so rapid changes occurring between samples are missed entirely (1). The reconstructed wave is a poor match for the original — higher frequencies in particular are lost, because they change too quickly to be captured by widely spaced samples (1), so the playback sounds muffled or distorted and may bear little resemblance to what was recorded (1).
Why this scores
Two distinct and independent sources of loss must be identified — gaps along the time axis and rounding along the amplitude axis. The strongest point in the second half is that high frequencies are lost first, because they are the components changing too rapidly for a low sampling rate to follow.

Key Formulae — Representing sound

The formulae you need to memorise for representing sound on the AQA GCSE 8525 paper, with every variable defined in plain English and a note on when to use it.

Sound file size in bits
$\text{File size (bits)} = \text{rate} \times \text{res} \times \text{secs}$
$rate$
Sampling rate — samples per second, in hertz
$res$
Sample resolution — bits per sample
$secs$
Duration in seconds (convert from minutes first)
When to use
Every sound file-size calculation. This is the specification's own formula. Divide the result by 8 for bytes, then by 1,000 per prefix step.
Example
44,100 × 16 × 30 = 21,168,000 bits ÷ 8 = 2,646,000 bytes
Duration conversion
$\text{secs} = \text{minutes} \times 60$
$minutes$
The duration as given in the question
$× 60$
Must be applied BEFORE substituting into the formula
When to use
Whenever a duration is given in minutes. Omitting this step makes the answer sixty times too small, and it usually carries its own mark.
Example
3 minutes = 3 × 60 = 180 seconds
Amplitude levels from resolution
$\text{levels} = 2^{\text{res}}$
$res$
Sample resolution in bits per sample
$levels$
The number of distinct amplitudes that can be recorded
When to use
Stating how precisely amplitudes can be stored — 8 bits gives 256 levels, 16 bits gives 65,536.
Example
16-bit resolution → 2¹⁶ = 65,536 amplitude levels
Stereo adjustment
$\text{stereo size} = 2 \times \text{mono size}$
$2$
Two channels — left and right
$Mono$
One channel, needing no adjustment
When to use
Only when a question states the recording is in stereo. AQA's formula assumes a single channel unless told otherwise.
Example
169,344,000 bits per channel × 2 = 338,688,000 bits

Key Definitions and Keywords — Representing sound

Definitions to memorise and the exact keywords mark schemes credit for representing sound answers — sharpened from recent examiner reports for the 2026 AQA GCSE 8525 sitting.

Analogue
Examiner keyword
A continuously varying signal that can take infinitely many values. Sound is analogue and must be converted to digital before a computer can store it.
Related: Digital, Sampling
Digital
Examiner keyword
A representation using discrete binary values. A digital recording stores a series of separate samples rather than a continuous wave.
Related: Analogue, Sample
Sampling
Examiner keyword
Measuring the amplitude of an analogue signal at regular intervals in order to create a digital version of it.
Related: Sample, Sampling rate
Sample
Examiner keyword
A measure of amplitude at a point in time. Each sample is stored as a binary value using the number of bits given by the sample resolution.
Related: Sampling, Amplitude, Sample resolution
Amplitude
Examiner keyword
The height of the sound wave at a given moment, corresponding to the loudness of the sound. It is the quantity each sample measures.
Related: Sample, Sample resolution
Sampling rate
Examiner keyword
The number of samples taken in one second, measured in hertz. 1 hertz means 1 sample per second. A higher rate captures the shape of the wave more accurately.
Example
CD quality uses 44,100 Hz.
Related: Sample resolution, Hertz
Sample resolution
Examiner keyword
The number of bits used per sample. A resolution of n bits allows 2ⁿ different amplitude levels, so a higher resolution records each amplitude more precisely.
Example
16-bit resolution gives 65,536 amplitude levels.
Related: Sampling rate, Amplitude
Hertz (Hz)
Examiner keyword
The unit of sampling rate, meaning samples per second. 44,100 Hz means 44,100 samples are taken each second.
Related: Sampling rate
Sound file size
Examiner keyword
Calculated as rate × res × secs, giving a result in bits which is then divided by 8 for bytes. This is the uncompressed size.
Related: Sampling rate, Sample resolution, Compression
Channel
Examiner keyword
A separate stream of audio. Mono uses one channel; stereo uses two, doubling the file size. AQA's formula assumes one channel unless stated otherwise.
Related: Sound file size
Approximation
Examiner keyword
The fact that a digital recording can never exactly match the original analogue signal, because the wave between samples is not recorded and each amplitude is rounded.
Related: Sampling, Sample resolution
Duration
Examiner keyword
The length of the recording in seconds. It must be converted from minutes before use in the file-size calculation, which is a frequent source of lost marks.
Related: Sound file size, Sampling rate

Common Mistakes and Misconceptions — Representing sound

The traps other students keep falling into on representing sound questions — taken from recent AQA GCSE 8525 examiner reports and mark schemes — and how to avoid them.

✕Confusing sampling rate with sample resolution — for example describing the sampling rate as '16 bits'.
AQA 8525 specification 3.3.7 defines sampling rate in hertz and sample resolution as the number of bits per sample.
Why it happens
Both are numeric settings that control recording quality and both increase file size, so under pressure the two names attach to the wrong quantities.
How to avoid it
Use the units as the anchor: sampling rate is in hertz (samples per second) and sample resolution is in bits (per sample). Rate acts along the time axis — how often — while resolution acts along the amplitude axis — how precisely.
✕Substituting a duration in minutes directly into the formula instead of converting to seconds.
AQA 8525 specification 3.3.7 defines the formula as rate × res × secs, with secs meaning the number of seconds.
Why it happens
The duration is given in minutes and the formula asks for a number, so the value is used as it appears — the units mismatch is easy to overlook when working quickly.
How to avoid it
The formula requires seconds, so multiply by 60 first: 3 minutes is 180 seconds. Substituting 3 makes the answer sixty times too small, and the conversion usually carries a mark of its own.
✕Giving the answer in bits when the question asked for bytes, kB or MB.
The ÷8 is awarded as a separate method mark in sound file-size questions.
Why it happens
The multiplication feels like the complete calculation, and the unit conversion reads as a formality rather than part of the answer.
How to avoid it
The formula produces bits. Divide by 8 for bytes, then by 1,000 for kB and again for MB. Sense-check the result — a three-minute song coming out at 127 MB rather than 16 MB is exactly a factor of eight out.
✕Ignoring the word 'stereo' and calculating for a single channel only.
Questions specifying stereo award a mark for the doubling, which is lost if the detail is overlooked.
Why it happens
The formula makes no mention of channels, so the detail in the question feels like background context rather than a required part of the calculation.
How to avoid it
Stereo means two channels, so the result must be multiplied by 2. Mono is one channel and needs no adjustment. Underline the word 'stereo' when you see it — AQA's formula assumes a single channel unless the question states otherwise.
✕Working backwards from a file size given in bytes without converting it to bits first.
Rearrangement questions asking for a missing duration, rate or resolution appear regularly and require this first step.
Why it happens
The size is stated in bytes so it seems natural to substitute it directly — the fact that the formula is defined in bits is easy to forget when rearranging.
How to avoid it
The formula gives bits, so a size given in bytes must be multiplied by 8 before rearranging. Note the direction reverses: forwards you divide by 8, backwards you multiply. Skipping this makes the answer eight times out.
✕Defining a sample vaguely, such as 'a small piece of the sound' or 'part of the recording'.
AQA 8525 specification 3.3.7 additional information: 'Students should understand that a sample is a measure of amplitude at a point in time.'
Why it happens
The everyday meaning of 'sample' is a small portion of something, so the technical definition never displaces the intuitive one.
How to avoid it
Use the specification's exact wording: a sample is a measure of amplitude at a point in time. Both halves matter — it is a measurement of amplitude, and it is taken at a single instant rather than over a period.

Representing Sound — AQA GCSE Computer Science 8525 (3.3.7)

Direction

What it does

Where it happens

Analogue → digital

Samples the wave for storage

When recording, e.g. a microphone input

Digital → analogue

Reconstructs a wave from the samples

When playing back through speakers

Rate

Quality

8,000 Hz (8 kHz)

Telephone quality

44,100 Hz (44.1 kHz)

CD quality

48,000 Hz

Professional audio

Resolution

Amplitude levels

Quality

8 bits

2⁸ = 256

Basic

16 bits

2¹⁶ = 65,536

CD quality

24 bits

2²⁴ ≈ 16.7 million

Professional

Sampling rate

Sample resolution

Measures

How often samples are taken

How accurately each is stored

Units

Hertz (samples per second)

Bits per sample

Axis

Time (horizontal)

Amplitude (vertical)

Increasing it

Captures the wave's shape better

Records each amplitude more precisely

The specification gives the formula verbatim:

$\text{File size (bits)} = \text{rate} \times \text{res} \times \text{secs}$

where rate = sampling rate (Hz), res = sample resolution (bits), secs = number of seconds.

The method

Identify rate, resolution and duration.
Convert the duration to seconds if given in minutes.
Multiply all three → size in bits.
Divide by 8 → bytes, then by 1,000 for kB, again for MB.
Sense-check against real-world scale.

Worked example

A 30-second recording is sampled at 44,100 Hz with a resolution of 16 bits. Calculate the file size in MB.

Step 1 — bits: $44{,}100 \times 16 \times 30 = 21{,}168{,}000 \text{ bits}$

Step 2 — bytes: $21{,}168{,}000 \div 8 = 2{,}646{,}000 \text{ bytes}$

Step 3 — megabytes: $2{,}646{,}000 \div 1{,}000{,}000 = 2.646 \text{ MB}$

Sense-check: CD-quality audio runs at roughly 5 MB per minute, so 30 seconds at around 2.6 MB is exactly right ✓

Converting minutes to seconds

A very common step — and a very common slip:

$3 \text{ minutes} = 3 \times 60 = 180 \text{ seconds}$

Always convert before substituting. Putting "3" into the formula instead of 180 makes the answer sixty times too small.

Working backwards

Rearrange as needed:

$\text{secs} = \frac{\text{size in bits}}{\text{rate} \times \text{res}} \qquad \text{res} = \frac{\text{size in bits}}{\text{rate} \times \text{secs}}$

A file of 1,411,200 bits was recorded at 44,100 Hz with 16-bit resolution. How long is it?

$\frac{1{,}411{,}200}{44{,}100 \times 16} = \frac{1{,}411{,}200}{705{,}600} = 2 \text{ seconds}$

Stereo

Rate

Effect on quality

Effect on size

Higher

Samples taken more often, so the shape of the wave is captured more accurately and the playback sounds closer to the original

Larger file, proportionally

Lower

Fewer samples, so detail between them is lost and the sound is less faithful

Smaller file

Resolution

Effect on quality

Effect on size

Higher

More bits per sample, so each amplitude is recorded more precisely with less rounding

Larger file, proportionally

Lower

Amplitudes rounded to fewer levels, producing a less accurate, potentially distorted sound

Smaller file

Rate

Resolution

Bits

Bytes

22,050 Hz

8 bits

1,764,000

220,500

44,100 Hz

8 bits

3,528,000

441,000

44,100 Hz

16 bits

7,056,000

882,000

Use

Typical settings

Why

Telephone speech

8 kHz, 8 bits

Speech is intelligible at low quality; bandwidth matters more

Music (CD)

44.1 kHz, 16 bits

High enough that most listeners cannot detect the difference

Professional recording

48 kHz+, 24 bits

Quality must survive later editing

Representing sound

Detailed Study Notes

At a glance

What you’ll learn

Sound is analogue and must be converted

Why conversion is necessary

Sampling — how the conversion works

The digital version is an approximation

The two conversions

Sampling rate and sample resolution

Sampling rate

Sample resolution

The distinction that matters

Calculating sound file size

The method

Worked example

Converting minutes to seconds

Working backwards

Stereo

How rate and resolution affect quality and file size

Effect of sampling rate

Effect of sample resolution

Worked comparison

Why the digital version is never perfect

Choosing appropriately

Quick recap

Memorise this

How it’s examined

Take this whole topic with you

1Explain why sound must be converted to digital (3 marks)

2Distinguish sampling rate from sample resolution (4 marks)

3Calculate a sound file size (4 marks)

4Handle a duration given in minutes (5 marks)

5Rearrange the formula to find a missing value (5 marks)

6Compare the effect of rate and resolution (5 marks)

7Justify settings for two different recordings (6 marks)

8Calculate a stereo recording's file size (6 marks)

9Explain why the digital version is an approximation (6 marks)

Sound file size in bits

Duration conversion

Amplitude levels from resolution

Stereo adjustment

Analogue

Digital

Sampling

Sample

Amplitude

Sampling rate

Sample resolution

Hertz (Hz)

Sound file size

Channel

Approximation

Duration

✕Confusing sampling rate with sample resolution — for example describing the sampling rate as '16 bits'.

✕Substituting a duration in minutes directly into the formula instead of converting to seconds.

✕Giving the answer in bits when the question asked for bytes, kB or MB.

✕Ignoring the word 'stereo' and calculating for a single channel only.

✕Working backwards from a file size given in bytes without converting it to bits first.

✕Defining a sample vaguely, such as 'a small piece of the sound' or 'part of the recording'.

Representing sound

Detailed Study Notes

At a glance

What you’ll learn

Sound is analogue and must be converted

Why conversion is necessary

Sampling — how the conversion works

The digital version is an approximation

The two conversions

Sampling rate and sample resolution

Sampling rate

Sample resolution

The distinction that matters

Calculating sound file size

The method

Worked example

Converting minutes to seconds

Working backwards

Stereo

How rate and resolution affect quality and file size