Sound: sampling rate and resolution

What you need to be able to do

Analogue and digital

A sound is a vibration: the air pressure goes up and down smoothly, and a microphone turns that into a smoothly changing voltage. Its height at any moment is its amplitude. A computer can only store a limited list of binary numbers, so the smooth wave has to be measured at moments in time and each measurement rounded to a number.

Sampling: turning the wave into numbers

1. The amplitude of the analogue sound wave is measured at set (regular) time intervals: each measurement is a sample.
2. Each sample is given the nearest binary value: the number of bits for each one is the sampling resolution.
3. The binary values are stored in sequence (in time order).
4. On playback, the values are turned back into a voltage, step by step: an approximation of the original wave.

Changing the sampling rate and resolution

A bigger file has knock-on effects in context: it takes longer to transmit (e.g. to a smartphone), uses more bandwidth, and fills the storage faster, so fewer recordings fit (June 2024, the video doorbell).

Sound file size

1. Change the rate to samples per second: 44.1 kHz = 44 100, 50 kHz = 50 000 (kilo = 1000 here).
2. Change the duration to seconds: 20 minutes = 1200 s.
3. Multiply rate × resolution × seconds → bits; ÷ 8 → bytes; then ÷ 1000s (kB, MB) or 1024s (KiB, MiB).

Mark-scheme focus

Sampling rate, then the effect of a higher sampling resolution (3 marks)

Credited

• (i) the number of samples taken per unit time / per second B1
• (ii) increases the number of bits per sample // larger range of values B1; which means that the file size increases B1
• or: makes the sound file more accurate // digital waveform closer to the analogue waveform // smaller quantisation errors B1

Not credited

• (i) “the rate of taking samples” (repeats the term)
• (ii) answers about taking more samples (that is the sampling rate)

Why

The report: in (ii) “a popular incorrect answer explained the effect of increasing the sampling rate … rather than the sampling resolution”.

From: 9618/11/O/N/22 Q1(d)

“Explain how an analogue sound wave is converted into digital data” (2 marks)

Credited

• Value / magnitude of the analogue wave is measured a set number of times each second / at set intervals B1
• Each sample is given the binary number and stored in sequence B1

Not credited

• “The sound is converted to binary” (how?)
• “It is sampled” with no mention of measuring at intervals

Why

The two marks are the two halves of the process: measure at intervals, then a binary value per sample in order.

From: 9618/13/O/N/24 Q6(c)(i)

“Explain the impact of changing the sampling resolution on the accuracy of a sound recording” (3 marks)

Credited

• Increase: the number of bits used for each sample is increased B1
• more values available to represent each sample // more amplitudes can be represented B1
• each binary amplitude is closer to the analogue amplitude // quantisation errors are reduced // the digital wave is closer to the original B1

Not credited

• “It sounds better” (not technical)
• Points about file size (the question asks about accuracy)
• Mixing increase and decrease in one answer

Why

Pick one direction (increase or decrease) and follow it through: more bits → more values → closer to the real amplitude.

Also accepted

The same three steps for a decrease, each reversed.

From: 9618/12/O/N/23 Q6(b)

Worked examples

Example 1 · Three sound terms (Paper 1, 3 marks) ★ Easy

1 Read it with me what is actually happening?

Three definitions given, you supply the term. Watch the wording: “how many times” vs “how many bits”.

• number of times … per time intervalhow often → rate
• number of bits … each measurementhow precisely → resolution
• before it is recordedcontinuous → analogue

2 What is it asking for? the target, part by part

• (b)three terms3

3 What do we know? everything the question gives us

4 Plan the route work backwards from what we want

FindKey ideaAlready have itWork out firstTrap

• Terms
  • Ratetimes per second
  • Resolutionbits per sample
  • Analoguecontinuous wave

Is “frequency” acceptable for the first?

The scheme's term is sampling rate: use it.

5 Now solve it have a go on paper first, then open the worked solution

1. the number of times the amplitude is measured per time interval → sampling rate B1
2. the number of bits used to store each amplitude measurement → sampling resolution B1
3. the type of sound wave before it is recorded → analogue B1

6 Check it does the answer make sense?

Rate ↔ times, resolution ↔ bits ✓.

Example 2 · Duration, rate and resolution: two effects each (Paper 1, 3 marks) ★★ Medium

1 Read it with me what is actually happening?

Each change gets two lines: one to a file-size box and one to an accuracy box.

• two lines from each changeone size effect + one accuracy effect
• durationlength of the recording

2 What is it asking for? the target, part by part

• (a)six lines3: 1 per change with both lines right

3 What do we know? everything the question gives us

4 Plan the route work backwards from what we want

FindKey ideaAlready have itWork out firstTrap

• Each change
  • Sizemore data? fewer bits?
  • Accuracycloser to the analogue wave?
  • Traplonger = more accurate

Does a longer recording sound more accurate?

No: every second is sampled the same way; there are just more seconds.

5 Now solve it have a go on paper first, then open the worked solution

1. increase the duration → file size gets bigger and no change to the accuracy B1
2. increase the sampling rate → file size gets bigger and accuracy improves B1
3. decrease the sampling resolution → file size gets smaller and accuracy worsens B1

6 Check it does the answer make sense?

Each change has exactly one size line and one accuracy line ✓.

Example 3 · Accuracy ticks, then a file size (Paper 1, 4 marks) ★★ Medium

1 Read it with me what is actually happening?

(a) uses the effects table. (b) is the size formula with two unit changes hidden in it: kHz → Hz and minutes → seconds.

• 50 kHz50 000 samples per second
• 20 minutes1200 s
• megabytes÷ 8 ÷ 1000 ÷ 1000

2 What is it asking for? the target, part by part

• (a)three ticks2
• (b)size in MB2

3 What do we know? everything the question gives us

• 40 → 60 kHz
• 20 → 40 min
• 24 → 16 bits
• 50 kHz, 16 bits, 20 min(b)

4 Plan the route work backwards from what we want

FindKey ideaAlready have itWork out firstTrap

• (b)MB
  • Hz50 × 1000
  • Seconds20 × 60
  • Multiply× 16 bits ÷ 8 ÷ 10⁶

What was the reported error?

Multiplying by 50 instead of 50 000.

How many bytes per sample?

16 bits = 2 bytes.

5 Now solve it have a go on paper first, then open the worked solution

1. (a) 40 → 60 kHz: accuracy increases; 20 → 40 minutes: does not change; 24 → 16 bits: decreases B2
2. (b) 50 000 × (20 × 60) × 16 bits = 960 000 000 bits M1
3. ÷ 8 = 120 000 000 bytes = 120 megabytes A1

6 Check it does the answer make sense?

2 bytes × 50 000 × 1200 = 120 million bytes ✓. (a): 2 marks for three ticks, 1 for one or two.

Example 4 · Describe how sound is represented; then why rate and resolution improve precision (Paper 1, 7 marks) ★★★ Hard

Why it is hard: the June 2023 report: (i)'s third point (limited binary values stored in sequence) was rarely given, and (ii) was answered with definitions instead of reasons.

1 Read it with me what is actually happening?

(i) is the sampling process in three steps. (ii) asks for reasons why each change improves precision: two reasons for the rate, two for the resolution.

• describe how … representedmeasure, give a binary number, store in sequence
• explain the reasons whycause → effect, not definitions

2 What is it asking for? the target, part by part

• (c)(i)the process3
• (c)(ii)2 reasons for rate + 2 for resolution4

3 What do we know? everything the question gives us

4 Plan the route work backwards from what we want

FindKey ideaAlready have itWork out firstTrap

• (ii) rate
  • Causesamples more often → smaller gaps
  • Effectdigital wave closer to analogue
• (ii) resolution
  • Causemore bits → more amplitude values
  • Effecteach sample closer, smaller quantisation error
• Trapjust defining rate and resolution

What did most candidates do wrong in (ii)?

The report: they “saw the two terms and simply wrote down the definitions without answering the question”.

5 Now solve it have a go on paper first, then open the worked solution

1. (i) The amplitude is recorded a set number of times a second B1; each amplitude is given a corresponding binary number B1; the binary numbers are saved in sequence B1
2. (ii) Rate: smaller ‘gaps’ in the sound wave // recorded more often B1; the digital waveform is closer to the analogue waveform // smaller quantisation errors B1
3. Resolution: more bits per sample // a wider range of amplitudes can be stored B1; each binary amplitude is closer to the analogue amplitude // smaller quantisation errors B1

6 Check it does the answer make sense?

Every (ii) point is a reason (cause → closer to the original), none is just a definition ✓.

Summary

• Analogue: continuously changing. Sampling: measure the amplitude at regular intervals, give each sample a binary value, store in sequence.
• Sampling rate = samples per second; sampling resolution = bits per sample.
• Higher rate or resolution: more accurate (closer to the analogue wave, smaller quantisation errors) and a bigger file. Longer duration: bigger file, same accuracy.
• Size (bits) = rate (Hz) × resolution (bits) × time (s).

Can I do this?

Quiz

Concept checks

Answer each one in your head first, then open it.

Exam-style practice

New questions, not used above. Do them on paper, then mark your work against the official mark scheme.