Sound Representation

Sound is a continuous, analogue phenomenon, yet computers store only discrete bits. This lesson explains how analogue-to-digital conversion (ADC) turns a sound wave into numbers through sampling, how sampling rate and sample resolution (bit depth) control fidelity, what the Nyquist idea tells us about the rate we must choose, how to calculate an audio file's size, and how a DAC turns the numbers back into sound.

---

Spec Mapping

This lesson addresses the H446 1.4.1 Data Types content on representing sound:

• Explain analogue-to-digital conversion: sampling the amplitude of a sound wave at regular intervals and storing each measurement in binary.
• Define sampling rate and sample resolution (bit depth) and describe how each affects the accuracy of the digital representation and the file size.
• State the relationship between sampling rate and the highest reproducible frequency (the Nyquist idea) and explain aliasing.
• Calculate the storage requirement of a sampled sound from rate, resolution, duration and channels.
• Describe the role of the ADC and DAC in the capture-and-playback pipeline.

(This is a paraphrase of the specification content, not a verbatim quotation.)

---

Analogue vs Digital Sound

Sound is a continuous wave of pressure variation in the air, which a microphone converts into a continuously varying voltage — an analogue signal that can take any value at any instant. A computer, however, can only store a finite set of discrete numbers, so the analogue signal must be digitised: measured at particular moments and rounded to particular values.

Aspect	Analogue	Digital
Signal	Continuous in both time and value	Discrete in both (sampled and quantised)
Fidelity	The original itself	An approximation of the original
Storage	Physical medium (vinyl groove, tape magnetism)	Binary numbers
Copying	Each copy adds noise/degrades	Bit-for-bit perfect copies

Digitising loses information in principle — the smooth wave is replaced by a staircase of samples — but if the rate and resolution are high enough, the approximation is indistinguishable to the human ear, while gaining the durability and perfect copyability of digital data.

The word staircase is worth holding onto. A digital recording is literally a sequence of flat steps: it holds each sampled value constant until the next sample. The two ways that staircase departs from the smooth original map exactly onto the two parameters — the width of each step is set by the sampling rate (more samples ⇒ narrower steps ⇒ better in time), and the height resolution of each step is set by the bit depth (more bits ⇒ finer steps ⇒ better in value). Improving fidelity therefore always means narrowing the steps in one or both directions, at the cost of more data.

The pay-off for accepting that approximation is substantial and examinable:

Advantage of digital	Why
Perfect copies	Bits can be duplicated exactly; an analogue copy (e.g. tape-to-tape) adds noise each generation
Robust transmission	Digital signals can be regenerated cleanly; an analogue signal accumulates noise
Easy editing	Samples are just numbers — cut, mix, apply effects, undo
Compression possible	Redundant or inaudible data can be removed (the next lesson)
Durable storage	No physical wear like a vinyl groove or a magnetised tape

---

Analogue-to-Digital Conversion (the ADC pipeline)

Sampling is measuring the amplitude (instantaneous height, i.e. loudness) of the analogue wave at regular time intervals and recording each measurement as a binary number. The full capture pipeline is:

flowchart LR
    A["Sound wave<br/>(air pressure)"] --> B["Microphone<br/>(wave to voltage)"]
    B --> C["Sample at fixed<br/>intervals"]
    C --> D["Quantise to nearest<br/>level (round)"]
    D --> E["Store each sample<br/>as binary"]

So digitisation is really two roundings, and you must keep them distinct:

• Sampling discretises time — we only look at the wave at set instants, controlled by the sampling rate.
• Quantisation discretises value — each measured amplitude is rounded to the nearest available level, controlled by the bit depth.

Parameter	Definition	Effect
Sampling rate	Samples taken per second (hertz, Hz)	Higher captures higher frequencies more faithfully
Sample resolution (bit depth)	Bits used to store each sample	Higher gives more amplitude levels, less quantisation noise
Duration	Length of the audio in seconds	Longer means more samples, larger file
Channels	1 = mono, 2 = stereo	More channels multiply the data

Worked Example — quantising actual samples

To see the two roundings in action, suppose a 3-bit ADC offers $2^{3}=8$23=8 amplitude levels, numbered $0$0–$7$7, evenly spaced across the input voltage range. At five successive sample instants the analogue wave has these true heights (on the same $0$0–$7$7 scale), and each is rounded to the nearest level:

Sample #	True value	Nearest level (stored)	Quantisation error
1	2.3	2	$-0.3$−0.3
2	4.8	5	$+0.2$+0.2
3	6.6	7	$+0.4$+0.4
4	3.1	3	$-0.1$−0.1
5	0.4	0	$-0.4$−0.4

The stored sequence is $2, 5, 7, 3, 0$2,5,7,3,0, written in binary as 010 101 111 011 000. The quantisation error column is the price of rounding: every stored value is up to half a level away from the truth. Add a fourth bit ($2^{4}=16$24=16 levels) and the steps halve, so the worst-case error halves too — directly less quantisation noise. This is the concrete meaning of "more bits give a more faithful amplitude".

---

Sampling Rate

The sampling rate is how often per second the amplitude is measured, in hertz. A higher rate places sample points closer together in time, so rapid changes in the wave — which correspond to high-frequency (high-pitched) sounds — are captured rather than missed.

Sampling rate	Quality	Typical use
8{,}000 Hz (8 kHz)	Telephone	Voice calls
22{,}050 Hz	AM-radio	Low-quality audio
44{,}100 Hz (44.1 kHz)	CD	Music CDs
48{,}000 Hz	Broadcast/DVD	Video production
96{,}000 Hz	Studio	Professional recording

Why a higher rate captures detail

Picture a wave that rises and falls quickly. If samples are taken far apart in time, the few points captured can be joined into many different curves — including a wrong, slower one — so the rapid wiggles are lost or misread. Taking samples closer together pins the curve down at more points, so the reconstruction follows the original faithfully. The number of samples grows in direct proportion to the rate:

$\text{samples} = \text{sampling rate} \times \text{duration (s)}$samples=sampling rate×duration (s)

So a 5-second clip at $44{,}100$44,100 Hz contains $44{,}100 \times 5 = 220{,}500$44,100×5=220,500 samples, whereas the same 5 seconds at the telephone rate of $8{,}000$8,000 Hz contains only $8{,}000 \times 5 = 40{,}000$8,000×5=40,000 samples — roughly five times fewer points describing the same wave, which is why phone-quality audio sounds dull and band-limited compared with CD.

---

Sample Resolution (Bit Depth)

Sample resolution, also called bit depth, is the number of bits used to store each sample. Because each bit doubles the number of distinct values, the number of amplitude levels follows the familiar power of two:

$\text{amplitude levels} = 2^{\,b} \quad\text{for a bit depth of } b \text{ bits}$amplitude levels=2bfor a bit depth of b bits

Bit depth	Levels $=2^{b}$=2b	Quality
8 bits	$2^{8}=256$28=256	Low — audible noise
16 bits	$2^{16}=65{,}536$216=65,536	CD quality
24 bits	$2^{24}=16{,}777{,}216$224=16,777,216	Studio quality
32 bits	$2^{32}\approx 4.3\text{ billion}$232≈4.3 billion	Professional mastering

Quantisation is the rounding of each measured amplitude to the nearest of these levels; the gap between the true value and the stored value is the quantisation error. With too few levels the staircase is coarse and the rounding errors are large and audible — a hiss known as quantisation noise. More bits mean finer steps, smaller errors and a smoother, quieter result. A neat way to remember the division of labour: the rate decides how faithfully pitch/frequency is captured, while the bit depth decides how faithfully loudness/amplitude is captured.

Bit depth and dynamic range

Bit depth also fixes the dynamic range — the span between the quietest and loudest sound the format can distinguish. Each extra bit doubles the number of amplitude levels, which means it can resolve a sound half as loud as before, adding roughly 6 dB of range per bit. (The figure comes from $20 \times \log_{10} 2 \approx 6$20×log10​2≈6 dB; you are not expected to derive it, only to know that more bits widen the gap between the softest and loudest representable signal.)

Bit depth	Levels $=2^{b}$=2b	Approx. dynamic range
8 bits	256	$\approx 8 \times 6 = 48$≈8×6=48 dB
16 bits	65,536	$\approx 16 \times 6 = 96$≈16×6=96 dB
24 bits	16,777,216	$\approx 24 \times 6 = 144$≈24×6=144 dB

The practical consequence: with only 8 bits the loudest sound is just 256 steps above the quietest, so soft passages sit close to the quantisation noise floor and the hiss is audible. With 16 bits the floor is pushed about 48 dB lower, so a CD can carry both a whisper-quiet intro and a full orchestra without the quiet parts being swamped by noise. So bit depth answers two linked questions — how finely each sample is resolved (quantisation error) and how wide a loud-to-quiet range the recording can span (dynamic range).

---

The Nyquist Idea and Aliasing

How fast must we sample? The Nyquist (Nyquist–Shannon) sampling theorem gives the answer:

$\text{sampling rate} \;\geq\; 2 \times \text{highest frequency in the sound}$sampling rate≥2×highest frequency in the sound

Intuitively, to capture a wave that goes up and down, you must catch at least its peak and its trough each cycle — two samples per cycle. Sample any slower and the reconstructed wave is wrong.

Rearranging the theorem gives an equally useful form — the highest frequency a given sample rate can reproduce is half the rate (this half-the-sample-rate value is called the Nyquist frequency):

$\text{highest reproducible frequency} = \frac{\text{sampling rate}}{2}$highest reproducible frequency=2sampling rate​

Worked Example — rate to highest frequency

A voice recorder samples at $8{,}000$8,000 Hz. What is the highest frequency it can faithfully reproduce?

$f_{\max} = \frac{8{,}000}{2} = 4{,}000 \text{ Hz}$fmax​=28,000​=4,000 Hz

This is fine for intelligible speech (most of which lies below 3.4 kHz, which is exactly why the telephone network historically used an 8 kHz rate) but it discards the brightness and sibilance of music, whose energy reaches up to 20 kHz. To capture a sound containing a 15 kHz component without aliasing, the rate must satisfy $\text{rate} \geq 2 \times 15{,}000 = 30{,}000$rate≥2×15,000=30,000 Hz, so $8{,}000$8,000 Hz is far too low and that component would alias.

Worked Example — required rate for an instrument

A cymbal produces overtones up to about $18{,}000$18,000 Hz. The minimum sampling rate to capture it faithfully is:

$\text{rate} \geq 2 \times 18{,}000 = 36{,}000 \text{ Hz}$rate≥2×18,000=36,000 Hz

so a $44{,}100$44,100 Hz CD rate (giving a Nyquist frequency of $22{,}050$22,050 Hz) comfortably captures it, whereas a $22{,}050$22,050 Hz rate (Nyquist frequency $11{,}025$11,025 Hz) would not — the 18 kHz overtones exceed its limit and alias into the audible band as spurious lower tones.

Why 44{,}100 Hz for CDs?

Human hearing reaches roughly 20{,}000 Hz (20 kHz). The Nyquist minimum is therefore:

$2 \times 20{,}000 = 40{,}000 \text{ Hz}$2×20,000=40,000 Hz

CD audio uses $44{,}100$44,100 Hz — a little above the minimum, giving headroom for the filters that remove frequencies above the hearing range. This is why the apparently odd figure 44.1 kHz is "CD quality". Its Nyquist frequency, $44{,}100 \div 2 = 22{,}050$44,100÷2=22,050 Hz, sits just above the limit of human hearing, so every audible frequency is captured with margin to spare.

Aliasing

If the rate is below twice the highest frequency, aliasing occurs: frequencies above the Nyquist limit are mis-captured and reappear as false lower-frequency tones that were never in the original, audible as distortion. To prevent it, an anti-aliasing (low-pass) filter removes too-high frequencies before sampling. Aliasing is the headline penalty for sampling too slowly, and a frequent exam discriminator.

Worked example — a tone aliasing. Suppose a system samples at $8{,}000$8,000 Hz (Nyquist frequency $4{,}000$4,000 Hz) but the input contains a $5{,}000$5,000 Hz tone — above the limit. The tone is undersampled and folds back, appearing as a false tone at $8{,}000 - 5{,}000 = 3{,}000$8,000−5,000=3,000 Hz, a pitch that was never present. The listener hears a spurious 3 kHz whistle. The fix is the anti-aliasing filter, which removes everything above 4 kHz before the samples are taken, so nothing can fold back. The everyday visual analogue is the "wagon-wheel effect" in film, where a fast-spinning wheel appears to turn slowly or backwards — the camera's frame rate is below twice the rotation rate, so motion aliases.

---

Calculating Audio File Size

The raw size of sampled audio is the product of all four factors:

$\text{file size (bits)} = \text{sampling rate} \times \text{bit depth} \times \text{duration (s)} \times \text{channels}$file size (bits)=sampling rate×bit depth×duration (s)×channels

with mono $=1$=1 channel and stereo $=2$=2. Convert with $\div 8$÷8 for bytes, then $\div 1024$÷1024 for KB and again for MB.

Worked Example — CD-quality stereo track

A 3-minute stereo track at CD quality ($44{,}100$44,100 Hz, 16-bit):

$\text{duration} = 3 \times 60 = 180 \text{ s}$duration=3×60=180 s

$\text{bits} = 44{,}100 \times 16 \times 180 \times 2 = 254{,}016{,}000 \text{ bits}$bits=44,100×16×180×2=254,016,000 bits

$\text{bytes} = \frac{254{,}016{,}000}{8} = 31{,}752{,}000 \text{ bytes}$bytes=8254,016,000​=31,752,000 bytes

$\text{KB} = \frac{31{,}752{,}000}{1024} \approx 31{,}007.8 \text{ KB}$KB=102431,752,000​≈31,007.8 KB

$\text{MB} = \frac{31{,}007.8}{1024} \approx 30.3 \text{ MB}$MB=102431,007.8​≈30.3 MB

About 30 MB for one song — which is exactly why music is normally stored compressed (MP3/AAC), foreshadowing the next lesson.

Worked Example — short mono voice clip

A 10-second mono voice memo sampled at $8{,}000$8,000 Hz with 8-bit resolution:

$\text{bits} = 8{,}000 \times 8 \times 10 \times 1 = 640{,}000 \text{ bits}$bits=8,000×8×10×1=640,000 bits

$\text{bytes} = \frac{640{,}000}{8} = 80{,}000 \text{ bytes} \approx 78.1 \text{ KB}$bytes=8640,000​=80,000 bytes≈78.1 KB

The contrast — 30 MB versus 78 KB — shows how dramatically rate, depth and channels drive size.