Skip to main content
EnglandMusic TechnologySyllabus dot point

Why do we measure sound level in decibels, and how does the logarithmic decibel relate to amplitude and power?

The decibel as a logarithmic ratio: the power formula and the amplitude (voltage) formula, dBFS and headroom, the relationship between decibel change and perceived loudness, and dynamic range.

A focused answer to the Edexcel 9MT0 decibel content, covering the decibel as a logarithmic ratio, the power and amplitude formulae, dBFS and headroom, how decibel changes map to perceived loudness, and dynamic range.

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

Edexcel wants you to treat the decibel as a logarithmic ratio, not a unit of loudness in itself. You must know the power formula and the amplitude (voltage) formula, when to use each, what dBFS means in a digital system, how decibel changes map to perceived loudness, and how dynamic range is expressed in decibels. This is the most calculation-heavy idea in the course, and Component 3 reliably asks for a decibel value.

The answer

Why the decibel is logarithmic

The quietest audible sound and the threshold of pain differ in pressure by a factor of about a million. A linear scale would be unwieldy; the logarithm compresses that range into roughly 00 to 120120 on the sound pressure level scale. The same logic applies to every gain, fader and meter in a studio.

The two formulae

Choosing the right formula is the single most common exam decision. Faders, microphone voltages and sample amplitudes are amplitude ratios, so they use 20log⁡1020\log_{10}. Acoustic power or intensity ratios use 10log⁡1010\log_{10}. Mixing them up is the classic error.

Useful reference points

These are worth memorising so you can sanity-check a calculation. If you raise a fader by 66 dB you have doubled the signal amplitude; if a track is 33 dB louder it carries twice the power; if a mix needs to sound twice as loud you need about 1010 dB, not double the number on the meter.

dBFS and headroom

In a digital system, level is measured in dBFS, decibels relative to full scale. The maximum representable level is 00 dBFS, and every real signal sits below it as a negative number. The space between the peak level and 00 dBFS is the headroom. Leaving several decibels of headroom protects against unexpected peaks, inter-sample overs and the need for further processing, all of which can push a signal into clipping.

Dynamic range

Examples in context

When a compressor shows 66 dB of gain reduction, it has roughly halved the amplitude of the peaks. When you read −3-3 dBFS on a master meter, the loudest sample is just below the ceiling, with little headroom left. When a converter spec quotes 120120 dB of dynamic range, it is telling you how far the quietest detail can sit below the loudest peak before disappearing into noise. The decibel turns all of these into a single, comparable scale.

Try this

Q1. State the decibel formula for an amplitude ratio. [1 mark]

  • Cue. dB=20log⁡10(A2A1)\text{dB} = 20\log_{10}\left(\frac{A_2}{A_1}\right).

Q2. By how many decibels does the level change if the power is doubled? [2 marks]

  • Cue. 10log⁡10(2)=+310\log_{10}(2) = +3 dB.

Q3. A 2424-bit recording offers about how much dynamic range, using roughly 66 dB per bit? [2 marks]

  • Cue. 24×6=14424 \times 6 = 144 dB.

Exam-style practice questions

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

Edexcel 9MT0/03 20184 marksThe output voltage of a microphone amplifier increases from 0.200.20 V to 0.400.40 V. Calculate the gain in decibels, and explain why the decibel is used to measure level rather than a simple ratio.
Show worked answer →

For a voltage (amplitude) ratio the decibel formula is 20log⁡10(V2V1)20\log_{10}\left(\frac{V_2}{V_1}\right).

20log⁡10(0.400.20)=20log⁡10(2)=20×0.301=6.020\log_{10}\left(\frac{0.40}{0.20}\right) = 20\log_{10}(2) = 20 \times 0.301 = 6.0 dB.

The decibel is used because hearing is logarithmic: equal ratios of amplitude (such as every doubling) are perceived as roughly equal steps in loudness, so a logarithmic scale matches perception and compresses the enormous range from the quietest audible sound to the threshold of pain into a manageable set of numbers.

Markers reward the 20log⁡1020\log_{10} formula (not 10log⁡1010\log_{10}, because this is a voltage ratio), the value 6.06.0 dB, and the link to logarithmic hearing.

Edexcel 9MT0/03 20213 marksA producer says a mix peaks at −6-6 dBFS. Explain what dBFS means, what the value −6-6 dBFS indicates, and why leaving headroom is good practice.
Show worked answer →

dBFS means decibels relative to full scale: in a digital system, 00 dBFS is the maximum level that can be represented before clipping, and all real signals sit below it as negative values. A peak of −6-6 dBFS means the loudest sample is 66 dB below the clipping point, so there is 66 dB of headroom.

Leaving headroom is good practice because it prevents clipping on unexpected peaks, leaves room for further processing and the mastering stage, and keeps the signal clean. A signal that reaches 00 dBFS risks inter-sample peaks and harsh digital distortion.

Markers reward dBFS defined against full scale, 00 dBFS as the clipping ceiling, −6-6 dBFS as 66 dB of headroom, and a sensible reason for keeping headroom.

Related dot points

Sources & how we know this