How do decibels work?

05 Aug, 2026

Fixing my understanding

What is a Decibel?

A decibel is a bit of a Russian doll you keep expanding. What are the units of dB? dB is itself a unit, but the ratio it describes is dimensionless - they're similar in that way to radians.

Your hearing is actually pretty insane when you consider the scale of things you hear. I'm often in an anechoic chamber, and I hear just fine. I also hear jet engines just fine. The problem is the ratio between those extremes is insane and insane ratios are annoying to work with directly. That's where logs, and decibels, come in.

Why we need logs: dynamic range

A very quiet anechoic chamber can sit below ~10 dB SPL of background noise. A concert can push something like 100 dB SPL. That's roughly a 90 dB difference. SPL means sound pressure level, measured relative to a reference pressure of 20 micropascals in air.

Power is proportional to the square of amplitude, provided the impedance and other relevant conditions stay the same.
Power is energy transferred per second. Sound intensity is the acoustic power per unit area reaching your ears - different from the electrical wattage rating of a 50,000-watt sound system.
Amplitude is the size of the wave's variation. In sound, pressure amplitude is the maximum change in air pressure caused by the vibration. In wireless, field amplitude is the peak strength of the field.

When a speaker pushes air, it creates a tight pocket of high pressure (compressing the air molecules in front of it). When the speaker snaps back, it creates a region of low pressure as nearby air expands. Low amplitude means your eardrum barely moves. And because power scales with the square of amplitude, doubling the amplitude means quadrupling the power - and it takes a correspondingly larger amount of power to produce that amplitude rise in the first place.

Concretely: a 100 dB difference represents a factor of 100,000 in amplitude, and 10,000,000,000 in power, under those same conditions. The dynamic range of human hearing is roughly 120–140 dB from the threshold of hearing to the threshold of pain, depending on frequency and the listener. Audio engineers use dynamic range to describe the ratio between the largest undistorted signal a system can handle and its noise floor (the level of background noise the system generates on its own) - and the general idea is the ratio between the largest and smallest usable signal levels.

So yes - the dynamic range of human hearing, and a lot of things we try to reason about, is large. Logarithms are what make that tractable.

The log, briefly

A logarithm answers one question: what power do I raise a certain number to, in order to get my answer?

$\log_{10} (100) = 2 \Leftrightarrow 10^{2} = 100$

$\ln (e^{3}) = 3 e^{3} \approx 20.086$

$\ln$ is just $\log_{e}$ - the same operation, base $e$ instead of base 10.

Decibels (and, briefly, Nepers)

Logs give you two common ways to measure a ratio like this. The one that matters for this post is the decibel. For amplitudes:

$level difference in dB = 20 \log_{10} (A_{2} / A_{1})$

For sound pressure levels, use RMS (root-mean-square) pressure rather than peak pressure.

The neper is worth mentioning as well, which does the same job using the natural log instead of log base 10:

$level difference in Np = \ln (A_{2} / A_{1})$

Both are useful for describing attenuation. With a constant attenuation coefficient and no geometric spreading, amplitude decays exponentially with distance:

$A (z) = A_{0} e^{- α z}$

where $A_{0}$ is the starting amplitude, $α$ is the amplitude attenuation constant, and $z$ is distance traveled. Sound spreading out from a source also loses amplitude through geometric spreading, which this equation doesn't include.

Divide both sides by $A_{0}$ and take $\ln$ :

$\ln (\frac{A (z)}{A_{0}}) = - α z$

As a sanity check: if an amplitude drops from 10 to 3.68, the starting amplitude divided by the ending amplitude is about $e$ , so:

$loss in Np = \ln (10 / 3.68) = \ln (2.717 \dots) \approx 1$

A reduction in amplitude by a factor of $e$ is, by definition, a loss of 1 neper. In decibels, that same drop is:

$loss in dB = 20 \log_{10} (10 / 3.68) \approx 8.69$

We're expressing the loss as a positive number here; the level change is negative.

So $1 Np \approx 8.69 dB$ - two units measuring the exact same ratio, just with a different log base and scaling.

Decibels use a 20 rather than a 10 in front of the log when comparing amplitudes because dB was originally defined in terms of power. When power is proportional to amplitude squared with the same proportionality constant, $\log (x^{2}) = 2 \log (x)$ , so that 2 gets folded into the constant, turning 10 into 20.

A decibel is one-tenth of a bel - $1 dB = 0.1 B$ - and a bel works out to $\frac{1}{2} \ln (10)$ nepers, about $1.151 Np$ .

Actually using dB

So imagine someone tells you their concert measured 100 dB SPL versus their usual show at 70 dB SPL. You may want to know how much more power is actually hitting their ears.

That's a 30 dB difference. Assuming the same acoustic conditions and receiving area, plug it into the power-ratio formula:

$level difference in dB = 10 \log_{10} (P_{2} / P_{1})$

$30 = 10 \log_{10} (P_{2} / P_{1})$

$3 = \log_{10} (P_{2} / P_{1})$

$P_{2} / P_{1} = 10^{3} = 1000$

A 30 dB increase is 1,000× the power. Easier things to remember:

10 dB = 10× power.
Every 3 dB approximately doubles power.
0 dB → 1×
3 dB → ≈2×
6 dB → ≈4×
9 dB → ≈8×
10 dB → 10×

The additive property of dB is kinda why everyone likes it: successive gains and losses multiply as ratios but add in dB. Pretty simple to work with attenuators etc.