I'm playing a pure tone on an iPad through a pair of "standard" earphones. From a programming perspective, the output level is in the range of $0$ to $1$ where $1$ means "full volume".

I've observed using a sound pressure level meter that when playing a pure tone at $1000 \text{Hz}$ with output level in $0$ to $1$ with step size $0.05$, that the relationship between output level and sound pressure level is logarithmic.

However, I'm not sure if this is correct or an issue with my use of a sound field sound pressure level meter (instead of a coupler).

Should doubling the output level double the sound pressure level? That is, should the relationship be linear? It makes intuitive sense to me that as the output level or voltage is increased, the earphone speaker will vibrate more, and thus the sound pressure will also increase.


1 Answer 1


As you say, pressure will be proportional to output level, if the output level is equivalent to voltage. But sound pressure level (SPL) is by convention expressed in the logarithmic decibel (dB) scale, with 0 dB referenced to some standard sound pressure level. The relationship is:

$$\text{SPL} = 20\log_{10}\left(\frac{A}{A_\text{ref}}\right) \text{ dB,}$$

with a reference output level $A_\text{ref}$ that would give an SPL reading of 0 dB. For measures of power such as $A^2$ the formula changes slightly:

$$\text{SPL} = 10\log_{10}\left(\frac{A^2}{A^2_\text{ref}}\right) \text{ dB.}$$

With $A$ and $A_\text{ref}$ non-negative, the equations are equivalent with the square inside the logarithm or outside it as a factor $2$.

  • 1
    $\begingroup$ (The 20 is just the ^2 brought out of the log10.) $\endgroup$
    – endolith
    Sep 28, 2017 at 18:25
  • $\begingroup$ @endolith thaks, incorporated into the answer. $\endgroup$ Sep 29, 2017 at 14:27

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