Unpacking the Decibel

Jonathan South Fundamentals July 17, 2019

Once you’ve dealt with them for a while, working with values in decibels (dB) becomes second nature, and warrants little further thought. Sometimes though, the process of getting to that stage can pose a few hurdles, some of which I was exposed to recently via an undergraduate course I had to re-take as part of my Masters. This isn’t intended to teach you everything about the decibel, but maybe clear up a bit of background an nomenclature.

It’s all in a name

I think once you see it, it’s a bit of a groan moment – the seemingly arbitrary mathematical definition of the decibel becomes glaringly obvious.

So Bell Labs – a pretty famous research laboratory, founded by a pretty famous guy Alexander Graham Bell – obviously had a lot to do with early telephony. One of the many reoccuring values of interest in these studies was measurements of power, such as electrical power. The researchers found that even within the one context, measurements they were taking had a HUGE range of values, differing by many orders of magnitude. So naturally, logarithms were used to “tame” some of these values into a more usable and comparable (luckily, human response is also roughly logarithmic).

A bel is the name they gave to the base 10 logarithm of a power-like quantity (I, here), scaled against some reference value. (Don’t forget, you can only take the logarithm of a dimensionless quantity).

$\log_{10} \left(\frac{I}{I_{ref}}\right)$

So that’s the bel, how about the deci? Well, what’s a milli to a meter? A micro to a gram? Your knowledge of metric prefixes should give you a deci-bel equal to

$10 \log_{10}\left(\frac{I}{I_{ref}}\right)$

We defined I as a power-like quantity – perhaps intensity or power – but what about other quantities? Typically we’ll be working with scalar amplitudes such as pressure or voltage, which relate to their power-like counterparts by the square of said amplitude. For example, with sound pressure

$\begin{aligned} SPL &= 10 \log_{10}\left(\frac{p^2}{p_{ref}^2}\right)\\ &= 2 \times 10 \log_{10}\left(\frac{p}{p_{ref}}\right) \\&= 20 \log_{10}\left(\frac{p}{p_{ref}}\right)\end{aligned}$

This is the other decibel “equation” you will see around commonly, and often the difficulty arises in knowing intuitively whether to use to 10x form or the 20x form. Remember – they stem from the same base definition of a bel, and the extra factor of 2 arises from some specifics of the measurement quantity.

Quick Hints

Some quick hints you should know by heart

Scaling of amplitudes via multiplication implies addition/subtraction of decibels, this is known as gain (sometimes a decrease in gain is referred to as attenuation)
10 fold increase/decrease in amplitude scale results in a $\pm$ 20dB of gain.
2 fold increase/decrease in amplitude scale results in a $\pm$ 6dB of gain. $\left(20 \log_{10} \left(2\right) \approx 6\right)$
Doubling/halving the number of sound sources changes SPL by $\pm$ 3dB $\left(10 \log_{10} \left(2\right) \approx 3\right)$
From the inverse square law, sound intensity of a point source remains constant over the surface area of a sphere, hence any doubling of distance from a point source results in SPL decreasing by 6dB $\left(10 \log_{10} \left(\frac{W}{4 \pi (2r)^2} \frac{4 \pi (r)^2}{W}\right) = 10 \log_{10} \left(\frac{1}{4}\right) \approx -6\right)$
There are similar arguments for infinite line sources, where sound power remains constant over the surface area of a cylinder, hence any doubling of distance from a line source results in SPL decreasing by 3dB $\left(10 \log_{10} \left(\frac{W}{k (2r)} \frac{k (r)}{W}\right) = 10 \log_{10} \left(\frac{1}{2}\right) \approx -3\right)$
As you get very far away from a physical sound source, they always start to look like a point source.
Real sound sources are usually somewhere between a point, a line and a plane, so you can use the above for back of the envelope approximations!