Crash course to decibels

First of all: decibels would be easy if they weren’t based on human perception.

Why decibels exist

Decibels are the best way of representing a change in sound pressure level, by a number, as perceived by humans. Decibel is also not an absolute value. To make things even worse, there are “different types” of decibels, or at least their zero-point changes between use scenarios. For this article we’ll stick with a relative dB. You’ll be able to read about the other uses of decibel in an upcoming article.

Stop the mathness

Fair. No one likes maths anyways. Why do we need such unnecessary calculations just to get a simple number out of a value? Because, as we learned, decibels represent our perception by using mathematics. And because we sense in a way that essentially makes no sense, the sensation is represented on a logarithmic scale. Logarithmic scale is the one where equal change between two points is smaller depending on how “high” up the scale we are. For example: $log(2/1)≈0.3$, but $log(3/2)≈0.18$. (Because you’re wondering what’s the difference, in real world, the difference between 1 and 2, and 2 and 3 would be equal — in logarithmic scale it’s not).

In the digital domain, an amplitude is represented, when normalized, with values between -1.0 and 1.0. These values represent what would be voltage on an analog circuit. If we never needed to listen to the audio we’re working with, we would never need to convert it to a logarithmic scale. But as soon as you start, for example, mixing a song, and the artist says “can you turn that down by 0.2?”, you wouldn’t know what that maps to in real world. Sure, you’d know the change in amplitude, so you know what the computer is doing, but you wouldn’t know whether the change is 0.8 dB or 21 dB. Put simply, decibels describe the perceived change in loudness. This matters because doubling perceived loudness doesn’t mean doubling voltage. It means doubling power, and power and voltage don’t scale the same way.

The mathness ensues

Let’s get back to numbers, since they seem to be the only point of truth in this whimsical world. Let’s dig deeper to the multiplier first: $2\times 10\times log(n)$. Why $2$ and why $10$?

The ten is there for making the unit the right magnitude. What we essentially get from the $log(n)$ calculation, is in bel, named after Alexander Bell, who wanted to have a unit for describing the signal loss over telephone circuits, and of course he had to name it after himself, because why not? To make it deci, we multiply it by ten. Put simply: the formula gives you the right answer with or without $\times 10$, but to make the units more workable, it’s often preferred to put the magnitude conversion right into the equation.

Two, on the other hand depends on what we’re trying to solve. For example, if we were calculating acoustical energy addition, we would use $\Delta dB = 10\times log({80\over{80}})≈3$. Note that we are not multiplying the result of the logarithm function by two, or twenty. This is because power is proportional to voltage squared. Or, put fancy, $P \alpha V^2$. This comes directly from the Ohm’s law, but first we need to understand what power actually is.

The universal definition of power

We know from the Ohm’s law, that $V=IR$. Rearrange to solve for current, we get $I={V\over R}$. Essentially, current ($I$) is the product we want from using electricity, and we know to get $I$, we need to $VR$.

To make that product $I$ tickle our senses, we are faced with power ($P$). To get there, we do the following: $P=V\times ({V\over R})$ (because we know $I$ is just $V\times R$). To simplify this, we can do the following: $P={V^2\over R}$. Now we know that to make $2P$, we need $\sqrt 2\times V$ and to make $4P$, we need $\sqrt{4/1}\times V$, which simplifies to $2V$, or as we saw it earlier, $V^2$.

In practice

Assume the voltage is 2 V and that gives us 10 W of power. To get 20 W of power, we would need $V = \sqrt {20/10}\times 2 ≈ 2.83$.

Back on track

All of this matters for audio calculations since the key concept with the proportionality between power and voltage is essential for understanding what’s actually happening when we “increase the volume by 3 dB”. Now you can say, that to get your speakers 3 dB louder, you need to give it $2P$ more power, or $\sqrt{2/1}\times V$ more voltage.