# Do We Really Need Another Article on the Decibel?

Last year I attended the Audio Engineering Society convention in Los Angeles. As I looked at the schedule of lectures and presentations, I rolled my eyes when I saw there were two one-hour lectures on the decibel.

Surely at a convention of audio engineers the information presented would be ground well covered by countless books and articles.

A few weeks later however I started thinking about the textbook definitions and even conversations with other engineers that showed there still is a bit of confusion about all this.

I then decided that yes, there does need to be one more article about the decibel.

### The Definition (Really?)

In a computer programming course I took a few years ago we were studying the time it took for a program to complete its computations, which, depending on how it’s programmed, can be described with logarithms. In introducing the concepts, the teacher surprisingly told the class “I have a Ph.D. in computer science and a master’s degree in mathematics, but logarithms have always made me uncomfortable.” So when I first read the Yamaha Sound Reinforcement Handbook by Gary Davis and Ralph Jones and found myself confused by what seemed like a simple definition for what a decibel is, I figured there must be something they knew that I didn’t. The following is what confused me.

We start with the definition of the Bel, the logarithm of an electrical, acoustic, or other power ratio.

That seems straightforward, but now we want to define the decibel. The Yamaha Sound Reinforcement Handbook and, to be fair, many other engineering texts I’ve seen, define the decibel as below:

This looks very familiar, and we’ve all taken the above expression on faith. But when we look at it a bit more closely, does it hold up? The prefix “deci” means 1/10, so when we affix that term to the Bel on the left side of the first equation, we are dividing that term by 10. Those who have learned elementary algebra know that whatever one does to one side of an equation must be done to the other side for the expression to remain true. So if we divided the left side by 10, why do we multiply the right side by 10?

In what I believe is one of the worst statements I’ve ever read in a technical publication, the authors of the Sound Reinforcement Handbook state: “If you are not familiar with logs, it may not make much sense to you that the right half of the Bel equation is multiplied by 10 to get a value that’s supposed to be 1/10 of a Bel. This is correct, though.”

In one fell swoop, the authors offer an incorrect definition, then stiff-arm the confused reader by telling him or her to just accept it. In reality, almost every text shows the definition of dB this way, but that doesn’t mean it’s accurate or clear. So what’s the problem? Let’s start with the original definition of the Bel and work from there.

Now, let’s divide just the right hand side by 10.

But we know we can’t arbitrarily manipulate one side of the equation. We either have to perform the same operation on the other side, or neutralize our action. We divided the right hand side by 10, so let’s undo that by multiplying the right hand side by 10.

Now, knowing that “deci” is the same as dividing by 10, let’s make that substitution.

The above expression is subtly but fundamentally different than that given in textbooks. The equation given in all of these texts comes from the fact that one can verbally define the decibel as 10 times the logarithm of the ratio of two power quantities. But when authors decide to format that statement as a pseudo-equation, which is what (1) is, the result is sloppy math and confusing writing.

So why do we even bother with the decibel? What was wrong with the Bel in the first place?

Scott Wilkinson, in his book Anatomy of a Home Studio, wrote “…Bels ‘compress’ power-ratio values too much to be useful in audio circuits. As a result, audio engineers use the decibel…”. Using Bels would almost always result in numerical values less than 10, often less than 1, in day-to-day calculations. Bumping the unit down by a factor of 10 let’s us increase the numbers by a factor of 10, and those numbers are then easier to deal with. In the end, it’s just about convenience.

### Half: -3 dB or -6 dB?

Within my favorite DAW, Cubase, there’s a Gain function that all DAWs have. When you apply the process, a box appears in which you can type in by how many dB you want to increase or decrease your signal. What Cubase shows that most other audio software programs don’t is a raw scale factor displayed in percent. So, for instance, with a 0 dB adjustment, the scale factor is presented as 100%. This makes sense, as adjusting a signal by 0 dB, also known as unity, is the same as multiplying the signal by 1. When I enter a 6 dB reduction (actually 6.02 dB), Cubase tells me this will be a 50% reduction. In other words, the ratio of the output to the input of the process will be ½. Let’s put this into our dB formula.

Crap. Now what’s wrong? We went through all that trouble of meticulously defining the decibel, and now Cubase is telling us something’s up. By our calculation a 50% reduction in level should be a 3 dB drop, but Cubase is saying it’s 6 dB. The key to this apparent contradiction lies in one word in the definition of the decibel: power.

The decibel is a ratio of power levels. Power isn’t a property that can be directly measured, but is calculated using properties that can. In electrical engineering, voltage, amperage, and resistance are measured and can then be used to determine power used in a system.

Let’s take a look at how this is done, understanding that the digital samples in our DAW represent the analog voltages in an audio signal.

For a gain function, we then have an input voltage Vin and an output voltage Vout. For the value of the resistance, we only have to mandate that the resistance is the same for both Vin and Vout. We can now plug (8) into (6):

We see that the resistance term, R, cancels out. The ratio of the square of the two voltage terms is the same as the ratio of the two voltage terms squared (read that while looking at the third and fourth terms in the series above). Lastly, by the property of logarithms, the value of any exponent of the argument (in this case, Vout/Vin) can be moved to the outside of the logarithm and used as a multiplier.

What this tells us is that when we’re dealing with voltage or current values in the analog domain or sample values in the digital domain, we have to use the multiplier 20 instead of 10 in order to use the decibel equation as it was intended: for power values. When we do, we see that all is right in the world and that Cubase was correct, of course.

### Conclusion (aka So What?)

If you’re reading this section then you just finished a lengthy article on the definition of a term you’ve probably had no trouble using for years. Will what you know now make your mixes sound better? Probably not. Do you now have golden ears? Nope. Then why did I write this article? I wrote it because the decibel is a fundamental term in audio engineering, and having a deep understanding of it enables you to understand other concepts such as gain, power, signal flow, and more. Beyond this, the practice of learning terms and ideas at their most fundamental and basic level gives us the power to make better, more informed decisions. When we take information presented to us on faith, we give up our ability to discern fact from fiction. We let those with more opinions than sense, or those with an agenda, to dominate the studio, the lab, and online forums. As engineers, these concepts are our tools, and we should learn them inside and out.

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Barak Shpiez has earned several degrees in composition, audio engineering, and electrical engineering. His music can be found on programming on MTV, The History Channel, and more. Barak has also worked for several concert venues running live sound and as an engineer for Line 6 and DTS.

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• Pierandrea Neonbug Petazzi

Thank you 🙂 Though it won’t make my mixes better, as you pointed out at the end, I confess that I too, notwithstanding a phd in physics, hadn’t really understood it that well until now. Thank you again

• Tyler Arndt

Good read, very well layed out. I’ve certainly mixed up doubling your distance to a sound source being -3db and that -6db is half the power ratio

• patandrae

Sir:
I’m a chemist by trade and I’ve had courses in math, physics, etc. I’m just as confused about your definition of the decibel as I would no doubt be confused by the textbooks you describe. It might have an easy explanation. I’m not sure. But in attempting to clarify the issue, you have shown equations that make little sense to me.
You give the definition of the decibel in equation 1. You then “return to the original definition” by introducing equation 3. Equations one and three look different to me. What is the “Bel” on the right side of the equation doing there? Is it a term in the equation? If so we can factor out Bel and obtain: 1 = log10(P1/P2). This seems quite wrong. If we take equation 1 and substitute 10 decibel = 1 Bel, we get 10 deciBel = log10(P1/P2) -or- deciBel = (1/10)log10(p1/P2). The math is right but the definition is wrong. You apparently understand the math but your definition and syntax of your equations are not clearing it up for me. Any way of explaining it another way?

• patandrae

OKAY! NOW I get it. And you’re correct. The definition of decibel (and Bel) as traditionally given, appear to me to be incorrect. Equation one should be: The DIFFERENCE in BELS between the speakers with power 1 and power 2 is: log(P2/P1) Plug in some values and obtain, for example: DIFFERENCE IN BELS = 30 BELS How many decibels is 30 Bels? 300 decibels. If you are to obtain the answer of 300 decibels, you need to multiply your equation by TEN. Ergo: the DIFFERENCE in DECIBELS of our two speakers = 10 log(P2/P1) = 300. It seemed clear to me once the definition is made clear and putting “dB =” on one side is misleading at best.

• patandrae

If you use the actual power ratios (say 50W/100W) you get -3.01. Wrong answer. So we use the equation you’ve shown with voltage values. WELL, the voltage ratio, if I’ve correctly calculated, would be approximately .707. The log is -.150 and multiplying by 20 gives -3.01. We will get the cubase answer only if it is the voltage that is multiplied by 50%, not the power (which would be using it as it was intended). Am I right? Tell me where I went wrong? No one will answer probably.

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