43

Prior to the invention of electronic calculators, multiplying and dividing long numbers was painful. Yet that was something engineers (including audio ones) had to do often.

Logarithms were invented in the 17th century to simplify calculations.

We have these fundamental rules of logarithms:

log(X * Y) = log(X) + log(Y)

And

log(X / Y) = log(X) - log(Y)

While rather innocent looking, they conceal great power: The ability to replace complex multiplication and division calculations with simple addition and subtraction ones.

There is a catch though - you needed to know the log value of numbers. But as the benefit outweighed the pain, soon books with log charts appeared showing the log value of numbers (my dad had one!).

So to multiply two long numbers, you find in the chart the log of both first, add them, and then search for the inverse log of your addition result.

Logarithmic rulers soon appeared, allowing quicker calculations:

With advances in electronics and audio recording technology it soon became clear that our perception of loudness is *exponential*, rather than linear.

That is, when reproduced via a speaker, the loudness change from 1V to 2V is not perceived as equal as the loudness change from 2V to 3V. The doubling of voltage is much closer to our loudness perception - 1V to 2V is roughly like 2V to 4V.

Log functions are exponential.

Now guess what? The engineers at Bell Laboratories decided to make the log an integral part of audio. They gathered what you may have gathered by now:

- Logs make calculations easier.
- Logs lend themselves better to the exponential nature of our hearing.

So they simply devised a unit called *Bell* (in honour of Graham Bell), which is nothing but a ratio between two measurements, logged:

Bell = log(m1 / m2)

Soon though, the realisation came that for audio applications the Bells always had an extra significant digit to the right of the decimal point. So they introduced another unit called *decibel*, which, after working out the maths of the original definition of the bell, resulted in:

dB = 10 * log(m1 / m2)

The decibel definition was devised for power measurements (watts), so we can look at it as a ratio between two power measurements:

dB = 10 * log(p1 / p2)

Once we throw into the equation voltages (where P = V^2 / R) and simplify it, the equation becomes:

dB = 20 * log(v1 / v2)

Which is most often written:

dB = 20log(v1 / v2)

Practically speaking, all dB equations are 20log, barring those dealing with power.

So far, dBs allow us to express the ratio between two measurements. But what if we measured the level of our signal and ask:

"How loud is this in dBs?"

The answer will be:

"Well, how loud compared to what?"

So the idea was to choose a reference, so we can express the level of something in dBs by measuring it and comparing it to a standard reference. In equation form:

dB = 20log(m / r)

So now we can develop various systems of dBs, each uses a different reference.

- The dBm system uses 1mW as a reference.
- The dBu system uses 0.775V as a reference (a number that was common in early electronic circuits).
- The dBV system uses 1V as a reference (a much easier number to deal with).

When digital audio emerged, it needed its own dB system, which brought the need to choose a reference. But there was a problem:

Digital audio can be 8 bits (sample values from 0 to 255) or 16 bit (0..65535) or 24 bit (0..16777216).

It seems like the common thing in all these bit depths is the zero, but we can't use it as a reference as we'll get:

dBFS = 20log(m / 0)

While this is sufficient to rule out 0 as a reference, there is another fact to consider: We know that during ADC a higher bit depth extends the dynamic range downwards not upwards. In other words, the loudest analog voltage an ADC supports always translates to the highest value the bits can represent. If the analog limit is 1V, in an 8 bit system it will get a sample value of 255, but in a 16 bit system it will get 65535.

This make the choice rather obvious - the reference for dBFS should be the highest sample value of the system bit depth. So:

dBFS = 20log(sample value / highest possible sample value)

Now in logs, if the numerator is smaller that the denominator, you always get a negative result; if the two are equal, you get 0. So in the case of 16 bits:

0dB = 20log(65535 / 65535)

and

-6dB = 20log(32768 / 65535).

Since the sample value can never exceed the highest sample value, dBFS values are always equal to, or smaller than 0.

All of this is sound, but there's a bit more.

As users, we don't really care about the dB level of the signal within the system, we care for the dB level compared to the standard operating level of the system (above which you may clip).

So audio meters don't show the dBu, dBV or dBFS level of the signal (some do, but as an extra). Instead they show 0dB as the standard operating level of the system. In analog this is known as 0dBVU (dB Volumatric Unit), and for digital as 0dBr (dB reference). For pro analog equipment 0dBVU is calibrated to +4dBu; for semi-pro equipment it is -10dBV. So regardless of the device itself, and the dB system used, dBVU always tells me how much below or above the standard operating level of the system the signal is.

DAWs are using 32 bit float internally, so the integer range of 0..65535 is aligned to the decimal range `0`

..`1.0`

; While 65535 is 0dBFS, `1.0`

is 0dBr. But floating point systems can go above `1.0`

, in fact, much higher than `1.0`

. Yet as at some point you'll have to go back to integers, the 0dB you see is aligned with 0dBFS.

10

What is 0 dB in digital audio?

0dB does actually represent the largest signal that a digital system can produce. Once a signal is in the digital domain it takes on a life of its own and 0dB remains full-scale. If you took a digital audio file (maybe from a CD or wav file) and loaded it onto another machine, the new machine would see the same binary numbers as the original machine so 0dB represents the biggest value numerically in the "digital world".

But what if you have different quantization levels i.e. 24bit instead of 16bit (CD audio)? Well, the biggest unsigned peak number on 16 bits is 65,536 and on 24 bits it is 16,777,216 BUT, for each 0dB is taken as their maximum value. The 24 bit "system" will have greater depth of clarity because it is representing an analogue signal with over 16 million digital levels whereas the 16 bit system is only representing an analogue signal with a bit over 65,000 digital levels.

what are the decibels being based on?

The decibel is one-tenth of 1 bel and years ago in Bell laboratories (clue in the name) they conducted listener tests to see how the average listener perceived sound levels of different amplitudes with loudness variations being introduced by longer cable lengths. The following quote is taken from **Wikipedia**

The decibel originates from methods used to quantify reductions in audio levels in telephone circuits. These losses were originally measured in units of Miles of Standard Cable (MSC), where 1 MSC corresponded to the loss of power over a 1 mile (approximately 1.6 km) length of standard telephone cable at a frequency of 5000 radians per second (795.8 Hz), and roughly matched the smallest attenuation detectable to the average listener. Standard telephone cable was defined as "a cable having uniformly distributed resistance of 88 ohms per loop mile and uniformly distributed shunt capacitance of .054 microfarad per mile" (approximately 19 gauge).[4] The transmission unit (TU) was devised by engineers of the Bell Telephone Laboratories in the 1920s to replace the MSC. 1 TU was defined as ten times the base-10 logarithm of the ratio of measured power to a reference power level.[5] The definitions were conveniently chosen such that 1 TU approximately equaled 1 MSC (specifically, 1.056 TU = 1 MSC).[6] In 1928, the Bell system renamed the TU the decibel.[7] Along with the decibel, the Bell System defined the bel, the base-10 logarithm of the power ratio, in honor of their founder and telecommunications pioneer Alexander Graham Bell.[8] The bel is seldom used, as the decibel was the proposed working unit.[9]

+1dB means it got louder by about 10%. -1dB means it got quieter by about 10%.

From other experiments it was determined that the human ear works logarithmically because each time a doubling of perceived sound level occured, the associated instrumentation showed that the power of the signal increased by ten times. The upshot of this is that a 10W amplifier/speaker is twice as loud as a 1W amp/spkr and that a 100W amp/spkr is four times louder than a 1W amp/spkr and twice as loud as the 10W rig.

So in engineering, 10dB now represents a ten-fold increase in the power of a signal.

In electronics, engineers use the term 0dBm and this refers to the power of 1mW. Other commonly used terms are dbuV where the u means micro and the V is voltage.

The 1929 paper from Bell Labs devising the decibel mentions nothing of the mentioned listener test. Nor am I aware that the 10dB as perceived doubling was researched until much later. The Fletcher-Munson curves, for instance, were only published in 1937. It'll be great to see a reference to the listening test mentioned. – Izhaki – 2013-09-13T04:03:01.760

@Izhaki - in case my schooling was wrong or I've become wrong over the years I've replaced the section you focussed on with a wiki quote. Bell labs did do average listener tests though. – None – 2013-09-13T10:37:29.010

1"0dB does actually represent the largest signal that a digital system can produce." No, "0 dBFS" does. "0 dB" is a relative gain. "0 dBFS" is an absolute level. – endolith – 2016-08-10T15:02:51.657

The question asked was what does 0 dB reference to. I just copied and pasted the words the guy used asking the question. I think it's clear enough. – Andy aka – 2016-08-10T15:08:56.947

6

Since the dB is a ratio as you say, "0 dB" just means "no change". In the context of digital sampling, I'd consider 'full scale' to be the reference point -- i.e. 255 in an 8-bit system -- making anything less a negative dB value. But unless it's specified, (as in dBa, dBm etc) it's just speculation and open to interpretation.

There may be a DWS standard or consensus, but that's not my world so I can't say.

5

It is the amount of change. 0dB is full signal, negative dBs are a reduction in signal (down to -infinity which is fully off) and +dB are when a signal has a gain applied. 0dB is also sometimes called Unity. This is consistent across digital mixing and analog mixing where anything below 0dB is reductive (the source signal is reduced) and anything above 0dB (when the board or software supports it) applies a gain to the signal via a pre-amp (in analog processing) or increasing the amplitude by the selected amount(in digital).

When metering, 0dB is often the either the maximum or ideal capability of the hardware or digital signal space you are working with. In the case of setting it to the optimum strength, there will be some headroom above the ideal signal level before hitting clipping. You should check documentation for the particular board or system you are using for the specific implementation.

2

Going to next level…

Sound pressure level (SPL), in dB, uses the threshold of human hearing as the reference level (denominator): A fluctuating pressure of 2x10^-5 Pascals. I’d have to dig for the citation, but that reference was determined by averaging the response of healthy individuals (ears) back in the 1930’s(???). So, 0 dB is the threshold of human hearing.

Sound power level is referenced to 10^-12 watts – the power that would produce 0 dB SPL when distributed over a sphere with a surface area of 1 m (r = 0.283 m)

1Yes, so basically, 0dB is the 'relative' reference point. If the dB unit has a suffix, like SPL or A, it is known as 'absolute' because the suffix indicates the reference point. **Note**: With SPL, you have to also indicate at what **distance from the source**, the level is measured. – Marc W – 2015-11-17T04:10:55.313

1@MarcW Yes, you need to specify distance if the source is a point source that gets quieter as you move away from it. If it's an ambient level in a room where it's the same everywhere, then you don't need to specify distance. The important thing to understand is that the sound pressure is measured at a point in space, so you need to specify "what point?" The SPL in the horn of a trumpet is different from the SPL 1 mile away. – endolith – 2016-08-10T15:20:52.760

2

Lots of lengthy answers here when all that is needed is a simple answer.

The 0dB that you refer to is actually 0dBFS which stands for dB with reference to "Full Scale".

0dBFS is the highest peak digital sample level. Anything below this is normal signal, therefore shown as a negative number.

-20dBFS is 20dB Below Full Scale.

1

It depends.

With a logarithmic scale like the decibel (or bel; 1B = 10dB), you usually need a reference point, and this reference point cannot be zero (otherwise the math involves division or multiplication by zero which doesn't get you anywhere). This reference point depends on what you're measuring. In audio, that's usually one of the following:

Voltage (potential): 0dBV = 1V, regardless of impedance. 0dBu = .7746V (the voltage inherent in a 600-ohm conductor dissipating one milliwatt of power). Home-quality audio equipment typically has a peak signal power handling of -10dBV (.316V), which itself is often marked as "0dB", while professional audio is typically calibrated to a much more powerful +4dBu (1.228V)

Wattage (power): 0dBm = 1mW. This is typically of value in wireless transmitters/receivers, to judge signal amplitude.

Sound Pressure: 0dBSPL = 20uPA (= .00002N/m^{2}). This reference point was based on one series of tests to determine the threshold of silence in human hearing. Other studies say humans with healthy auditory systems may be able to differentiate sounds as faint as -10dB. These curves are calibrated for different kinds of sound environments; OSHA has two common "weightings", A and C. A is based on the average spectrum of industrial noise, while C is based on the spectrum common in amplified audio. Exposure limits are officially defined for and measured using A-weighting.

0

It's really simple. It's the inverse ratio. So, 0 is no less. -3dB is the ratio for 3 dB less than 0.

Decibels are used to describe a ratio, rather than a definite amplitude. To figure out negative decibels, you do the same thing you'd do with positive decibels, except with a negative number.

To create a 3dB increase, you'd multiply the sound's power by 10^(3/10), or 1.9952623149688796013524553967395 (approximately twice the original amplitude).

To create -3dB change, you'd multiply the sound's power by 10^(-3/10), or 0.50118723362727228500155418688495 (which is approximately 1/2 of the original sound).

So, in short, it's just the inverse. 3dB is about twice the amplitude, and -3dB is about half of the amplitude.

I'm confused as to why this was downvoted. It answers the question, and it's correct. Granted, it may not answer my question completely, but it does address something that would be too long to fit in a comment. – Cole Johnson – 2013-09-17T02:05:17.670

@Cole Johnson: I didn't vote, but your question was: "what are the decibels being based on?" This is not addressed at all. Additionally, the post's structure is unclear and repetitive. – None – 2013-09-17T08:06:12.497

for one thing, you've mixed up amplitude and power. power is proportional to amplitude squared, so +6 dB increases amplitude by 2× and power by 4×. – endolith – 2016-08-10T15:17:49.433

Really good answer Izhaki and I'm looking into the dB and listener tests to see where I may have got my schooling (many years ago) mistaken. +1 – None – 2013-09-13T10:16:45.390

To be honest, the comment I've left you made me doubt the accuracy of my own statement - that the choice of logs for dBs had something to do with our loudness perception. Although I'm pretty sure the exponential nature of our loudness perception was known at the time of the gramophone, I'm not sure it is something Bell engineers accounted for directly when they devised the Bell and decibel. But I haven't made it up - pretty sure I've read it somewhere, so I'm also checking the accuracy of my answer. – Izhaki – 2013-09-13T10:35:32.713

Well I doubted mine so I've amended it to give a little more historical credibility. Interesting subject though. – None – 2013-09-13T10:41:38.260

2It should be noted that the

FSindBFSstands for "full scale", i.e. dBFS = "decibels relative to full scale". Also, you need some way to represent negative values, so 16-bit audio actually ranges from -32768 to +32767, and dBFS is calculated by first dividing by 32768 (or 32767), not 65535. – None – 2013-09-13T15:48:46.613@chirlu, all your points are valid; as for the second one - sticking to positive numbers makes the explanation easier, it also saves the need to discuss amplitude vs magnitude. – Izhaki – 2013-09-13T16:27:07.303

1ockquote>

"Since the sample value can never exceed the highest sample value, dBFS values are always equal to, or smaller than 0."

Not true. They can exceed 0 when audio is represented in floating-point format. And it is even quite common when decoding some lossy audio files to floating-point (which is required to avoid introducing quantization noise again). – None – 2013-09-15T11:03:42.857

Since 0dBFS integer is mapped to 1.0 in floating point, and in floating point you can represent numbers bigger than 1.0, you can say that "In floating point, you can exceed the number representing 0dBFS in integers". But by definition, 0dBFS corresponds to the highest possible sample value, which in floating point is 3.40282*e^38 (in Binary 0 11111110 11111111111111111111111); This value is 771dB above 1.0 (0dBFS in integer, or 0dBr in your DAW). But you still can't go above that max value. – Izhaki – 2013-09-15T11:53:57.533

It might help to point out that dB = 10 * log(p1 / p2) → 10 * log(v1^2/v2^2) → 2 * 10 * log(v1/v2) – endolith – 2016-08-10T15:01:47.400

1"0dB = 20log(65535 / 65535)" Actually, 16-bit audio is signed, so it goes from -32768 to +32767, so max amplitude is 32768. – endolith – 2016-08-10T15:25:37.973

@endolith I noticed that, but then I also noticed that it was already mentioned in chirlu's comment. – Marc W – 2016-08-10T16:58:52.273

This is a fantastic explanation, thank you so much! But can we get one more bit of info? If 0 dBVU is +4dBu or -10dBV in analogue equipment, what's 0dBVU in my DAW? Or is there no such thing, and the meters in the DAW are actually showing 0dBFS? – Charles Wood – 2019-01-26T16:32:09.007

1@CharlesWood That's a good question. In theory you shouldn't see something like dBVU on a DAW - being a more analog thing. If you do, then most likely 0dBVU is calibrated to 0dBr (ie 0dBFS if you bounce to integer). But, and it's a big BUT, dBVU meters on a DAW mimic the scale and ballistics of those in the analog domain, and too many engineers are terribly accustomed to these because they are everywhere on analog consoles. Being average, rather than peak meters, these reveal the approximate loudness of the sounds as they near your 0VU. Hope it makes sense. – Izhaki – 2019-01-27T23:50:13.977

Regarding multiplying by 10 to change Bells to decibels, this is normal, nothing special about dB. For example, a deciliter is 1/10 of a liter. So 1.5 liters is 15 deciliters. You get the number of deciliters (15 in this example) by multiplying by ten. Same for dB. – Matt – 2020-05-09T13:57:47.127

@Matt You are right. 0.5 metres is 5 decimetres; thus 0.5 Bells is 5 decibels. I'll have to dig deep to find that equation from the original Bell Labs paper - I'm pretty sure in it you'll find that if you divide the power by 10 you get the x10 in decibel equation; but I'll have to dig MUCH deeper to find out if it was that or just a simple application of a deci- prefix. Anyhow, I've deleted the related paragraph and thank you so much for pointing it out. – Izhaki – 2020-05-09T20:59:26.343