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Could someone explain in simple terms (let's say, limited to a high school calculus vocabulary) why decibels are measured on a logarithmic scale?

(This isn't homework, just good old fashioned curiousity.)

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see also Sound pressure @wikipedia – Tobias Kienzler Mar 10 '11 at 9:53
For that matter, why do we say "40 deciBels" instead of 4 Bels? I assume Bels are also logarithmic? – barrycarter Mar 10 '11 at 16:29
Ever heard from Decimeters? – Georg Mar 13 '11 at 11:30
up vote 17 down vote accepted

I don't know anything about the history of the Bel and related measures.

Logarithmic scales--whether for audio intensities, Earthquake energies, astronomical brightnesses, etc--have two advantages:

  • You can look at phenomena over a wide ranges of scales with numbers that remain conveniently human-sized all the time. An earthquake you can barely detect and one that causes a regional disaster both fit between 1 and 10. Likewise the stillness of an audio-dead room and the pain of an amp turned up to 11 fit between 10 and 130.
  • Fractional measures are converted into differences which most people find easier to compute quickly. Three decibels reduction is always the same fractional difference; the EEs get a lot of mileage out of this.

These scales may seem very artificial at first, but if you use them they will become second nature.

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+1 "turned up to 11" :) – Ben Mar 10 '11 at 4:15
I believe the unit bel is named in honor of Alexander Graham Bell – Mark Rovetta Aug 4 '13 at 4:40

Human senses, nearly all, work in a manner and obey Weber–Fetcher law, that response of the sense machinery is logarithm of an input. It is true at least for hearing, but also for eye sensitivity, temperature sense etc. And of course, in areas where it works normally. Because in extreme, there are other processes such as pain, etc.

So as in a cause of hearing, what you experience is the logarithm of power of a sound wave, by "biological, natural, hear sense construction. So, it is natural to use logarithmic units.

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This is the correct answer, as far as senses are concerned! +1 – Georg Mar 10 '11 at 10:08

It's just because sounds that the human ear is capable of hearing range over a very large range of amplitudes. If you talked about the power delivered to the ear, rather than the log of the power delivered to the ear, you would need to use numbers like $10^{12}$ to talk about airplane engines. So, rather than deal with that, we use logarithims, so that most of the numbers we deal with when talking about sounds vary over reasonable number ranges.

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Thanks! The other answer was a bit more detailed though. – Ben Mar 10 '11 at 4:17

Not only is the human ear (and other human senses) capable of observing signals over many orders of magnitude, we also perceive these signals more or less on a logarithmic scale. Take the 80 dB of a room full of loud conversation as an example. We don't perceive this as a thousand times louder than the 50 dB of a washing machine, nor a hundred times quieter than the 100 dB of a jackhammer. (Examples from Wikipedia.) Therefore, the decibel scale is not just useful for calculating 3 dB increases or keeping the numbers on human-sized scales, it also approximates the way our senses work.

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Correct me if i'm wrong, but doubling a source gives +3 dB, so (100-80)/3 = about seven times as loud, not a hundred. See… – Cees Timmerman Dec 12 '13 at 10:58
That's incorrect, the decibel scale is logarithmic: +10 dB gives 10x the sound pressure (and +3 dB gives about 2x, as you say) so +20 dB (100-80) gives 100x. In other words, $2^{(100-80)/3} \approx 100$. – ptomato Dec 13 '13 at 4:39

Human ear can detect 10^13 units of intensity. By using logarithmic scale, you get a scale of 130 db.

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It's a historical accident which has left us a lasting pain in the ass. There's no reason to express sound as decibels: writing the pressure level in Pascals in scientific notation is just as convenient as writing a level in decibels relative to some reference pressure level. And it's often MORE convenient for calculating certain things (i.e. a noise floor expressed in pascals-per-root-Hz makes sense; dB-per-root-Hz or dB-per-Hz makes no sense)

That being said, there are a few possible reasons for using dB:

1) For folks that don't know scientific notation.

2) To match human physiology/psychology, since human perception of sound has a roughly logarithmic response to the pressure level.

3) For attenuators or amplifiers, you can just add things (this is more common when dealing with RF). If I put a 10 dB attenuator in series with a 5 dB attenuator I get a 15 dB attenuator. And if I put a +2 dB power level of RF into a 15 dB attenuators, I get a -13 dB power out. And even though I don't like dB, I do have to admit that's pretty convenient.

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""It's a historical accident which has left us a lasting pain in the ass."" Right, the much more intelligent Neper scale based on Napiers (natural) logarithms, did not florish. – Georg Mar 11 '11 at 20:01
Yet, "dBc/Hz" is a common measure of sideband power. – nibot Mar 11 '11 at 20:29
@nibot Is it? My understanding is that dBc/Hz is typically used as a measure of phase noise. It's nonsensical as a measure of amplitude (or power) noise because I can't get the noise in dB by multiplying by the bandwidth in Hz (due to the logarithmic nature of the dB scale) – Anonymous Coward Mar 11 '11 at 21:54

While most of the answers received so far emphasize the use of decibels as a measure of the loudness of sound, it is important to note that dB are most commonly used in RF engineering and can be used in describing any wave phenomenon that transports energy.

The utility of dB comes from two properties:

  1. The logarithmic scale applies conveniently to situations where you have a huge dynamic range. For example, in an RF system, you might have signals with powers ranging from picowatts to megawatts.
  2. Secondly, the logarithmic scale converts multiplication to addition, which is convenient.

A quantity in dB always represents a ratio. An absolute power level in "dB" is always given with respect to some reference value. For example, signal power in RF engineering is often given in "dBm", which are decibels with respect to 1 mW:

$$ \mathrm{[dBm]} = 10\ \log_{10} \frac{P}{\left(1 \mathrm{\ mW}\right)}$$

(Sadly, the "m" in "dBm" refers to "mW" !)

Now, suppose an initial signal with power 7 dBm is followed by an amplifier with gain 2 (approx 3 dB). The power at the output will be 7 dBm + 3 dB = 10 dBm.

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