# Decibels, acoustic intensity and sound perception

I know the formula: $$I[dB] = log_{10}\left(\frac{I}{I_0}\right)$$ with $$I$$ being the acoustic intensity in $$[W/m^2]$$ and $$I_0 = 10^{-12} [W/m^2]$$ being the human threshold of hearing. Also, I am told that "the sound intensity is doubled every 3 dB". But since this measurement is the logarithm of a ratio, I don't get it:

• What physical value is doubled exactly?
• Is it related to the "sound perception" that we have? If it is, I find it strange because I really don't have the feeling that the sound is "twice as loud" when I set my speaker to 3 more dB.
• If the answer to the previous question is no, what is the actual relation between that formula and the "feeling of loudness"?

What physical value is doubled exactly?

It is the sound intensity, as in the quote. The sound intensity is a specific physical quantity, representing the power carried along by the sound wave per unit area, measured in watts per meter-squared (W/m$$^2$$). The way to think about this is that if there is a "spherical" speaker emitting sound waves, and the amount of energy emitted per second is $$P\times1$$ second (so that $$P$$ is the power), then that amount of energy gets spread out over a larger and larger sphere, centered on the source, as the wave moves away from the source. Dividing the power by the surface area of that sphere gives a sort-of "density" of power. This quantity is relevant because it is the one that determines how "loud" a sound is.

Note that if we increase the intensity by a factor of 2, then $$\textrm{SIL}_{\textrm{new}} - \textrm{SIL}_{\textrm{old}} =(10\,\textrm{dB})\log_{10}\left(\frac{I_{\textrm{new}}}{I_0}\right) -(10\,\textrm{dB})\log_{10}\left(\frac{I_{\textrm{old}}}{I_0}\right) =(10\,\textrm{dB})\log_{10}\left(\frac{I_{\textrm{new}}}{I_{\textrm{old}}}\right) =(10\,\textrm{dB})\log_{10}2\approx3\,\textrm{dB}\,.$$

Is it related to the "sound perception" that we have? If it is, I find it strange because I really don't have the feeling that the sound is "twice as loud" when I set my speaker to 3 more dB.

It is related to the sound perception but in a complicated way. "Doubling the intensity" does not translate into "perceived as twice as loud". I'll put the basics below.

If the answer to the previous question is no, what is the actual relation between that formula and the "feeling of loudness"?

So, here we go.

The human hearing apparatus does not respond the same to all frequencies. It is easier for us to hear sounds at frequencies near 1000 - 5000 Hz than it is to hear sounds with frequencies significantly lower or higher than this. An accumulation of experiments have lead to a set of standard equal loudness curves, shown below (from Wikipedia).

The way to understand the figure is as follows. The frequency of a pure tone is on the horizontal axis, and the sound pressure level (viz., sound intensity level or decibel level) is on the vertical axis. A single point in the space therefore represents a particular pure tone played at a particular frequency and particular intensity. The red curves represent notes that humans perceive as being the same loudness. (You can do this exercise yourself.) What the curves show is that if you have a low-frequency sound playing, you have to play it at a higher intensity than a sound played at, say 1000 Hz. Similarly for the high-frequency sounds.

The red curves are also known as equal-phon curves: we come up with a new quantity, called the phon, which acts like the decibel scale but takes into account this difference in the frequency response of our hearing apparatus. The phon scale is set so that a change of 10 phons equals a change of 10 decibels (roughly) for 1000 Hz sounds.

Now, this gets us closer to the subjective experience of sound, and the final step is that studies also show if a sound increases by 10 phons, than we perceive that sound as being twice as loud. So finally, there is the sone scale, which is a scale that doubles every time the number of phons increases by 10.

So! - the relationship between the bare physical facts of the sound wave---i.e., its intensity, which is in a sense a measure of how "spread-out" the energy is in the wave---and the subjective experience of the loudness of the sound is somewhat complicated. But, we can trace back through all of the relationships above to make the following statement:

Roughly speaking, for sounds played at 1000 Hz, if the intensity increases by a factor of 10, we perceive the sound as being twice as loud. (This is because doubling the perceived loudness corresponds to doubling the number of sones, which corresponds to increasing the phon level by 10, which corresponds to increasing the dB level by 10 (because the sound is at 1000 Hz!), which corresponds to increasing the intensity by a factor of 10.

However, at other frequencies, this relationship is not quite the same, and one needs to trace through this sequence of relationships again.

• Very nice and complete answer. One mystery solved ! And something to be smarter in conversations when people pull out the famous 3 dB thing. Jan 12 at 23:05
• @c.leblanc Note that "the sound intensity is doubled every 3 dB" is correct: that follows directly from the definition of the decibel scale because $10\log_{10}(2)$ is approximately 3. It's not the same as the perceived loudness doubling, though. I wasn't sure if this was a question of yours or not, so I'll add it into the post if it is. Jan 12 at 23:30
• The fact is that I have been surrounded a lot by speeches confusing the two (sound intensity and perception), I've always been perplexed at this level, seems like I had the right to be :) Indeed I was wondering about that too (the approximation of 3), thank you ! Jan 12 at 23:47