One-third octave explained

A one-third octave is a logarithmic unit of frequency ratio equal to either one third of an octave (1200/3 = 400 cents: major third) or one tenth of a decade (3986.31/10 = 398.631 cents: M3).^[1] An alternative (unambiguous) term for one tenth of a decade is a decidecade.^{[2] [3]}

Definitions

Base 2

ISO 18405:2017 defines a "one-third octave" (or "one-third octave (base 2)") as one third of an octave, corresponding to a frequency ratio of

2 ^{1/3} .A one-third octave (base 2) is precisely 400 cents.

Base 10

IEC 61260-1:2014 and ANSI S1.6-2016 define a "one-third octave" as one tenth of a decade, corresponding to a frequency ratio of

10 ^{1/10} . This unit is referred to by ISO 18405 as a "decidecade" or "one-third octave (base 10)".^[4]

One decidecade is equal to 100 savarts (approximately 398.631 cents).

See also

• Decibel
• Octave band
• Pseudo-octave
• Tritonic scale

Further reading

• A terminology standard for underwater acoustics and the benefits of international standardization. . Ainslie . Michael A. . Halvorsen . Michele B. . Robinson . Stephen P. . 2021-11-09 . January 2022 . 0364-9059 . 1558-1691 . . . 47 . 1 . 179–200 . 10.1109/JOE.2021.3085947 . 243948953 . free . https://web.archive.org/web/20221220112042/https://ieeexplore.ieee.org/stamp/stamp.jsp?tp=&arnumber=9607022<!--https://ieeexplore.ieee.org/stamp/stamp.jsp?tp=&arnumber=9607022--> (22 pages)

Notes and References

1. https://books.google.com/books?id=1x_RvffW-hcC&dq=one+third+octave+base+2+10&pg=PA1325 Malcolm J. Crocker, Handbook of Acoustics (1997)
2. von Benda-Beckmann, A. M., Aarts, G., Sertlek, H. Ö., Lucke, K., Verboom, W. C., Kastelein, R. A., ... & Ainslie, M. A. (2015). Assessing the impact of underwater clearance of unexploded ordnance on harbour porpoises (Phocoena phocoena) in the Southern North Sea. Aquatic Mammals, 41(4), 503.
3. ISO 18405 Underwater Acoustics - Terminology (International Organization for Standardization, Geneva, 2017)
4. (This makes sense as, if we want one third of an octave, the ratio will be f 2 /f 1=2 ^{1/3} , and if we log10 both members of equation we have, log{(f 2 /f 1 )} = log{( 2 ^{1/3} )} - > log(f 2 /f 1 ) = log( 2 )* 1 / 3 , which is approximately 0,1.