Standard Octave Bands

The “standard” centre frequencies for 1/3 octave bands are based upon the Preferred Numbers. These date from the 19th century when Col. Charles Renard (1849–1905) was given the job of improving captive balloons used by the military to observe enemy positions. This work resulted in what are now known as Renard numbers. Preferred Numbers were standardised in 1965 in British Standard BS2045:1965 Preferred Numbers and in ISO and ANSI versions in 1973. Preferred numbers are not specific to third octave bands. They have been used in wide range of applications including capacitors & resistors, construction industry and retail packaging.

In BS2045 these preferred numbers are called the R5, R10, R20, R40 and R80 series. The ‘R’ is used to acknowledge the original work of Renard in the 1870’s. The relationship is

Preferred Series NoR10R20R40R80
1/N Octave1/31/61/121/24

The basis of audio fractional octave bands is a frequency of 1000Hz. There are two ISO and ANSI approved ways in which the exact centre frequencies may be defined. One scheme is the base2 method where the ratio between 2 exact centre frequencies is given by 2^(1/N) with N as 3 for 1/3 octaves and so on. The other method is the base 10 method where the ratio is given by 10^(3/[10N]). This ratio may also be written as 2^(3/[10Nlog2]). For nearly all practical purposes both ratios are the same but tones at band edges can be interesting and may appear to be in different octave bands. The base 2 one is simpler to use (and is often favoured by non-engineering programmers!), but the base 10 one is actually sounder numerically.

One very good reason for using base 10 is that all the exact centre frequencies are the same for each decade. This is not the case for the base 2 frequencies.

As an example (using base 2) the theoretical centre frequency of the 1/3 octave below 1000 is found by dividing by 2^(1/3). This is 793.7005… . Using base 10 the corresponding centre frequency is 794.3282… . In both cases the nearest preferred frequency is 800Hz so that is what the band is called. When working out the edge band frequencies for a 1/3 octave then these are respectively

upper = centre * 21/6
lower = centre / 21/6

where the centre frequency is the exact one not the preferred one. For (1/N)th octave the relationship is simply

upper = centre * 21/2N
lower = centre / 21/2N

If we use the base 2 method and find the centre frequency of the third octave band 10 steps below we get 99.21257… Hz, but with base 10 we get exactly 100.0Hz. If we continue further down to 10Hz and 1Hz then the base 2 centre frequencies are 9.84313…Hz and 0.97656…Hz respectively. The base 10 values are at 10Hz and 1Hz of course. The point to notice is that theses low centre frequencies now differ by approximately (1/24)th of an octave between the two methods.

Generally in audio work we are not too concerned about the very low frequencies. It does explain, however, why the standards use the 1kHz rather than the logical 1Hz as the reference centre frequency. If the 1Hz was used as the reference centre frequency then there would be serious discrepancies between the two schemes at 1kHz, which is very important acoustically. It is also interesting to note that third octave band numbering does use 1Hz as the reference point. We have 1Hz = 100 is third octave band 0, 10Hz = 101 is band 10, 100Hz = 102 is band 20, 1000Hz = 103 is band 30 and so on.

 1 1.6 2.5 4 6.3 1.03 1.65 2.58 4.12 6.5 1.06 1.7 2.65 4.25 6.7 1.09 1.75 2.72 4.37 6.9 1.12 1.8 2.8 4.5 7.1 1.15 1.85 2.9 4.62 7.3 1.18 1.9 3 4.75 7.5 1.22 1.95 3.07 4.87 7.75 1.25 2 3.15 5 8 1.28 2.06 3.25 5.15 8.25 1.32 2.12 3.35 5.3 8.5 1.36 2.18 3.45 5.45 8.75 1.4 2.24 3.55 5.6 9 1.45 2.3 3.65 5.8 9.25 1.5 2.36 3.75 6 9.5 1.55 2.43 3.87 6.15 9.75

Preferred Values 1Hz to 10Hz, 1/24th Octave
The R80 table above gives the 1/24th octave preferred frequencies. For 1/12th skip one to get 1.0, 1.06, 1.12 etc. For 1/6 skip three to give 1.0, 1.12, etc. For 1/3 then skip seven to get 1.0, 1.25 and so on.

So for 1/3 octave bands we obtain the well known sequence: 1.0, 1.25, 1.6, 2.0, 2.5, 3.15, 4.0, 5.0, 6.3, 8.0

References

A. Van Dyck. Preferred numbers. Proceedings of the Institute of Radio Engineers, volume 24, pages 159-179 (February 1936)

British Standard BS2045:1965 Preferred Numbers

ISO 3-1973, Preferred Numbers – Series of Preferred Numbers.

ISO 17-1973, Guide to the Use of Preferred Numbers and of Series of Preferred Numbers.

ISO 497-1973, Guide to the Choice of Series of Preferred Numbers and of Series Containing More Rounded Values of Preferred Numbers.

ANSI Z17.1-1973, American National Standard for Preferred Numbers.

Preferred Number on Wikipedia

Preferred Numbers from Sizes.com

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Dr Colin Mercer

Chief Signal Processing Analyst at Prosig
Dr Colin Mercer was formerly at the Institute of Sound and Vibration Research (ISVR), University of Southampton where he founded the Data Analysis Centre. He then went on to found Prosig in 1977. Colin retired as Chief Signal Processing Analyst at Prosig in December 2016. He is a Chartered Engineer and a Fellow of the British Computer Society.

17 thoughts on “Standard Octave Bands”

1. Dr Colin Mercer Post author

Hi
The band numbers are only defined for third octaves as explained in the text. The real central body for these is ANSI rather than ISO. See ANSI S1.11 -2004 “Specification for Octave Band and Fractional Octave Band Analog and Digital Filters”. The preferred values are OK for 1/1 to 1/3 the octave bands, but for narrower octave bands there now exists a way of computing the preferred frequencies without reference to a table look up.

I feel an explanatory article is needed.

1. richard

Thank you for this information.
I wonder if you could direct me to the documentation (or Matlab code) for your statement —-
“there now exists a way of computing the preferred frequencies without reference to a table look up.”

thank you
richard

2. amit bhola

Dr Colin,

I have a basic doubt. Why only ‘one-third’ ? Why each octave is not divided into 4 or 5 subdivisions to make ‘one-fourth’ or ‘one-fifth’ octave filter ? What is the significance behind 3 ?

Regards
Amit B

1. amit bhola

My query meant why 1/3 octave is so common in analysis but not the further octaves? Is there any significant meaning behind 1/3 divisions?

1. Dr Colin Mercer Post author

Amit
The origin of 1/3 octaves being so popular is that in a general sense they are close in bandwidth to the way our hearing distinguishes between different frequencies. If you ever get into psychoacoustics there is a unit there called a Bark whose bandwidth is similar to 1/3 octaves.

That is in an approximate sense 1/3 octaves match our hearing.

1. Yulik Yagudin

I’d say that this system also directly refers to a musical scale. In a equal tempered chromatic scale the smallest step – a semitone – is exactly a 1/12 octave. So 1/3 octave is equal to a major third.

3. khaleel

Dear Sir,

We were analyzed the effect of AHU room wall acoustic insulation (50mm thick – 32 kg/m3 density) and we found from our observation that there is negligible sound transmission loss of STC 3 only.
And also as per the ASHRAE 2011 Standard adding acoustical absorption in the AHU room will reduces the reverberant sound maximum of 4 dB @ high frequency only.

Whether can we remove the AHU room lining? (Usually we are lining any two critical walls of the AHU room as per the client requirement)

As per our analysis AHU with return mixing box with 900mm length of low static return duct silencer will provide effective noise reduction than lining the AHU room walls.

Anticipating your valuable suggestion for the same

4. Collin B. Sower

Dr. Mercer,
I am looking for an algorithm to convert discrete vibration analysis data into 1/3 octave. In other words, if given a data set from 0-250 Hz, with vibration readings recorded every 0.3125 Hz, how would one mathematically create a 1/3 octave analysis?

1. Dr Colin Mercer Post author

Collin

As well as the frequency scaling, it is also important to do the correct amplitude scaling. I will deal with the amplitude stuff first. This is because we have to deal with properly summing the data in each third octave band.

Third Octave data is usually shown as a spectrum level on a dB scale, that is if you data on a linear scale in terms of say V, then you have to square it into $V^2$ units. Similarly if you have your data as a spectral density in units of $(V^2/Hz)$ you need to multiple it by the effective noise bandwidth to get to $V^2$ unit. If your data is calculated at frequencies $k*df$ (in your case $df$ is 0.3125 Hz) then it is assumed you have a scaled rectangular block of ‘energy’ of amplitude $A_k$, that is in units of $V^2$, running from frequency $(k-0.5)df$ to $(k+0.5)df$

Now each 1/3 octave runs from $F_{low}$ to $F_{high}$, where these are the exact band edges in Hz of the particular band you are working on. So in principle you find all those scaled values in the frequency range $F_{low}$ to $F_{high}$ and sum them together. For the two frequency components at the edges $F_{low}$ and $F_{high}$ you need to add in only that part in the current third octave band. This is a bit fiddly but necessary.

We now have a sum of squares for a particular 1/3 octave band. Do NOT divide by number of values you summed, we need the sum of squares in the frequency domain. You can decide to take the square root so the dB conversion is

$20 log_{10}\left(\frac{\sqrt{A}}{dBRef}\right)$

or

$10log_{10}\left(\frac{A}{dBref^2}\right)$

Obviously repeat for each 1/3 octave. You can find the exact band edge frequencies by knowing the nominal centre frequency, computing the exact centre frequency and from that the exact band edges for the integration. Each third octave value is plotted against its nominal or preferred centre frequency, not against the exact centre frequency.

This method is quite accurate, but the frequency edge slopes of each band do not match ANSI standards but the results are nevertheless very good.

Colin Mercer

5. PKE Kumar

Dear Prof. Colin
I am having a doubt in 1/3 octave analysis. For a sound signal, I am having the Power spectrum (PSD vs Frequency plot). I want the 1/3 octave representation of the signal. I order to do the 1/3 octave analysis, I divide the frequency into n no of bins. After dividing it into bins I have to get a single amplitude value for each bin. What is the procedure to get a single amplitude value for each octave bin.

1. Dr Colin Mercer Post author

Hi Kumar
The response immediately above your question should give you the basic scheme. Also be very careful about what is meant by “psd” as some people have this in “units Squared” rather than the correct units of “units squared/Hz”. I have also seen some in “units/sqrt(Hz)”. One way or another you need to form the sum of squares, which is different from the square of the sum.

6. Collin B. Sower

Dr. Mercer,

The acoustic algorithm I understand. My situation, however, is dealing with vibration. Obviously, vibration is not a logrithmic measure, as acoustics are. The other defining difference is the lack of a reference such as is used in acoustics.

Are you aware of the mathematical algorithm that will take discrete vibrational data and convert it to 1/3 octave?

1. Dr Colin Mercer Post author

Hi
Firstly there is no difference in the algorithm in converting into 1/N th octave bands for any type of signal.

What is different however is the reference level. If you have an acceleration in $m/sec^2$ then the reference level is $10^{-6} m/sec^2$; if it is a velocity then the reference is $10^{-9} m/sec$; if it is a displacement it is $10^{-12} m$. That is there is a factor of 1000 between each type of vibration. At around 159Hz all three types give the same dB level. Also acceleration measurements in g use a reference of $9.80665*10^{-6}$. That is the g values are effectively converted to $m/sec^2$. A curious by product is that the velocity reference is sometimes taken as $10^{-8}$ because 9.80665 is close to 10 and the US Navy uses $10^{-5}$ for acceleration for the same reason – that is they assume acceleration measurements are only ever in g – so beware!! It can be even more entertaining if you measure in $feet/sec^2$ or even $inches/sec^2$. To be certain and consistent effectively convert to metric

Please see the table below for the usual reference levels.

Note if you have computed a sum of squares then the dB level is
$10 log\frac{Sum-of-squares}{dB-ref^2}$
or you can use $dB = 20 log \frac{\sqrt{Sum-of-squares}}{dB-Ref}$

Incidentally did you know that the original definition of a Bel was the change in voltage level over a mile of ‘standard’ telegraph wire. Reference level for Volts is unity

A useful Table

Sound Energy Level : $L_w$ | $10 log\frac{W}{W_0}$ | $10^{-12}J$

Sound Energy Density Level : $L_E$ $10 log\frac{E}{E_0}$ | $10^{-12} J/m^3$

Sound Intensity Level : $L_I$ | $10 log\frac{I}{I_0}$ | $10^{-12} W/m^2$

Sound Power Level : $L_{Pac}$ | $10 log\frac{P}{P_0}$ | $10^{-12} W$

Sound Pressure Level : $L_p$ – in air | $20 log\frac{p}{p_0}$ | $2 x 10^{-5} Pa$

Sound Pressure Level : $L_p$ – liquids and solids | $20 log\frac{p}{p_0}$ | $1 x 10^{-6} Pa$

Force Level : $L_F$ | $20 log\frac{F}{F_0}$ | $10^{-6} N$

Vibration Acceleration Level : $L_a$ | $20 log\frac{a}{a_0}$ | $10^{-6} m/s^2$

Vibration Velocity Level : $L_v$ | $20 log\frac{v}{v_0}$ | $10^{-9} m/s$

Vibration Displacement Level : $L_d$ | $20 log\frac{d}{d_0}$ | $10^{-12} m/s$

Voltage Level : $L_u$ | $20 log\frac{v}{v_0}$ | $1 Volt$

People sometimes wonder where $2*10^{-5}$ comes from in SPL in air. Well if you square $2*10^{-5}$ and multiply by the product of standard air density and speed of sound you will get $10^{-12}$ to a first order approximation, which is sound power for a spherical wave – so there.

Colin

1. Collin B. Sower

Dr. Mercer,

First of all, thank you for such a quick and comprehensive response. The presence of a reference measurement within the world of vibration is an epiphany of enormous proportion! If I understand the table correctly, the reference for acceleration (a sub naught) is 10e-6 m/s^2?

Pending your answer to the reference value question, I will attempt to apply the algorithm you provided to my data sets–one being a “narrowband”/discrete test performed with a delta f of 0.3125 and the second being a 1/3 octave generated by the software suite.

Would it be too much of an imposition to share those results with you to ensure that I am performing the operation correctly?

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