Fundamentals of Acoustics and Noise

Unit 4
Frequency analysis,
Frequency bands,
Decibel scales,
Descriptors for time varying noise levels Fundamentals of Acoustics and Noise: Unit 4 – Frequency Analysis, Decibel Scales, Special Descriptors

4-1

Contents
Frequency Analysis of Sound Pressure Signals

Constant Proportion Bandwidth Frequency Bands

Constant Bandwidth Frequency Bands

Decibel Scales
Descriptors for Time Varying Noise Levels

Equivalent Continuous Sound Level

Sound Exposure Level

Percentile Exceeded Sound Level

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Fundamentals of Acoustics and Noise: Unit 4 – Frequency Analysis, Decibel Scales, Special Descriptors

Frequency Analysis of Sound Pressure Signals
A microphone is constructed to produce a voltage proportional to the sound pressure of interest.
These voltage signals, which are functions of time, are referred to as sound pressure signals. It is necessary to have knowledge of the frequency composition of these signals. Fourier Series and the Fourier Transform can be used to mathematically relate functions in the time and frequency domains. The use of these relationships in the practical frequency analysis of signals will not be discussed in detail in this course. A less mathematical and more physical description of elementary frequency analysis is given in this section. The purpose of an elementary frequency analysis is to determine how the “strengths” of the components of the sound pressure are distributed as a function of frequency.
Consider first a pure tone. At a particular point in space, the time history of the sound pressure of the pure tone is given by: p(t ) = P cos 2πft

(4.1)

The elementary frequency analysis of this pure tone shows that at a frequency of f, it has a strength or amplitude (peak value) of P. It is often more convenient to measure the strength of a component by its root mean square value, prms, which here is given by P / 2 . The time domain representation of the root mean square value of a pure tone is shown in Figure 4.1, and the frequency domain representation of this pure tone is shown in Figure 4.2.

Figure 4.1

Time domain representation of a pure tone.

Fundamentals of Acoustics and Noise: Unit 4 – Frequency Analysis, Decibel Scales, Special Descriptors

4-3

Figure 4.2

Frequency domain representation of a pure tone.

When the sound is a simple pure tone, its magnitude may be represented by its amplitude, which in this case is also the same as its peak value. The average value of the sound pressure over any period of more than a few cycles will be zero because the positive half-cycles will cancel the negative pressure half-cycles. For more complex waveforms, such as harmonic, transient or random noise, the expression of magnitude is not as simple, but the time-averaged value is still zero.
A commonly used expression of magnitude is the RMS value of the sound pressure. This gives a non-zero average, corresponding to the square root of the mean (average) of the square of the pressure. Figure 4.3 Processing stages of a complex sound pressure signal to determine its RMS value.

Consider now a complex sound pressure obtained by the superposition of two harmonically related pure tones of frequencies mf and nf. p(t ) = Pm cos 2πmft + Pn cos 2πnft

(4.2)

The frequency domain representation of this acoustic pressure is shown in Figure 4.4.
It is evident that the strengths of the individual components of frequencies mf and nf, in terms of their RMS values are pmrms = Pm / 2 and pnrms = Pn / 2 .

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Fundamentals of Acoustics and Noise: Unit 4 – Frequency Analysis, Decibel Scales, Special Descriptors

Figure 4.4

Frequency domain representation of a complex acoustic pressure.

A measure of the total strength of the sound pressure is useful, and the RMS value is again appropriate. The total RMS value of the sound pressure whose time history is given by
Equation (4.2) can be found by squaring Equation (4.2) and averaging it over a time equal to l/f, resulting in:

(

2
2
2 prms = pm rms + pn rms

)

(4.3)

It is of interest to consider the significance of this result in terms of sound energy. An expression for the sound energy density (energy per unit volume) of one dimensional plane acoustic waves associated with a simple harmonic variation of the sound pressure can be written in terms of the RMS sound pressure as:
E =

2 p rms ρc 2

(4.4)

2
The sound energy density associated with the component whose frequency is mf is pm rms ρc 2
2
and for the component whose frequency is nf the sound energy density is pn rms ρc 2 .

Sound energy and sound energy density are scalar quantities and so the sound energy densities associated with the two components can be added to give a total sound energy density of
2
2 pm rms + pn rms ρc 2 . The total sound energy density associated with the total sound pressure is

(

)

2 given by prms ρc 2 . Equating these two expressions for the sound energy density leads to
Equation (4.3).

It can also be concluded, from energy considerations, that, even if the frequencies are not harmonically related, provided that they are all different, the total RMS value of a sound pressure which is composed of N simple harmonic components of different frequencies whose
RMS values are p1 rms , p2 rms , ...., p N rms is given from:

(

2
2
2 prms = p12rms + p2 rms +....+ p N rms

)

Fundamentals of Acoustics and Noise: Unit 4 – Frequency Analysis, Decibel Scales, Special Descriptors

(4.5)

4-5

Constant Proportion Bandwidth Frequency Bands
Many commonly occurring sound pressures are composed of a large number of components and it is not feasible to try to individually describe all these components. Instead, the RMS values of the sound pressures in frequency bands are used to present frequency domain information. The most widely used frequency bands for analysing sound pressure signals are the octave bands.
The word octave means a doubling of frequency. The ideal octave band filter allows frequency components greater than a lower frequency f l and less than an upper frequency f u = 2 f l to pass through the filter unimpeded while other components are completely blocked. The centre frequency f c of the pass band is geometrically defined by: fc fu
=
fl fc (4.6)

fc and f u = 2 f c . The
2
centre frequencies of adjacent bands are then an octave apart. The preferred octave bands have centre frequencies of 31.5, 63, 125, 250, 500, 1000, 2000, 4000, 8000 and 16 000 Hz. The audible frequency range from about 20 Hz to 20 000 Hz is therefore covered in 10 octave bands. Use of the result f u = 2 f l in this equation leads to the results f l =

Usually, octave band sound pressures are obtained from measured sound pressures by amplifying the voltage signal obtained from the microphone, passing it into octave band electrical filters and then measuring the RMS value of the filtered voltage.
Octave bands are quite wide in a frequency sense. The 1000 Hz octave band extends from
1000 / 2 = 707 Hz to 1000 × 2 = 1414 Hz , that is, it is 707 Hz wide. The widths of octave bands are in fact equal to the low frequency limit of the band.
Frequently, it is desirable to make analyses in narrower bands, and hence one third octave bands are used. One third octave bands, as the name implies, are one third of an octave wide in a frequency sense. Three one third octave bands are contained in every octave band. This is done
1
by setting the upper frequency limit of the band f u to 2 3 f l where f l is the lower frequency limit of the band. The centre frequency of the band f c is again defined geometrically by f f f Equation (4.6), that is, c = u , and so f l = 1 c6 and f u = 21/ 6 f c . The preferred one third
2/
fl fc octave bands have centre frequencies of 16, 20, 25, 31.5, 40, 50, 63, 80, 100, 125, 160, 200, ...
Hz.
Actual octave and one third octave band filters do not provide the characteristics of the ideal filters discussed so far. The pass bands of the ideal and typical octave filters are shown in
Figure 4.5.

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Fundamentals of Acoustics and Noise: Unit 4 – Frequency Analysis, Decibel Scales, Special Descriptors

Figure 4.5

Ideal and typical octave pass bands.

The octave and one third octave bands which have been discussed so far are examples of constant proportion bands. The widths of the bands are proportional to the centre frequencies of the bands. Frequently, narrower constant proportion bands are used and a 6% bandwidth is commonly used. The bandwidth percentage is the ratio of the bandwidth to the band centre frequency, expressed as a percentage.

% bandwidth =

bandwidth (Hz )
× 100% centre frequency (Hz )

(4.7)

Example 4.1
A complex sound pressure is known to be composed of five individual pure tones. The frequencies and amplitudes of the individual components are given in the following table.

Frequency
(Hz)
100
150
160
210
250

Amplitude
(Pa)
2
4
6
2
1

Determine the total RMS value of the complex sound pressure, and the RMS value of the components in the ideal 125 Hz octave band.

Fundamentals of Acoustics and Noise: Unit 4 – Frequency Analysis, Decibel Scales, Special Descriptors

4-7

Solution:
The RMS values of the individual components in ascending order of frequency are 1.414,
2.828, 4.242, 1.414 and 0.707 Pa. Application of Equation (4.5) gives:

(

prms = 1.414 2 + 2.8282 + 4.242 2 + 1414 2 + 0.707 2
.

)

1

2

Pa = 5523 Pa
.

The ideal 125 Hz octave band extends from 125 / 2 = 88.4 Hz to 125 × 2 = 176.78 Hz . Thus the first three of the tabulated frequency components are in this band. Application of Equation
(4.5) gives:

(

p rms = 1.414 2 + 2.828 2 + 4.242 2

)

1

2

Pa = 5.291 Pa

Exercise 1
1.

Two complex sound pressure signals have RMS values p1 rms and p2 rms . The smaller signal has an RMS value of 0.1 times that of the larger signal. What error will be introduced in using the RMS value of the larger signal to estimate the total RMS value of the combined signals?

Exercise 2
1.

4-8

Determine the percentage bandwidths of octave and one third octave band filters.

Fundamentals of Acoustics and Noise: Unit 4 – Frequency Analysis, Decibel Scales, Special Descriptors

Constant Bandwidth Frequency Bands
It is sometimes an advantage to analyse sound pressure signals into narrow frequency bands of constant bandwidth and not constant proportion bandwidths. Reasonable estimates of the power spectral densities of sound pressure signals can be obtained with constant bandwidth analyses.
The power spectral density of a sound pressure signal at a frequency of f can be interpreted as the mean square value of the sound pressure signal after it has been filtered by an ideal pass band filter which is 1 Hz wide and is centred at a frequency of f .
The power spectral density is a function of f and is generally denoted S ( f ) . The contribution to the mean square value of a sound pressure signal from components between two frequencies f 1 and f 2 is given by the integral of S ( f ) between f 1 and f 2 .
If the power spectral density is relatively uniform between two frequencies the sound pressure is said to be “white” between these frequencies.
The mean square value associated with this white portion of the sound pressure is simply given by the product of the power spectral density and the frequency range.
Mean square value of white noise = S ( f ) × ( f 2 − f1 )

(4.8)

Example 4.2
A narrow band frequency analyser with a constant bandwidth of 10 Hz is used to analyse a sound pressure signal. It is found that when the centre frequency is 1000 Hz the measured RMS value is 0.1 Pa. Determine the power spectral density at this frequency.

Solution:
The mean square value in this frequency band is (0.1)2 Pa2. The band is 10 Hz wide and so the power spectral density at this frequency is given by:
(01) 2 Pa 2 10 Hz = 0.001 Pa 2 Hz
.

Exercise 3
1.

The power spectral density of a sound pressure varies linearly between 1000 Hz and
5000 Hz. At 1000 Hz the power spectral density is 0.001 Pa2/Hz and at 5000 Hz the power spectral density is 0.0001 Pa2/Hz. Determine the RMS value of the sound pressure between these frequency limits.

Fundamentals of Acoustics and Noise: Unit 4 – Frequency Analysis, Decibel Scales, Special Descriptors

4-9

Decibel Scales
Quantities such as sound pressure, sound intensity and sound power vary over wide ranges and it is convenient to express these quantities on a logarithmic scale. The decibel scale, which is based on logarithms to base 10 is widely used. It is convenient in discussing the decibel scale to begin by considering its application to describing sound powers.
A sound source which is radiating W watts of sound power has a sound power level of LW relative to a reference sound power of Wref watts defined by:
LW = 10 log10

W
Wref

(4.9)

LW has the “units” of dB re Wref . It is international practice to set Wref to 10-12 watts. Suppose, for example, that W is 1 watt. LW is then 120 dB re 10-12 watts.
Sound intensity, being the sound power flow through a unit area is closely related to sound power. The intensity level LI for a sound wave whose intensity at a specified point is I , is defined by:
LI = 10 log10

I
I ref

(4.10)

LI has the “units” of dB re I ref . It is international practice to set I ref to l0-12 watts/m2.
Sound intensity is proportional to the square of the sound pressure ( I ∝ p 2 ). Hence, the sound pressure level Lp corresponding to a sound pressure p can be defined by:
L p = 10 log10

2 p rms
2
p ref

(4.11)

Equation (4.11) can be written as:
Lp = 20 log10

prms pref (4.12)

Lp has the “units” of dB re pref . It is international practice to set pref to 2×10-5 Pa. Suppose,

for example, that p is 2 Pa. Lp is then 100 dB re 2×10-5 Pa. Often, the “re 2×10-5 Pa” term is not included as it is considered to be implied. Further, this reference pressure is taken to be the
RMS value. It is of interest that the threshold of hearing at 1000 Hz for a young listener with good hearing occurs at a sound pressure of approximately 2×10-5 Pa, that is, pref corresponds to the threshold of hearing.
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Fundamentals of Acoustics and Noise: Unit 4 – Frequency Analysis, Decibel Scales, Special Descriptors

The reference quantities for sound power, sound intensity and sound pressure have been selected so that the corresponding levels are inter-related under certain conditions. An expression for the intensity of a travelling plane wave can be expressed in terms of the RMS value of the sound pressure as:

I =

P2 p2 = rms ρc 2 ρc

(4.13)

When ρc is 415 rayls and I is set to the reference sound intensity of 10-12 watts/m2, the corresponding sound pressure is 2.057×l0-5 Pa. This value is close to the reference sound pressure of 2×l0-5 Pa. Thus LI is approximately equal to Lp for a travelling wave.
Other quantities can be expressed on the decibel scale. For example, the calibration constant of a measurement microphone can be expressed on a decibel scale, as can a voltage. Suppose that a voltage signal has a value of V volts. This voltage can be expressed in terms of a level relative to some reference voltage V0 by use of an equation similar to Equation (4.12):
LV = 20log10

V
V0

(4.14)

LV has units of dB re V0 .

Care must be used in adding and subtracting quantities expressed on decibel scales. The safest procedure to follow when quantities expressed on decibel scales must be added or subtracted is to express the quantities in physical units and then these quantities can be added or subtracted in the appropriate manner. Sound powers and intensities, when expressed in the physical units of watts or watts/m2 can be added or subtracted directly. However, sound pressures, when expressed in the physical units of Pa must be added or subtracted by adding or subtracting the mean square pressures and not the root mean square pressures.

Example 4.3
The calibration constant of a microphone is known to be 0.001 V/Pa. Express this in dB re l V/Pa.

Solution:
The calibration constant in dB re l V/Pa = 20log10

0.001
= − 60 dB re l V/Pa.
1.0

Fundamentals of Acoustics and Noise: Unit 4 – Frequency Analysis, Decibel Scales, Special Descriptors

4-11

Example 4.4
An octave band analysis of a sound pressure gives the following results.
Octave Band Centre Frequency
(Hz)
31.5
63
125
250
500
1000
2000
4000
8000
16000

Sound Pressure Level ( Lp )
(dB re 2×10-5 Pa-2)
80
90
100
90
100
90
90
110
80
60

Determine the RMS value of the total sound pressure and the corresponding sound pressure level. Solution:
Application of Equation (4.11) allows the mean square sound pressures in each of the octave bands to be found. The mean square sound pressures in each octave band are given in the following table.
Octave Band Centre Frequency
(Hz)
31.5
63
125
250
500
1000
2000
4000
8000
16000

Mean Square Sound Pressure
2
( prms ) (Pa2)
0.04
0.4
4
0.4
4
0.4
0.4
40
0.04
0.0004
∑ = 49.6804 (Pa)2

These mean square sound pressures can then be summed by use of Equation (4.5).
The total RMS sound pressure is prms = 49.6804 = 7.05 Pa.
The corresponding sound pressure level can be found by application of Equation (4.12) to be:
L p = 20 log10

4-12

prms pref = 20 log10

7.05
2 × 10

−5

= 110.9 dB re 2×10-5 Pa.

Fundamentals of Acoustics and Noise: Unit 4 – Frequency Analysis, Decibel Scales, Special Descriptors

Exercise 4
1.

At a point midway between two typewriters the sound pressure level when both typewriters are working is 86 dB. What will be the sound pressure and sound pressure level at this point when only one typewriter is working?

2.

A compound 10m x 10m contains 10 barking dogs. At a distance of 50 m from the centre of the compound the sound pressure level with the dogs barking is found by measurement to be 60 dB. An additional 40 dogs are introduced into the compound.
What is the increase in the sound pressure level due to the additional dogs?

3.

Two sound pressure levels differ by x dB. Derive, as a function of x, an expression for the quantity which must be added to the larger sound pressure level to give the sound pressure level of the combined sounds.

Descriptors for Time Varying Noise Levels
So far it has been assumed that the sound is steady, and it can be represented by the overall sound pressure level Lp , or the A-weighted sound pressure level L A .
Frequently the sound is not steady and this causes difficulties. The first attempt to overcome this problem is to use the Fast (F) or Slow (S) time constants on the Sound Level Meter. The time constant “F ” is designed to approximate the response of the ear.
There are a number of more sophisticated measures which attempt to quantify sound which is not steady by a single number. These are the equivalent continuous sound levels ( Leq , LAeq ), and the sound exposure levels ( SEL , LAE ).

Equivalent Continuous Sound Level
Sound power is proportional to p 2 . Consider a sound pressure p(t ) whose level fluctuates in some way over a time T. It is useful to find an equivalent sound pressure whose level is constant over the time T such that the total sound energy (power × time) associated with the equivalent sound pressure is the same as that associated with the fluctuating sound pressure. It is thus necessary to have
T

p × T = ∫ p 2 (t )dt
2
eq

0

⇒

2 peq 2 pref 1
=
T

 p 2 (t ) 
∫  pref dt
 2 
0

T

Fundamentals of Acoustics and Noise: Unit 4 – Frequency Analysis, Decibel Scales, Special Descriptors

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⇒

⇒

Leq ,T

2
 peq 
1

= 10log10 2  = 10log10 
p 
T
 ref 


1
Leq ,T = 10 log10 
T

n

∑ T × 10

Leq ,Ti / 10

i

i =1

T

∫
0

p 2 (t )  dt 
2
pref 






(4.15)

(4.16)

Usually, the A-weighted equivalent level LAeq is of interest. It is found by A-weighting p(t ) and so it is given by
1
LAeq ,T = 10log10 
T


⇒

1
LAeq ,T = 10 log10 
T

2 p A (t )  dt 
2
 pref 

T

∫
0

n

∑ T × 10 i =1

L Aeq ,Ti / 10

i

(4.17)





(4.18)

LAeq is used as a rational means of obtaining a single number to describe a sound of fluctuating

level as shown, for example, in Figure 4.6.

Figure 4.6 Leq as a measure of fluctuating level.
Leq measurements are used to describe fluctuating machinery noise, eg. the above Lp vs time

plot could be noise produced by a machine which works on a cycle.
Leq (or more particularly the LAeq ) measurements are useful in assessing long-term noise

exposure associated with occupational and environmental problems.
Often for these purposes, LAeq is measured over a time T and so LAeq is written as LAeq ,T .
For occupational purposes T is commonly 8 hours, but for environmental purposes different values of T are used. Thus LAeq ,9 might be measured from 22:00 to 07:00 to describe the noise levels during sleep times.

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Fundamentals of Acoustics and Noise: Unit 4 – Frequency Analysis, Decibel Scales, Special Descriptors

Sound Exposure Level
The Sound Exposure Level ( SEL , also denoted LE ) is similar in form to Leq , but is used for discrete noise events. It is defined as the constant sound pressure level which, if maintained for a period of one second, would deliver the same noise energy to the receiver as the original event itself. Thus t2 p 2 × T0 = ∫ p 2 (t )dt t1 T0 = 1 second

where

t 2 − t1 = a stated time interval long enough to encompass all significant sound of the event

⇒

⇒

SEL = 10log10

t2

1
T0

p 2 (t ) dt 2 pref ∫

t1

1
SEL = 10log10 
T
 0

n

(4.19)

∑ T × 10

Leq ,Ti / 10

i

i =1






(4.20)

Usually the pressures are A-weighted in which case LAE is used for sound exposure level,
LAE

⇒

1
= 10log10
T0

t2

∫

t1

1
LAE = 10log10 
T
 0

2 p A (t ) dt 2 pref n

∑ T × 10 i =1

i

(4.21)

L Aeq ,Ti / 10






(4.22)

It can be seen that either SEL or LAE gives a measure of the total energy of a single noise event, whereas either Leq or LAeq gives a measure of the average sound power of a sound.
LAE is useful for describing transient noise events such as vehicle drive-by noise and aircraft fly-overs. Figure 4.7 shows an example of a measurement beginning and ending in background noise.
LAE does not depend on the time of measurement whereas LAeq does.

Fundamentals of Acoustics and Noise: Unit 4 – Frequency Analysis, Decibel Scales, Special Descriptors

4-15

Figure 4.7 Comparison between LAE and LAeq for a transient noise event.

It is also useful to be able to calculate an LAeq value applying for a given period from data on individual noise events which may occur in the period. Examples could be when different types of vehicles drive by or if a number of different types of machines are used in, say, an eight hour period. The LAE value and number of occurrences of each event must be known. For a single event occurring in a time period T (where T is in seconds), it can be shown that LAeq is related to LAE by:

LAE = LAeq ,T + 10 log

⇒

LAeq = L AE − 10 log

T
T0

where T0 = 1 second

T
T0

where T0 = 1 second

(4.23)

(4.24)

For n events with individual sound exposure levels given by L AEi

n

LAeq = 10 log ∑ 10 i =1

4-16

L AEi
10

− 10 log

T
T0

where T0 = 1 second

(4.25)

Fundamentals of Acoustics and Noise: Unit 4 – Frequency Analysis, Decibel Scales, Special Descriptors

Percentile Exceeded Sound Level
When noise is fluctuating, the energy averaged Leq alone does not indicate the level of annoyance as no information is given concerning the degree of fluctuation or the maximum and minimum values and their duration. Statistical analysis of the noise is a convenient way of quantifying the temporal variation of the noise level over the measurement period.
The common statistical descriptor of fluctuating noise levels is the Percentile Exceeded Sound
Level L N which is the noise level exceeded for N% of the time over which the measurement was made. Figure 4.8 shows how L N is related to the noise signal. To be rigorous, the notation
LN ,T should be used where T specifies the time of measurement. Usually the sound is Aweighted and the notation used is LAN ,T .
Often LA 90,T is used to estimate the residual background noise level, that is, L90 represents the noise level exceeded for 90% of the time.

LA10,T is used to estimate the maximum levels, that is, L10 represents the noise level exceeded for 10% of the time.

Figure 4.8 Percentile exceeded level related to noise signal.

Fundamentals of Acoustics and Noise: Unit 4 – Frequency Analysis, Decibel Scales, Special Descriptors

4-17

Exercise 5
1.

A sound level meter is set on the ‘linear’ scale and has been used to obtain the following data for a sound source totally in the 1000 Hz octave band: time period
0 seconds to 5 seconds

meter reading
Lp = 70 dB re 2×10-5 Pa

5 seconds to 15 seconds

Lp = 110 dB re 2×10-5 Pa

15 seconds to 30 seconds

Lp = 90 dB re 2×10-5 Pa

30 seconds to 100 seconds

Lp = 70 dB re 2×10-5 Pa

a)

Determine Leq and SEL .

b)

Determine Leq and SEL if the final time period were from 30 seconds to 200 seconds (still with Lp = 70 dB re 2×10-5 Pa). Hence comment on which measure of sound is the more appropriate.

Exercise 6
1.

Consider a noise event lasting for ten seconds. As shown on the graph below, the noise level increases from 60 dB(A) re 2×10-5 Pa to 70 dB(A) re 2×10 5 Pa during the fourth second and again increases from 60 dB(A) re 2×10-5 Pa to 90 dB(A) re 2×10 5 Pa during the eighth second.
LA [dB(A) re 2*10-5 Pa]

100
90
80
70
60
50
40
0

1

2

3

4
5
6
7
time [seconds]

8

9

10

11

Determine:
(a) L Aeq for the ten second period
(b)
(c)
(d)
(e)
(f)

4-18

LAE for the first five second period
LAE for the second five second period
LAE for the ten second period
LA 90 %, 10 seconds
LA 10 %, 10 seconds

Fundamentals of Acoustics and Noise: Unit 4 – Frequency Analysis, Decibel Scales, Special Descriptors