37.3 Levels and Decibels

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Sound pressure and power are commonly expressed in terms of decibel levels. This allows us to use a

logarithmic rather than a linear scale. It provides the distinct advantage of allowing accurate

computations using small numerical values, while accommodating a wide range of numerical values.

37.3.1 Sound Power Level

Sound power level describes the acoustical power radiated by a given source with respect to the

international reference of 10212 W. The sound power level, LW ; is defined as

LW ¼ 10 log

W

Wre

􀀏 􀀐

ðdBÞ ð37:7Þ

where W denotes sound power in question and Wre ¼ 10212 W (reference).

Example 37.2

Determine the sound power level of a small ventilation fan that generates 10 W of sound power.

Solution

LW ¼ 10 log

W

Wre

􀀏 􀀐

¼ 10 log

10

10212

􀀏 􀀐

¼ 130 dB

37.3.2 Sound Pressure Level

Sound pressure levels are expressed in decibels, as are sound power levels. The sound pressure level, Lp; is

defined as

Lp ¼ 10 log

p􀀊2

p2

re

􀁻 !

¼ 20 log

p􀀊

pre

􀀏 􀀐

ðdBÞ ð37:8Þ

where p􀀊 denotes root-mean-square (RMS) sound pressure in question Pa or N/m2 and

pre ¼ 20 £ 1026 Pa ¼ 0.0002 mbar. The pressure of 20 £ 1026 Pa has been chosen as a reference because

it has been found that the average young adult can perceive a 103 Hz tone at this pressure. This reference

is often referred to as the threshold of hearing at 103 Hz.

Example 37.3

Giving Lp ¼ 50 dB for the Aeolian tone of 200 Hz, determine the RMS pressure of the tone.

Solution

Given Lp as 50 dB, then p􀀊 is determined by using Equation 37.8.

50 ¼ 20 log

p􀀊

pre

􀀏 􀀐

then

p􀀊 ¼ 1050=20 pre ¼ 316:2pre

p􀀊 ¼ 6:32 £ 1023 Pa ¼ 0:0632 mbar

Note that this value is very small, contradicting the magnitude of the sensory impression of the

human ear.

Sound Levels and Decibels 37-3

© 2005 by Taylor & Francis Group, LLC

37.3.3 Overall Sound Pressure Level

The sound pressure level is defined assuming “pure tone” sound. However, practically any real sound

contains various components of pure tone sound. Let us consider a set of n components of pure tone,

denoted by

p1ðtÞ ¼ a1 sinð2pf1t þ f1Þ

.. .

pnðtÞ ¼ an sinð2pfnt þ fnÞ

8>>><

>>>:

ð37:9Þ

pðtÞ ¼ p1ðtÞ þ · · · þ pnðtÞ ð37:10Þ

If Lp of pðtÞ is evaluated in RMS pressure, p􀀊; we have

p􀀊 ¼ lim

T!1

1

T

ðT

0

p2ðtÞdt

􀀒 􀀓1=2

¼ lim

T!1

1

T

ðT

0 ðp1ðtÞ þ · · · þ pnðtÞÞ2dt

􀀒 􀀓1=2

ð37:11Þ

Since

lim

T!1

1

T

ðT

0

piðtÞpjðtÞdt ¼ 0; i – j

is valid, p􀀊 is obtained as

p􀀊 ¼

h

p1ðtÞ2 þ · · · þ pnðtÞ2

i1=2

¼

h

p􀀊21

þ · · · þ p􀀊2

n

i1=2

ð37:12Þ

where

p􀀊2i

; piðtÞ2 ; lim

T!1

1

T

ðT

0

p2i

ðtÞdt ¼

1

2

a2i

ð37:13Þ

Let us define Lpi ; 10 logð􀀊p2i

=p2

reÞ ði ¼ 1; 2; …; nÞ: Then the overall sound pressure level, Lp; of pðtÞ is

expressed by

Lp ; 20 log

p􀀊

pre ¼ 10 log

1

p2

re ðp􀀊21 þ · · · þ p􀀊2

nÞ

or Lp is expressed by Lpi ði ¼ 1; 2; …; nÞ as follows:

Lp ¼ 10 logð10Lp1 =10 þ 10Lp2 =10 þ · · · þ 10Lpn =10Þ ð37:14Þ

Example 37.4

Determine the overall sound pressure level of the combination of three pure tones, the sound pressure

levels of which are expressed by

Lp1 ¼ 60 dB ðf1 ¼ 250 HzÞ; Lp2 ¼ 65 dB ðf2 ¼ 500 HzÞ; Lp3 ¼ 55 dB ðf3 ¼ 1000 HzÞ

Solution

We have 10Lp1 =10 ¼ 106; 10Lp2 =10 ¼ 106:5; and 10Lp3 =10 ¼ 105:5: Then the overall level, Lp; is determined by

using Equation 37.14 as follows:

Lp ¼ 10 logð106 þ 100:5 £ 106 þ 1020:5 £ 106Þ

¼ 10 log106ð1 þ 100:5 þ 1020:5Þ ¼ 60 þ 10 log4:479 ¼ 66:5 dB

Note that the sum of 65, 60 and 55 dB is just 66.5 dB.

37-4 Vibration and Shock Handbook

© 2005 by Taylor & Francis Group, LLC

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