3 Presentation3 5 dB Arithmetic

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dB ArithmeticdB Arithmetic

Basic SPL CalculationsBasic SPL Calculations

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Decibels (dB) scale Decibels (dB) scale

Definition: Definition:

A linear numbering scale used to A linear numbering scale used to define a logarithmic amplitude define a logarithmic amplitude scale, thereby compressing a scale, thereby compressing a wide range of amplitude values wide range of amplitude values to a small set of numbers. to a small set of numbers.

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Our ear detect Sound Intensity, I.Our ear detect Sound Intensity, I.

We always measure Sound We always measure Sound Pressure, p (Pa – N/mPressure, p (Pa – N/m22).).

Relationship between Sound Relationship between Sound Intensity, I and Pressure, pIntensity, I and Pressure, p

I I (pressure) (pressure)22

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Our ear detects sound pressure as Our ear detects sound pressure as low as low as 2 x 102 x 10-5-5 N/m N/m22..

The highest (before pain) is 200 The highest (before pain) is 200 N/mN/m22..

______ 200 N/m______ 200 N/m22

}} 101066 N/m N/m22

Units of Units of differencedifference

_______ 2 x 10_______ 2 x 10-5-5 N/m N/m22

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Bell (Alexander Graham) took logs Bell (Alexander Graham) took logs to reduce the range by a factor of to reduce the range by a factor of 1010

Bel = log ( I / IBel = log ( I / Irefref ) )

= log ( p / p= log ( p / prefref))22

= log (= log (actual sound pressureactual sound pressure / / reference sound reference sound

pressurepressure))22

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But the scale was not fine enough, so he But the scale was not fine enough, so he developed a tenth of this scale or deciBeldeveloped a tenth of this scale or deciBel

dB = 10 log ( p / pdB = 10 log ( p / prefref))22

= 20 log ( p / p= 20 log ( p / prefref))

where pwhere prefref = 2 x 10 = 2 x 10-5-5 N/m N/m22

Note that dB is not a unit !!!!Note that dB is not a unit !!!!

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Sound IntensitySound Intensity In sound field where waves emitting In sound field where waves emitting

from a source refer to “intensity” of from a source refer to “intensity” of sound at a point.sound at a point.

IntensityIntensity, I is the amount of energy , I is the amount of energy passing through unit area per unit time passing through unit area per unit time and is expressed in Watts / mand is expressed in Watts / m22

I I (pressure) (pressure)22

I I (p) (p)22

(I / I(I / Irefref) = (p / p) = (p / prefref))22

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but,but, dB = 10 log (p / p dB = 10 log (p / prefref))22

= 10 log (I / I= 10 log (I / Irefref))

where Iwhere Irefref = threshold intensity, = threshold intensity, measured in measured in

Watt / mWatt / m22

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For plane wave,For plane wave,

Intensity, I = pIntensity, I = p2 2 / / cc

WhereWhere = density = density

c = velocity of soundc = velocity of sound

c = 410 rayls in air at normal c = 410 rayls in air at normal temperatures and pressures. temperatures and pressures.

= characteristic acoustic impedance = characteristic acoustic impedance of air.of air.

so, threshold intensity = [ (2x10so, threshold intensity = [ (2x10-5-5))22 / 410 ] W/m / 410 ] W/m22

= 10= 10-12-12 W/m W/m22

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Pressure in dB is called Sound Pressure Pressure in dB is called Sound Pressure Level (SPL) and can be written as,Level (SPL) and can be written as,

SPL = 20 log (actual sound pressure / SPL = 20 log (actual sound pressure / ref. ref. sound sound pressure)pressure)

= 10 log (I / I= 10 log (I / Irefref))

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ExampleExample

The RMS pressure of a sound is 200 The RMS pressure of a sound is 200 Pa (N/mPa (N/m22). What is the sound ). What is the sound pressure level (SPL) ?pressure level (SPL) ?

SPL = 20 log [ (200) / (2x10SPL = 20 log [ (200) / (2x10-5-5) ]) ] = 20 log [10= 20 log [1077]] = 140 dB= 140 dB

Note: 200 Pa or 140 dB is the peak Note: 200 Pa or 140 dB is the peak action level of the Noise at Work action level of the Noise at Work Regulations 1989.Regulations 1989.

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ExampleExample

What is the intensity of a sound whose RMS What is the intensity of a sound whose RMS is 200 Pa ?is 200 Pa ?

I = pI = p22 / / cc = (200)= (200)22 /410 /410

= 97.8 W/m= 97.8 W/m22

Sound Intensity level of the sound = 10 log Sound Intensity level of the sound = 10 log (I / I(I / Irefref))

= 10 log [ (97.8) / (10= 10 log [ (97.8) / (10-12-12) ) ]]

= 10 log 10= 10 log 101414

= 140 dB= 140 dB

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ExampleExampleWhat is the sound pressure level in What is the sound pressure level in decibels of a sound whose decibels of a sound whose intensity is 0.01 W/m2 ?intensity is 0.01 W/m2 ?

SPL = 10 log [ 0.01 / 10SPL = 10 log [ 0.01 / 10-12-12 ] ] = 10 log 10= 10 log 101010

= 100 dB= 100 dB

Intensity can also be defined as the Intensity can also be defined as the amount of sound energy (or sound amount of sound energy (or sound power) that is delivered to the ear.power) that is delivered to the ear.

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ExampleExample

Suppose p = 2 x 10Suppose p = 2 x 10-1-1 N/m N/m22. What . What is the decibel level?is the decibel level?

dB = 20 log [ (2x10dB = 20 log [ (2x10-1-1) / (2x10) / (2x10-5-5) ]) ]

= 10 log [ 10= 10 log [ 1044 ] ]

= 80 dB= 80 dB

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……………………CONTINUECONTINUE

If the sound pressure is double, If the sound pressure is double, what is the new dB value?what is the new dB value?

dB = 20 log [ (2x2x10dB = 20 log [ (2x2x10-1-1) / (2x10) / (2x10-5-5) ]) ]

= 10 log [ 2x10= 10 log [ 2x1044 ] ]

= 86 dB= 86 dB

When we double the sound When we double the sound pressure, 6 dB is added.pressure, 6 dB is added.

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Addition of Sound LevelAddition of Sound Level

If we add 2 or more sound If we add 2 or more sound pressure levels together, we add pressure levels together, we add the intensities and convert back the intensities and convert back to dB.to dB.

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ExampleExample

Two machines each have 90 dB Two machines each have 90 dB sound level. What is the sound level. What is the resultant dB when both resultant dB when both machines are working together?machines are working together?

90 dB90 dB = = 10 log (I10 log (I9090 / I / Irefref))

9 dB 9 dB = = log (I log (I9090 / I / Irefref))

101099 = = II9090 / I / Irefref

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II9090 = 10 = 1099 x I x Irefref …………………………machine 1machine 1

AlsoAlso

II9090 = 10 = 1099 x I x Irefref …………….machine …………….machine 22

Adding,Adding, I = I = [I[I9090 + I + I9090]]

= [ 2x10= [ 2x1099 x I x Irefref ] watts/m ] watts/m22

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dB = 10 log (I / IdB = 10 log (I / Irefref))

where where

IIrefref = 10 = 10-12-12 watts/m watts/m22

dBdB = 10 log [(2x10 = 10 log [(2x1099 x I x Irefref) / I) / Irefref]]

= 10 log [(2x10= 10 log [(2x1099]]

= 93 dB= 93 dB

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Averaging decibelsAveraging decibels

The average of a number of The average of a number of decibels may be found from the decibels may be found from the following equation:following equation:

LLAVAV = 10 log [ (10 = 10 log [ (10L1/10L1/10 + 10 + 10L2/10L2/10 + + …….+ …….+ 10 10Ln/10Ln/10) / n ]) / n ]

= 10 log [(10= 10 log [(10L1/10L1/10 + 10 + 10L2/10L2/10 + ……. + …….+ + 10 10Ln/10Ln/10)] - 10 log n)] - 10 log n

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wherewhere

n = number of different n = number of different soundssounds

LLAVAV = average of the = average of the decibelsdecibels

LL11 = first SPL in dB = first SPL in dB

LL22 = second SPL in dB = second SPL in dB

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Time varying noiseTime varying noise

The level of many noises varies with The level of many noises varies with time, for eg. traffic sounds. It is not time, for eg. traffic sounds. It is not easy to find a measure which can easy to find a measure which can accurately quantify with a single accurately quantify with a single number what is heard. Statistical number what is heard. Statistical measurement are made in dB(A) measurement are made in dB(A) because that corresponds because that corresponds approximately to the response of the approximately to the response of the ear.ear.

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LL1010 = sound level in dB(A) which is = sound level in dB(A) which is exceeded for 10% of the time. exceeded for 10% of the time.

LL5050 = sound level in dB(A) which is = sound level in dB(A) which is exceeded for 50% of the time. exceeded for 50% of the time.

LL9090= sound level in dB(A) which is = sound level in dB(A) which is exceeded for 90% of the time. exceeded for 90% of the time.

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Equivalent continuous noise Equivalent continuous noise level, Llevel, Leqeq

LLeqeq is the sound pressure level of is the sound pressure level of a steady sound that has, over a a steady sound that has, over a given period, the same energy as given period, the same energy as the fluctuating sound. It is an the fluctuating sound. It is an average and is measured in average and is measured in dB(A).dB(A).

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LLeqeq = 10 log {(t = 10 log {(t11 x 10 x 10L1/10L1/10 + t + t22 x 10 x 10L2/10L2/10 + + tt33 x 10 x 10L3/10L3/10 + ..........+ t + ..........+ tnn x 10 x 10Ln/10Ln/10) / T}) / T}

WhereWhere

tt11 = time at L = time at L11 dB(A) dB(A)

tt22 = time at L = time at L22 dB(A) dB(A)

tt33 = time at L = time at L33 dB(A) dB(A)T = total time over which T = total time over which

the the LLeqeq is required. is required.

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ExampleExample

Calculate the LCalculate the Leqeq over an eight-hour over an eight-hour day for a worker exposed to the day for a worker exposed to the following noise levels and duration.following noise levels and duration.

dB(A)dB(A) Time (hour)Time (hour)

9494 33

8989 22

9898 0.50.5

8383 2.52.5

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LLeqeq = 10 log {(3 x 10 = 10 log {(3 x 109.49.4 + 2 x 10 + 2 x 108.98.9 + 0.5 x + 0.5 x

10109.89.8 + 2.5 x 10 + 2.5 x 108.38.3) / 8}) / 8}

= 92 dB(A)= 92 dB(A)

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Sound PowerSound Power

Total sound power in watts is equal Total sound power in watts is equal to the intensity in watts / mto the intensity in watts / m22 multiplied by the area in mmultiplied by the area in m22. .

Since Intensity is directly Since Intensity is directly proportional to the square of the proportional to the square of the sound pressure, so the sound sound pressure, so the sound power level (SWL) (or Intensity power level (SWL) (or Intensity level) is proportional to the square level) is proportional to the square of sound pressure level.of sound pressure level.

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SWL = 10 log (SWL = 10 log (actual Intensityactual Intensity / / ref. ref.

IntensityIntensity))

= 10 log (W= 10 log (Wactualactual / W / Wrefref))

where where

WWactualactual is the sound power is the sound power in watts and Win watts and Wrefref is the reference is the reference power of 10power of 10-12-12 watts. watts.

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ExampleExample

Determine the sound power level Determine the sound power level of 0.001 watts.of 0.001 watts.

SWL SWL = 10 log (0.001 / 10= 10 log (0.001 / 10-12-12))

= 10 log (10= 10 log (10-3-3) + 120) + 120

= 90 dB re 10= 90 dB re 10-12-12 W W

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Sound Power and Sound Sound Power and Sound IntensityIntensity

The sound intensity from a point source The sound intensity from a point source of sound radiating uniformly into free of sound radiating uniformly into free space can be found from the power space can be found from the power output and the distance from the output and the distance from the source, r.source, r.

Intensity, I = [(sound power W (watts)) / Intensity, I = [(sound power W (watts)) / 44rr22 ] ]

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If the sound is produced at If the sound is produced at ground level, assuming that the ground level, assuming that the ground is perfectly reflecting, ground is perfectly reflecting, then the energy is only radiated then the energy is only radiated into a hemisphere instead of a into a hemisphere instead of a complete sphere. Intensity complete sphere. Intensity becomes,becomes,

I = W / 2I = W / 2rr22

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ExampleExample

Calculate the intensity and SPL Calculate the intensity and SPL of a sound at a distance of 10 m of a sound at a distance of 10 m from a uniformly radiating from a uniformly radiating source of 1 watt power.source of 1 watt power.

Intensity, I = Intensity, I = W / 4W / 4rr22

= = 1.0 / 41.0 / 4 (10) (10)22

= = 7.95 x 107.95 x 10-4-4 W/m W/m22

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SPL = 10 log [(7.95 x 10SPL = 10 log [(7.95 x 10-4-4) / 10) / 10-12-12]]

= 10 log 7.95 x 10= 10 log 7.95 x 1088

= 89 dB= 89 dB

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Relation between Sound Power Relation between Sound Power Level (SWL), Sound Pressure Level (SWL), Sound Pressure Level (SPL) and distance from Level (SPL) and distance from source to receiver ‘r’ in a free source to receiver ‘r’ in a free field ie with no reflections except field ie with no reflections except the ground.the ground.

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SWL = 10 log (W / WSWL = 10 log (W / Wrefref))(1)(1)

WWrefref = 10 = 10-12-12 Watts Watts

SPL = 10 log (p / pSPL = 10 log (p / prefref))22

(2)(2)

pprefref = (2x10 = (2x10-5-5) N/m) N/m22

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Total power, W = [ pTotal power, W = [ p22 / / c] x area of spherec] x area of sphere

= [ p= [ p22 / / c] x 4c] x 4rr22 (3) (3)

Substitute (1) and (2) into (3),Substitute (1) and (2) into (3),

SPL = SWL – 20 log r – 11SPL = SWL – 20 log r – 11 (for spherical (for spherical radiation)radiation)

SPL = SWL – 20 log r – 8 (for hemispherical SPL = SWL – 20 log r – 8 (for hemispherical radiation) radiation)

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ExampleExample

How many dB’s is the result of How many dB’s is the result of doubling the intensity?doubling the intensity?

SWLSWL11 = 10 log ( I / I = 10 log ( I / Irefref))

= 10 log ( I / 10= 10 log ( I / 10-12-12))

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Double,Double,

SWLSWL22 = 10 log ( 2I / I = 10 log ( 2I / Irefref))

= 10 log ( 2I / 10= 10 log ( 2I / 10-12-12))

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Suppose,Suppose, I = 1 watt/mI = 1 watt/m22

SWLSWL11 = 120 dB = 120 dB

SWLSWL22 = 123 dB = 123 dB

ie: 3 dB increase.ie: 3 dB increase.

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dB dB Summation…..Summation…..

Other Method…Other Method…

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dB Summation….. Other dB Summation….. Other Method…Method…

50 dBA + 50 dBA = 100 dBA ???50 dBA + 50 dBA = 100 dBA ???

………………………………..WRONG !WRONG !

50 dBA + 50 dBA = 50 dBA + 50 dBA = 5353 dBA dBA

…………. . THAT’S RIGHT !THAT’S RIGHT !

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dB SummationdB Summation

Cannot add up the Cannot add up the numbersnumbers………… .as the dBs are ………… .as the dBs are “ratios”.“ratios”.

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dB SPLdB SPL11=Pressure 1, dB =Pressure 1, dB SPLSPL22=Pressure 2,=Pressure 2,

New Pressure= Pressure 1 + New Pressure= Pressure 1 + Pressure 2.Pressure 2.

New SPL = 20 log (New SPL = 20 log (New pressureNew pressure / / Ref Ref

pressurepressure))

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Use of ChartsUse of Charts Use of TablesUse of Tables Exact CalculationsExact Calculations

Use of charts and tables are Use of charts and tables are convenient, and for quick rule of convenient, and for quick rule of thumb estimates.thumb estimates.Use of calculators for accurate Use of calculators for accurate calculations.calculations.

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By CalculationsBy Calculations

dBdB11 = 10 log (Pressure = 10 log (Pressure1 1 / Ref pressure)/ Ref pressure)22

dBdB22 = 10 log (Pressure = 10 log (Pressure2 2 / Ref pressure)/ Ref pressure)22

1010dB1/10 dB1/10 = (Pressure= (Pressure1 1 / Ref pressure)/ Ref pressure)22

1010dB2/10dB2/10 = (Pressure = (Pressure2 2 / Ref pressure)/ Ref pressure)22

(New Pr(New Pr / Ref pr.)/ Ref pr.)22 = (Pr = (Pr1 1 / Ref pr)/ Ref pr)2 2 + (Pr+ (Pr2 2 / Ref / Ref pr)pr)2 2

= 10= 10dB1/10dB1/10 + 10 + 10dB2/10dB2/10

New dB = 10 log (New PressureNew dB = 10 log (New Pressure / Ref / Ref pressure)pressure)22

= 10 log (10= 10 log (10dB1/10 dB1/10 + 10+ 10dB2/10 dB2/10 ).).

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By CalculationsBy Calculations

Nett for 2 SPL = 10 log ( 10 Nett for 2 SPL = 10 log ( 10 dB1/10dB1/10 + 10 + 10 dB2/10dB2/10).).

3 SPL = 10 log (103 SPL = 10 log (10dB1/10 dB1/10 + 10+ 10dB2/10 dB2/10 + + 1010dB3/10dB3/10).).

n SPL = 10 log (10n SPL = 10 log (10dB1/10 dB1/10 + 10+ 10dB2/10 dB2/10 + + 1010dB3/10dB3/10

+……+ 10+……+ 10dBn/10dBn/10).).

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Use of ChartsUse of Charts

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To use the curve proceed as follows:1. Calculate the difference, ∆ L, between the two sound pressure levels.2. Use the curve to find L+.3. Add L+ to the highest level to get Lt, the total level.

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Use of ChartsUse of Charts Example :Example :

Difference between two Difference between two levels “6 levels “6 dBA”dBA”

Add to highest level “1 dBA”.Add to highest level “1 dBA”.

55 dBA + 49 dBA = ?55 dBA + 49 dBA = ?

Difference 6 dBADifference 6 dBA

Nett = 55 + 1 = 56 dBA.Nett = 55 + 1 = 56 dBA.

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Addition of Two SourcesAddition of Two Sources

If the levels differ If the levels differ byby

The following The following should be should be

added to the higher added to the higher

0 or 1 dB0 or 1 dB 3 dB3 dB

2 or 3 dB2 or 3 dB 2 dB2 dB

4 to 9 dB4 to 9 dB 1 dB1 dB

10 dB or over10 dB or over 0 dB0 dB

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Subtraction of Two Subtraction of Two SourcesSources

If the levels differ If the levels differ byby

The difference is The difference is the higher level the higher level

minusminus

More than 10 dBMore than 10 dB 0 dB0 dB



3 dB3 dB 3 dB3 dB

2 dB2 dB 5 dB (approx)5 dB (approx)

1 dB1 dB 7 dB (approx)7 dB (approx)

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ExamplesExamplesNotes page 16Notes page 16 1.1. 78 dB + 81 dB = ?78 dB + 81 dB = ?

Difference between 2 levels = 81-78 = 3 Difference between 2 levels = 81-78 = 3 dBdB

Value to be added to higher level = 2 dBValue to be added to higher level = 2 dB(From table / graph)(From table / graph)

Resulting level = 2 + 81 = 83 dBResulting level = 2 + 81 = 83 dB

By calculationsBy calculations Result = 10 log (10 Result = 10 log (10 7.87.8 + 10 + 10 8.18.1 ) = 82.8 dB ) = 82.8 dB

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Examples…..Examples…..

2.2. 86 dB + 92 dB = ?86 dB + 92 dB = ?Difference between 2 levels = 92-86 = 6 Difference between 2 levels = 92-86 = 6

dBdBValue to be added to higher level = 1 dBValue to be added to higher level = 1 dBResulting level = 1 + 92 = 93 dBResulting level = 1 + 92 = 93 dB

By calculationsBy calculations Result = 10 log (10 Result = 10 log (10 86/1086/10 + 10 + 10 92/1092/10 ) )

= 10 log (10 = 10 log (10 8.68.6 + 10 + 10 9.29.2 ) ) = 93.0 dB= 93.0 dB

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3.3. Fan 85 dBA as measured at worker’s Fan 85 dBA as measured at worker’s position.position.

Fan to be added at first fan location, 84 Fan to be added at first fan location, 84 dBA. dBA. Find Find new noise levels.new noise levels.

85 dBA + 84 dBA = ?85 dBA + 84 dBA = ?Difference between 2 levels = 85-84 = 1 Difference between 2 levels = 85-84 = 1

dBAdBAValue to be added to higher level = 3 dBAValue to be added to higher level = 3 dBAResulting level = 3 + 85 = 88 dBAResulting level = 3 + 85 = 88 dBA

By calculationsBy calculations Result = 10 log (10 Result = 10 log (10 8.58.5 + 10 + 10 8.48.4 ) = 87.5 dBA ) = 87.5 dBA

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4. A manufacturer’s data sheet shows 4. A manufacturer’s data sheet shows that a compressor model A01 has an that a compressor model A01 has an overall sound power level of 100 dB(A) overall sound power level of 100 dB(A) and model A02 an overall sound power and model A02 an overall sound power level of 92 dB(A). The measured noise level of 92 dB(A). The measured noise level at a worker’s position with both level at a worker’s position with both compressors operating was 87 dB(A). compressors operating was 87 dB(A). The compressors are located together. The compressors are located together. What is the likely noise level at the What is the likely noise level at the worker’s position if model A02 is shut worker’s position if model A02 is shut off? Accuracy of manufacturer’s data off? Accuracy of manufacturer’s data can be assumed. can be assumed.

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Model A01 = 100 dBAModel A01 = 100 dBA Model A02 = 92 dBAModel A02 = 92 dBA

Noise at worker’s position = 87 dBANoise at worker’s position = 87 dBAWhat is the noise level with Model A02 off ?...What is the noise level with Model A02 off ?...

Difference between 2 levels = 100-92 = 8 dBADifference between 2 levels = 100-92 = 8 dBAValue to be subtracted from higher level = 1 dBAValue to be subtracted from higher level = 1 dBA

Resulting level = 87-1 = 86 dBAResulting level = 87-1 = 86 dBA

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Practical Example…..Practical Example…..

Two noise sources operating simultaneously with total Two noise sources operating simultaneously with total SPL of 95 dBA.SPL of 95 dBA.

With one noise source switched off , the remaining SPL With one noise source switched off , the remaining SPL is 91 dBA. is 91 dBA.

What is the SPL of the noise source that’s switched off ?What is the SPL of the noise source that’s switched off ? Original total SPL = 95 dBAOriginal total SPL = 95 dBA

Difference in levels was 4 dBA, meaning that 2 Difference in levels was 4 dBA, meaning that 2 dBA dBA to be to be deducted from the higher original deducted from the higher original level.level.Resulting level = 95-2 = 93 dBAResulting level = 95-2 = 93 dBA

Check Sum :Check Sum :91 dBA + 93 dBA = 95 dBA91 dBA + 93 dBA = 95 dBA

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Practical UsagePractical Usage

The subtraction of noise The subtraction of noise source(s) and noise levels from source(s) and noise levels from an existing total measured level an existing total measured level is often used to determine the is often used to determine the individual noise contribution of a individual noise contribution of a particular noise source(s).particular noise source(s).

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Octave Band SummationOctave Band Summation Respective octave band values are Respective octave band values are

summed up to obtain the overall level.summed up to obtain the overall level.

Frequency31.

5 63 125 250 500100

0200

0400

0 8000

(Hz)

SPL Linear dB 73 63 50 48 46 40 35 30 22

Overall dB Overall dB = 73 dB + 63 dB +50 dB + ..= 73 dB + 63 dB +50 dB + ..…….+ 22 dB …….+ 22 dB

= 10 log( 10 = 10 log( 10 7.37.3 +10 +10 6.36.3 +10 +10 5.05.0 +10 +10 4.8 4.8 +10 +10 4.64.6 +10 +10 4.04.0+10 +10 3.53.5+10 +10 3.03.0+10 +10 2.22.2 ) )

= 73 dB= 73 dB

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Octave Band SummationOctave Band Summation If the individual octave bands SPL are in If the individual octave bands SPL are in

linear,linear, the overall summation result in the the overall summation result in the overalloverall Linear SPLLinear SPL..

If the individual octave bands SPL are in If the individual octave bands SPL are in “A-“A-weighted”weighted” , the overall summation result in , the overall summation result in the the “overall A-weighted SPL”“overall A-weighted SPL”..

To convert linear SPL into “A-weighted” SPL To convert linear SPL into “A-weighted” SPL requires “A-weighted” adjustments in the requires “A-weighted” adjustments in the individual frequency bands according to the individual frequency bands according to the “A-weighted curve”.“A-weighted curve”.

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A-weighted SPL A-weighted SPL ConversionsConversions

Consider a sound with linear octave band Consider a sound with linear octave band values as tabulated below. Find the overall values as tabulated below. Find the overall linear SPL and overall A-weighted SPL.linear SPL and overall A-weighted SPL.

Frequency31.

5 6312

525

050

0100

0200

0400

0800

0 Overall

Calculated

SPL Linear 73 63 50 48 45 40 36 30 22 73

A Correction -39

-26 -16 -9 -3 0 1 1 -1

SPL dBA 34 37 34 39 42 40 37 31 21 47

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A-weighted Octave Band A-weighted Octave Band SummationSummation

Respective octave band values after A-weighted Respective octave band values after A-weighted corrections are summed up to obtain the overall level.corrections are summed up to obtain the overall level.

Frequency31.5 63 125 250 500

1000

2000

4000 8000

(Hz)

SPL dBA 34 37 34 39 42 40 36 31 21

Overall dBAOverall dBA = 34 dBA+37 dBA+39 dBA+ = 34 dBA+37 dBA+39 dBA+ ….+ 21 dBA ….+ 21 dBA

= 10 log( 10 = 10 log( 10 3.43.4 +10 +10 3.73.7 +10 +10 3.43.4 +10 +10 3.9 3.9 +10 +10 4.24.2 +10 +10 4.04.0+10 +10 3.63.6+10 +10 3.13.1+10 +10 2.12.1 ) )

= 47 dBA= 47 dBA

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66

67

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Cannot add up the Cannot add up the numbersnumbers………… .as the dBs are ………… .as the dBs are “ratios”.“ratios”.

Have to use anti-log to convert the Have to use anti-log to convert the dB (ratio) back to pressures, ….. and dB (ratio) back to pressures, ….. and add up the pressure. The dB of the add up the pressure. The dB of the resulting pressure is then obtained resulting pressure is then obtained as a ratio to the reference pressure. as a ratio to the reference pressure.

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3 Presentation3 5 dB Arithmetic