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It is worth noting that a doubling or halving of the sound pressure causes a 6dB
change in the sound pressure level.
Example: -
If the sound pressure doubles in value as an example from 1Pa to 2Pa then the
sound pressure level will increase by 6dB: -
Lp
20 log
⋅
1
20 10
6
−
⋅
⋅
dB
:=
Lp
94dB
:=
Lp
20 log
⋅
2
20 10
6
−
⋅
⋅
dB
:=
Lp
100dB
:=
Although an increase of 6dB represents a doubling of the sound pressure, an
increase of approximately 10dB is required before the sound is perceived as
being twice as loud. Approximately 3dB in sound pressure level is the smallest
change we can detect.
If however the sound power doubles or halves then the sound power level
change will be 3dB. This is commonly referred to as 3dB doubling.
Example: -
If the sound power doubles in value as an example from 1W to 2W then the
sound power level will increase by 3dB: -
L
w
10 log
1
10
12
−
⋅
dB
:=
L
w
110 dB
⋅
:=
L
w
10 log
2
10
12
−
⋅
dB
:=
L
w
113 dB
⋅
:=
It is worth noting that a doubling or halving of the sound power level causes a
3dB change in the sound pressure level.
As decibels are logarithmic it is not possible to directly add or subtract decibel
values.
Use the equation below to add or subtract decibel values (for subtractions
simply replace the plus sign with a minus sign): -
10 log
⋅
10
A
10
10
B
10
+
⋅
dB
Example: -
To add 60dB and 65dB replace A and B with the 60 and 65 and then calculate
the equation to give: -
60dB
65dB
+
66.2dB
:=
