Basic Documentation

Sound Measurement Parameters
Again, there are no units associated with decibels since they are a comparison between two
values, (or more scientifically, a ratio between different magnitudes). Also, decibels are used
for different parameters besides sound power level. Decibels are also used to express sound
pressure level, which is discussed below, and as we are keenly aware, is a different sound
parameter than the sound power level.
Sound Pressure Level
As previously stated, sound pressure is concerned with the effect that a specific sound power
level has on a receiver that is usually some distance away from the sound source. Recall in
our analogy about covering the surface of an expanding sphere with a fixed quantity of paint,
a receiver is like a limited area of the sphere, it will receive only a small portion of the paint
(sound power).
Therefore, a receiver is only exposed to a portion of the total sound power. In other words,
the effect of the sound power becomes less and less (is attenuated more and more) on the
receiver. Therefore, the sound power level and sound pressure level are different parameters
and cannot be used interchangeably. However, decibels also are used to express the ratios
of the relative sound intensity or loudness at the receiver.
The basic unit of acoustic pressure is the Pascal (Pa). (One PSI is equivalent to 6,895
Pascals.) Even though a Pascal is a very small unit of pressure measurement, the specific
values of Pascals that are encountered with sound pressure are so small and vary over such
a wide range that the “decibel” approach is applied to express sound pressure levels in a
more practical manner.
The basic formula to determine a specific sound pressure level in decibels is:
Lp = 10 x Log (P ÷ Pref)
2
Where:
Lp = the sound pressure level in dB.
P = the pressure of a specific sound at the receiver in Pascals.
Pref = the reference pressure and is always.
2 x 10
-5
Pascals is approximately the sound pressure on an eardrum at the hearing
threshold.
A person speaking in a normal voice, about three feet away from a listener, will produce a
sound pressure of around 0.02 Pa. Using this formula, let’s determine the sound pressure
level in dB that the listener would experience.
Lp = 10 x Log (2 x 10
-2
÷ 2 x 10
-5
)2
= 10 x Log (10
3
)
2
= 10 x Log (10
6
)
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