Page 59 - The Master Handbook Of Acoustics

34 CHAPTER TWO

observed. The convention for Equation 2-3 is that sound power is pro-
2
portional to (sound pressure) . The voltage-level gain of an amplifier
in decibels is 20 log (output voltage/input voltage), which holds true
regardless of the input and output impedances. However, for power-
level gain, the impedances must be considered if they are different. If
it is a line amplifier with 600-ohm input and output impedances, well
and good. Otherwise, a correction is required. The important lesson is
to clearly indicate what type of level is intended, or else label the gain
in level as “relative gain, dB.” The following examples illustrate the
use of the decibel.

Example: Sound-Pressure Level
A sound-pressure level (SPL) is 78 dB. What is the sound pressure, p?
– 6
78 dB = 20 log p/(20 × 10 )
– 6
log p/(20 × 10 ) = 78/20
– 6
p/(20 × 10 )=10 3.9
– 6
p = (20 × 10 ) (7,943.3)
p = 0.159 pascals
Remember that the reference level in SPL measurements is 20 µPa.
Example: Loudspeaker SPL

An input of 1 watt produces a SPL of 115 dB at 1 meter. What is the
SPL at 20 ft (6.1 meters)?
SPL = 115 – 20 log (6.1/1)
= 115 – 15.7
= 99.3 dB
The assumption made in the 20 log 6.1 factor is that the loudspeaker is
operating in a free field and that the inverse square law is operating.
This is a reasonable assumption for a 20-foot distance if the loud-
speaker is remote from reflecting surfaces.
An Electro-Voice “constant directivity” horn Model HP9040 is rated at
a sound pressure level of 115 dB on axis at 1 meter with 1 watt into 8 ohms.
If the input were decreased from 1 watt to 0.22 watts, what would be the
sound-pressure level at 1 meter distance?
SPL = 115 – 10 log (0.22/1)
= 115 – 6.6
= 108.4 dB