Fig. 9.1 Frequency response for the A-, B-, and C-weighting

A-weighted sound pressure level (spectrum) can be calculated by adding the unweighted sound pressure level and A-weighted gain or loss as shown below:

Lp½dBA& ¼ Lp½dB& þ A weighting

Therefore, A-weighted spectrum is given by:

(33.8, 49.3, 72.8, 80, 47.2) dB at (125, 250, 500, 1000, 2000) Hz, respectively

(c) A-weighted SPL [dBA]:

¼10 log ð2399 þ 85114 þ 1:9E7 þ 10:0E7 þ 52481Þ ¼ 10 log ð11:9E7Þ ¼ 80:8 ½dBA&

Therefore, the A-weighted SPL ¼ 80.8 [dBA].

9.5Homework Exercises

Exercise 9.1

The mathematical expression for a level is:

Exercise 9.2

The sound pressure levels measured in the following five octave bands are:

(a)Determine the unweighted sound pressure level.

(b)Determine the A-weighted spectrum of the octave bands.

(c)Calculate the A-weighted sound pressure levels.

(d)What is the PSIL and communication voice level at 1.2 m between a speaker and a listener?

(Answers): (a) 90.7 [dB]; (b) (81.3, 68.8, 82.0, 66.2, 51.0) [dBA]; (c) 84.9 [dBA];

(d) 73 [dB], shouting.

Exercise 9.3

The sound pressure levels measured in five of the octave bands are given below:

(a)Determine the unweighted sound pressure level.

(b)Determine the A-weighted spectrum of the octave bands.

(c)Calculate the A-weighted sound pressure levels.

(d)What is the PSIL and communication voice level at 1.5 m from a listener?

(Answers): (a) 71.5 [dB]; (b) (18.8, 36.3, 61.8, 70.0, 61.2) [dBA]; (b) 71.1 [dBA];

(c) 65 [dB], very loud voice.

Exercise 9.4

Given the time domain function p(t) below:

p(t) ¼ 0.02 cos (2π 500 t) + 0.04 sin (2π 1000 t)+ 0.06 cos (2π 1500 t) + 0.08 sin (2π 2000 t)[Pa]

(a)Determine the total (combined) sound pressure level (SPL) using:

(i)The summation of the squares of the RMS pressures.

(ii)The SPLs of all individual frequencies.

(iii)The SPL spectrum of octave bands.

(b)Determine the A-weighted sound pressure level (SPL) based on the sound level conversion table.

(Answers): (a) i-iii 71.76 [dB]; (b) 72.72 [dBA].

Exercise 9.5

Given the sound pressure function p(t):

pðtÞ ¼ 0:03 cos ð2π 400 tÞ þ 0:04 sin ð2π 400 tÞ

þ0:06 cos ð2π 1200 tÞ þ 0:08 sin ð2π 1200 tÞ þ0:09 cos ð2π 1600 tÞ þ 0:12 sin ð2π 1600 tÞ½Pa&

(a)Calculate the unweighted sound pressure level.

(b)Calculate the A-weighted sound pressure level.

(c)Calculate the preferred speech interference level (PSIL).

(d)Determine the communication voice level at 0.6 m from a listener.

Hint: Combine the harmonic waves of the same frequencies into one harmonic wave function using the four equivalent forms.

(Answers): (a) 76.41[dB]; (b) 77.08[dBA]; (c) 70.14 [dB]; (d) raised voice.

Exercise 9.6

Given the sound pressure function p(t):

pðtÞ ¼ 0:05 cos ð800πtÞ þ 0:10 cos ð2400πtÞ þ 0:15 cos ð3200πtÞ½Pa&

(a)Calculate the unweighted sound pressure level.

(b)Calculate the A-weighted sound pressure level based on the sound level conversion table.

(c)Calculate the preferred speech interference level (PSIL).

(d)Determine the communication voice level at 0.6 m from a listener.

Answers): (a) 76.41[dB]; (b) 77.08[dBA]; (c) 70.14 [dB]; (d) raised voice.

Exercise 9.7

The sound pressure levels measured in five of the octave bands are given below:

(a)Determine the unweighted total sound pressure level.

(b)Calculate the A-weighted sound pressure level.

(c)Calculate the preferred speech interference level (PSIL).

(d)Determine the communication voice level at 1 m from a listener.

(Answer): (a) 71.52 [dB]; (b) 71.08; (c) 65 [dB]; (d) raised voice.