Engineering Acoustics. Malcolm J. Crocker

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Название Engineering Acoustics
Автор произведения Malcolm J. Crocker
Жанр Техническая литература
Серия
Издательство Техническая литература
Год выпуска 0
isbn 9781118693827



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4.5 Human auditory field envelope.

      4.3.2 Loudness Measurement

      The way in which the brain interprets the neural pulses is still a matter for research. However, various experiments have been conducted on groups of people to determine people's average sensation of loudness, etc. We should stress that no one's hearing is exactly the same as any other and hence we must find statistical responses.

Schematic illustration of equal loudness contours in which the contours join the sound pressure levels of different frequency pure tones that are judged to be equally loud. The numbers on each contour are the loudness levels in phons.

      We see from the equal loudness contours that the ear is most sensitive to sound at about 4000 Hz. This is mainly due to a quarter wavelength resonance in the ear canal at this frequency. The increase in sensitivity at 12 000 Hz is mainly caused by diffraction effects around the head.

      The equal loudness contours are not flat as the frequency is changed. Notice at 1000 Hz we can just detect sounds of 4 dB, while at 100 Hz they must be 25 dB. Thus the intensity must be 21 dB higher at 100 Hz than 1000 Hz for us to detect a pure tone. This represents a 100‐fold increase in intensity or a 10‐fold increase in sound pressure amplitude. As the sound pressure level is increased, the equal loudness contours became flatter. A 40 dB tone at 1000 Hz appears equally as loud as a 51 dB tone at 100 Hz, an 11 dB increase. A 70 dB tone at 1000 Hz is as loud as a 75 dB tone at 100 Hz, only a 5 dB increase.

      where S is loudness (sone) and P is loudness level (phon).

Graph depicts the relationship between the loudness and the loudness level of a sound.

      Example 4.1

      Example 4.2

      Given three pure tones with the following frequencies and sound pressure levels: 100 Hz at 60 dB, 200 Hz at 70 dB, and 1000 Hz at 80 dB:

      1 Calculate the total loudness in sones of these three pure tones.

      2 Find the sound pressure level of a single 2000 Hz pure tone which has the same loudness as all the three pure tones combined.

      Solution

      1 The loudness level in phons of each tone is found from Figure 4.6 and the corresponding loudness in sones is found from Eq. (4.1). Then we have that100 Hz @ 60 dB has P = 50 phons and S = 2 sones200 Hz @ 70 dB has P = 70 phons and S = 8 sones1000 Hz @ 80 dB has P = 80 phons and S = 16 sonesTherefore, the total loudness of these three pure tones is 2 + 8 + 16 = 26 sones.

      2 The total loudness of the combined tones (26 sones) corresponds to a loudness level of P = 10log2(26) + 40. Since log2(26) = ln(26)/ln(2) = 4.7, then P = 87 phons.Now, from Figure 4.6 we observe that a pure tone of 2000 Hz and loudness level of 87 phons has a sound pressure level of 82 dB.

      So far we have discussed the loudness of pure tones. However, many of the noises we experience, although they may contain pure tones, are predominantly broadband. Similar loudness rating schemes have been worked out for broadband noise. In 1958–1960, Zwicker [12], taking into account masking effects (see later discussion in Section 4.3.3), devised a graphical method to compute the loudness of a broadband noise. It should be noted that Zwicker's method can be used both for diffuse and free field conditions and for broadband noise