By David Howard, Jamie Angus

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**Example text**

Calculate the sound intensity level (SIL) at a distance of 2 m from the loudspeaker when it is mounted on 1, 2 and 3 mutually orthogonal boundaries. 18 except for the addition of the term for the directivity, Q. The presence of 1, 2 and 3 mutually orthogonal boundaries the sphere to a hemisphere, half hemisphere and quarter hemisphere, which corresponds to a Q of 2, 4 and 8, respectively. As the only difference between the result with the boundaries is the term in Q, the sound intensity level at 2 m can be calculated as: SIL1 boundary = SIL2 m + 10 log10(Q) = 93 dB + 10 log10(2) = 93 dB + 3 dB = 96 dB SIL2 boundaries = SIL2 m + 10 log10(Q) = 93 dB + 10 log10(4) = 93 dB + 6 dB = 99 dB SIL3 boundaries = SIL2 m + 10 log10(Q) = 93 dB + 10 log10(8) = 93 dB + 9 dB = 102 dB From these calculations we can see that each boundary increases the sound intensity at a point by 3 dB, due to the increased directivity.

2 The effect of phase The phase, which expresses the starting value of the individual sine wave components, also affects the waveshape. 48 shows what happens to a square wave if alternate partials are subtracted rather than added, and this is equivalent to changing the phase of these components by 180°. 48. However, although the time domain waveform is radically different the frequency domain is very similar, as the amplitudes are identical, only the phase of some of the harmonics have changed.

As in the case of diffraction, it is the size of the object with respect to the wavelength that determines how much, and in what way, the energy is scattered. 44. 43. 43 Scattering from an object at low frequencies. 44 Scattering from an object at high frequencies. diffraction, the size at which the object becomes significant is when it is about two thirds of a wavelength (2⁄3λ). Thus objects smaller than this will tend to scatter the energy in all directions whereas objects bigger than this will be more directional.