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Excess attenuation – barriers

Excess attenuation – barriers. Noise screens (1). A noise screen causes an insertion loss  L :  L = (L 0 ) - (L b ) (dB) where L b and L 0 are the values of the SPL with and without the screen.

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Excess attenuation – barriers

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  1. Excess attenuation – barriers

  2. Noise screens (1) • A noise screen causes an insertion lossL: • L = (L0) - (Lb) (dB) • where Lb and L0 are the values of the SPL with and without the screen. • In the most general case, there arey many paths for the sound to reach the receiver when the barrier is installed: • diffraction at upper (B) and side(C,D)edges of the screen • passing through the screen (SA), • reflection over other surfaces present in proximity (building, etc. - SEA). 

  3. Noise screens (2): the MAEKAWA formulas • If we only consider the enrgy diffracted by the upper edge of an infinitely long barrier we can estimate the insertion loss as: • L= 10 log (3+20 N) for N>0 (point source) • L = 10 log (2+5.5 N) for N>0 (linear source) • where N is Fresnel number defined by:  • in which  is the wavelength and  is the path difference among the diffracted and the direct sound.

  4. Maekawa chart

  5. Noise screens (3): finite length • If the barrier is not infinite, we need also to consider its lateral edges, each with its Fresnel number (N1, N2), and we have: • L = Ld - 10 log (1 + N/N1 + N/N2) (dB) • Valid for values of N, N1, N2 > 1. • The lateral diffraction is only sensible when the side edge is closer to the source-receiver path than 5 times the “effective height”. heff R S

  6. Noise screens (4) • Analysis: • The insertion loss value depends strongly on frequency: • low frequency  small sound attenuation. • The spectrum of the sound source must be known for assessing the insertion loss value at each frequency, and then recombining the values at all the frequencies for recomputing the A-weighted SPL. SPL (dB) No barrier – L0 = 78 dB(A) Barrier – Lb = 57 dB(A) f (Hz)

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