Waves in fluid and solid media - Building Acoustics - page 77

1
1

+ R

p

Z = ρ c (3.68)

g 0 0

− R

p

and furthermore

Z

g

− 1

ZZ −

ρ c

g 0

00

R = = where Z = ρ c . (3.69)

p 0 0 0

Z

ZZ +

g

g 0

+ 1

ρ c

00

Inserting this expression into Equation (3.62) we arrive at the absorption factor for
normal incidence expressed as

Z
Z

⎧⎫
⎨ ⎩⎭

g

4Re

0

α = . (3.70)

2

Z Z

⎧⎫

g g

+ 2Re + 1

⎨⎬
⎩⎭

Z Z

0 0

For most simple illustration using these equations we may assume that the boundary
“surface” is dividing two different gases. As a thought experiment we shall have an
infinite long tube containing the gases, which are separated by a massless membrane.
Letting the gases be air and helium, having at 20°C a characteristic impedance of 415
and 170 Pa⋅s/m, respectively, this “surface” will give a reflection factor | R p | equal to 0.42
using Equation (3.69) and an absorption factor α equal to 0.82 by using Equation (3.70).
Another example, which may be more interesting, is the boundary between air and
water. Setting the density and sound speed for water as 1000 kg/m 3 and 1500 m/s,
respectively; we arrive at a characteristic impedance of 1.5⋅10 6 Pa⋅s/m. This implies that
we obtain a pressure reflection factor approximately equal to 0.9995 and an absorption
factor of about 1.1⋅10 -3 . A water surface is therefore, practically speaking, a totally
reflecting surface or a nearly “infinitely hard” surface.

Some special cases of the equations above may be listed:
•

An “infinitely hard” surface, i.e. Z g ⇒ ∞ gives | R p | =1, δ = 0 and α = 0.

• A “soft” surface, denoted a pressure release surface , i.e. Z g ⇒ 0 gives | R p | =1, δ

= π and α = 0.

• A totally absorbing surface, i.e. Z g = Z 0 gives | R p | = 0, δ = 0 and α = 1.

As is apparent from Equation (3.70) , the impedance of the boundary surface
uniquely determines the absorption factor but the opposite is not true. Representing the
absorption factor by parametric curves in a Cartesian coordinate system, using the real
part of the impedance as abscissa and the imaginary part as ordinate, it is relatively easy
to show that the curves are circles. The circles have their centres in ( x 0 ,0) and the their
radii will be ( x 0 2 - 1) 1/2 , where x 0 = (2/α) - 1. This is shown in Figure 3.11 having an
elliptical form due to a difference in scale on the two axes. Finally, Figure 3.12 maybe
illustrate in a better way how the impedance components should be adjusted to achieve a
high absorption factor.