Sound absorbers - Building Acoustics - page 204

− j kx j kx

px () =⋅ Ae +⋅ Be

− ⋅ )

(5.78)

1

(

− j kx j kx

and vx ( ) = Ae ⋅ Be .

Z

c

The quantities A and B will be determined by the boundary conditions on each side of the
layer. On the left-hand side, where x is equal zero, we get

pAB =+

1

1

(5.79)

(

− )

and v = A B .

1

Z

c

On the right-hand side, where x is equal to d , the pressure is given by

− j kd j kd

p =⋅ A e +⋅ B e = + ⋅ ( A B ) cos kd −⋅ j ( A − ⋅ B ) sin kd .

(5.80)

2

Hence, using the Equations (5.79)

p =⋅ p cos kd −⋅ j Z v cos kd .

(5.81)

2 1 c 1

Correspondingly, the particle velocity on the output side will be

p

1

(5.82) v =⋅ v cos kd − ⋅ j sin kd .

2 1

Z

c

We may now cast the Equations (5.81) and (5.82) into the form sought after

⎡ cos kd j ⋅ Z sin kd ⎤

c

p p

⎡⎤ ⎡⎤

⎢ ⎥

⎥

1 2

= .

(5.83) sin kd

⎢⎥ ⎢⎥

⎢ ⋅

j cos kd

v v

⎣⎦ ⎣⎦

1 2

⎢ Z ⎥

⎣ ⎦

c

As an example, we shall assume that the layer is placed on to an infinitely hard wall,
which implies that v 2 is equal to zero or that the load impedance Z L in Equation (5.76) is
infinite. The input impedance will then be

p

1

Z ==−⋅ j Z t ( kd ) ,

(5.84)

1 c

v

1

which was expected from our formerly derived result (see Equation (5.50) ).
Equation (5.83) may, however, be put into a general form. First, it is common to
substitute the wave number by the propagation coefficient Γ, which is given by j⋅ k .
Second, we shall not assume normal incidence but introduce oblique incidence giving the
wave vector propagating through the material an angle ϕ with the normal. The result will
be