Building Acoustics

162 Building acoustics

{}

{}

{ }

{}

r1 r 2

12

i1 i 2

F(, F(,

(,) , ( ,) ,

F(, F(,

pxt pxt

Rx f Rx f

pxt px t

== (5.6)

and a transfer function H 12 for the total pressure in these two positions:

{}

{}

{ }

{}

2i2r2

12

1i1r 1

.

F ( ,) F ( ,) ( ,)

()

F(,) F (,) (,)

px t p x t p x t

Hf

px t p x t p x t

+

==

+

(5.7)

x

pi(x 1 ,t)

pr(x 1 ,t)

pi(x 2 ,t)

pr(x 2 ,t)

Figure 5.5 Wave components in a standing wave tube.

Correspondingly, for the pressure in the incident and reflected wave, respectively,
we may define transfer functions

(^) []

{}

{}

[]

{ }

{}

i2 r2

(^12) ir 12
i1 r1
.

F(,) F (,)

() , ()

F(,) F (,)

pxt p xt

Hf Hf

pxt p xt

== (5.8)

From Equations (5.6) to (5.8) we get by eliminating R(x 2 ,f):

[ ]

[]

(^1212) i
1
(^12) r 12

(,).

HH

Rx f

HH

−

=

−

(5.9)

We have assumed plane wave propagation only and we may then write

(^) [He 12 ]ir=−⋅−j(kxx12 21)and [H e 12 ] =j(kxx^2121 ⋅−), (5.10)
where k 12 and k 21 are wave numbers for the incident and reflected wave, respectively.
Furthermore, assuming no energy losses in the tube between these two positions, we may
write k 12 = k 21 = k 0 and Equation (5.9) will become
0
0
j
12
(^1) j
12

(,) ,

kd

kd

He

Rx f

eH

− −

=

−

(5.11)