Building Acoustics

Sound transmission in buildings. Flanking transmission. 339

the prospective absorber. Similar conditions are imposed on the receiving side, hence we
may write
ττ τS=⋅S,pl S,a and τ τ τR= ⋅R ,pl R,a. (9.14)

There is a major problem using transmission factor data, normally determined for diffuse
field conditions, in this situation where the sound field on the receiving and sending side,
respectively, are far from diffuse. Another problem is how the power transmitted into the
plenum spreads out. One part will be transported in the direction of the partition, another
goes in the direction of the backing wall, if there is one, and in turn is reflected back. The
ratio of these two parts is characterized by the quantity sS, a ratio that for lack of better
alternatives is put equal to 0.5. Both parts will be attenuated during propagation in the
direction of the partition, attenuation is assumed to take place exponentially as seen in
the expression

Wx eS,h()∝ −mxS⋅, (9.15)

where mS is the power attenuation coefficient (m-1) in the plenum of height h. With an
absorber the plenum could be considered as a rectangular duct lined on one side with an
absorber, and we may use routines for finding the complex propagation coefficient Γ or
the complex wavenumber k ́ in such a duct (see e.g. Mechel (1976)). The attenuation
coefficient is then found from

m=⋅ Γ=−⋅2Re{ } 2Im ',{k} (9.16)

where Re and Im denote the “the real part of” and “the imaginary part of”, respectively.
All power contributions are integrated over the length LS to arrive at an expression
for the total power passing over to the plenum on the receiver side. A similar derivation
is carried out for the receiving side except for taking account of the attenuation partly
caused by transmission through the ceiling and into the receiver room. Hence, the
attenuation coefficient here is expressed as

(^) RRRR,
s
mm
h
τ
′ =+ (9.17)
where sR is the ratio mentioned above, applied to the receiver side of the plenum. Now
assuming that all sidewalls in the plenum are totally reflecting, i.e. the reflection factor is
equal to 1.0, we get

SRSR R(^2 SS)(^2 RR)

SS R R

cl^11 mL mL

ss L

ee

mL m L h

.

ττ

τ =⋅−−− − ′

⋅

(9.18)

The other extreme situation, assuming the sidewalls are totally absorbing, gives the same
expression, however without the factor 2 in the exponential terms. In the case of minor
attenuation in the plenum (mSLS, mRLR << 1), also putting sS = sR = 0.5, we arrive at the
following very simple expression for the transmission factor

(^) cl^2 S R R.
4

L

h

τεττ= (9.19)