Reverberation time

in non-diffuse rooms

www.odeon.dk

in non-diffuse rooms

Jens Holger Rindel

Odeon A/S

Contents

1. Theory: Mean free path and reverberation time

2. Reverberation in rooms with non-diffuse sound field -

Room acoustics 2

non-diffuse sound field -example

1. Theory: Mean free path and reverberation time

Room acoustics 3

reverberation time

A sound wave represented by a ray

Mean free path,

3D sound field: lm = 4V/ S

V : Volume

S : Surface area

Room acoustics 4

Mean free path,

2D sound field: lm = π Sx /U

Sx : Area

U : Perimeter

NB: It is assumed that all surfaces have the same

absorption coefficient αm

A sound wave represented by a ray

Room acoustics 5

Energy of sound

wave is reduced by

(1- αm) after each

reflection

Sound pressure after n reflections

)1ln(2

0

2

0

2 e)1()( mnn

m pptpα

α−⋅

⋅=−⋅=

m

i

i lntcl ⋅=⋅=∑t

l

cm

mptp⋅−⋅

⋅=

)1ln(2

0

2 e)(α

Total path:

⇒⋅=⇒=−622 10)( ptpTtRevTime - 60 dB decay:

Room acoustics 6

60

)1ln(6 )1ln()10ln(6e10

60

Tl

cm

m

Tl

cm

m⋅−⋅=⋅−⇒=

⋅−⋅

−

α

α

m

m

m

m

c

l

c

lT

αα ⋅

⋅≈

−⋅−

⋅=

8.13

)1ln(

8.1360

General equation for

reverberation time

⇒⋅=⇒=−62

0

2

60 10)( ptpTtRevTime - 60 dB decay:

c : Speed of sound

Vl

4=3D:

General:m

m

m

m

c

l

c

lT

αα ⋅

⋅≈

−⋅−

⋅=

8.13

)1ln(

8.1360

Reverberation time equations

3.55 V⋅ V⋅3.55

Room acoustics 7

S

Vlm

4=

U

Sl x

m

π=

xm ll =

3D:

2D:

1D:

)1ln(

3.55

mSc

V

α−⋅⋅−

⋅

m

x

c

l

α⋅

⋅8.13

mSc

V

α⋅⋅

⋅3.55

m

x

Uc

S

α⋅⋅

⋅4.43

)1ln(

4.43

m

x

Uc

S

α−⋅⋅−

⋅

)1ln(

8.13

m

x

c

l

α−⋅−

⋅

Eyring and Sabine equations (3D)

)1ln(

3.5560

mSc

VT

α−⋅⋅−

⋅=Eyring:

More correct for very high absorption, αm → 1

Special case for 3D diffuse field

Room acoustics 8

Sabine:mSc

VT

α⋅⋅

⋅=

3.5560

Approximately the same as Eyring for αm < 0,3

More correct for very high absorption, αm → 1

2. Reverberation time in non-diffuse rooms – example

Room acoustics 9

rooms – example

Example: Rectangular room

Room acoustics 10

Uneven distribution of absorption – the preconditions for the

reverberation equations of Sabine and Eyring are violated

Results according to theory

Direction lm (m) αm T60 (s)

3-dim. (Sabine)3-dim. (Eyring)

2-dim. (horizontal)

5.75.7

10.5

0.300.30

0.10

0.760.63

4.21

Room acoustics 11

1-dim. (length)1-dim. (width)1-dim. (height)

20105

0.100.100.45

8.024.010.45

lm : Mean free pathαm : Mean absorption coefficientT60 : Reverberation time

The scattering coefficient, s

(1-s)1Ratio of

reflected

Room acoustics 12

s

reflected

energy in

non-specular

directions

Simulation with ray tracing

Low scattering:

s = 0.01

Room acoustics 13

T30 = 1,69 s @ 1 kHz

Simulation with ray tracing

Normal scattering:

s = 0.05

Room acoustics 14

T30 = 1,40 s @ 1 kHz

Simulation with ray tracing

High scattering:

s = 0.50

Room acoustics 15

T30 = 0,71 s @ 1 kHz

Simulation with ray tracing

Very high scattering:

s = 1.00

Room acoustics 16

T30 = 0,71 s @ 1 kHz

Conclusion

• 1-D and 2-D sound fields have longer RT and 3-D sound fields have shorter RT (due to different mean free path)

• The classical equations for reverberation time are based on 3-D sound field and even distribution of absorption

• With uneven distribution of absorption the degree of scattering is most important (RT varies from 1.7 s to 0.7

Room acoustics 17

• With uneven distribution of absorption the degree of scattering is most important (RT varies from 1.7 s to 0.7 s in one example)

• The decay curve is not a straight line in case of low scattering and uneven distribution of absorption, i.e. different results for different evaluation range (20 dB or 30 dB)