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Hargreaves, JA and Cox, TJ

Title A time domain boundary element method for compliant surfaces

Authors Hargreaves, JA and Cox, TJ

Type Conference or Workshop Item

URL This version is available at: http://usir.salford.ac.uk/id/eprint/19380/

Published Date 2008

USIR is a digital collection of the research output of the University of Salford. Where copyright  permits, full text material held in the repository is made freely available online and can be read,  downloaded and copied for non­commercial private study or research purposes. Please check the  manuscript for any further copyright restrictions.

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1

University of Salford, UK

Architectural and building acoustics Digital signal processing Outdoor sound propagation

Environmental acoustics Virtual acoustic prototypes Acoustics of porous materials

6* 2001 RAE

(3)

2

A time domain Boundary Element Method

for compliant surfaces

(4)

3

• BEM overview

• Welled obstacle model

• Numerical results

(5)

4

• Scattering of sound by an obstacle

• Wave method

• Only the surface of the obstacle is modelled

• The obstacle’s acoustic properties are stated

as surface properties

– Typically as surface impedance for harmonic excitation

• Time domain BEM permits transient broadband

excitation

(6)

5

Animation to illustrate mutual interaction of scatterers, and the time iterating process by which the BEM solves the scattering problem.

(7)

6

(8)

7

d

<<

λ

φin

φ

o

u

t

Consider an obstacle whose surface contains a well with cross-section small w.r.tλ

(9)

8

( )

t in

(

t d c

)

out =

ϕ

−2

ϕ

d

φin

φ

o

u

t

(10)

9

( )

t in

(

t d c

)

out =

ϕ

−2

ϕ

d

( )

t

( )

t p x, =−

ρ

0

ϕ

& x,

( )

x,t

( )

x,t

v =∇

ϕ

(11)

10

( )

t in

(

t d c

)

out =

ϕ

−2

ϕ

(12)

11

( )

t in

(

t d c

)

out =

ϕ

−2

ϕ

d

( )

( )

i kd

in

out Φ e

Φ

ω

=

ω

2

(13)

12

( )

t in

(

t d c

)

out =

ϕ

−2

ϕ

d

( )

( )

Z

( )

i c

( )

kd

V P n t cot 0

ρ

ω

ω

ω

= = −

( )

( )

i kd

in

out Φ e

Φ

ω

=

ω

2

(14)

13

( )

t in

(

t d c

)

out =

ϕ

−2

ϕ

d

( )

( )

Z

( )

i c

( )

kd

V P n t cot 0

ρ

ω

ω

ω

= = −

( )

t v

( ) ( )

t z t

pt = − n

( )

( )

i kd

in

out Φ e

Φ

ω

=

ω

2

(15)

14

( )

t in

(

t d c

)

out =

ϕ

−2

ϕ

d

( )

( )

Z

( )

i c

( )

kd

V P n t cot 0

ρ

ω

ω

ω

= = −

( )

t v

( ) ( )

t z t

pt = − n

( )

( )

i kd

in

out Φ e

Φ

ω

=

ω

2

However, However, a found by inverse discrete Fourier transform of Z) is

(16)

15

in in

t

(17)

16

( )

t

[

( )

t

(

t

d

( )

c

)

]

p

t

x

,

=

ρ

0

ϕ

&

in

x

,

+

ϕ

&

in

x

,

2

x

in

in t

(18)

17

( )

t

[

( )

t

(

t

d

( )

c

)

]

p

t

x

,

=

ρ

0

ϕ

&

in

x

,

+

ϕ

&

in

x

,

2

x

in

in t

( )

[

( )

t

(

t d

( )

c

)

]

c t

vn x, = 1

ϕ

&in x, −

ϕ

&in x, −2 x

(19)

18

( )

=

∫∫

(

( )

∗ ⋅∇

(

)

(

)

( )

)

S

n t

s y,t

ϕ

x,t nˆ g R,t g R,t v x,t dx

ϕ

( )

t

[

( )

t

(

t d

( )

c

)

]

pt x, =−

ρ

0

ϕ

&in x, +

ϕ

&in x, −2 x in

in t

( )

[

( )

t

(

t d

( )

c

)

]

c t

vn x, = 1

ϕ

&in x, −

ϕ

&in x, −2 x

(20)

19

( )

=

∫∫

(

( )

∗ ⋅∇

(

)

(

)

( )

)

S

n t

s y,t

ϕ

x,t nˆ g R,t g R,t v x,t dx

ϕ

( )

t

[

( )

t

(

t d

( )

c

)

]

pt x, =−

ρ

0

ϕ

&in x, +

ϕ

&in x, −2 x in

in t

( )

[

( )

t

(

t d

( )

c

)

]

c t

vn x, = 1

ϕ

&in x, −

ϕ

&in x, −2 x

(21)

20

( )

=

∫∫

(

( )

∗ ⋅∇

(

)

(

)

( )

)

S

n t

s y,t

ϕ

x,t nˆ g R,t g R,t v x,t dx

ϕ

( )

t

[

( )

t

(

t d

( )

c

)

]

pt x, =−

ρ

0

ϕ

&in x, +

ϕ

&in x, −2 x in

in t

( )

[

( )

t

(

t d

( )

c

)

]

c t

vn x, = 1

ϕ

&in x, −

ϕ

&in x, −2 x

φ

in

is the fundamental unknown

You might notice that this means that φinis the fundamental surface unknown; all other surface and scattered quantities can be calculated from it.

(22)

21

( )

=

∫∫

(

( )

∗ ⋅∇

(

)

(

)

( )

)

S

n t

s y,t

ϕ

x,t nˆ g R,t g R,t v x,t dx

ϕ

Back to the KIE:

(23)

22

( )

=

∫∫

(

( )

∗ ⋅∇

(

)

(

)

( )

)

S

n t

s y,t

ϕ

x,t nˆ g R,t g R,t v x,t dx

ϕ

(

)

(

)

R c R t t R g π δ 4 , = − x y− = R

g(R,t) is the free-space time domain greens function, which describes how a instantaneous point source (monopole) at t= 0 radiates to an observer at distance

(24)

23

( )

=

∫∫

(

( )

∗ ⋅∇

(

)

(

)

( )

)

S

n t

s y,t

ϕ

x,t nˆ g R,t g R,t v x,t dx

ϕ

(

)

(

)

R c R t t R g π δ 4 , = − x y− = R

(25)

24

( )

=

∫∫

(

( )

∗ ⋅∇

(

)

(

)

( )

)

S

n t

s y,t

ϕ

x,t nˆ g R,t g R,t v x,t dx

ϕ

(26)

25

( )

t i

( )

t s

( )

t

t y,

ϕ

y,

ϕ

y,

ϕ

= +

( )

=

∫∫

(

( )

∗ ⋅∇

(

)

(

)

( )

)

S

n t

s y,t

ϕ

x,t nˆ g R,t g R,t v x,t dx

ϕ

(27)

26

( )

t i

( )

t s

( )

t

t y,

ϕ

y,

ϕ

y,

ϕ

= +

( )

=

∫∫

(

( )

∗ ⋅∇

(

)

(

)

( )

)

S

n t

s y,t

ϕ

x,t nˆ g R,t g R,t v x,t dx

ϕ

(28)

27

( )

t cv

( )

t pt x, =

ρ

0 n x,

This is because the last piece in the puzzle that lets us solve for surface sound from incident sound, the boundary condition, is written in terms of total sound.

(29)

28

(30)

29

Rigid plate:

(31)

30

Virtual well:

(32)

31

(33)

32

580 elements 400 elements

31% fewer 52% faster

1m sq body, 0.5m deep, 0.1m wells on front face. Equivalent mixed and well

(34)

33

Point Source

o

b

s

ta

c

le

100m

(35)

34

TD

FD

FD

( )

t in

(

t d c

)

out =

ϕ

−2

ϕ

( )

i c

( )

kd

Z

ω

=

ρ

0 cot

Here we have scattered sound magnitude versus position. We have the time domain model, and two different frequency domain models, all very different implementations of different meshes & boundary conditions. To the right of the figure, the source side of the obstacle, an interference pattern is seen and all models agree very well. Inside the obstacle there is good cancellation.

(36)

35

Scope?

(37)

36

(

)

(

ω

)

( )

ω

ω

Z V

P

n t

=

− ,

,

x x

Pt Vn

( )

t v

( ) ( )

t z t

pt x, = − n x, ∗

(38)

37

φ

in

φ

out

( )

t in

( )

t w

( )

t

out x, =

ϕ

x, ∗

ϕ

(

)

(

)

( )

( )

( )

1

1 , , 0 0 + − = = Φ Φ c Z c Z W in out

ρ

ω

ρ

ω

ω

ω

ω

x x

(39)

38

Pt Vn Pt Vn Pt Vn Pt Vn Pt Vn Pt Vn Pt Vn

(

)

( )

( )

1

cos 1 cos , 0 0 + − = = c Z c Z R P P i r

ρ

ω

θ

ρ

ω

θ

θ

ω

P i Pr

θ

(

)

(

ω

)

( )

ω

ω

Z V P n t = − , , x x

(40)

39

(

)

(

)

( )

( )

ω

ω

θ

θ

ω

θ

ω

W W R R − + = − + 1 1 cos , 1 , 1 P i Pr

θ

Φin Φout Φin Φout Φin Φout Φin Φout Φin Φout Φin Φout Φin Φout

(

)

(

ω

)

( )

ω

ω

W Φ Φ in out = , , x x

(41)

40

p i

θ

φin φout φin φout φin φout φin φout φin φout φin φout φin φout

( )

t in

( )

t w

( )

t

out x, =

ϕ

x, ∗

ϕ

( )

,t p

( ) ( )

,t r t,

θ

pr x = i x

pr

From which it’s easy to transfer to the time domain.

So in principle we have here a time domain model of locally reaction compliant obstacles which only makes the same assumptions as the generally accepted use of surface impedance.

(42)

41

• Well mouths can be modelled by a surface impedance • Equivalent formulation proposed in time domain

– Based on causal relation between incoming & outgoing waves • Shown to be accurate and stable

• Could be the basis of a time domain model of locally reacting materials

– Is this a realistic way of representing the response of a material?

(43)

42

University of Salford, UK

Architectural and building acoustics Digital signal processing Outdoor sound propagation

Environmental acoustics Virtual acoustic prototypes Acoustics of porous materials

6* 2001 RAE

References

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