Acoustic diffraction by an Oscillating strip. This problem is basically solved by a technique called...

Acoustic diffraction by Acoustic diffraction by an Oscillating stripan Oscillating strip

This problem isbasically solved by a technique

calledWiener Hopf

technique

Scheme of Presentation Scheme of Presentation This presentation includes: This presentation includes:

1.1. Introduction of Introduction of OriginatorsOriginators of this Technique of this Technique

2.2. A general tour of A general tour of Wiener Hopf techniqueWiener Hopf technique

4.4. Applications of Wiener-Hopf techniqueApplications of Wiener-Hopf technique

5.5. Acoustic diffraction by an Oscillating stripAcoustic diffraction by an Oscillating strip

a)a) GeometryGeometry

b)b) Formation of the ProblemFormation of the Problem

c)c) Solution of the problemSolution of the problem

d)d) Graphical / Numerical ComputationGraphical / Numerical Computation

Introduction of

Originators of this

Technique

A general tour ofWiener Hopf

technique

Application ofApplication of Wiener-Hopf technique Wiener-Hopf technique

Acoustic diffraction by Acoustic diffraction by an Oscillating stripan Oscillating strip

What is an acoustic Plane What is an acoustic Plane Wave?Wave?

A wave coming from a source A wave coming from a source located at an infinite distance with located at an infinite distance with

constant frequency and amplitude. constant frequency and amplitude.

How it looks like?How it looks like?

GeometryGeometry

Strip GeometryStrip Geometry

Formulation Formulation of the of the

ProblemProblem

We consider the scattering of plane acoustic We consider the scattering of plane acoustic wave from an oscillating strip occupying the wave from an oscillating strip occupying the space at and is oscillating space at and is oscillating in a direction perpendicular to the screen in a direction perpendicular to the screen with velocity , where is a with velocity , where is a periodic function of time whose periodic function of time whose generalized Fourier seriesgeneralized Fourier series is given by is given by

l x 0 y 0

u0 f1t f1tt

f1t n n Cnein 0t.

Here, the Fourier coefficients are given Here, the Fourier coefficients are given

byby

Cn

Cn 1T0

f1te in 0tdt,

with non zero fundamental frequency

0 2T0

0.

Assume the continuity of the velocity across the boundary as given in P. A Cannel paper

ty u0 ft, l x 0,

where the total velocity potential satisfies the wave equation

t

2

x2 2

y2 t 1

c2

2 t t2

.

For convenience, we write

t i,

y 0, x 0,

iwhere φ is the diffracted field and is the incident field given by

i exp ik 1x cos 0 y sin 0 i 1 t, 0 0 ,

Where is the frequency and is the speed of the sound.

k 1 1c , 1 c

We need to solve the following boundaryWe need to solve the following boundary

value problem:value problem:

2

x2 2

y2 1

c2

2 t2

,

y x, 0, t ik 1 sin 0e ik1x cos 0 i 1t u0 f1t l x 0,

x, 0 , t x, 0 , t,y x, 0 , t

y x, 0 , t,, x l, x 0, y 0.

Scheme of SolutionScheme of Solution

1.1. Fourier temporal transformFourier temporal transform of the of the ProblemProblem is taken to is taken to convert it from convert it from time domaintime domain to to frequency domainfrequency domain

2.2. Then Then Fourier Integral transformFourier Integral transform is taken to switch the is taken to switch the problem to problem to Complex domainComplex domain from from Spatial domainSpatial domain

3.3. Problem is now solved using Wiener-Hopf technique in Problem is now solved using Wiener-Hopf technique in Complex domainComplex domain

4.4. Inverse Fourier transform Inverse Fourier transform is taken bring back the problem is taken bring back the problem in in Spatial domainSpatial domain

5.5. Then Then Inverse Temporal transformInverse Temporal transform is taken to bring it back is taken to bring it back to to time domaintime domain

Solution of the ProblemSolution of the Problem

x,y,

x,y, tei tdt,

x,y, t 12

x,y, e i td .

Define the temporal Fourier transform pair as

We need to solve the following boundaryWe need to solve the following boundary

value problem:value problem:

2

x2 2

y2 1

c2

2 t2

,

y x, 0, t ik 1 sin 0e ik1x cos 0 i 1t u0 f1t l x 0,

x, 0 , t x, 0 , t,y x, 0 , t

y x, 0 , t,, x l, x 0, y 0.

Taking the temporal Fourier transform, the above take the form as follows:

2

x2 2

y2 k 2 x,y, 0,

y x, 0, 2 ik 1 sin 0e ik1x cos 0 1 u0 f , l x 0,

x, 0 , x, 0 , ,y x, 0 ,

y x, 0 , , x l, x 0, y 0,

where

k c k 1 ik 2 with k 2 0 and

f 2 Cn n 0.

,y,

x,y, ei xdx,

x,y, 12

,y, e i xd ,

Define Spatial Fourier transform over the variable as follows:x

with

,y, ,y, 1 ,y, e i l ,y, ,

,y, 0

x,y, ei xdx,

1 ,y, l

0

x,y, ei xdx,

,y,

0

x,y, ei xdx.

1 , 0, 2 k1 sin 0 1

k1 cos 01 exp i k 1 cos 0l

u 0f i 1 exp i l

, l x 0,

Transforming above equations into -plane,

x, 0 , x, 0 , ,y x, 0 ,

y x, 0 , , x l, x 0, y 0,

where 2 2 k 2 with Re 0 in the strip Imk Im Imk.

d 2

dy2 2 ,y, 0,

,y, A , e y , if y 0,

A , e y, if y 0,

The Solution of after using the continuity of Ψ′ across y=0 is given by

d 2

dy2 2 ,y, 0,

,y, ,y, 1 ,y, e i l ,y, ,Now using above and following equation,

we have

A , 0 , 1 , 0 , e i l , 0 , ,

and

A , 0 , 1 , 0 , e i l , 0 , .

Adding and subtracting above two equations Adding and subtracting above two equations we have we have First Wiener HopfFirst Wiener Hopf equation: equation:

2S e i l 2 , 0, 2J1 , 0, 0,

2S , 0 , , 0 , ,

2J1 , 0, 1 , 0 , 1 , 0 , ,

2 , 0, 2 , 0 , , 0 , , 0 , ,

where

andandA J2 , 0, ,

whereJ2 , 0, 1

2 1 , 0 , 1 , 0 , ,

,y, A , e y , if y 0,

A , e y, if y 0,

Now, solving again

,y, ,y, 1 ,y, e i l ,y, ,and

x, 0 , x, 0 , ,y x, 0 ,

y x, 0 , , x l, x 0, y 0,

and

A , 0, 1

, 0, e i l , 0, .

Using the continuity of across y 0

We arrive at the Second Wiener Hopf equation

e i l , 0, J2 , 0,

, 0, 2 k 1 sin 0 1G u 0f

i 1 exp i l,where

G 1 exp i k1 cos 0l k1 cos 0

.

equations mentioned above in red boxes

and after solving

We get

Solution of above W.E equation, as laid down in B. Noble’s book, we Solution of above W.E equation, as laid down in B. Noble’s book, we arrive at the following value of the arrive at the following value of the

A 1 A exp i k 1 cos 0l k 1 cos 0 k 1 cos 0

A R2 exp i l

exp i lA kT C2

iu0 f exp i l

0

exp i lT Tk k iu0 f k1

2kT2k 1

k 1

0

exp i lT Tk k iu0 f

k1 2kT2k

A

k 1 cos 0 k 1 cos 0

A eilk1 cos 0R1 iu0 f

0

T Tk k iu0 f

k1 2kT2k

1 k

1 0

T Tk k iu0 f

k1 2kT2k

AT kC1. 5.44

A

x,y, 12

,y, e i xd 12

A , e |y | i xd .

The function x,y, can be obtained by taking the inverse Fourier transform

Now, x,y, can be broken up as follows:

x,y, sepx,y, intx,y, ,

sepx,y, 12

1

A exp i k1 cos 0l k1 cos 0 k1 cos 0

A k1 cos 0 k1 cos 0

where

iu 0f exp i l 0

iu 0f 0

e |y | i xd ,

intx,y, 12

1 A R2 exp i l A kT C2 exp i l

T Tk k iu 0f exp i l

k1 2kT2k

1 k

1 0

T Tk k iu 0f exp i l

k1 2kT2k

A eilk1 cos 0R1 T Tk k iu 0f

k1 2kT2k

1 k

1 0

T Tk k iu 0f

k1 2kT2k

AT kC1 e |y | i xd .

Here sepx,y, consists of two parts each representing the diffracted field produced by the edges at x 0

and x l, respectively, as though the other edge were absent while intx,y, gives the interaction of

one edge upon the other.

Far Field Approximation Far Field Approximation

The far field approximation can found asymptotically. We arrive at the following results

Graphical Graphical DiscussionDiscussion

Case 1Case 1We consider the case where the We consider the case where the

angle of incidence to the oscillating angle of incidence to the oscillating strip (strip (with constant frequencywith constant frequency) is) is

different say 30different say 30oo, 45, 45oo and 90 and 90oo

Amplitude of the separated field for different values of angle of incidence

Case 2Case 2We consider the case where the We consider the case where the frequency of the incidence ray is frequency of the incidence ray is

different and the strip is oscillating different and the strip is oscillating with constant frequency. with constant frequency.

Amplitude of the separated field for different values of wave frequency

Case 3Case 3We consider the case where the We consider the case where the frequency of the incidence ray is frequency of the incidence ray is

constant and the strip is oscillating constant and the strip is oscillating with different frequencies. with different frequencies.

Amplitude of the separated field for different values of strip frequency

Respected Gentlemen Respected Gentlemen I take pride to introduce aI take pride to introduce a

New ideaNew idea inin

Wiener Hopf technique Wiener Hopf technique in the in the

Strip GeometryStrip Geometry

To understand this, we need to To understand this, we need to have the idea of the half plane and have the idea of the half plane and

the strip geometries. the strip geometries. Lets have a trip to these. Lets have a trip to these.

Half Plane GeometryHalf Plane Geometry

Strip GeometryStrip Geometry

DDiscussioniscussionIf we gradually increase the value of If we gradually increase the value of

the strip length the strip length ll. . PleasePlease

Imagine what will happenImagine what will happen??

DiscussionDiscussion

The Strip geometry supposed to be The Strip geometry supposed to be converted into Half plane geometry.converted into Half plane geometry.

This phenomenon must be This phenomenon must be implemented to check the validity of implemented to check the validity of

the results obtained for the Strip the results obtained for the Strip geometry both Mathematically and geometry both Mathematically and

Graphically.Graphically.

Mathematical VerificationMathematical Verification

Mathematical VerificationMathematical Verification

Graphical VerificationGraphical Verification

Amplitude of the separated field for Amplitude of the separated field for different values of strip frequency for different values of strip frequency for

l 101

Amplitude of the separated field for different values Amplitude of the separated field for different values of strip frequency forof strip frequency for

l 1014

Amplitude of the separated field for different values Amplitude of the separated field for different values of of stripstrip frequency for frequency for l 1022 .

Bashir’s Half Plane resultBashir’s Half Plane result

Amplitude of the diffracted field for different values of Amplitude of the diffracted field for different values of half planehalf plane frequency frequency

Amplitude of the separated field for different

values of Oscillating Strip frequency for l 1022 .

Amplitude of the diffracted field for

different values of Half Plane frequency

Question/AnswerQuestion/Answer

SessionSession