(kx i t i kx i kx ik x ik ∫∫( 2

Problem 1

MASSACHUSETTS INSTITUTE OF TECHNOLOGY DEPARTMENT OF OCEAN ENGINEERING

13.811 Advanced Structural Dynamics and Acoustics

Second Half - Problem Set 2 Solution

Using equation (2.52) from text and substituting the expression for p(x,y,0),

( ) P kx , ky =

Poe-it

(ei2kx + e-i2kx )e-ikx xe-ik y ydxdy

- -

=

Poe-it

(e-i(k x -2k ) x + e-i(2k +k x ) x )e-ik y ydxdy

- -

Using equation (1.5) from text,

( ) P kx , ky = 4 [ 2Poe-it ( kx - 2k ) + ( kx + 2k ) ] ( ky )

(1)

Using equation (2.50) and substituting equation (1) above,

1

p(x, y,z) =

=

= Where

I 1

=

[ ] 1

4 2

[4

- -

2 Poe -it

(kx

- 2k) + (kx

+ 2k)

(k y )eikz z ]eikx x eik y y dk x dk y

[ ] Poe-it (k x - 2k ) + (k x + 2k ) eikx xdk x (k y )ei(kz z+k y y)dk y

-

-

[ ] Poe-it (k x - 2k ) + (k x + 2k ) eikx xdk x I1

(2)

-

(k y )ei(kzz+ky y)dk y

-

Using sifting property of the delta function integral (equation (1.37) of text) and recognizing that k z = k 2 -k x 2 -k y 2 ,

2

I 1

=

ei k2 -kx2 z

(3)

Thus, substituting equation (3) above into (2) above,

p(x, y,z) =

[ ] Poe-it

(k x - 2k ) + (k x + 2k ) eikx xei

dk k 2 -kx 2 z x

-

Again, using sifting property of the delta function integral (equation (1.37) of text),

p(x, y,z) =

[ ] Poe-it ei2kxei -3kz + e-i2kxei -3kz

= Poe-ite- 3kz cos(2kx)

p Figure 1. shows the normalized pressure field 20log dB for normalized distance 0 kx 10 and 0 kz 10.

Po

3

p Figure 1. Normalized pressure field 20log in dB

Po for normalized distance 0 kx 10 and 0 kz 10.

4

Problem 2

Using equation (2.61) and let z = z' = 0,

.

W (kx , k y ,0)

=

kz ock

P(kx , k y ,0)

(1)

By recognizing the definition in equation (2.56) and substituting (1) above into the expression,

P(kx , k y ) P(kx , k y ,0)

P(

kx,ky )

ock

kz

.

W(

kx , ky ,0)

(2)

Using equation (2.50) and observing the form of the inverse 2-D Fourier Transform expression given in equation (1.17)

[ ( ) ] p(x,

y,

z

)

=

Fx

F -1 y

-1

P

kx , ky

eikz z

[ ( )] p(x,

y,0)

=

F F -1 -1 xy

P

kx

,

ky

(3)

5

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