Prof. David R. JacksonECE Dept.

Spring 2014

Notes 13

ECE 6341

1

Plane Wave ExpansionThe goal is to represent a plane wave in cylindrical coordinates as a series of cylindrical waves (to help us do scattering problems).

Generating function: (Schaum’s Outline Eq. (24.16))

0 ( )jkx

E E x

e

1

2 ( )t

ntn

n

e J t

( , )

x

y

cosx

2

Plane Wave Expansion (cont.)

Let

j

k

t je

2 cos1 1

j

j j jt je je je jt je

1

22 cos

2( ) ( )cos

)( )(t

tk j jkx

njnnj k

nj eJ kee e e

Hence

3

Plane Wave Expansion (cont.)

Generalization:

where

1( )x z zjk x jk z jk z jn

nnn

e e e J k ej

1

2 2 2x zk k k k

or1

( )jkx jnnn

n

e J k ej

“Jacobi-Anger Expansion”

4

Note: the plane wave and the scattered field must have the same kz.

Alternative Derivation

Multiple by e-jmf and integrate over f [0, 2p]

Note that

Let

( )jkx jnn n

n

e a J k e

2

0

2 ,

0,jm jn m n

e e dm n

2

0

2 ( )jkx jmm me e d a J k

Hence

5

Note: The plane-wave field on the LHS is finite on the z axis.

Alternative Derivation

2

0

2 ( )jkx jmm me e d a J k

Hence

2

0

1

2 ( )jkx jm

mm

a e e dJ k

2cos

0

1

2 ( )jk jm

mm

a e e dJ k

or

6

Note: It is not obvious, but am should be a constant (not a function of ).

Alternative Derivation (cont.)

Hence

or

Identity (adapted from Schaum’s Mathematical Handbook Eq. (24.99)):

2 ( cos )0

2( )j x m

mme d J x

j

1 2( )

2 ( )m mmm

a J kJ k j

1m m

aj

7

1( )jkx jn

nmn

e J k ej

We then have

Scattering by CylinderA TMz plane wave is incident on a PEC cylinder.

iH

iE

qi

( )0ˆ x zi j k x k z

yH y H e cos

sinx i

z i

k k

k k

TMz

8

k

x

z

a

Scattering by Cylinder (cont.)

Let( )

1x zj k x k zi

zA A e

( )1

1

1x z

ii zy

j k x k zx

AH

x

jk A e

To find A1:

9

Scattering by Cylinder (cont.)

Hence

or

0 1y x

jH k A

1 0

1y

x

A j Hk

i sz z zA A A

For denotea

10

Scattering by Cylinder (cont.)

where

To solve for , first put into cylindrical form (Jacobi-Anger identity):szA i

zA

1

1( )zjk zi jn

z nnn

A A e J k ej

2 2 2 2 2sin cosz i i xk k k k k k k

21

1( )zjk zs jn

z n nnn

A A e a H k ej

Assume the following form for the scattered field:

11

Scattering by Cylinder (cont.)

At

Both will be satisfied if

a 0

0zE

E

22

2

2

1

1 1

zz z

z

AE k A

j z

AE

j z

, , 0zA a z

12

Scattering by Cylinder (cont.)Hence

This yields

, , , ,z zs iA a z A a z

2n n nJ k a a H k a

(2)

nn

n

J k aa

H k a

or

Wethen have

1

2

2

1( )

( )z

z

njk zs jnnn

n n

J k aA e A H k e

j H k a

13

Scattering by Cylinder (cont.)

Note: We were successful in solving the scattering problem using only a TMz scattered field. This is because the cylinder

was perfectly conducting. For a dielectric cylinder, the scattered field must have BOTH Az and Fz.

14