EM scattering 4 new - Sharifee.sharif.edu/~emscattering_ms/Lecture 4.pdfScattering from cylindrical...

Electromagnetic scattering

Graduate CourseElectrical Engineering (Communications)1st Semester, 1390-1391Sharif University of Technology

Scattering from cylindrical objects 2

Contents of lecture 4

� Contents of lecture 4:

• Scattering from cylindrical objects� Introduction

� Scalar waves in cylindrical coordinates

� Vector wave equation

� Vector solutions in cylindrical coordinates

� Expansion of plane waves

� Scattering by perfectly conducting infinitely long cylinder

� Scattering width

� Scattering by a dielectric cylinder


Introduction

� Scattering from flat, layered media is an example of an exactly solvable problem

� Another class of exactly solvable systems is the scattering by cylinders of infinite length

� In case of dielectric cylinders they are solvable for a constant dielectric constant

ik


Introduction

� The system has rotational symmetry but, of course, the incident plane wave is not rotationally symmetric

� Nonetheless, we can still analyze the problem using the ‘natural’solutions of the wave equation in cylindrical coordinates

� First, consider scalar waves in a homogeneous medium

x

y

z

� �

� �2 2 0k ��

2 22

2 2 2

1 10k

z

� � ��

� � � � ��

� � � ��

In cylindrical coordinates

2 2k � ��


Scalar waves in cylindrical coordinates

� Try solutions of the type

� �( , , ) ( ) exp zz f jm jk z� � � � ��

� General solution:

� �2

2 2 2 22

( ) ( )( ) 0

d f dfk m f

d d ��

� � � ��

� � � �

2 2zk k k� � �

� � � �, ( )zm k m mf AJ k BY k� ��

� � � � � �, ( , , ) expzm k m m zz AJ k BY k jm jk z� ��


Scalar waves in cylindrical coordinates

� Alternatively, one can represent the solution in terms of Hankel

functions of the 1st and 2nd kind

� � � �(1,2) ( )m m mH k J k jY k� � ��

� � � � � �(1) (2), ( , , ) expzm k m r m r zz AH k BH k jm jk z� � � � � ��

� Hankel functions preserve the ‘wave’ picture. For large radial

distances they represent inward and outward moving waves

� � � �(1,2) 2 : ( ) exp / 4mmH k j jk jk� ��

� ��


Vector wave equation

� These were the solutions in a homogeneous medium

� At the first sight the scattering problem for a cylinder can now be

easily solved: just solve the Bessel and Hankel waves outside

and inside the cylinder (if dielectric) and match them at the

interface

� But there are two problems:

• We have to deal with a vector problem and a vector wave equation: EM fields are vector fields and have a polarization.

• Incident wave is not a ‘cylindrical’ wave, but a normal plane wave

� These facts make the analysis ‘a bit’ more complicated!



� Remember vector wave equation in a homogeneous medium

� � � �2 2 2 0k k�� E E E E E

� Instead of trying to move to cylindrical coordinates and directly

solve the equation, consider the following general theorem

� Assume that �(r) is a solution of the ‘scalar’ wave equation solved in any coordinate system

� Also consider a known vector field a(r), such that

( ) 0�� a r



� Now, let us introduce the new vector fields

� �1( ) ( ) ( )k

�� M r r a r 1( ) ( )k

� ��N r M r

� It can then be shown that the 1st vector field is a solution of the

vector wave equation if

� �2 0� �� a a� For the derivation use

� � � ��

2

2 2

� �

� � � �

� � � �

��

��

��

a a

a a a a

a a a a



� The 2nd vector field is then automatically a solution as well

� These two vector fields are linearly independent from each other

� They both have zero divergence

� Every solution of the vector wave equation in a homogeneous

medium can be written as a combination of these vectors,

generated by different solutions of the scalar wave equation

� One cannot generate more functions by taking more curls. Note

that

1( ) ( ) ( )k

k�� N r M r M r



� Solution 1:

� �1( ) ( ) ( ) ( )k

�� E r M r r a r

0 0

1 1( ) ( ) ( ) ( )

j

j j�� H r E r M r N r

� Solution 2:

0

1( ) ( ) ( )

j

j�� H r N r M r

1( ) ( ) ( )

k� � ��E r N r M r



� Example: choose the scalar function to be a solution of the wave

equation (Helmholtz equation) in the Cartesian system

� This would be a simple scalar plane wave

� � � �2 2 0 expk j� �� k r� Take, as an example, a constant a(r), i.e., ˆ( ) �a r x

� �ˆ ˆ( ) expj j� � � � �M r k x k r

� � � �ˆ ˆ ˆ( ) exp j� � � � � �N r k k x k r xk̂M

N

� These correspond to two independent polarizations


Vector waves in cylindrical coordinates

� For cylindrical objects, a proper choice is to use the constant

vector field ˆ( ) �a r z

� This field satisfies all the requirements and results in

� �1 1ˆ ˆk k

� �� M z z

� � 22 2

22

1 1ˆ ˆ

1ˆ

k k z

kk z

� � �

� �

��

N z z

z



� We next use cylindrical coordinates

, , ,,

ˆˆ ˆz z zz

m k m k m km k z

� � ��

� � �

� � ��

� � �z

, ,, ,

,,

1 ˆ ˆˆ

ˆˆ

z z

z z

z

z

m k m km k m k

m km k

k k k

jm

k k

� ��

� � ��

��

� ��

� ��

� ��

M z

, 2, , ,2

1 ˆˆ ˆ zz z z

m k zm k z m k m k

mkjk k

k ��

� ��

��

N z



� Let us see what kind of fields they represent. 1st solution:

,, ,

( )ˆˆexp( ) ( ) zz z

m km k z m k

dfjmjm jk z f

k kd

��

� ��

� � � � ��

E M

� This field has no vertical component (TEz wave). It has an elliptic

polarization in the x-y plane.

� Its corresponding magnetic field is

,

2,

, ,2 2 2

1exp( )

( ) ˆˆ ˆ ( ) ( )

z

z

z z

m k z

m kz zm k m k

j jjm jk z

j

df kjk k mf f

k d k k�

��

��

� �

� � ��

� ��

H E N

z



� The 2nd solution for the electric field is:

� Here, the electric field does have a vertical component.

� It again has elliptic polarization, but in a plane rotated around

the radial unit vector.

,

2,

, ,2 2 2

exp( )

( ) ˆˆ ˆ ( ) ( )

z

z

z z

m k z

m kz zm k m k

jm jk z

df kjk k mf f

k d k k�

�

��

� �

� � � �

� ��

E N

z



� The corresponding magnetic field is

� �, ,,

,

1 1

( )ˆˆ exp( ) ( )

z z

z

z

m k m k

m kz m k

j

j j k

dfj jmjm jk z f

k kd

��

��

� � �

� � ��

� ��

� �

H E M M

� Now the magnetic field lies in the horizontal plane. It has no

vertical component (TMz wave)

� Note also that in both cases

, , 0 0z zm k m k� � � � �M N E H



� Example: m = 0 corresponds to the cylindrically symmetric fields

,0,

( )ˆ exp( ) zz

m kk z

dfjk z

kd

�

�� M

2,

0, ,2 2

( )ˆ ˆexp( ) ( )z

z z

m kzk z m k

df kjkjk z f

k d k��

�

� ��

� �� N z

� Example: corresponds to solutions uniform along z0zk �

,0 ,0ˆ exp( ) ( )m mjm f� �� N z

,0,0 ,0

( )ˆˆexp( ) ( ) mm mdfjm

jm fk kd

��

� ��

� � ��

M

k k� �



� For later use, let us consider the vector functions when

� �(2), ( )zm k mf r H k� ��

(2) 21 ( ) exp4

mm

jk H k j jk

k� � ��

��

� ��

� �

� This choice is useful for representing scattered wave since it

satisfies the radiation condition

� Consider the functions at a large distance from the center.

Asymptotic relation for the Hankel function



� The TE solution has the asymptotic behavior:

, 0

exp( )ˆ z

mz

m k

j k jk jm jk zC

k k� �

�

� �

�

� � �� E M

0, 0

exp( )ˆ ˆ

z

mzz

k

k j k jk jm jk zkjC

k k k k� � �

�

� �

� � �

� � � � ��

�H N z

0

2exp

4

jC j

��

� ��

1k� � ��



� At large distance these are

conical waves propagating

with the wave vector

for every angle �. They all

make an angle

with the z-axis

� They behave as TEM waves:

fields are perpendicular to

each other and to the wave

vector in all directions

ˆ

ẑˆ

ˆ ˆzk k� � z

ˆ ˆzk k� � zH

E

ˆ ˆzk k� � zWave front

�

�

sin /zk k� �



� Magnetic field has no �

component, electric field has

only a � component

� Again the electric and magnetic

fields are orthogonal

� At a large distance these wave

behave as TEM waves

ˆ

ẑˆ

ˆ ˆzk k� � zH

E

� Remember: the above analysis of the asymptotic behavior is

only valid for solutions based on Hankel function of the 2nd kind



� The behavior of the TM solution at

large distance is similar to TE but

now the electric field has no �

component and magnetic field only

has a � component

ˆ

ẑˆ

ˆ ˆzk k� � z

H

E

� �0, 0exp( )

ˆ ˆz

mzz

k

k j k jk jm jk zkj C

k k k k� � �

�

� �

�

� � � � ��

�E N z

, 0

exp( )ˆ z

mz

m k

j k jk jm jk zj jC

k k� �

�

� �

� � �

� � � � ��

� ��H M


Expansion of a plane wave

� So far we have analyzed the ‘natural’ solutions to the vector

wave equation in cylindrical coordinates. We expect them to be

useful when treating scattering by cylindrically symmetric objects

� We have seen that asymptotically, these solutions behave as

TEM waves at large distance

� But the actual problem we are interested is not the scattering of

these waves, but the scattering of a simple plane wave such as

� �0( ) expi i ij� � �E r E k r

i k�k



� To be able to use the cylindrical

vector solutions, we first have to

expand the plane wave into these

functions

� Let us write

ik

� � � �, , , , , ,, , cos , sin ,i i x i y i z i i i i i zk k k k k k� �� k

x

y

z

i�,ik �

ik,i zk

2 2 2, ,i i zk k k� � �



� �, ,cosi i i i zk k z�� k rx

y

z

�

r

�

� � � �exp cos ( ) ( ) expm mm

jz j J z jm� ��

��

� � � ��

� We next use the following relationship

from the theory of Bessel functions

� � � � � � � �, ,exp exp ( ) expmi i z m i im

j jk z j J k jm� � � ��

��

� � � � � � �� k r

� Remember that

� � � �, , cos , sin ,x y z z� � � �� r



� Remember that

x

y

z

�

r

�

� � � �,,

exp ( ) ( , , )expi z

m Ji m k i

m

j j z jm� � � ��

��

� � � ��k r

� � � �,, , ,( , , ) exp

i z

Jm k m i i zz J k jm jk z��

� is one solution of the scalar wave

equation with the Bessel function of the

first kind

� This is the expansion of a scalar plane wave for an incident

wave with . What about the vector plane wave?, ,i z ik �



� 1st result:

� � � �,,

1ˆ exp ( ) ( , , ) exp

i z

m Ji m k i

m

j j z jmk

� � ��

��

�� z k r M

� �

� � � � � �

,

, , ,

,

, , ,

,, , ,

1 ˆˆˆ

ˆˆ exp

i z

i z i z i z

Jm kJ J J

m k m k m k

im i m i i z

jm

k k k

kjmJ k J k jm jk z

k k�

� �

��

� �

� � ��

��

�

� ��

� �

M z

� � � �mmdJ z

J zdz

� �



� The first vector relation now follows:

� � � � � �,,ˆˆ exp ( ) ( , , ) expi zm Ji i m k im

j j j z jm� � ��

��

� � � � � ��z k k r M

� For the 2nd vector relation take the curl of the above equation

� � � � � �,,1 ˆˆ exp ( ) ( , , ) expi zm Ji i m k im

j j j z jmk

� � ��

��

� �� z k k r N

, ,, ,

1i z i z

J Jm k m kk

� ��N M



� Explicit form of N:

� �

� � � � � �

,, ,

2, , ,,

, , ,2 2 2

exp

ˆˆ ˆ

i z

Jm k i z

i z i ii zm i m i m i

jm jk z

jk k kmkJ k J k J k

k k k� �

� � �

�

� � ��

� � �

� ��

� ��

N

z

� � � � � �,,ˆ ˆˆ exp ( ) ( , , ) expi zm Ji i i m k im

j j z jm� � ��

��

� �� z k k k r N

� The 2nd vector relation now becomes



� Now, we define the unit vector system

ˆˆˆ ˆ ˆˆˆˆ

ii i i i

i

��

�

z kv h k kz k

ˆˆˆˆˆ

ii

i

��

�

z khz k

ˆik

ẑ

ˆˆ i�z k

ˆik

ẑ

� �ˆ ˆˆ i i� �z k k



� Let us see what these vectors are:

� � ˆˆˆ ˆsin ,cos ,0ˆˆ

ii i i i

i

� ��

� � � ��

z khz k

,,

ˆ ˆ ˆ ˆˆ ˆ ˆii zi i i i i ikk

k k�� v h k k ρ z

� � � �, , , , , ,, , cos , sin ,i i x i y i z i i i i i zk k k k k k� �� k

� �, , ,1ˆ cos , sin ,i i i i i i zk k kk � �� k

� �ˆ cos ,sin ,0i i i� ��ρ

x

y

z

i�,ik �

ik,i zk

ˆiρ

̂i

These cylindrical unit vectors are defined with respect to ki. They have a subscript i. Do not confuse them with the cylindrical unit vectors for position!



� We thus have the relations

� � � �

� �,

,,

ˆ ˆexp exp

( ) ( , , ) expi z

i i i i

m Jm k i

mi

j j

kj j z jm

k ��

�

��

� � � � � �

� ��

h k r k r

M

� � � �

� �,

,,

,,

ˆ ˆ ˆexp exp

( ) ( , , ) expi z

ii zi i i i

m Jm k i

mi

kkj j

k k

kj z jm

k

�

�

� � ��

��

� ��

� �

� ��

v k r ρ z k r

N



� For any polarization of the incident plane wave:

� � � � � � � �0 0 0ˆ ˆ ˆ ˆ( ) exp expi i i i i i i i i ij j� �� E r E k r E h h E v v k r

, ,, ,( ) ( , , ) ( , , )

i z i z

J Ji m m k m m k

m m

u z v z� � � ��

��

� �� E r M N

� � � �

� � � �

0

,

0

,

ˆ ( ) exp

ˆ ( ) exp

mm i i i

i

mm i i i

i

ku j j jm

k

kv j jm

k

�

�

�

�

� � � �

� � � �

E h

E v

Related to TE: horizontal electric field

Related to TM: horizontal magnetic field


Exact solution of the problem

� We now consider the scattering of a plane wave by an infinitely

long, perfectly conducting cylinder

ik

� When the incident wave hits the

cylinder, surface currents (and

charges) are induced.

� These currents create the ‘scattered’

field. At any point, the total electric

field is

� �0( ) expi i ij� � �E r E k r

( ) ( ) ( )i s� �E r E r E r



� We saw how the incident plane wave can be represented in

terms of cylindrical vector solutions

ik

� The scattered field (outside the

cylinder) can also be expanded in

terms of those solutions

� But: for the scattered field we should

use vectors with the right condition at

the infinity � ��



� We should use the Hankel function of the 2nd kind for these

waves which satisfy the radiation condition (behave as

outgoing waves at infinity)

� Also, since the system is uniform in z-direction, kz is the same

as that of the incident wave: . Of course the wave will

be scattered along different angles (different �’s)

, ,, ,( ) ( , , ) ( , , )

i z i z

H Hs m m k m m k

m m

a z b z� � � ��

��

� �� E r M N

, ,, ,

1ˆ

i z i z

H Hm k m kk

�� M z , ,, ,1

i z i z

H Hm k m kk

� ��N M

,z i zk k�



� The vector functions are explicitly given by

� �

� � � �,, ,

,(2) (2), ,

exp

ˆˆ

i z

Hm k i z

im i m i

jm jk z

kjmH k H k

k k�

� �

�

� ��

� � �

� ��

M

� �

� � � � � �

,, ,

2, , ,,(2) (2) (2)

, , ,2 2 2

( ) exp

ˆˆ ˆ

i z

Hm k i z

i z i ii zm i m i m i

jm jk z

jk k kmkH k H k H k

k k k� �

� � �

�

� � ��

� � �

� ��

N r

z

� Note that we have used 2 2 2 2, ,z i z ik k k k k k� ��



� Zero tangential component of total electric field

a

zEE�

( , , ) ( , , ) 0zE a z E a z��

� The next step is to choose the coefficients of the expansion

such that the condition at the surface of the cylinder is satisfied

� 1st Result:

� ��

� � � ��

,

(2),

,0

(2), ,

ˆ ( ) exp

m i

m m

m i

m i mi i i

i m i

J k ab v

H k a

J k akj jm

k H k a

�

�

�

� �

�

� �

� � �E v



� 2nd result:

� ��

� � � ��

,

(2),

,0

(2), ,

ˆ ( ) exp

m i

m m

m i

m i mi i i

i m i

J k aa u

H k a

J k akj j jm

k H k a

�

�

�

� �

�

��

��

� � ��

E h

� Note that a TE incident wave excites TE polarized waves

(horizontal polarization), and a TM wave excites TM polarized

waves (vertical polarization)


Exact solution of the problem (TE polarization)

� TE scattering (horizontal electric field):

� ��

� � � ��

,

,

,

,

,

,(2),

,0,(2)

, ,

( ) ( , , )

( , , )

ˆ ( ) exp ( , , )

i z

i z

i z

TE Hs m m k

m

m m i Hm k

m m i

m i m Hi i i m k

mi m i

a z

u J k az

H k a

J k ajkj jm z

k H k a

�

�

�

� �

� �

� �

� � �

�

��

�

��

�

��

�

��

�

��

�

�

�

�

E r M

M

E h M



� Explicit form in components

� � � �

� ��

� � � �

,0,

,

, (2),(2)

,

expˆ( )

( ) exp

i zTEs i i

i

mm i

m i im m i

jk zE

k

j J k am H k jm

H k a

��

��

�

�

� � ��

��

��

��

r E h

� � � ��

� ��

0, ,

, (2),(2)

,

ˆ( ) exp

( ) exp

TEs i i i z

mm i

m i im m i

E j jk z

j J k aH k jm

H k a

�

��

�

� � ��

��

� � � �

��

r E h



� The magnetic field in the TE case

� �� ,

,,(2)

,

( ) ( , , )i z

m m iTE Hs m k

m m i

u J k ajz

H k a

�

�

� ��

�

��

��

��H r N

� In components:

� � � ��

� ��

0 ,, ,

, (2),(2)

,

ˆ( ) exp

( )exp

TE i zs i i i z

mm i

m i im m i

jkH jk z

k

j J k aH k jm

H k a

�

��

�

�

� � ��

��

� ��

� ��

� � ��

r E h



� � � �

� ��

� � � �

,0 ,,

,

, (2),(2)

,

expˆ( )

( )exp

i zTE i zs i i

i

mm i

m i im m i

jk zkH

k k

j J k am H k jm

H k a

��

��

�

� �

� � ��

��

��

� �

��

r E h

� � � ��

� � � � � �

,0, ,

, (2),(2)

,

ˆ( ) exp

( )exp

iTEs z i i i z

mm i

m i im m i

kH jk z

k

j J k aH k jm

H k a

�

��

�

�

� � ��

��

� ��

� ��

� ��

r E h


Exact solution of the problem (TM polarization)

� For the TM case we have

� ��

� � � ��

,

,

,

,

,

,(2),

,0,(2)

, ,

( ) ( , , )

( , , )

ˆ ( ) exp ( , , )

i z

i z

i z

TM Hs m m k

m

m i Hm m k

m m i

m i m Hi i i m k

mi m i

b z

J k av z

H k a

J k akj jm z

k H k a

�

�

�

� �

� �

� �

� � �

�

��

�

��

�

��

�

� �

� � �

�

�

�

E r N

N

E v N




� � � �

� ��

0 ,, ,

, (2),(2)

,

ˆ( ) exp

( )exp

TM i zs i i i z

mm i

m i im m i

jkE jk z

k

j J k aH k jm

H k a

�

��

�

� � ��

��

� ��

� � � ��

r E v

� � � �

� ��

,0 ,,

,

, (2),(2)

,

expˆ( )

( )exp

i zTM i zs i i

i

mm i

m i im m i

jk zkE

k k

j J k am H k jm

H k a

��

��

�

�

� � ��

��

��

��

r E v




� � � �

� ��

,0, ,

, (2),(2)

,

ˆ( ) exp

( )exp

iTMs z i i i z

mm i

m i im m i

kE jk z

k

j J k aH k jm

H k a

�

��

�

� � ��

��

� ��

� ��

� ��

r E v

� The magnetic field

� �� ,

,

,(2),

( ) ( , , )i z

m iTM Hs m m k

m m i

J k ajv z

H k a�

�

� ��

�

��

� � �H r M



� � � �

� ��

,0,

,

, (2),(2)

,

expˆ( )

( ) exp

i zTMs i i

i

mm i

m i im m i

jk zH

k

j J k am H k jm

H k a

��

��

�

� �

� � ��

��

��

��

r E v

� � � �

� ��

0, ,

, (2),(2)

,

ˆ( ) exp

( ) exp

TMs i i i z

mm i

m i im m i

jH jk z

j J k aH k jm

H k a

�

��

�

�

� � ��

��

� � � �

��

r E v


Scattering by a perfectly conducting infinite cylinder

� Remember that this is the full solution everywhere. We are

actually interested in the scattering in the far field limit.

� We again use the asymptotic relationship

� � � �(2)0

exp exp2( 1) exp

4

m m

m

j jk j jkjH k jC

k k

� ��

� �

� ��

� � �

� ��

� �

� We find that in the far field limit the fields are parallel to:

,ˆ( 1) TEs ik � � � �E

,,, ˆ ˆ( 1)

ii zTMs i

kkk

k k�

� ��

� �E z

Same as θ̂



� More specifically:

� � � �

� ��

, 0 , ,, 0

,

,

(2),

exp( )ˆ ˆˆ

exp

i i i zTM i zs i i

i

m i

im m i

k jC jk jk zk

k k k

J k ajm

H k a

� �

�

�

�

�

�

� ��

��

� � ��

� �

��

E z E v

� �

� ��

� �

00

,

(2),

exp( )ˆ ˆ

exp

zTEs i i

m i

im m i

jC jk jk z

k

J k ajm

H k a

�

�

�

�

�

�

� ��

��

� ��

��

E E h



� The interesting point here is the behavior of far field as

function of the angle � , which is governed by the functions

� ��

� �,(2)

,

expm i im m i

J k ajm

H k a

�

�

� ��

��

��

� Remember that

(2) (2)( 1) ( 1)m mm m m mH H J J� ��

� ��

,

(2),

expm i im m i

J k ajm

H k a�

�

� ��

��

��

TE

TM


TE Scattering by a thin conducting cylinder

� To get some insight let us consider the limit of a thin cylinder

, 1ik a ka� � �

� For small arguments (sufficient to consider positive or zero m’s)

� � � �1

0 1( ) 2 ( 1)!

m

m m m m

m zJ z J z J z

z m

�

� �� 0 2

zJ z� ��

� �(2)02j

H zz�

� �� (2) (2) (2)0 1 12 !

( )m

m m m m

m j mH z H z H z

z z��



� To the lowest order we have for the TE case

� � � �

� �

20,0

,

, ,

ˆˆ( , , )

4

exp( ) 1 2cos

i i iTEs

i

i i z i

k aCz

k

jk jk z

�

�

�

��

�

� � �

��

� � � ��

E hE

� ��

� �

� � � �

,

(2),

2

,

exp

1 2cos4

m i

im m i

i

i

J k ajm

H k a

k aj

�

�

�

� �

��

�

��

��

� � ��

�

�



� Consider the scattered electric field. It’s phase does not

depend on angle. It’s amplitude is given below

, , ,ˆ( , , ) ( ) exp( )TE TEs s i i zz E jk jk z� �� E

� � � � � �20

,0,

,

ˆ( ) 1 2cos

4

i i iTEs i

i

k aCE

k

��

�

��

�

��

E h�

i� ��ik

x

y

� � i�

,TEsE �



� Note that scattering is largest when

� �cos 1i i� � � � ��

i� � ��

x

y

� � i�

,TEsE �

� This is the backward scattering case

� It is zero when 2i�

� ��


Numerical experiments

� Let us return to the exact result for TE case and again

focus on the azimuthal component of the electric field

� We would like to compare this with the asymptotic result

� This a difficult function to compute because of ‘bad’ behavior of

Bessel functions of large order

� � � ��

� ��

0, ,

, (2),(2)

,

ˆ( ) exp

( ) exp

TEs i i i z

mm i

m i im m i

E j jk z

j J k aH k jm

H k a

�

��

�

� � ��

��

� � � �

��

r E h



� So let us consider the exact numerical result for the azimuthal

component of the electric field

� We plot this function when

� �, , , ,0 ,0, ,i i i x i z i i xk k k k�� k

� � � �0 0, , , ,ˆ ˆ( ) exp expi i y i i y i x i zE j jk x jk z� � � � � �E r y k r E y

0 0,

ˆ ˆ ˆˆ i i i i i yE� � � �h y E h

x

y

,i xk

iE

,TEsE �

�



� For a thin cylinder we have the amplitude profile

, 0.25ik a� �

,i xk

iE / 3�



� It is also instructive to look at constant phase fronts, at which

, 0.25ik a� �

( ) 0phase Q �



� So in the far field, the phase fronts are almost circular in the x-

y plane: there is no angle-dependence of the phase of the

scattered field, as found from the thin-film approximation

� Also the amplitude is largest in case of backscattered waves,

as found from the same approximation

� The amplitude becomes nearly zero when

� Again this is in line with what we found before

� So this approximation is quite accurate

3i�

� ��


TM Scattering by a thin conducting cylinder

� For the TM case from a thin cylinder we have

, 1ik a ka� � �

� For small arguments (consider positive or zero m’s)

� � 1! 2

m

m

zJ z

m� ��

� � �(2)02

lnj

H z z�

�� (2)02 ( 1)!m

m m

j mH z

z��

��

� Lowest order in : ,ik a�

� ��

,

(2), ,

exp2ln

m i

im m i i

J k a jjm

H k a k a�

� �

��

�

��

��



� Resulting far scattered field for a thin conducting cylinder

� � � �

,,, , ,

0 0,

,,

ˆ ˆ( , , ) exp( )

ˆ2 ln

iTM TM i zs s v i i z

TMs v i i

ii

kkz E jk jk z

k k

CE

k ak

��

��

� � �

��

� ��

� �

� �

E z

E v

� In the lowest order approximation there is no angle

dependence for the TM scattered wave!

� The amplitude in all directions is the same



� Let us compare the scattering strength in the two cases

� ��

20,0

,, ,

ˆ( ) 1 2cos

4

i i iTEs i

i i

k aCkE

k k

�

��

��

�

��

E h

� �0 0

,

, ,

ˆ2 ln

TMs v i i

i i

CE

k k a� �

��

� �E v

� For equal horizontal and vertical components of the incident

field we have

� � � �2, , ,,,

( ) 3ln

2

TEs

i iTMis v

E kk a k a

kE

��

�

��



� This ratio is quite small for thin cylinders

� But this result is to be expected: a

vertically polarized incident electric field

has a component along the ‘wire’ and

easily induces electric currents along the

wire. These currents, in turn, generate the

scattered field.

� A horizontally polarized incident wave has

no longitudinal components and cannot

excite such currents

TEiE

TMiE


TE Scattering by a thick conducting cylinder

� Now, let us investigate the other limit, that of a thick conducting

cylinder which satisfies

2 2, , 1i i zk a a k k� � � �

� Remember: scattered electric field

� � � �� ,0 ,0

(2), ,

exp( )ˆ ˆ expm ii zTEs i i imi m i

J k ajC jk jk zjm

k H k a

��

� �

��

�

�

��

�� E E h

� �� ,

,

,(2),

( ) ( , , )i z

m m iTE Hs m k

m m i

u J k az

H k a

�

�

� ��

��

��

��E r M

� Far field limit



� ��

� �

� � � �

, ,,(2)

,

,

exp exp2 4

( ) exp

m i ii i

m m i

mm i i

m

J k a k a jjm j jk a

H k a

j J k a jm

� ��

�

�

� ��

� �

�

��

�

��

� � ��

��

�

�

�

� � � �cos exp cos ( ) ( ) expm mm

j jz j J z jm� � ��

��

��

� Next, use the relation

� Let us approximate the denominator

� �(2) (2) 1 21 ( ) exp4

mm m

jz H z jH z j jz

z

��

� � ��

� � �



� � � ��

0

, , ,

ˆ ˆ( , , ) cos

exp ( ) cos

TEs i i i

i i i i z

az

jk a jk a jk z� �

� � � ��

� � �

� � �

� ��

E E h�

� The TE far scattered field becomes

� Remember that this result is also based on the far-field behavior

of the vector solutions (used here), which in turn, was based on

the asymptotic behavior of the Hankel function when

1k��


Numerical results for a thick cylinder

� So let us again look at some numerical results for the azimuthal

component of the electric field

� We again plot this function when

� �, , , ,0 ,0, ,i i i x i z i i xk k k k�� k

� � � �0 0, , , ,ˆ ˆ( ) exp expi i y i i y i x i zE j jk x jk z� � � � � �E r y k r E y

0 0,

ˆ ˆ ˆˆ i i i i i yE� � � �h y E h

x

y

,i xk

iE

,TEsE �

�



� The amplitude profile is shown below

, 10i xk a �

,i xk

iE



� Phase fronts:

, 10i xk a �,i xk

iE



� Note that in front of the cylinder the scattered wave looks

normal, it has a circular phase front

� But behind the cylinder, at a not too far distance, the phase

fronts are flat!

� Besides, its amplitude is much larger than the waves scattered

back (to the left).

� But this is just a misconception. To understand this point, one

should realize that the total field is

( ) ( )TEi s�E r E r



� Let plot the � component of

the total field

� The amplitude of the

scattered wave behind the

cylinder is large because it

has to partially cancel the

incident wave, to form a

shadow region



� Nevertheless, the numerical results are very different from the

asymptotic result we found earlier if we only demand

� To understand why, consider the exact electric field

� � � ��

� ��

0, ,

, (2),(2)

,

ˆ exp

( ) exp

TEs i i i z

mm i

m i im m i

E j jk z

j J k aH k jm

H k a

�

��

�

� � ��

��

� � � �

��

E h

� Asymptotic behavior of Hankel function depends on whether

argument is larger, smaller, or comparable to order m

1k��



� For a thin cylinder only small m’s contribute (|m|=0,1)

� Then, far field asymptotic is valid since

, 1ik a� �

� ��

,

(2),

m i

m i

J k a

H k a

�

�

�

�

m

, 1ik a� �

, 10ik a� �

, 0.25ik a� �

, 1ik � � �

� �2 (2), , ,,

2( ) exp / 4mi m i i

i

k m H k j jk jk� � ��

� � � ��

� � ��

� The correct asymptotic result is



� But, for a thick cylinder, all orders with have a

contribution so that large orders are also important

� The used asymptotic is then only accurate when

,im k a��

� �2

2

, ,,

2i i

i

ak k a� �

�

��

��

� This is similar to the ‘definition’ of the far field region (why?)

� So our earlier result (and the earlier discussion of the far-field

behavior of vector wave solutions) is, strictly speaking, only

applicable in this limit



� Similar discussion applies to TM scattering: whereas in the far

field limit the scattered wave behaves as TEM waves with a

conical wave front (circular in x-y plane), the behavior outside

this limit can be totally different

� One can have flat wave fronts, a shadow region, etc

� In particular, note that when the wavelength is very short

compared to the dimensions of the object, the far field result is

only valid so far away that it may be useless for practical

applications


Scattering cross section (scattering width)

� We saw in the beginning how a scattering cross section is

defined for a finite scatterer in terms of the scattered power

� An infinite cylinder, however, is not a finite object

� The field radiated by sources inside the cylinder (scattered

field) does not drop as , but as

� We have to reformulate our definition of scattering cross

section for infinite cylindrical objects

1/ r 1/ �



� Again consider an incident wave

ik

� �0 expi i iE j� � �E e k r

� Now, the scattered far field

is always of the type

� � � �0 , ,ˆ ˆ,

exps i

s s i i z

fE jk jk z��

� � �k k

E esk

Note: scattering amplitude may depend on the polarization of the incident wave!



� Corresponding far field Poynting

vectors are

ik

sk

�

20

ˆ2

i

i i��E

S k

� � 22 20 0ˆ ˆ,ˆ ˆ

2 2

s is i

s s s

f

� � ��

k kE ES k k

� It is important to bear in mind that

, ,ˆ ˆ ˆi i zsk k

k k�� k ρ z

,i zk

,ik �



� The Poynting vector is independent of z, it is

the same for all vertical coordinates

� We, therefore, cannot define the total

scattered power since it is infinite

� But, we can define the scattered power per

unit length by integration over the angle

� �2 2

2,

0 0

ˆ ,iL s i ik

P d S f dk

� �� S ρ

�



� Remember that

ik,i zk

,ik �i k�ki�

, cosi ik

k� ��

� �2

2,

0

,iLW ii

kPf d

S k

��

� Total scattering width is

now defined as



ik

� As an example consider the TE case

� �0 ˆ ˆ ˆ exp i i i i iE j� � � �E h k r h ̂i

sk

ˆ

� The scattered far field:

� �

� ��

� �

0 ,0

,

,

(2),

exp( )ˆ ˆ

exp

i zTEs i i

i

m i

im m i

jC jk jk z

k

J k ajm

H k a

�

�

�

�

�

�

� ��

��

� ��

��

E E h



� The scattering amplitude

� � � ��

� �2

,

(2), ,

2, expm iTE i i

mi m i

J k af jm

k H k a

�

� �

� � � ��

�

��

��

� Scattering width for TE waves:

� � � ��

22

2 ,,

(2)0 ,

4, m iiTE TEW s i s

m m i

J k akf d

k k H k a

��

�

� � � ��

��

��

��



� TM case:

� �0ˆ expi i iE j� � �E v k r

,, ̂

ˆˆ ˆ ˆii zi i i ikk

k k�� v k ρ z

ik

ˆiv

� Scattered field:

� � � �

� ��

, 0 , ,, 0

,

,

(2),

exp( )ˆ ˆˆ

exp

i i i zTM i zs i i

i

m i

im m i

k jC jk jk zk

k k k

J k ajm

H k a

� �

�

�

�

�

�

� ��

��

� � ��

� �

��

E z E v

sk



� Scattering amplitude:

� � � �� 2

2 ,

(2), ,

2, expm iTM i i s

mi m i

J k af jm

k H k a�

� �

� � � ��

�

��

� ��

� Scattering width:

� � � ��

22

2 ,,

(2),0

4, m iiTM TMW i

m m i

J k akf d

k k H k a

��

�

� � � ��

��

� � ��



� Example: thin cylinder

� ��

� � � �2

24 8,

, ,(2),

4 3

4m iTE

W i im m i

J k ak a O k a

k kH k a

��

�

��

�

��

��

��

� ��

2 22

4,

,(2), ,

4 1

lnm iTM

W im m i i

J k aO k a

k kH k a k a�

��

��

�

��

� ��

� ��



� Numerical results for the TE case

TEWk�

,ik a�



� Plotted in a different way

2

TEW

a

�

,ik a�



� For the TM case

,ik a�

TMWk�


Scattering by an infinitely long dielectric cylinder

� So far we discussed a (perfectly) conducting cylinder

� What about a dielectric cylinder? We can actually use the

same machinery to solve the problem

ik

� Incident plane wave:

,

,

,

,

( ) ( , , )

( , , )

i z

i z

Ji m m k

m

Jm m k

m

u z

v z

� �

� �

�

��

�

��

�

�

�

�

E r M

N



� For the scattered wave (outside cylinder) we use the same

expansion as before

ik,

,

,

,

( ) ( , , )

( , , )

i z

i z

Hs m m k

m

Hm m k

m

a z

b z

� �

� �

�

��

�

��

�

�

�

�

E r M

N

� Inside the cylinder we use solutions based

on Bessel function of 1st kind (why?)

, ,, ,( ) ( , , ) ( , , )

i z i z

J Jc m m k m m k

m m

c z d z� � � ��

��

� �� E r M N



� Next step is to match the fields at the boundary:

� �ˆ ˆi s c� � � �E E E

� This again leads to algebraic equations for

coefficients which can be solved

� Note that inside the cylinder

a

zEE�

� �ˆ ˆi s c� � � �z E E z E

2 2 2 2, , , 0 i c d i z d dk k k k k� � � ��

��

EM scattering 4 new - Sharifee.sharif.edu/~emscattering_ms/Lecture 4.pdfScattering from cylindrical...

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Transcript of EM scattering 4 new - Sharifee.sharif.edu/~emscattering_ms/Lecture 4.pdfScattering from cylindrical...