Cylindrical Radiation and Scattering - ECE/CISweile/ELEG648Spring06/...Scattering from a Circular...

Cylindrical RadiationScattering

Cylindrical Radiation and Scattering

Daniel S. Weile

Department of Electrical and Computer EngineeringUniversity of Delaware

ELEG 648—Radiation and Scattering in CylindricalCoordinates

D. S. Weile Cylindrical Radiation and Scattering


Outline

1 Cylindrical RadiationSources of Cylindrical RadiationGreen’s Function and Far FieldWave Transformations

2 ScatteringScattering from a Circular CylinderScattering from a Wedge2.5-D Problems



Outline





Sources of Cylindrical RadiationGreen’s Function and Far FieldWave Transformations

Outline






Two-Dimensional Sources

Two-dimensional radiation is created by sourcesindependent of z.The simplest such source (akin to a Hertzian Dipole) is aninfinite filament.Such a current radiates TMz .

Consider a filamentary current I on the z-axis. Since itsradiation is

1 Independent of φ,2 Independent of z, and3 Outwardly traveling,

the magnetic vector potential must be of the form

Az = µCH(2)0 (kρ)




Radiation from a Filament

Now, we must have

limρ→0

2π∫0

Hφρdφ = I

NowH =

1µ∇× A

implies

Hφ = −C∂

∂ρ

[H(2)

0 (kρ)]= kCH(2)

1 (kρ).

For small ρ, we have

H(2)1 (kρ)→ kρ

2+

jπ

2kρ




Radiation from a Filament

Substituting this into our equation,

kC limρ→0

2π∫0

(kρ2

+jπ

2kρ

)ρdφ = I

2jCπ

2π = I

so finally

C =14j

and we have the

Magnetic Vector Potential

Az =µI4j

H(2)0 (kρ)




Fields of a FilamentGiven Az we can easily compute

Filamentary Fields

Ez =−k2I4ωε

H(2)0 (kρ)

Hφ =kI4j

H(2)1 (kρ)

Using large argument approximations, we can find

Far Fields

Ez = −ηkI

√j

8πkρe−jkρ

Hφ = kI

√j

8πkρe−jkρ




Observations

Near the source, E and H are not in phase and have a complexrelationship.

Far from the source, E and HAre in phase,Have ratio η, andDecrease as ρ−

12 .

The total radiation (per unit length) must be independent ofradius (which can be proven directly). Therefore, we can use thefar field expression to compute the power:

P = −2π∫

0

EzH∗φρdφ = −2π∫

0

(−ηkI

√j

8πkρe−jkρ

)(kI∗√−j

8πkρejkρ

)ρdφ

=ηk2|I|2

8πkρ2πρ =

ηk4|I|2




Two-Dimensional Notation

To proceed further, it is useful to define

ρ = xux + yuy

ρ′ = x ′ux + y ′uy

The distance between these points is

|ρ− ρ′| =√

(ρ− ρ′) · (ρ− ρ′)

=√ρ2 + (ρ′)2 − 2ρ · ρ′

=√ρ2 + (ρ′)2 − 2ρρ′ cos(φ− φ′)




Radiation Due to a Filament

The radiation due to a current I at the origin we have seen is

Az =µI4j

H(2)0 (kρ).

Therefore, if the current is located at ρ′, we have

Filamentary Radiation

Az(ρ) =µI4j

H(2)0 (k |ρ− ρ′|)

Similarly, a filamentary magnetic current K at ρ′ radiates

Fz(ρ) =εK4j

H(2)0 (k |ρ− ρ′|)




Outline






Green’s Function

If a current density Jz(ρ) is independent of z, we can think of itas a bundle of filaments with current Jz(ρ)dS. Then we havethe

General Formulas for Two-Dimensional Radiation

Az(ρ) =µ

4j

∫∫Jz(ρ

′)H(2)0 (k |ρ− ρ′|)dS′

Fz(ρ) =ε

4j

∫∫Mz(ρ

′)H(2)0 (k |ρ− ρ′|)dS′

From here, the fields can be computed in the usual way.




Far FieldsIn the far field, we can make the standard approximation

|ρ− ρ′| → ρ− ρ′ cos(φ− φ′)

Similarly, for x →∞,

H(2)0 (x)→

√2jπx

e−jx .

Combining these (and remembering to use a simpler approximationin the denominator) we have

The Far Field

Az(ρ) = µe−jkρ√8jπkρ

∫∫Jz(ρ

′)ejkρ′ cos(φ−φ′)dS′

Fz(ρ) = εe−jkρ√8jπkρ

∫∫Mz(ρ

′)ejkρ′ cos(φ−φ′)dS′




Far Fields

We can compute the far fields using the definitions of Aand F.The results are similar to three-dimensional results.

Relationship Between E and H

Eφ = ηHz Ez = −ηHφ

Far Electric Fields

Eφ = −jωηFz

Ez = −jωAz




Outline






Plane Wave Expansion

We often want to express plane waves in cylindrical wavefunctions. The result must be

Finite at the origin, and2π-periodic.

Thus

e−jx = e−jρ cosφ =∞∑

n=−∞anJn(ρ)ejnφ.

To find the an, use orthogonality

2π∫0

e−jρ cosφe−jmφdφ =∞∑

n=−∞anJn(ρ)

2π∫0

ej(n−m)φdφ





Using orthogonality

2π∫0

e−jρ cosφe−jmφdφ = 2πamJm(ρ)

The integral on the right-hand side is well-known

Jn(x) =jn

2π

2π∫0

e−jx cosφe−jmφdφ





Substituting this in, we find

am = j−m

and finally

The Plane Wave Expansion of Cylindrical Waves

e−jx = e−jρ cosφ =∞∑

n=−∞j−nJn(ρ)ejnφ

How might this formula be modified for waves travelling in otherdirections?




The Addition Theorem

We are also interested in expanding the field of a filamentwith respect to a different center.We have seen that a filamentary current I located at ρ = ρ′

radiates a field with Az = µψ with

ψ(ρ, φ) =I

4πH(2)

0 (k |ρ− ρ′|).

We can think of the current generating this field as acurrent sheet, confined to the ρ = ρ′ cylinder, with

Jz(φ) =Iδ(φ− φ′)

ρ

where the denominator ensures that∫ 2π

0Jz(φ)ρdφ = I





We can expand the field inside and outside the tube of current:

ψ(ρ, φ) =

∞∑

n=−∞a−n Jn(kρ)ejnφ for ρ < ρ′

∞∑n=−∞

a+n H(2)

n (kρ)ejnφ for ρ > ρ′

Since Ez ∝ ψ, ψ must be continuous at ρ′. Thus

a−n Jn(kρ′) = a+n H(2)

n (kρ′).

This is solved if we let

a−n = H(2)n (kρ′)an

a+n = Jn(kρ′)an





We now have

ψ(ρ, φ) =

∞∑

n=−∞anH(2)

n (kρ′)Jn(kρ)ejnφ for ρ < ρ′

∞∑n=−∞

anJn(kρ′)H(2)n (kρ)ejnφ for ρ > ρ′

Now

Hφ = −∂ψ∂ρ

=

−k

∞∑n=−∞

anH(2)n (kρ′)J ′n(kρ)ejnφ for ρ < ρ′

−k∞∑

n=−∞anJn(kρ′)H

(2)′n (kρ)ejnφ for ρ > ρ′





From boundary conditions, we have

Hφ(ρ = ρ′+)− Hφ(ρ = ρ′−) = Jz

We thus have

−k∞∑

n=−∞an[Jn(kρ′)H

(2)′n (kρ′)− H(2)

n (kρ′)J ′n(kρ′)]ejnφ = Jz

The standard Wronskian formula gives

Jn(kρ′)H(2)′n (kρ′)− H(2)

n (kρ′)J ′n(kρ′) =

−2jπkρ





Therefore, plugging in,

2jπρ

∞∑n=−∞

anejnφ =Iδ(φ− φ′)

ρ

We can now find the an using orthogonality:

jπ

∞∑n=−∞

an

2π∫0

ejnφe−jmφdφ = I

2π∫0

δ(φ− φ′)e−jmφdφ

4jam = Ie−jmφ′

am =I4j

e−jmφ′





We thus find that the field of a filament can be expanded asAz = µψ with

ψ(ρ, φ) =

I4j

∞∑n=−∞

H(2)n (kρ′)Jn(kρ)ejn(φ−φ′) for ρ < ρ′

I4j

∞∑n=−∞

Jn(kρ′)H(2)n (kρ)ejn(φ−φ′) for ρ > ρ′

Equating our earlier expression, this gives


H(2)0 (k |ρ− ρ′|) =

∞∑

n=−∞Jn(kρ)H

(2)n (kρ′)ejn(φ−φ′) for ρ < ρ′

∞∑n=−∞

Jn(kρ′)H(2)n (kρ)ejn(φ−φ′) for ρ > ρ′



Scattering from a Circular CylinderScattering from a Wedge2.5-D Problems

Outline






TM Scattering by a Circular Cylinder

Consider a PEC cylinder of radius a.Let it be excited by a z-polarized incident wave

E iz = E0e−jkx = E0e−jkρ cosφ

Using our plane wave expansion, we may write

E iz = E0

∞∑n=−∞

j−nJn(kρ)ejnφ

The total field is of course

Ez = E iz + Es

z ;

it must vanish on the surface of the cylinder.




TM Scattering from a Circular Cylinder

We may expand the scattered field as

Esz = E0

∞∑n=−∞

j−nanH(2)n (kρ)ejnφ

The total field on the surface of the cylinder is thus

Ez = E0

∞∑n=−∞

j−n[Jn(ka) + anH(2)

n (ka)]

ejnφ = 0

Thereforean = − Jn(ka)

H(2)n (ka)




TM Scattering from a Circular Cylinder

We may expand the scattered field as

Esz = E0

∞∑n=−∞


The total field on the surface of the cylinder is thus

Ez = E0

∞∑n=−∞

j−n[Jn(ka) + anH(2)

n (ka)]

ejnφ = 0

Thereforean = − Jn(ka)

H(2)n (ka)




TM Scattering from a Circular Cylinder, Current

From here we can find any information about the scattering wewant. For instance

Jz = Hφ|ρ=a =1

jωµ∂Ez

∂ρ

∣∣∣∣ρ=a

Now

1jωµ

∂Ez

∂ρ

∣∣∣∣ρ=a

=E0

jωµ

∞∑n=−∞

j−n[J ′n(ka) + anH(2)′

n (ka)]

ejnφ

=−E0

jωµ

∞∑n=−∞

j−n[H(2)

n (ka)J ′n(ka) + Jn(ka)H(2)′n (ka)

] ejnφ

H(2)n (ka)




TM Scattering from a Circular Cylinder, Current

Using the Wronskian relation, we find

The Surface Current

Js(φ) =−2E0

ωµπa

∞∑n=−∞

j−nejnφ

H(2)n (ka)

For a thin wire, the first term dominates and we can even write

I = 2πE0

jωµ log ka




TM Scattering from a Circular Cylinder, Scattered Field

Of course, the exact scattered field is given above:

Esz = E0

∞∑n=−∞


Using the far field formula for Hankel Functions, we find the

Scattered Far Field

Esz = E0e−jkρ

√2πkρ

∞∑−∞

anejnφ

We may treat the other polarization in the same way.




Scattering due to Filamentary Excitation

We can also consider the scattering due to a current I locatedat ρ′. In this case, the incident field is

E iz =

−k2I4ωε

H(2)0 (k |ρ− ρ′|)

=−k2I4ωε

∞∑n=−∞

H(2)n (kρ′)Jn(kρ)ejn(φ−φ′) for ρ < ρ′

We can write the scattered field in the form

Esz =−k2I4ωε

∞∑n=−∞

cnH(2)n (kρ′)H(2)

n (kρ)ejn(φ−φ′)




Scattering due to Filamentary Excitation

From the preceding, it is obvious that

cn = − Jn(ka)

H(2)n (ka)

The final solution is thus the

Total Field from Filamentary Scattering

Ez =k2I4ωε

∞∑n=−∞

H(2)n (kρ>)

[Jn(kρ<) + cnH(2)

n (kρ<)]

ejn(φ−φ′)

Hereρ< = min(ρ, ρ′) ρ> = max(ρ, ρ′)




Observations

Our answer is symmetrical with respect to the substitutionρ↔ ρ′ and φ↔ φ′. Why?

The coefficients cn are the same as the an from the lastproblem. Why? How can we assure they are the same forevery problem?It is often said this problem is a generalization of the planewave scattering problem. Why? How can we recover theplane wave solution from this one?How else might we approach the solution to this problem?




Observations

Our answer is symmetrical with respect to the substitutionρ↔ ρ′ and φ↔ φ′. Why?The coefficients cn are the same as the an from the lastproblem. Why?

How can we assure they are the same forevery problem?It is often said this problem is a generalization of the planewave scattering problem. Why? How can we recover theplane wave solution from this one?How else might we approach the solution to this problem?




Observations

Our answer is symmetrical with respect to the substitutionρ↔ ρ′ and φ↔ φ′. Why?The coefficients cn are the same as the an from the lastproblem. Why? How can we assure they are the same forevery problem?

It is often said this problem is a generalization of the planewave scattering problem. Why? How can we recover theplane wave solution from this one?How else might we approach the solution to this problem?




Observations

Our answer is symmetrical with respect to the substitutionρ↔ ρ′ and φ↔ φ′. Why?The coefficients cn are the same as the an from the lastproblem. Why? How can we assure they are the same forevery problem?It is often said this problem is a generalization of the planewave scattering problem. Why?

How can we recover theplane wave solution from this one?How else might we approach the solution to this problem?




Observations

Our answer is symmetrical with respect to the substitutionρ↔ ρ′ and φ↔ φ′. Why?The coefficients cn are the same as the an from the lastproblem. Why? How can we assure they are the same forevery problem?It is often said this problem is a generalization of the planewave scattering problem. Why? How can we recover theplane wave solution from this one?

How else might we approach the solution to this problem?




Observations

Our answer is symmetrical with respect to the substitutionρ↔ ρ′ and φ↔ φ′. Why?The coefficients cn are the same as the an from the lastproblem. Why? How can we assure they are the same forevery problem?It is often said this problem is a generalization of the planewave scattering problem. Why? How can we recover theplane wave solution from this one?How else might we approach the solution to this problem?




Outline






Scattering from a Wedge

Now we look at filamentaryscattering from a wedge.The wedge is composed oftwo half planes φ = α andφ = 2π − α.The filament carriescurrent I and is located at(ρ′, φ′).

x

y

α

ρ� φ�




A Different Approach

We can write the

Total Electric Field

Ez =∑ν

aνJν(kρ<)H(2)ν (kρ>) sin[ν(φ′ − α)] sin[ν(φ− α)]

The dependence on ρ and φ is clear enough.How can I just write a dependence on ρ′ and φ′?Since Ez(α) = Ez(2π − α) = 0,

ν = νm =mπ

2(π − α)




Expansion of the Current

The current can be thought of as a cylindrical sheet current(A/m) with an impulsive distribution:

Jz = Iδ(φ− φ′)

ρ′

This current can be expanded in a Fourier series in the usualway:

Jz =I

(π − α)ρ′∞∑

m=1

sin νm(φ′ − α) sin νm(φ− α)

From boundary conditions, this current is related to the field by

Jz = Hφ(ρ′+)− Hφ(ρ

′−)




Scattering from a Wedge

From the usual equations we can find

Hφ =

k

jωµ

∞∑m=1

aνm H(2)νm (kρ′)J ′νm

(kρ) sin νm(φ′ − α) sin νm(φ− α) ρ < ρ′

kjωµ

∞∑m=1

aνm H(2)′νm (kρ′)Jνm(kρ) sin νm(φ

′ − α) sin νm(φ− α) ρ > ρ′

Using the Wronskian, we can write

Jz = − 2ωµπρ′

∞∑m=1

aνm sin νm(φ′ − α) sin νm(φ− α)

Equating the two expressions for Jz we find

aνm = − ωµπI2(π − α)




Plane Wave Scattering from a Wedge

We can generalize our solution to plane wave scattering bytaking the limit as the limit as the source recedes. The originalincident field was

E iz =−k2I4ωε

H(2)0 (k |ρ− ρ′|)

Using the large argument approximation and the far fieldapproximation of |ρ− ρ′| we write

E iz = −ωµI

4

√2jπkρ′

e−jkρ′ejkρ cos(φ−φ′)





This can be written as

E iz = E0ejkρ cos(φ−φ′)

where

E0 =−ωµI

4

√2jπkρ′

e−jkρ′

Why is E0 dependent on ρ?

For large ρ′ our solution is

Ez =

√2jπkρ′

∞∑m=1

anjνmJνm(kρ) sin νm(φ′ − α) sin νm(φ− α)





This can be written as

E iz = E0ejkρ cos(φ−φ′)

where

E0 =−ωµI

4

√2jπkρ′

e−jkρ′

Why is E0 dependent on ρ? For large ρ′ our solution is

Ez =

√2jπkρ′

∞∑m=1

anjνmJνm(kρ) sin νm(φ′ − α) sin νm(φ− α)





Plugging in for I and an we find

The Total Field

Ez =2πE0

π − α

∞∑m=1

jνmJνm(kρ) sin νm(φ′ − α) sin νm(φ− α)




Outline






2.5 Dimensions?

2.5 dimensions is a “term of art,” not an actual physicaldescription.The geometry of the problem is assumed two-dimensional.The source, however, may be three dimensional.The problems are attacked by superposition (i.e. FourierMethods).

Suppose (∂2

∂x2 +∂2

∂y2 +∂2

∂z2 + k2)ψ = 0




The General Approach

Since the boundaries are independent of z, we can define ψ interms of

A Superposition

ψ(x , y , z) =1

2π

∞∫−∞

ψ(x , y , kz)ejkzzdkz

Substituting into the Helmholtz equation, we find that[∂2

∂x2 +∂2

∂y2 + (k2 − k2z )

]ψ = 0

We define (as usual) k2ρ = k2 − k2

z .




A Filament

Suppose we have a filament of current I(z).We of course have a TMz field, Az = µψ.The wave function can be written in the form

ψ =1

2π

∞∫−∞

f (kz)H(2)0 (kρρ)dkz .

Here, then, the transform of the function is

ψ = f (kz)H(2)0 (kρρ).




A Filament

The azimuthal magnetic field is

Hφ = −∂ψ∂ρ

= −kρf (kz)H(2)′0 (kρρ).

Now, by Ampère’s law,

limρ→0

∮Hφd` = I

For small ρ,

Hφ = kρf (kz)d

dx

(2jπ

lnγx2

)x=kρ

=2jπρ

f (kz)




A Filament

Thuslimρ→0

∮Hφd` = 2πρ

2jπρ

f (kz) = I

or

f (kz) =I4j

Finally, we find the

2.5-D Filament Solution

ψ =1

8πj

∞∫−∞

I(kz)H(2)0

(ρ

√k2 − k2

z

)ejkzzdkz




Another Way

We can always make a line current out of dipoles. We thus have

The Spatial Approach

ψ =

∞∫−∞

I(z ′)e−jk√ρ2+(z−z′)2

4π√ρ2 + (z − z ′)2

dz ′

Now, suppose we choose I(z ′) = I`δ(z ′). Then

ψ = I`e−jkr

4πr.

Also, we have I(kz) = I`, since the Fourier Transform of thedelta function is unity.




A New Identity

Plugging this transform into the 2.5-D solution,

ψ =I`

8πj

∞∫−∞

H(2)0

(ρ

√k2 − k2

z

)ejkzzdkz

Setting these two expressions to each other, we find

An Identity

ejkr

r=

12j

∞∫−∞

H(2)0

(ρ

√k2 − k2

z

)ejkzzdkz


Date post:	10-Jun-2018
Category:	Documents
Upload:	duonganh
View:	230 times
Download:	3 times

Cylindrical Radiation and Scattering - ECE/CISweile/ELEG648Spring06/...Scattering from a Circular...

Documents