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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
Cylindrical RadiationScattering
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
D. S. Weile Cylindrical Radiation and Scattering
Cylindrical RadiationScattering
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
D. S. Weile Cylindrical Radiation and Scattering
Cylindrical RadiationScattering
Sources of Cylindrical RadiationGreen’s Function and Far FieldWave Transformations
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
D. S. Weile Cylindrical Radiation and Scattering
Cylindrical RadiationScattering
Sources of Cylindrical RadiationGreen’s Function and Far FieldWave Transformations
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)
D. S. Weile Cylindrical Radiation and Scattering
Cylindrical RadiationScattering
Sources of Cylindrical RadiationGreen’s Function and Far FieldWave Transformations
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
D. S. Weile Cylindrical Radiation and Scattering
Cylindrical RadiationScattering
Sources of Cylindrical RadiationGreen’s Function and Far FieldWave Transformations
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)
D. S. Weile Cylindrical Radiation and Scattering
Cylindrical RadiationScattering
Sources of Cylindrical RadiationGreen’s Function and Far FieldWave Transformations
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
D. S. Weile Cylindrical Radiation and Scattering
Cylindrical RadiationScattering
Sources of Cylindrical RadiationGreen’s Function and Far FieldWave Transformations
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
D. S. Weile Cylindrical Radiation and Scattering
Cylindrical RadiationScattering
Sources of Cylindrical RadiationGreen’s Function and Far FieldWave Transformations
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′)
D. S. Weile Cylindrical Radiation and Scattering
Cylindrical RadiationScattering
Sources of Cylindrical RadiationGreen’s Function and Far FieldWave Transformations
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′|)
D. S. Weile Cylindrical Radiation and Scattering
Cylindrical RadiationScattering
Sources of Cylindrical RadiationGreen’s Function and Far FieldWave Transformations
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
D. S. Weile Cylindrical Radiation and Scattering
Cylindrical RadiationScattering
Sources of Cylindrical RadiationGreen’s Function and Far FieldWave Transformations
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.
D. S. Weile Cylindrical Radiation and Scattering
Cylindrical RadiationScattering
Sources of Cylindrical RadiationGreen’s Function and Far FieldWave Transformations
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′
D. S. Weile Cylindrical Radiation and Scattering
Cylindrical RadiationScattering
Sources of Cylindrical RadiationGreen’s Function and Far FieldWave Transformations
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
D. S. Weile Cylindrical Radiation and Scattering
Cylindrical RadiationScattering
Sources of Cylindrical RadiationGreen’s Function and Far FieldWave Transformations
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
D. S. Weile Cylindrical Radiation and Scattering
Cylindrical RadiationScattering
Sources of Cylindrical RadiationGreen’s Function and Far FieldWave Transformations
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
D. S. Weile Cylindrical Radiation and Scattering
Cylindrical RadiationScattering
Sources of Cylindrical RadiationGreen’s Function and Far FieldWave Transformations
Plane Wave Expansion
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
D. S. Weile Cylindrical Radiation and Scattering
Cylindrical RadiationScattering
Sources of Cylindrical RadiationGreen’s Function and Far FieldWave Transformations
Plane Wave Expansion
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?
D. S. Weile Cylindrical Radiation and Scattering
Cylindrical RadiationScattering
Sources of Cylindrical RadiationGreen’s Function and Far FieldWave Transformations
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
D. S. Weile Cylindrical Radiation and Scattering
Cylindrical RadiationScattering
Sources of Cylindrical RadiationGreen’s Function and Far FieldWave Transformations
The Addition Theorem
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
D. S. Weile Cylindrical Radiation and Scattering
Cylindrical RadiationScattering
Sources of Cylindrical RadiationGreen’s Function and Far FieldWave Transformations
The Addition Theorem
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′
D. S. Weile Cylindrical Radiation and Scattering
Cylindrical RadiationScattering
Sources of Cylindrical RadiationGreen’s Function and Far FieldWave Transformations
The Addition Theorem
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
D. S. Weile Cylindrical Radiation and Scattering
Cylindrical RadiationScattering
Sources of Cylindrical RadiationGreen’s Function and Far FieldWave Transformations
The Addition Theorem
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′
D. S. Weile Cylindrical Radiation and Scattering
Cylindrical RadiationScattering
Sources of Cylindrical RadiationGreen’s Function and Far FieldWave Transformations
The Addition Theorem
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
The Addition Theorem
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′
D. S. Weile Cylindrical Radiation and Scattering
Cylindrical RadiationScattering
Scattering from a Circular CylinderScattering from a Wedge2.5-D Problems
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
D. S. Weile Cylindrical Radiation and Scattering
Cylindrical RadiationScattering
Scattering from a Circular CylinderScattering from a Wedge2.5-D Problems
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.
D. S. Weile Cylindrical Radiation and Scattering
Cylindrical RadiationScattering
Scattering from a Circular CylinderScattering from a Wedge2.5-D Problems
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)
D. S. Weile Cylindrical Radiation and Scattering
Cylindrical RadiationScattering
Scattering from a Circular CylinderScattering from a Wedge2.5-D Problems
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)
D. S. Weile Cylindrical Radiation and Scattering
Cylindrical RadiationScattering
Scattering from a Circular CylinderScattering from a Wedge2.5-D Problems
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)
D. S. Weile Cylindrical Radiation and Scattering
Cylindrical RadiationScattering
Scattering from a Circular CylinderScattering from a Wedge2.5-D Problems
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
D. S. Weile Cylindrical Radiation and Scattering
Cylindrical RadiationScattering
Scattering from a Circular CylinderScattering from a Wedge2.5-D Problems
TM Scattering from a Circular Cylinder, Scattered Field
Of course, the exact scattered field is given above:
Esz = E0
∞∑n=−∞
j−nanH(2)n (kρ)ejnφ
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.
D. S. Weile Cylindrical Radiation and Scattering
Cylindrical RadiationScattering
Scattering from a Circular CylinderScattering from a Wedge2.5-D Problems
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′)
D. S. Weile Cylindrical Radiation and Scattering
Cylindrical RadiationScattering
Scattering from a Circular CylinderScattering from a Wedge2.5-D Problems
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′)
D. S. Weile Cylindrical Radiation and Scattering
Cylindrical RadiationScattering
Scattering from a Circular CylinderScattering from a Wedge2.5-D Problems
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?
D. S. Weile Cylindrical Radiation and Scattering
Cylindrical RadiationScattering
Scattering from a Circular CylinderScattering from a Wedge2.5-D Problems
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?
D. S. Weile Cylindrical Radiation and Scattering
Cylindrical RadiationScattering
Scattering from a Circular CylinderScattering from a Wedge2.5-D Problems
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?
D. S. Weile Cylindrical Radiation and Scattering
Cylindrical RadiationScattering
Scattering from a Circular CylinderScattering from a Wedge2.5-D Problems
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?
D. S. Weile Cylindrical Radiation and Scattering
Cylindrical RadiationScattering
Scattering from a Circular CylinderScattering from a Wedge2.5-D Problems
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?
D. S. Weile Cylindrical Radiation and Scattering
Cylindrical RadiationScattering
Scattering from a Circular CylinderScattering from a Wedge2.5-D Problems
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?
D. S. Weile Cylindrical Radiation and Scattering
Cylindrical RadiationScattering
Scattering from a Circular CylinderScattering from a Wedge2.5-D Problems
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
D. S. Weile Cylindrical Radiation and Scattering
Cylindrical RadiationScattering
Scattering from a Circular CylinderScattering from a Wedge2.5-D Problems
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�
D. S. Weile Cylindrical Radiation and Scattering
Cylindrical RadiationScattering
Scattering from a Circular CylinderScattering from a Wedge2.5-D Problems
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)
D. S. Weile Cylindrical Radiation and Scattering
Cylindrical RadiationScattering
Scattering from a Circular CylinderScattering from a Wedge2.5-D Problems
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
′−)
D. S. Weile Cylindrical Radiation and Scattering
Cylindrical RadiationScattering
Scattering from a Circular CylinderScattering from a Wedge2.5-D Problems
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)
D. S. Weile Cylindrical Radiation and Scattering
Cylindrical RadiationScattering
Scattering from a Circular CylinderScattering from a Wedge2.5-D Problems
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′)
D. S. Weile Cylindrical Radiation and Scattering
Cylindrical RadiationScattering
Scattering from a Circular CylinderScattering from a Wedge2.5-D Problems
Plane Wave Scattering from a Wedge
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)
D. S. Weile Cylindrical Radiation and Scattering
Cylindrical RadiationScattering
Scattering from a Circular CylinderScattering from a Wedge2.5-D Problems
Plane Wave Scattering from a Wedge
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)
D. S. Weile Cylindrical Radiation and Scattering
Cylindrical RadiationScattering
Scattering from a Circular CylinderScattering from a Wedge2.5-D Problems
Plane Wave Scattering from a Wedge
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)
D. S. Weile Cylindrical Radiation and Scattering
Cylindrical RadiationScattering
Scattering from a Circular CylinderScattering from a Wedge2.5-D Problems
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
D. S. Weile Cylindrical Radiation and Scattering
Cylindrical RadiationScattering
Scattering from a Circular CylinderScattering from a Wedge2.5-D Problems
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
D. S. Weile Cylindrical Radiation and Scattering
Cylindrical RadiationScattering
Scattering from a Circular CylinderScattering from a Wedge2.5-D Problems
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 .
D. S. Weile Cylindrical Radiation and Scattering
Cylindrical RadiationScattering
Scattering from a Circular CylinderScattering from a Wedge2.5-D Problems
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).
D. S. Weile Cylindrical Radiation and Scattering
Cylindrical RadiationScattering
Scattering from a Circular CylinderScattering from a Wedge2.5-D Problems
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)
D. S. Weile Cylindrical Radiation and Scattering
Cylindrical RadiationScattering
Scattering from a Circular CylinderScattering from a Wedge2.5-D Problems
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
D. S. Weile Cylindrical Radiation and Scattering
Cylindrical RadiationScattering
Scattering from a Circular CylinderScattering from a Wedge2.5-D Problems
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.
D. S. Weile Cylindrical Radiation and Scattering
Cylindrical RadiationScattering
Scattering from a Circular CylinderScattering from a Wedge2.5-D Problems
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
D. S. Weile Cylindrical Radiation and Scattering
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