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202 Chapter 7

Elementary Acoustic and Electromagnetic Edge Waves

Functions f ,

g and f hp, ghp are singular in the directions of the incident

and reflected plane waves (ϕ = π + ϕ0, ϕ = π ϕ0, ϕ = 2α π ϕ0), but their difference is finite. For instance,

 

 

(1)

γ0, π

ϕ0) = −

1

cot

 

ϕ0

+

 

1

 

cot

 

ϕ0

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

Fh

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

2n

 

 

 

 

n

 

 

4

2

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

1

 

cot

 

 

 

π ϕ0

 

 

 

 

 

ε(α

 

 

 

 

 

 

ϕ

 

)

1

 

 

 

cot

π ϕ0

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

2n

 

 

 

 

 

 

 

 

 

 

+

0

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

n

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

4

 

 

 

 

 

 

2

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

+

 

2n

 

 

 

 

 

 

 

 

 

n

 

 

 

 

 

 

 

 

 

 

 

 

 

 

0

 

4

 

 

 

 

 

 

 

 

 

2

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

1

 

cot

 

 

 

α π

 

 

 

 

 

 

ε(α

 

 

 

 

 

ϕ

)

 

1

cot

 

α π

,

 

(7.173)

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

(1)

γ

, π

ϕ

)

= −

1

cot

 

ϕ0

 

+

 

1

cot

ϕ0

 

1

 

 

cot

π ϕ0

 

 

 

 

 

 

 

 

Fs

 

 

 

0

 

 

0

 

2n

 

 

 

 

 

 

 

n

 

 

 

4

 

 

 

 

 

 

 

 

2

 

 

 

2n

 

 

 

 

 

 

 

 

n

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

ε(α

 

 

ϕ

)

1

cot

π ϕ0

 

 

 

 

 

 

 

 

1

 

 

 

 

cot

α π

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

+

4

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

n

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

0

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

2

 

 

 

 

 

 

 

2n

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

+

 

 

 

 

 

0

 

 

4

 

 

 

 

 

 

 

 

 

 

 

 

 

 

2

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

ε(α

 

 

ϕ

)

1

cot

α π

,

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

(7.174)

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

(1)

γ

, 2α

π

ϕ

)

= −

1

cot

 

ϕ0 π )

 

 

+

 

1

 

cot

ϕ0 π )

 

 

 

Fh

2n

 

 

 

 

 

 

 

 

 

 

 

 

 

 

4

 

 

 

 

 

 

0

 

 

 

 

0

 

 

 

 

 

 

 

 

 

 

 

 

 

 

n

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

2

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

+

 

1

 

cot

 

 

 

α π

 

 

 

 

1

cot

α π

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

2n

 

 

 

 

 

 

 

4

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

n

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

2

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

1

 

cot

 

 

 

α ϕ0

 

+

 

 

1

cot

α ϕ0

,

 

 

 

 

 

 

 

 

(7.175)

 

 

 

 

 

 

 

 

 

 

 

 

 

 

2n

 

 

 

 

 

 

 

 

 

4

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

n

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

2

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

and

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

(1)

γ

, 2α

π

γ

)

= −

1

cot

ϕ0 π )

 

+

 

 

1

cot

ϕ0 π )

 

 

2n

 

 

 

 

 

 

 

4

 

 

 

Fs

 

0

 

 

 

 

0

 

 

 

 

 

 

 

 

 

 

 

 

 

 

n

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

2

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

1

 

cot

 

 

 

 

α π

 

+

 

 

1

cot

α π

 

 

 

 

 

1

cot

α ϕ0

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

2n

 

 

 

 

 

 

4

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

n

 

 

 

 

 

 

 

 

 

 

 

 

 

 

2

 

 

 

 

 

 

 

 

 

 

2n

n

 

 

 

 

 

 

 

 

 

 

 

 

 

+

1

cot

α ϕ0

.

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

(7.176)

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

4

 

 

 

 

 

 

 

 

 

2

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

F(1)

These equations clearly show that functions h,s are free from the grazing singularity. Indeed, for the grazing directions of the incident wave (ϕ0 = π , ϕ0 = α π ), these functions have the finite values

Fh(1)

γ0, 0) = Fh(1)

1

 

π

+

1

 

 

 

α

 

γ0, α) = −

 

cot

 

 

tan

 

 

(7.177)

n

n

4

2

and

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

Fs(1)γ0, 0) = Fs(1)γ0, α) = −

1

tan

α

.

(7.178)

 

 

 

 

4

2

It also follows from these equations that functions Fh,s(1) are equal to zero when α = 2π and the wedge transforms into the half-plane. This result is the obvious consequence

TEAM LinG

7.9 Improved Theory of Elementary Edge Waves 203

of the general expressions (7.163) and (7.164) and the identities Vt = V thp, Ut = U thp, which are valid in the case α = 2π .

To find the total field scattered by finite objects, one should also calculate the contribution generated by the uniform component distributed over the finite elementary

strips (0 ≤ ξ1,2 l). In the far zone (R

 

 

kl2), it is determined by the integrals

 

dζ

 

eikR l

 

duh1(0) =

 

ik sin γ0 sin ϑ sin ϕ

 

 

 

0 jh1(0)1, ϕ0)eikξ1 cos β1 dξ1

(7.179)

4π

 

R

and

 

 

 

 

 

 

 

 

 

 

 

 

 

dζ

eikR

l

 

 

dus1(0) = −

 

sin γ0

 

 

 

0 js1(0)1, ϕ0)eikξ1 cos β1 dξ1.

(7.180)

 

4π

R

 

 

Replacing ξ1, β1, ϕ, ϕ0 here with ξ2, β2, α ϕ, α ϕ0, one obtains the equations associated with the field from strip 2 (0 ≤ ξ2 l). These integrals are easily calculated in closed form. We show only those results that relate to the grazing incidence (ϕ0 = π ):

(0)

= u0eikζ cos γ0

dζ eikR sin γ0 sin ϑ sin ϕ

)eikl(1−cos β1) − 1* ,

(7.181)

duh1

 

 

 

 

 

 

 

 

 

 

 

 

4π R

1 − cos β1

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

sin γ0

 

 

 

 

 

 

 

 

 

(0)

= u0eikζ cos γ0

dζ eikR

 

 

2

 

 

ei3π/4

 

kl(1−cos β1)

2

dus1

 

 

 

 

 

 

 

π

0

 

 

eit dt.

2π

R

1

cos β1

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

(7.182)

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

For the grazing scattering direction (β1 = 0, ϕ = 0), it follows from these equations that

duh1(0) = 0

 

 

 

 

 

 

(7.183)

and

 

 

 

 

 

 

 

 

 

 

 

dζ eikR

 

 

 

 

 

 

dus1(0) = u0eikζ cos γ0

 

 

2kl

 

 

 

 

 

sin γ0

 

 

ei3π/4.

(7.184)

2π R

π

As was expected, the field duh,s(0) is also free from the grazing singularity.

7.9.2Electromagnetic EEWs

This theory is based on the paper by Ufimtsev (2006b).

Asymptotic expressions for electromagnetic EEWs established in Section 7.8 possess the grazing singularity. Careful analysis reveals the reason for this singularity. As in

TEAM LinG

204 Chapter 7 Elementary Acoustic and Electromagnetic Edge Waves

(0)

the case of acoustic waves, it turns out that the original definitions of quantities j and

j(1) are not adequate for actual surface currents induced under the grazing incidence (ϕ0 = π or ϕ0 = α π ). In this case, for instance, the component jn(1), normal to the edge, does not vanish away from the edge, but instead it transforms into a plane wave

there. Also, the component jn(0) includes the absent reflected wave. Therefore, one

(0,1)

should modify the original definitions of j to avoid the grazing singularity.

(0) hp

With this purpose we introduce a new uniform component j j identical to the electric surface current induced on the illuminated side of the tangential perfectly

(1)

conducting half-plane. A new nonuniform component of the surface current j is the

difference j(1) = j(t) j hp, where j(t) is the total current induced on the tangential perfectly conducting wedge.

(1)

EEWs Radiated by the Nonuniform Component j

The incident electromagnetic wave (7.129) undergoes diffraction at a scattering per-

(1)

fectly conducting object (Fig. 8.1) and creates there a surface current j . This current radiates diffracted EEWs. Far from the diffraction point ζ (kR 1), these waves can be presented again in the form of Equations (7.130) and (7.131) with new functions

(1) (1)

F and G . These new functions can be found in the same way as described in Section 7.8 and in the papers by Butorin et al. (1987) and Ufimtsev (1991), with a simple modification based on the replacement (7.159). We provide here only the final results.

(1)

The field generated by j is described as

dE

(1)

 

 

dζ

(1), ϕ)

eikR

and

dH(1)

 

R

 

dE

(1)

/Z0

 

=

 

 

 

R

= [

×

]

 

 

2π E

 

 

 

 

 

 

 

with

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

(1), ϕ)

= [

E0t (ζ ) (1), ϕ)

+

Z0H0t (ζ )

(1), ϕ)

eikφi(ζ ),

E

 

 

 

 

 

F

 

G

 

]

 

 

 

where

(7.185)

(7.186)

Fϑ(1), ϕ) = [Ut 1, ϕ0) Uthp1, ϕ0)] sin ϑ

 

+ [Ut 2, α ϕ0) ε(α ϕ0)Uthp2, α ϕ0)] sin ϑ ,

(7.187)

Fϕ(1), ϕ) = 0,

(7.188)

Gϑ(1), ϕ) = [sin γ0 cos ϑ cos ϕ − cos γ0 sin ϑ cos σ1][Vt 1, ϕ0) Vthp1, ϕ0)]

− [sin γ0 cos ϑ cosϕ) − cos γ0 sin ϑ cos σ2]

 

× [Vt 2, α ϕ0) ε(α ϕ0)Vthp2, α ϕ0)],

(7.189)

TEAM LinG

7.9 Improved Theory of Elementary Edge Waves 205

and

Gϕ(1), ϕ) = −[Vt 1, ϕ0) Vthp1, ϕ0)] sin ϕ sin γ0

− [Vt 2, α ϕ0) ε(α ϕ0)Vthp2, α ϕ0)] sinϕ) sin γ0. (7.190)

It is supposed here that 0 < ϕ0 π .

In the directions ϑ = π γ0 associated with the diffraction cone, these expres-

sions are simplified as

 

 

 

Fϑ(1)γ0, ϕ) = −

 

1

[ f (ϕ, ϕ0, α) f hp, ϕ0)],

Fϕ(1), ϕ) = 0,

 

sin γ0

 

 

 

 

 

(7.191)

1

 

 

 

Gϑ(1)γ0, ϕ) = 0,

Gϕ(1)γ0, ϕ) =

 

[g(ϕ, ϕ0, α) ghp, ϕ0)],

sin γ0

 

 

 

 

 

(7.192)

with functions f , g and f hp, ghp defined above in Section 7.9.1. Comparison with Equations (7.167) and (7.168) reveals the following relationships between the electromagnetic and acoustic EEWs:

Fϑ(1)

1

 

Fs(1)γ0, ϕ)

 

γ0, ϕ) = −

 

 

(7.193)

sin γ0

and

 

 

 

 

 

 

 

1

Fh(1)γ0, ϕ).

 

Gϕ(1)γ0, ϕ) =

 

(7.194)

sin γ0

According to Section 7.9.1 these functions are free from the grazing singularities as well as from the singularities in the directions of the incident and reflected rays.

(0) hp

Field Radiated by the Uniform Component j j

hp

Here we investigate the field radiated by the current j induced on the finite elementary strips (0 ≤ ξ1,2 l) belonging to the finite plane faces of a scattering object (Fig. 7.3). In this investigation we apply Cartesian coordinates x, y, z and x , y , z associated with faces 1 and 2, respectively. Axes x and x belong to faces 1 and 2, respectively, and they are parallel to tangents τ1 and τ2. Utilizing the known solution

TEAM LinG

206

Chapter 7 Elementary Acoustic and Electromagnetic Edge Waves

 

 

 

 

 

 

 

for the half-plane diffraction problem, one finds the current on strip 1,

 

 

jx(0,1) jxhp,1 = 2H0zeikζ cos γ0 eikξ1(cos2 γ0sin2 γ0 cos ϕ0)

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

× 1 −

eiπ/4

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

2

 

 

 

,

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

·

 

 

sin γ0 cos

ϕ0

 

eit

 

 

dt

 

 

 

 

 

 

 

 

 

 

 

 

(7.195)

 

 

 

π

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

2 1

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

2

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

jz(,10) jzhp,1 = −2eikζ cos γ0 eikξ1 cos

2

γ0

 

1

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

sin γ0

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

× Y0E0z sin ϕ0eik

ξ1 sin

2

γ0 cos ϕ0

eiπ/4

 

 

 

 

 

 

 

2

 

 

 

 

 

 

 

 

·

 

 

sin γ0 cos

ϕ0

eit

dt

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

π

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

2 1

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

2

 

 

 

 

 

 

 

 

 

ϕ

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

sin

 

0

 

 

 

ei3π/4

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

2

 

eikξ1 sin2 γ0

 

 

 

sin ϕ0eikξ1 sin2 γ0 cos ϕ0

 

 

 

 

+ sin γ0

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

2π kξ1

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

cos ϕ0eikξ1 sin

2

 

γ0 cos ϕ0

eiπ/4

 

 

 

 

 

 

 

 

 

2

 

 

+ H0z cos γ0

 

 

 

 

 

·

 

sin γ0 cos

ϕ0

eit

dt

 

 

 

 

 

 

 

 

 

 

 

π

 

 

 

 

 

 

 

 

 

 

 

 

 

 

2 1

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

2

(7.196)

 

 

 

cos

20

 

 

 

ei3π/4

eikξ1 sin2 γ0

 

 

 

cos ϕ0eikξ1 sin2 γ0 cos ϕ0

 

 

 

 

 

 

 

ϕ

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

+ sin γ0

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

(0)

 

 

 

 

2π kξ1

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

where Y

1/Z

 

 

is the admittance of free space. The current components j

x ,2

and

(0)

0 =

 

0

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

jz,2

on strip 2 are determined by Equations (7.195) and (7.196) with the replacements

 

 

 

 

 

H0z → −H0z,

 

 

 

ξ1 ξ2,

 

 

 

ϕ0 α ϕ0.

 

 

 

 

 

(7.197)

 

The field radiated by these currents is determined (in terms of the retarded vector-

potential dA) as

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

dEϕ = −Z0 dHϑ = ikZ0[−dAx,1 sin ϕ + ε(α ϕ0)dAx ,2 sinϕ)],

(7.198)

and

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

dEϑ = Z0 dHϕ = ikZ0{−[dAz,1 + ε(α ϕ0)dAz,2] sin ϑ

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

+ [dAx,1 cos ϕ + ε(α ϕ0)dAx ,2 cosϕ)] cos ϑ ]}.

(7.199)

It is supposed here that 0 < ϕ0 π . For the far zone (R

 

kl 2), one can use the

approximation

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

dA

 

=

 

 

dζ

 

 

sin γ

eikR

 

 

l j

 

 

 

 

 

)eikξ1,2 cos β1,2 dξ

 

 

.

 

 

 

(7.200)

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

1,2

 

4π

 

 

0

 

R

0

1,2

 

 

1,2

 

 

 

 

 

 

 

 

 

 

 

 

 

1,2

 

 

 

 

 

 

 

TEAM LinG

7.9 Improved Theory of Elementary Edge Waves 207

(0)

After substitution of j 1,2 into Equation (7.200), this leads to the following expressions:

 

 

1

H0zeikζ cos γ0

dζ

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

eikR

 

 

 

 

 

 

dAx,1 = −

 

 

 

 

[B3, ϕ0) B1, ϕ0)] sin γ0

 

 

,

 

 

 

(7.201)

ik

2π

R

 

 

 

 

 

1

eikζ cos γ0

dζ eikR

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

dAz,1 = −

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

ik

2π

 

 

 

R

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

Y0E0z

B3, ϕ0) sin ϕ0

 

 

 

 

 

 

 

 

 

ϕ

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

B2(ϕ) sin

 

0

 

 

 

 

 

 

 

 

 

 

 

2

 

 

 

B1, ϕ0) sin ϕ0

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

+

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

sin γ0

 

ϕ0

 

 

 

 

 

 

 

 

 

 

 

×

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

B1, ϕ0) cos ϕ0 ,

 

 

 

H0z cos γ0 B3, ϕ0) cos ϕ0

 

 

B2(ϕ) cos

2

 

 

+

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

+

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

sin γ0

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

(7.202)

where

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

B

, ϕ

)

=

eikl (ϕ,ϕ0) − 1

,

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

(7.203)

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

1

 

0

 

, ϕ0)

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

eiπ/4

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

2

 

 

 

kl (ϕ)

2

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

B2(ϕ) =

 

 

 

 

0

 

 

 

 

eit

 

 

dt,

 

 

 

 

 

 

 

 

 

 

 

 

(7.204)

 

(ϕ)

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

π

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

B3

, ϕ0)

 

1

 

 

 

 

 

e

ikl (ϕ,ϕ0) eiπ/4 ∞

 

 

 

 

 

 

 

e

it

2

dt

 

1

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

=

, ϕ0)

 

 

 

 

 

 

 

π

 

 

2kl

sin γ0 cos

ϕ0

 

 

 

 

2

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

2

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

ϕ

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

+ B2(ϕ) sin γ0 cos

 

0

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

(7.205)

 

 

 

 

 

 

2

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

with

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

, ϕ0) = cos2 γ0 − sin2 γ0 cos ϕ0 − cos β1,

 

 

 

 

(7.206)

 

 

 

 

 

 

 

(ϕ) = 1 − cos β1.

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

(7.207)

Components dAx ,2 and dAz,2 are found from dAx,1 and dAz,1, respectively, with

replacements H0z → −H0z, ϕ α ϕ, ϕ0 α ϕ0, and β1 β2.

We then substitute the above equations for the vector dA into Equations (7.198) and (7.199) and obtain the field expressions in the form of Equations (7.130) and (7.131):

dE(0)

 

 

dζ

(0), ϕ)

eikR

,

dH

(0)

 

R

 

dE

(0)

/Z0,

(7.208)

=

 

 

 

R

 

= [

×

]

(0), ϕ)

2π E

 

 

 

 

 

 

 

 

= [

E0t (ζ ) (0), ϕ)

+

Z0H0t (ζ ) (0), ϕ)

eikφi (ζ ),

(7.209)

E

 

F

 

 

 

G

 

 

]

 

 

 

 

TEAM LinG

208 Chapter 7

Elementary Acoustic and Electromagnetic Edge Waves

 

 

where

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

Fϑ(0) = F1ϑ , ϕ0, ϑ ) + ε(α ϕ0)F1ϑ ϕ, α ϕ0, ϑ ),

Fϕ(0) = 0,

(7.210)

 

Gϑ(0) = G1ϑ , ϕ0, ϑ ) ε(α ϕ0)G1ϑ ϕ, α ϕ0, ϑ ),

 

 

(7.211)

 

Gϕ(0) = G1ϕ , ϕ0, ϑ ) + ε(α ϕ0)G1ϕ ϕ, α ϕ0, ϑ ).

 

 

(7.212)

Here,

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

B3, ϕ0) sin ϕ0

 

 

 

 

 

ϕ

 

 

 

 

sin ϑ ,

 

 

 

 

 

 

(ϕ)

sin

0

 

 

 

 

F

1ϑ , ϕ0

, ϑ )

=

+

B2

2

 

 

B1, ϕ0) sin ϕ0

sin γ0

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

(7.213)

 

 

 

 

B3, ϕ0) cos ϕ0

 

 

 

 

 

 

ϕ

 

 

B1, ϕ0) cos ϕ0 cos γ0 sin ϑ

 

 

 

 

 

 

(ϕ)

cos

 

0

 

 

G

1ϑ , ϕ0

, ϑ )

=

+

B2

2

 

 

 

 

 

 

 

 

 

 

 

sin γ0

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

− [B3, ϕ0) B1, ϕ0)] sin γ0 cos ϑ cos ϕ,

 

 

(7.214)

G1ϕ , ϕ0, ϑ ) = [B3, ϕ0) B1, ϕ0)] sin γ0 sin ϕ.

 

 

 

(7.215)

Notice that functions B1,2,3 are finite when = 0 and = 0. In particular, for the grazing incidence (ϕ0 = π ) and for the grazing scattering (ϑ = π γ0, ϕ = 0), they are equal to

 

eiπ/4

 

 

1

 

 

B1 = ikl, B2 =

2kl

 

 

,

B3 =

 

kl.

(7.216)

2

π

As expected, the field generated by the uniform component of the surface current is also free from the grazing singularity.

Thus, the asymptotic theory developed in Sections 7.9.1 and 7.9.2 is valid for all directions of incidence and scattering. It is well suited for calculation of bistatic scattering in the case when both planar faces of the edge are illuminated by the incident wave (α π ϕ0 π ). For other incidence directions ϕ0, one can apply the original theory presented in Sections 7.1 to 7.8.

Here it is pertinent to mention the alternative approach (Michaeli, 1987; Breinbjerg, 1992; Johansen, 1996) for elimination of the grazing singularity. The uniform and nonuniform components of the surface current are defined there according to the original PTD, and the grazing singularity is eliminated by truncation of

elementary strips (0 ≤ x1,2 l). Compared to this approach, a distinctive feature of

(1)

the present theory is as follows: It introduces a new nonuniform scattering source j that generates an elementary edge wave regular in all scattering directions. In other words, it allows the extraction of the fringe component from the total field in a pure explicit form.

TEAM LinG

7.10 Some References Related to Elementary Edge Waves 209

7.10 SOME REFERENCES RELATED TO ELEMENTARY EDGE WAVES

The investigation of EEWs has a long history. In Kirchhoff’s approach, the EEWs were first discovered by Maggi (Maggi, 1888; Bakker and Copson, 1950). The same result was rediscovered by Rubinowicz (1917). A similar approach to electromagnetic EEWs was developed by Kottler (1923).

Attempts to define EEWs more strictly (on the basis of the Sommerfeld (1896, 1935) exact solution of the wedge diffraction problem) were first undertaken by Bateman (1955) and Rubinowicz (1965). However, their expressions for EEWs satisfy the Dirichlet or Neuman boundary conditions everywhere at the faces of the canonical tangent wedge. For this reason, these EEWs predict incorrect values for the diffracted field at those parts of the virtual tangent wedge that are extended outside the real scattering object (as shown in Fig. 7.10 by dotted lines). Infact, according to this definition of EEWs, the infinite plane areas of free space (outside the scattering object) formally become perfectly reflecting.

The same drawback exists in another theory of EEWs suggested in by Tiberio et al. (1994, 1995, 2004). In PTD a similar shortcoming occurs only at the extensions of infinitely narrow elementary strips (Fig. 7.3). As PTD is a source-based theory, this shortcoming can be completely removed by the truncation of elementary strips (Johansen, 1996).

The directivity patterns of electromagnetic EEWs can be interpreted as equivalent edge currents (EECs). The EECs introduced in the work of Knott and Senior (1973) and Knott (1985) are based on GTD and are valid only for the directions of the diffracted rays. The EECs based on PTD are applicable for arbitrary scattering directions (Michaeli, 1986, 1987; Breinbjerg, 1992; Johansen, 1996). Notice that the untruncated EECs developed by Michaeli (1986) are in complete agreement with the EEWs derived in Section 7.8. However, his truncated EECs (Michaeli, 1987) are not free from some shortcomings, which were overcome by Johansen (1996). The paper by Johansen (1996) also contains additional references related to the EEC concept.

Another interpretation of the directivity pattern of EEWs is the so-called incremental length diffraction coefficient (ILDC). The term ILDC was introduced by Mitzner (1974), who determined the ILDCs on the basis of PTD. The ILDC concept was further developed in the work of Tiberio et al. (2004).

Figure 7.10 Perfectly reflecting solid prism of finite size (solid and dashed lines) and infinite faces of the virtual tangent wedge (dotted lines).

TEAM LinG

210 Chapter 7 Elementary Acoustic and Electromagnetic Edge Waves

Notice also the asymptotic theory for plane screens (Wolf, 1967), which is similar to PTD and the method of matched asymptotic expansions (Tran Van Nhieu, 1995, 1996), which lead to the PTD ray asymptotics.

PROBLEMS

7.1Start with Equations (7.3) and (7.4) and derive Equations (7.12) and (7.13) for the scattering surface sources js and jh on strip 1.

7.2Start with Equations (7.3) and (7.4) and derive Equations (7.16) and (7.17) for the scattering surface sources js and jh on strip 2.

7.3Start with Equations (7.19) and (7.20), use Equations (7.12) and (7.13), and verify the field expressions (7.25) and (7.26).

7.4Start with Equations (7.19) and (7.20), use Equations (7.16) and (7.17), and verify the field expressions (7.29) and (7.30).

7.5Start with Equation (7.37) for the pole σ ( p), verify its form (7.42), (7.43), and retrace

the trajectory of this pole in the complex plane (η), when the argument p changes from

−∞ to +∞.

7.6Apply the Cauchy residue theorem to the integral (7.35) and verify its transformation into the form (7.64).

7.7Apply the Cauchy residue theorem to the integral (7.36) and verify its transformation into the form (7.65).

7.8Apply the stationary phase technique to the integrals (7.64) and (7.65) and prove the asymptotics (7.89), (7.90) for EEWs.

7.9Verify that the directivity patterns Fs,h(1) of EEWs are always real functions, although their arguments σ1,2 can be complex quantities.

7.10Explain why the function Fs(1) equals zero for the grazing incidence (ϕ0 = 0 or ϕ0 = α).

7.11Functions Ut 1, ϕ0), Vt 1, ϕ0), as well as functions U01, ϕ0), V01, ϕ0), are singular at the point σ1 = ϕ0. Show that their differences, functions U = Ut U0 and V = Vt V0 remain finite there. Prove Equations (7.106) and (7.107).

7.12Show that for the scattering directions ϑ = π γ0, 0 ≤ ϕ α (which belong to the diffraction cone, outside the wedge), functions Fs(1), Fh(1) transform into functions f (1), g(1), respectively. Prove Equations (7.115).

7.13

Show that for the scattering directions ϑ = π γ0, α ϕ ≤ 2π (which belong to

 

the diffraction cone, inside the wedge), the total field of EEWs equals zero. Prove

 

Equations (7.116). Explain why this happens.

7.14

Prove that the incident wave

Ezinc = E0zeikz cos γ0 eikr sin γ0 cosϕ0),

Hzinc = H0zeikz cos γ0 eikr sin γ0 cosϕ0)

TEAM LinG

Problems 211

generates on elementary strip 1 (Fig. 7.3) the nonuniform currents

jx(1)

jz(1)

Hints:

 

1

D eik1x cos η[w−1+ ϕ0) + w−1ϕ0)]dη,

=

2α H0zeikz cos γ0

=eikz cos γ0 × +H0z cos γ 0eikz cos γ0

2α sin γ0

×eik1x cos η[w−1+ ϕ0) + w−1ϕ0)] cos η dη

D

,

Y0E0z eik1x cos η[w−1+ ϕ0) w−1ϕ0)] sin η dη .

D

Represent the field excited by the incident wave around the wedge in the form

 

=

 

=

E

E(x, y)eikz cos γ0 ,

H

H(x, y)eikz cos γ0 .

Use the Maxwell equations and express components Er,ϕ (x, y), Hr,ϕ as functions of components Ez(x, y), Hz(x, y). See Equation (5.4) in Ufimtsev (1962).

According to Chapter 4,

Ez = E0zeikz cos γ0 [v(k1r, ϕ ϕ0) v(k1r, ϕ + ϕ0)],

Hz = H0zeikz cos γ0 [v(k1r, ϕ ϕ0) + v(k1r, ϕ + ϕ0)].

 

 

 

= ˆ × [

 

 

]

 

 

go

 

H

 

, where H is the total field

Then define the current j(1) as j(1)

n

 

Hgo

 

around the wedge and H

 

is its geometrical optics part. The x-axis is shown in

 

Figure 7.3.

 

 

 

 

 

 

 

 

(1)

7.15 Use the same manipulations as those in Problem 7.14 and find the current j on elementary strip 2 (Fig. 7.3). Show that its components are determined by the equations:

j(1) = − 1 H eikz cos γ0 x 2α 0z

×eik1x cos η [w−1+ α ϕ0) + w−1α + ϕ0)]dη,

 

 

D

(1)

= −

eikz cos γ0

jz

 

 

2α sin γ0

 

×

+H0z cos γ0 D eik1x cos η [w−1+ α ϕ0) + w−1α + ϕ0)] cos η dη

 

+ Y0E0z D eik1x cos η [w−1+ α ϕ0) w−1α + ϕ0)] sin η dη, .

The x -axis is shown in Figure 7.3.

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