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Ufimtsev P. Fundamentals of the physical theory of diffraction (Wiley 2007)(348s) PEo

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272 Chapter 13 Backscattering at a Finite-Length Cylinder

Figure 13.2 Backscattering at the finite cylinder according to the PO approximations PO-1 (Equation (13.12)) and PO-2 (Equation (13.13)).

negligible in the case when d = 3λ, as is clearly seen in Figure 13.3. Because of that we use in the following calculations the simplest approximation (13.12). There is another reason in favour of using Equation (13.12). It matches an analogous type of approximation for the field (13.22), (13.23) generated by the nonuniform/fringe source j(1). This field is investigated in the following section.

Figure 13.3 Backscattering at the finite cylinder according to the PO approximations PO-1 (Equation (13.12)) and PO-2 (Equation (13.13)).

TEAM LinG

13.1 Acoustic Waves 273

13.1.2 Backscattering Produced by the Nonuniform Component j (1)

There are three types of nonuniform components j(1) of the scattering sources on a finite cylinder. The edge/fringe component jfr(1) concentrates near the edges. The component jcr(1) associated with creeping waves concentrates near the shadow boundary on the cylindrical part of the object and exponentially attenuates away from this boundary. The third component jdif(1) is caused by the transverse diffusion of the wave field between the adjacent rays reflected from the cylindrical surface. This component jdif(1) exists on the illuminated part of the cylindrical surface away from the shadow boundary. Among these components of j(1), a main contribution to the backscattering is provided by the fringe component jfr(1) and this contribution is investigated here. Notice also that the field generated by jdif(1) is comparable with that produced by jfr(1), but this situation occurs only in the direction of the specular rays reflected from the cylindrical surface. This topic is considered in the next chapter.

First we analyze the field generated by the fringe component located near the left

'

edge (z = −l, x2 + y2 = a). According to Equation (8.3) it is determined as

(1)left

= u0

a

i2kl cos ϑ eikR

2π

(1)

(ϕ )e

i2ka sin ϑ sin ϕ

 

.

 

us,h

 

e

 

 

 

Fs,h

 

dϕ

(13.16)

2π

 

R

0

 

In the direction ϑ = π , functions Fs,h(1) transform into functions f (1) and g(1) as shown in Equations (7.120) and (7.121). Recall that these functions are defined in Section 4.1. Hence for this direction,

us(1)left

 

f (1)

 

eikR

 

 

 

uh(1)left

 

= u0a g(1)

ei2kl

 

,

= π ).

(13.17)

R

For other directions ϑ , which satisfy the condition 2ka sin ϑ

1, the integral

in Equation (13.16) is evaluated asymptotically by the stationary-phase technique. There are two stationary points (ϕst,1 = π/2 and ϕst,2 = 3π/2) in the integrand of Equation (13.16). At these points, functions Fs,h(1) also transform into functions f (1) and g(1). The resulting asymptotic approximations for the field (13.16) are given as

(1)left

 

a

i2kl cos ϑ eikR

1

 

 

us

= u0

 

 

e

 

 

 

 

 

 

2

 

R

 

 

 

π ka sin ϑ

 

 

 

× [ f (1)(1)ei2ka sin ϑ iπ/4 + f (1)(2)ei2ka sin ϑ +iπ/4]

(13.18)

and

 

 

 

 

 

 

 

 

 

 

 

 

 

(1)left

= u0

a

i2kl cos ϑ eikR

1

 

 

uh

 

e

 

 

 

 

 

 

2

 

R

 

 

 

π ka sin ϑ

 

 

 

×

[g(1)(1)ei2ka sin ϑ iπ/4 + g(1)(2)ei2ka sin ϑ +iπ/4].

(13.19)

TEAM LinG

274 Chapter 13 Backscattering at a Finite-Length Cylinder

Here, the functions f (1)(1), g(1)(1) relate to the stationary point 1 (ϕst,1 = π/2) and

the functions f (1)(2), g(1)(2) relate to the stationary point 2 (ϕst,2 = 3π/2). These points are shown in Figure 13.1.

In order to construct the field approximations in the entire region π/2 ≤ ϑ π , we apply the idea suggested in Sections 6.1.4 and 13.1.1. In otherwords, we substitute the asymptotics

 

ei2ka sin ϑ iπ/4

J0

(2ka sin ϑ ) + iJ1

 

 

 

 

 

 

(2ka sin ϑ ),

(13.20)

 

π ka sin ϑ

ei2ka sin ϑ +iπ/4

J0

(2ka sin ϑ ) iJ1

 

 

 

 

 

(2ka sin ϑ )

(13.21)

 

π ka sin ϑ

into Equations (13.18) and (13.19) and then extend the obtained field expressions to the entire region π/2 ≤ ϑ π . It turns out that the resulting expressions

 

a

eikR

0f (1)(1)[J0(2ka sin ϑ ) + i J1(2ka sin ϑ )]

 

us(1)left = u0

 

 

 

ei2kl cos ϑ

 

 

2

R

 

+ f (1)(2)[J0(2ka sin ϑ ) i J1(2ka sin ϑ )]1

(13.22)

and

 

 

 

 

a

eikR

0g(1)(1)[J0(2ka sin ϑ ) + i J1(2ka sin ϑ )]

 

uh(1)left = u0

 

ei2kl cos ϑ

 

 

2

R

 

+ g(1)(2)[J0(2ka sin ϑ ) i J1(2ka sin ϑ )]1

(13.23)

exactly transform into Equation (13.17) when ϑ π . Therefore, these expressions can be considered as appropriate approximations for the scattered field in all directions

π/2 ≤ ϑ π .

The contribution of the right edge (z = +l,

 

x2 + y2

= a) to the field in the

region π/2 < ϑ < π is described by the

expression

 

 

 

 

 

 

 

'

 

 

 

 

 

(1)right

 

a

i2kl cos ϑ eikR

 

2π

(1)

 

 

i2ka sin ϑ sin ϕ

 

 

 

us,h

= u0

 

e

 

 

 

Fs,h (ϕ )e

 

 

dϕ

,

(13.24)

2π

 

R

π

 

 

analogous to Equation (13.16). Its asymptotic approximation found by the stationaryphase technique is determined as

uh

 

=

2

g

 

(3)

us(1)right

 

 

 

 

f (1)

(3)

 

 

 

u0 a

 

(1)

 

(1)right

 

 

 

 

ei2ka sin ϑ +iπ/4

ei2kl cos ϑ

eikR

,

(13.25)

 

 

 

 

π ka sin ϑ

 

 

R

 

where 2ka sin ϑ 1 and functions f (1)(3), g(1)(3) relate (ϕst,3 = 3π/2) shown in Figure 13.1. With the application

to the stationary point 3 of Equation (13.21), this

TEAM LinG

13.1 Acoustic Waves 275

expression is extended to all directions π/2 < ϑ < π and provides the following approximations to the field (13.24):

us(1)right = u0

a

f (1)(3)

[J0(2ka sin ϑ ) iJ1(2ka sin ϑ )]ei2kl cos ϑ

 

eikR

 

(13.26)

2

 

 

R

and

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

a

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

eikR

 

 

uh(1)right = u0

 

 

g(1)(3) [J0(2ka sin ϑ )

iJ1(2ka sin ϑ )]ei2kl cos ϑ

 

 

 

 

 

.

(13.27)

2

 

R

Thus, in the first approximation, the total field produced by the component

jfr(1) equals

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

a eikR

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

us(1)

= u0

 

 

 

 

 

 

 

 

 

0f (1)(1) [J0(2ka sin ϑ ) + iJ1(2ka sin ϑ )]ei2kl cos ϑ

 

 

 

 

 

 

 

 

 

 

2

 

 

R

 

 

 

 

 

 

 

 

 

 

 

+ [ f (1)(2)ei2kl cos ϑ + f (1)(3)ei2kl cos ϑ ][J0(2ka sin ϑ ) iJ1(2ka sin ϑ )] ,

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

(13.28)

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

1

and

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

a eikR

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

uh(1)

= u0

 

 

 

 

 

0g(1)(1) [J0(2ka sin ϑ ) + iJ1(2ka sin ϑ )]ei2kl cos ϑ

 

 

 

 

 

 

 

 

 

 

2

 

R

 

 

 

 

 

 

 

 

 

 

 

+ [g(1)(2)ei2kl cos ϑ + g(1)(3)ei2kl cos ϑ ][J0(2ka sin ϑ ) iJ1(2ka sin ϑ )] .

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

(13.29)

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

1

The functions f (1) and g(1) are determined according to Section 4.1 as

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

sin

π

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

1

 

 

 

 

 

 

 

1

 

 

 

 

 

 

1

 

cos ϑ

 

 

f (1)(1)

 

 

 

 

 

n

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

,

 

(13.30)

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

π

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

=

 

 

 

n

 

 

 

cos

 

 

 

 

1

cos

 

π

 

cos

π − 2ϑ

 

 

+

2 sin ϑ

 

 

 

 

 

 

 

 

 

 

 

 

 

n

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

n

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

n

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

sin

π

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

1

 

 

 

 

 

 

 

1

 

 

 

 

 

 

1

 

cos ϑ

 

 

g(1)(1)

 

 

 

 

 

 

n

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

,

 

(13.31)

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

π

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

=

 

 

 

n

 

 

 

cos

 

 

 

 

1

+ cos

 

π

 

cos

π − 2ϑ

 

 

2 sin ϑ

 

 

 

 

 

 

 

 

 

 

 

 

 

n

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

n

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

n

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

sin

π

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

1

 

 

 

 

 

 

1

 

 

 

 

 

 

1 cos ϑ

 

 

 

1 sin ϑ

f (1)(2)

 

 

 

 

 

 

n

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

,

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

π

 

 

 

 

 

π

 

 

2ϑ

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

=

 

 

 

n

 

 

cos

 

 

 

− 1

cos

 

 

− cos

 

 

 

2 sin ϑ

 

2 cos ϑ

 

 

 

 

 

 

 

 

 

 

n

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

n

n

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

(13.32)

 

 

 

 

 

 

 

 

 

 

sin

π

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

1

 

 

 

 

 

 

1

 

 

 

 

 

1

 

cos ϑ

 

 

 

1 sin ϑ

g(1)(2)

 

 

 

 

 

 

n

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

,

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

π

 

 

 

 

 

π

 

 

2ϑ

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

=

 

 

 

n

 

 

cos

 

 

 

− 1

+ cos

 

 

− cos

 

 

 

+

2 sin ϑ +

 

2 cos ϑ

 

 

 

 

 

 

 

 

 

 

n

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

n

n

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

(13.33)

TEAM LinG

276 Chapter 13

 

 

Backscattering at a Finite-Length Cylinder

 

 

 

 

 

 

 

 

 

 

 

 

 

 

sin

π

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

1

 

 

 

 

 

 

 

 

 

1

 

 

 

 

 

 

1

 

 

 

sin ϑ

 

 

 

f (1)(3)

 

n

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

,

(13.34)

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

π

 

 

 

 

 

n

 

 

n

 

 

 

 

 

 

=

 

 

 

 

 

 

 

 

 

n

 

 

cos

 

 

 

 

 

+

 

 

 

 

 

 

 

 

 

and

 

n

 

 

 

cos

 

 

 

 

1

 

π

 

 

cos

π + 2ϑ

 

 

 

 

 

2 cos ϑ

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

sin

π

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

1

 

 

 

 

 

 

 

 

 

1

 

 

 

 

 

 

 

1

 

 

sin ϑ

 

 

g(1)(3)

 

n

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

,

(13.35)

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

π

 

 

 

 

 

n

 

 

n

 

 

 

 

 

=

 

 

 

 

 

 

 

 

 

n

 

 

+ cos

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

n

 

 

 

cos

 

 

 

1

π

 

cos

π + 2ϑ

 

 

 

 

2 cos ϑ

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

with n = 3/2.

Although certain terms in these functions are singular in the directions ϑ = π/2 and ϑ = π , these singularities always cancel each other, and the functions f (1), g(1) remain finite. In the direction ϑ = π/2 they have the values

 

 

 

 

 

 

 

 

 

 

 

1

sin

π

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

1

 

 

π

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

f (1)(1) = 0,

 

 

 

f (1)(2) = f (1)(3) =

n

 

 

 

 

n

 

 

 

+

 

cot

(13.36)

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

cos

π

 

− 1

2n

 

n

 

 

 

 

 

 

 

 

 

 

 

 

n

 

 

 

 

 

 

 

 

 

 

 

 

 

and

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

2

 

sin

 

π

 

 

 

 

 

 

1

sin

 

π

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

1

 

 

 

π

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

g(1)(1) =

n

 

 

 

 

n

 

g(1)(2) = g(1)(3) =

 

 

 

n

 

 

 

 

n

 

 

 

 

 

 

 

 

,

 

 

 

 

 

 

 

 

cot

 

, (13.37)

 

 

π

 

 

 

 

 

 

π

 

 

 

2n

n

cos

 

 

 

− 1

 

 

 

 

 

cos

 

 

 

− 1

 

 

 

 

 

 

 

n

 

 

 

 

 

n

 

 

 

 

 

 

and in the direction ϑ = π they are determined as

 

 

 

 

 

 

 

1

sin

π

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

1

 

 

 

π

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

f (1)(1) = f (1)(2) =

 

 

n

 

 

 

n

 

 

+

 

cot

,

 

 

 

 

f (1)(3) = 0

 

 

 

 

 

(13.38)

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

cos

π

− 1

2n

n

 

 

 

 

 

 

 

 

 

 

 

 

 

 

n

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

and

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

1

sin

 

π

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

2

sin

 

π

 

 

 

 

 

 

 

 

 

 

 

 

 

 

1

 

 

 

 

π

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

g(1)(1) = g(1)(2) =

n

 

n

 

 

 

 

 

 

 

g(1)(3) =

 

n

 

n

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

cot

 

,

 

 

 

 

 

 

.

(13.39)

 

 

π

 

 

 

 

2n

n

 

 

 

 

 

π

 

 

cos

 

 

− 1

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

cos

 

 

− 1

 

n

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

n

 

We also obtain the expressions for the functions f (1)(4) and g(1)(4) related to the

stationary point 4 (Fig. 13.1), which

becomes visible in the directions ϑ

=

π/2 and

 

 

(1)

 

 

 

 

 

 

(1)

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

ϑ = π . For both directions, functions f

 

 

(4) and g

 

(4) have the same values,

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

2

 

sin

 

π

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

f (1)(4) = 0

 

 

and

 

 

 

g(1)(4) =

n

 

n

.

 

 

 

 

 

 

 

 

(13.40)

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

π

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

cos

 

 

 

− 1

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

n

 

 

 

 

 

 

 

 

 

As the contribution from point 4 equals zero for the soft cylinder, the approximation (13.28) can be used in the entire region π/2 ≤ ϑ π . In the case of

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13.1 Acoustic Waves 277

expression (13.29) for the hard cylinder, it is valid (strictly speaking) for directions π/2 + 0 ≤ ϑ π − 0 , when point 4 is invisible. However, the contribution of point 4 for the hard cylinder is ka (or kl) times less than the PO field in the direction ϑ = π (or ϑ = π/2) and it can be neglected in the case of large cylinders.

A quantitative influence of the field us,h(1) on the backscattering is illustrated graphically in the next section.

13.1.3 Total Backscattered Field

The total field is the sum

us,h(t) = us,h(0) + us,h(1),

(13.41)

where us(0) = −uh(0) and the terms uh(0), us,h(1) are determined by Equations (13.12), (13.28), and (13.29). Utilizing these approximations, we have calculated the normal-

ized scattering cross-section (13.14) and demonstrated the individual contribution of each term in Equation (13.41). Recall that the field us,h(0) is produced by the uniform scattering source js,h(0) and represents the PO approximation. The field us,h(1) is produced

by the nonuniform source js,h(1) concentrated near the edges and is denoted below as the fringe component of the backscattering. The numerical results are presented in the following for two sets of geometrical parameters of the cylinder: (a) d = 2a = λ, L = 2l = 3λ; and (b) d = 3λ, L = 9λ. Here, d is the diameter and L is the length of the cylinder.

Figure 13.4 Backscattering at a soft cylinder. According to Equation (13.46), the PO curve here also displays the backscattering of electromagnetic waves (with Ex -polarization) from a perfectly conducting cylinder.

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278 Chapter 13 Backscattering at a Finite-Length Cylinder

Figure 13.5 Backscattering at a hard cylinder. According to Equation (13.70), the PO curve here also displays the backscattering of electromagnetic waves (with Hx -polarization) from a perfectly conducting cylinder.

An interesting observation follows from Figures 13.4 to 13.7. Most of the maximums in the soft fringe field are located in the vicinity of the angular positions of the minimums of the PO field. The opposite situation is observed for the hard fringe field: its maximums are positioned near the maximums of the PO field. This observation explains why the minimums of the field scattered by soft cylinders are not as deep as those for the case of hard cylinders.

Figure 13.6 Backscattering at a soft cylinder. According to Equation (13.70), the PO curve here also displays the backscattering of electromagnetic waves (with Ex -polarization) from a perfectly conducting cylinder.

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13.2 Electromagnetic Waves 279

Figure 13.7 Backscattering at a hard cylinder. According to Equation (13.46), the PO curve here also displays the backscattering of electromagnetic waves (with Hx -polarization) from a perfectly conducting cylinder.

13.2ELECTROMAGNETIC WAVES

The original PTD of electromagnetic waves scattered from a finite perfectly conducting cylinder was published in the work of Ufimtsev (1958a, 1962). Below, we present in brief a revised version based on the concept of EEWs.

13.2.1 E-Polarization

The incident wave is defined as

Exinc = E0x eik(z cos γ +y sin γ ),

Eyinc = Ezinc = Hxinc = 0.

(13.42)

The uniform component (1.97) of the induced surface current is determined by

jx(0)disk = 2Y0E0x cos γ eikl cos γ eikρ sin γ sin ψ

 

and

 

 

jy(0)disk = jz(0)disk = 0

(13.43)

on the left base of the cylinder (Fig. 13.1), and by

 

jx(0)cyl = −2Y0E0x sin γ sin ψ eik(z cos γ +a sin γ sin ψ ), jy(0)cyl = 2Y0E0x sin γ cos ψ eik(z cos γ +a sin γ sin ψ ),

and

TEAM LinG

280 Chapter 13 Backscattering at a Finite-Length Cylinder

jz(0)cyl = 2Y0E0x cos γ cos ψ eik(z cos γ +a sin γ sin ψ )

(13.44)

on the cylindrical part of the surface (−l z l , π ψ ≤ 2π ). Here, Y0 = 1/Z0 is the admittance of free space (vacuum).

(0) (0)

The field Ex generated by the current j is found with the help of Equa-

m m =

tions (1.92) and (1.93), where one should drop off the terms Aϕ,ϑ , because j

−[ˆ ×

E

] =

 

(1)

 

 

 

 

 

n

 

0 due to the boundary condition on a perfectly conducting surface. The

 

 

 

(1)

 

 

= −

) and right (z

=

l)

noununiform (fringe) currents j concentrate near the left (z

 

 

edges. The field Ex

radiated by these currents is calculated in accordance with the

theory developed in Section 7.8. The total scattered field is the sum

 

 

 

 

 

Ex = Ex(0)disk + Ex(0)cyl + Ex(1)left + Ex(1)right .

(13.45)

One can show that

 

 

 

 

 

 

 

 

 

 

Ex(0)disk = us(0)disk ,

Ex(0)cyl = us(0)cyl.

(13.46)

The quantities us(0)disk and us(0)cyl are defined in Section 13.1.1, where one should set u0 = E0x . Therefore, the PO curves in Figures 13.4 and 13.6 for the backscattering of acoustic waves from a soft cylinder also display the backscattering of electromagnetic

waves from a perfectly conducting cylinder.

The fields Ex(1)left and Ex(1)right are calculated by the integration of EEWs, which are the functions of the local spherical coordinates with the origin at an edge point x = a cos ψ , y = a sin ψ . To avoid the possible confusion with the basic coordinates R, ϑ , ϕ of the observation point, we re-denote the local coordinates as r, θ , φ. The necessary preliminary work is to define the local coordinates in terms of the basic coordinates ϑ , ψ .

First, notice that one can use the following approximations

rleft = R + a sin ϑ sin ψ + l cos ϑ ,

rright = R + a sin ϑ sin ψ l cos ϑ ,

 

 

 

Rˆ

 

 

(13.47)

rleft

rright

y sin ϑ

z cos ϑ

(13.48)

ˆ

≈ ˆ

 

= −ˆ

+ ˆ

 

for the observation point (x = 0, y = −R sin ϑ , z = R cos ϑ ) in the far zone (R

ka2,

Rkl2). Then, we introduce the unit vectors

 

 

 

 

 

θ

= ˆ

 

+ ˆ

 

 

+ ˆ

 

,

 

φ

= ˆ

φ

+ ˆ

φ

 

+ ˆ

φ

z,

 

(13.49)

 

 

 

 

 

 

x θ

y θ

y

z θ

z

 

 

x

y

y

z

 

 

 

 

 

 

ˆ

 

 

x

 

 

 

 

 

 

ˆ

 

 

x

 

 

 

 

 

and find their components from the equations

 

 

 

 

 

 

 

 

 

 

 

 

 

 

Rˆ

·

ˆ

=

 

ˆ

·

[ ˆ

× ˆ] =

 

 

ˆ

·

 

ˆ

= −

(

 

 

 

 

 

 

 

 

 

ˆ

=

(13.50)

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

sin2 ϑ cos2 ψ ,

 

ˆ

× ˆ

 

 

θ

 

0,

θ

 

R

t

 

0,

t

 

 

θ

 

1

 

 

φ

 

R

θ ,

TEAM LinG

13.2 Electromagnetic Waves 281

where ˆt = xˆ sin ψ yˆ cos ψ is the tangent to the edge. According to these equations,

θx = −

 

 

 

 

sin ψ

θy =

 

 

cos ψ cos2 ϑ

 

 

 

 

 

 

 

 

,

 

 

 

 

,

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

1 − sin2 ϑ cos2 ψ

1 − sin2 ϑ cos2 ψ

 

 

 

 

cos ψ sin ϑ cos ϑ

 

'

 

 

 

 

 

 

θz =

 

'

 

,

 

 

 

 

 

 

(13.51)

 

 

 

 

 

 

 

 

 

 

 

 

 

 

'

1

 

 

sin2 ϑ cos2 ψ

 

 

 

 

 

 

 

 

 

 

 

cos ψ cos ϑ

 

 

 

 

 

sin ψ cos ϑ

 

 

φx = −

 

 

 

 

 

,

φy = −

 

 

 

 

,

'

 

 

 

 

 

1 − sin2 ϑ cos2 ψ

1 − sin2 ϑ cos2

ψ

φz = −

sin ψ sin ϑ

 

 

 

'

 

 

 

 

 

 

 

 

.

 

 

 

 

 

 

 

(13.52)

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

1 − sin2 ϑ cos2 ψ

 

 

 

 

 

 

 

The angle θ is

defined by the equation

 

 

 

 

 

 

 

 

 

 

 

'

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

Rˆ · ˆt = cos θ = sin ϑ cos ψ .

(13.53)

In order to define the angles φ and φ0, one should note that they are measured from the illuminated face of the edge in the plane perpendicular to the tangent ˆt to the edge.

By projecting the vectors R

y sin ϑ

 

z cos ϑ and Q

 

ki

 

 

y sin γ

z cos γ

 

 

 

 

 

 

 

 

 

 

 

 

 

 

ˆ = −ˆ

 

 

 

 

 

+ ˆ

 

 

 

 

 

 

 

ˆ

= −ˆ

= −ˆ

− ˆ

on this plane, one obtains

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

sin φ = −

 

 

 

 

 

 

 

 

cos ϑ

 

 

 

,

 

 

 

cos φ =

 

 

 

 

 

 

 

 

sin ϑ sin ψ

(13.54)

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

'

 

 

 

 

 

 

 

'

 

 

 

 

and

1 − sin2 ϑ cos2 ψ

 

 

 

1 − sin2 ϑ cos2 ψ

 

sin φ0 =

 

 

 

 

 

 

 

 

cos γ

 

 

,

 

 

 

cos φ0 =

 

 

 

 

 

 

 

 

sin γ sin ψ

(13.55)

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

1

sin2 ϑ cos2 ψ

 

 

1

sin2 ϑ cos2 ψ

for the left edge'(z

 

 

l), and

 

 

 

 

 

 

 

 

 

 

 

'

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

= −

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

cos ϑ

 

 

 

 

 

 

and sin φ = −

 

 

 

 

 

 

sin ϑ sin ψ

 

 

,

 

 

 

cos φ = −

 

 

 

 

 

 

 

 

 

 

 

(13.56)

 

 

 

 

 

 

 

 

 

 

 

 

 

 

'

 

 

 

 

 

 

'

 

 

1 − sin2 ϑ cos2 ψ

 

 

1 − sin2 ϑ cos2 ψ

 

sin φ0 = −

 

 

 

 

 

 

sin γ sin ψ

 

 

 

 

,

 

 

 

cos φ0 =

 

 

 

 

 

 

 

 

 

cos γ

 

 

 

 

 

(13.57)

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

'

 

 

 

 

 

 

1 − sin2 ϑ cos2 ψ

 

 

 

1 − sin2 ϑ cos2 ψ

for the right

edge (z l).

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

'

 

 

=

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

Now, according to Section 7.8, one obtains

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

a

 

 

eikR

 

2π

0sin ψ Fθ(1), θ , φ) · θx

 

 

 

 

 

 

 

 

Ex(1)left = E0x

 

 

 

 

ei2kl cos ϑ

 

 

 

 

 

 

 

 

 

 

 

2π

 

R

0

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

(1)

, θ , φ)

·

θ

 

G(1), θ , φ)

·

φ

 

ei2ka sin ϑ sin ψ dψ ,

 

 

+ cos

ϑ cos ψ [Gθ

 

 

 

 

 

 

 

 

x +

φ

 

 

 

 

 

 

 

 

 

 

 

x

]1

 

 

 

 

(13.58)

Ey(1,z)left = Hx(1)left

= 0,

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

(13.59)

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