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

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6.3 Scattering at Cones: Focal Field

139

In the limiting case when ω π/2,

 

 

 

 

 

 

 

σ (0) = π a2(ka)2.

(6.114)

Together with the contribution from the nonuniform scattering sources j(1)

, the

total backscattering cross-section equals

s,h

 

 

 

σs

=

π a2

i

tan2 ω(1

ei2kl )

 

 

 

 

 

 

 

2ka

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

2

 

 

 

 

π

 

 

 

1

 

 

 

 

 

 

 

 

 

 

 

 

 

 

n

 

1

 

 

 

 

n

 

 

 

 

 

 

 

 

 

 

 

 

 

 

n

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

sin

 

 

 

 

 

 

 

 

 

π

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

ei2kl

 

 

n %

 

n &

 

 

 

 

 

 

 

1

 

cos

π

 

 

 

 

 

 

 

2ω

 

 

 

 

and

 

 

 

cos

 

 

 

 

 

 

 

 

 

 

 

 

cos

 

 

 

 

 

 

 

 

 

 

π a2

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

σh

=

i

 

tan2 ω(1

ei2kl )

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

2ka

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

1

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

1

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

2

 

 

 

π

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

n

 

 

 

 

 

n

 

 

 

 

 

 

 

 

 

 

 

 

 

n

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

sin

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

ei2kl

 

 

 

 

 

 

 

 

 

 

 

 

 

 

π

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

+ n %

 

n &

 

 

 

 

 

 

 

1

+ cos

π

 

 

 

 

 

 

 

2ω

 

 

 

 

 

 

 

 

cos

 

 

 

 

 

 

 

 

 

 

cos

 

 

 

 

 

 

 

 

 

with n = 1 + + )/π .

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

In the limiting case when ω π/2,

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

2

 

 

 

 

 

 

 

 

 

=

 

 

 

 

 

 

 

+

 

2

 

 

 

 

π

 

 

 

 

 

+ n

 

n

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

ika

 

 

 

 

 

sin

 

 

 

 

 

 

 

 

1

 

 

 

 

π

 

 

 

 

 

 

 

 

 

 

 

 

σs

 

 

 

π a2

n

n

 

 

 

 

 

 

 

 

cot

 

,

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

π

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

n

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

and

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

cos

 

 

 

1

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

2

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

2

 

 

 

 

 

 

 

 

=

 

 

 

 

 

 

 

+

 

 

 

 

 

π

 

 

n

 

 

 

n

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

sin

 

 

 

 

 

 

 

 

 

 

1

 

 

 

π

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

n

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

σh

 

 

 

π a2

ika

 

 

 

 

 

n

 

 

 

 

 

 

 

 

 

cot

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

π

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

n

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

cos

 

 

 

 

 

 

1

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

2

2

(6.115)

(6.116)

(6.117)

(6.118)

with n = (3/2) + ( /π ).

Finally when both ω and are equal to π/2, and the cone transforms into the disk,

σs = π a2 |ika − 1|2

(6.119)

and

 

σh = π a2 |ika + 1|2 .

(6.120)

The normalized backscattering cross-section (6.112) of the cone was numerically analyzed as the function of three variables: the length l, the angle ω and the angle .

TEAM LinG


140 Chapter 6 Axially Symmetric Scattering of Acoustic Waves at Bodies of Revolution

Figure 6.10 Backscattering at a cone: dependence on the cone length l. According to Equation (6.95), the PO curve also represents the scattering of electromagnetic waves from a perfectly conducting cone.

The results are presented in the decibel scale as follows:

For the calculation of Equation (6.112) as a function of the length l, the variables were set as

ω = 45, = 90, a = l tan ω = l, 10 ≤ kl < 30.

In this case, 1.5λ < l < 4.8λ and 3λ < 2a < 9.6λ. The results are demonstrated in Figure 6.10. As is seen here, the data for the hard cone are higher than those for the soft cone. The difference between them is about 15 dB. The PO data are approximately in the middle.

For the calculation of Equation (6.112) as a function of the angle ω, the variables were set as

ka = 3π , 10ω ≤ 90, = 90.

In this case, 2a = 3λ, 0 ≤ l ≤ 8.5λ. The results are plotted in Figure 6.11. A big difference can be observed between the soft and hard data here, at about 40 dB for narrow cones.

For the calculation of Equation (6.112) as a function of the angle (Fig. 6.9), the variables were set as

ω = 10, ka = 3π , kl 17π , 0≤ ≤ π ω = 170.

In this case, 2a = 3λ and l 8.5λ. The results are plotted in Figure 6.12. The PO approximation does not depend on the angle , which is why it is represented here by a straight horizontal line. The difference between the soft and hard data is about 42–57 dB. The influence of the cone base shape approaches 11 dB for the soft cone and 16 dB for the hard cone.

TEAM LinG


6.4 Backscattered Focal Fields 141

Figure 6.11 Backscattering at a cone: dependence on the vertex angle ω. According to Equation (6.95), the PO curve also represents the scattering of electromagnetic waves from a perfectly conducting cone.

Figure 6.12 Backscattering at a cone: dependence on the angle . According to Equation (6.95), the PO curve also represents the scattering of electromagnetic waves from a perfectly conducting cone.

6.4 BODIES OF REVOLUTION WITH NONZERO GAUSSIAN CURVATURE: BACKSCATTERED FOCAL FIELDS

This section studies symmetrical scattering at bodies of revolution whose illuminated side is an arbitrary smooth convex surface with nonzero Gaussian curvature. A generatrix of such a surface and related geometry are shown in Figure 6.13.

TEAM LinG


142 Chapter 6 Axially Symmetric Scattering of Acoustic Waves at Bodies of Revolution

Figure 6.13 Generatrix of the body of revolution.

The incident plane wave (6.1) propagates in the positive direction of the z-axis, which represents the symmetry axis of a scattering object. We use two systems of coordinates: cylindrical coordinates ρ, ϕ, z and spherical coordinates r, ϑ , ϕ. The generatrix is given as a function ρ = ρ(z). It is assumed that d2ρ/dz2 =0 for the illuminated side (0 ≤ z l) of the object. This condition ensures that the Gaussian curvature of this surface is not zero. We also utilize the following denotations related to the edge points (ρ = a): dρ/dz = tan ω for the illuminated side (z = l − 0) and dρ/dz = − tan for the shadowed side (z = l + 0). The shadowed side is an arbitrary smooth surface with 0 ≤ ≤ π ω. In the limiting case = π ω, the scattering object is an infinitely thin (but still perfectly reflecting) screen ρ = ρ(z) with 0 ≤ z l.

The principal radii of curvature of the scattering surface are determined according to the differential geometry (Bronshtein and Semendyaev, 1985)

1 = {

 

+

|[ρ (z)|

 

 

 

2 =

'

+ [

 

]

 

 

 

 

 

1

 

ρ (z)]2

}3/2

 

 

 

 

 

 

 

 

 

,

 

R

 

 

 

 

 

and

R

 

ρ(z) 1

 

ρ

(z)

2

(6.121)

where ρ = dρ(z)/dz, ρ = d2ρ(z)/dz2. The radius R1 relates to the normal section of the surface by the plane (ρ, z). The radius R2 relates to the orthogonal normal section. As ρ = tan θ with ω θ π/2, the principal radii can be represented as

R1 =

1

 

R2 =

ρ(z)

 

 

and

 

.

(6.122)

|ρ (z)| cos3 θ

cos θ

The radius R2 is shown in Figure 6.13. The Gaussian curvature is determined by

kG = R1R2

=

ρ(z)

cos4 θ .

(6.123)

1

 

ρ (z)

 

 

The above expressions for R1, R2, and kG become indefinite at the point z = 0. In order to disclose these indefinitenesses, one can use the alternative expressions

 

 

 

1 + [z (ρ)]2

 

3/2

 

 

 

 

 

 

 

 

 

R

 

 

 

and

R

 

 

ρ

1 + [z (ρ)]2

,

(6.124)

=

5

z (ρ)

6

 

 

=

 

'

z (ρ)

1

 

 

 

2

 

 

TEAM LinG


6.4 Backscattered Focal Fields 143

where z = z(ρ), z = dz(ρ)/dρ, and z = d2z(ρ)/dρ2. It follows from these equations that at the point ρ = z = 0

1

 

R1 = R2 = z (ρ) .

(6.125)

The equality of the two principal radii means that the vertex point ρ = z = 0 of the scattering surface is an umbilic point.

6.4.1PO Approximation

This is the field radiated by the uniform components (1.31) of the scattering sources

 

 

 

js(0) = iu02knzeikz

and

jh(0) = u02eikz.

(6.126)

According to the differential geometry (Bronshtein and Semendyaev, 1985),

 

 

 

 

cos ψ

sin ψ

 

 

ρ (z)

 

 

nˆ = xˆ

 

 

+ yˆ

 

 

zˆ

 

 

 

(6.127)

 

'

 

'

 

'

 

or

1 + [ρ (z)]2

1 + [ρ (z)]2

1 + [ρ (z)]2

 

 

 

nˆ = xˆ cos θ cos ψ + yˆ cos θ sin ψ zˆ sin θ .

(6.128)

Here, we use the letter ψ for the polar coordinate of the scattering surface and retain the letter ϕ for the polar coordinate of the field point.

The scattered field in the far zone is determined by Equations (1.16) and (1.17),

where one should set

 

 

 

and ds = ρ(z)'

 

dzdψ ,

r sin ϑ = ρ(z),

r cos ϑ = z

 

1 + (ρ )2

(6.129)

 

mˆ = r = xˆ sin ϑ cos ϕ + yˆ sin ϑ sin ϕ + zˆ cos ϑ .

(6.130)

Due to the axial symmetry of the scattered field, it is sufficient to calculate the field only in the meridian plane ϕ = π/2. Taking this into account, one has

m

n

 

 

sin ϑ sin ψ ρ

(z) cos ϑ ,

 

ˆ · ˆ =

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

ik '

 

 

 

 

 

 

 

 

 

 

 

 

 

 

1 + [ρ (z)]2

 

2π

 

 

 

 

 

 

 

 

eikR

 

l

 

 

 

u(0)

=

u

 

 

 

 

 

 

 

 

eikz(1−cos ϑ )ρ(z)ρ (z)dz

eikρ(z) sin ϑ sin ψ dψ

0 2π

 

 

 

 

s

 

 

 

R

 

0

 

 

 

 

0

and

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

uh(0) = −u0

ik

eikR

l

 

 

 

 

 

 

 

0 eikz(1−cos ϑ )ρ(z)dz

 

 

 

2π

 

R

 

 

 

 

 

 

 

 

 

2π

eikρ(z) sin ϑ sin ψ [sin ϑ sin ψ ρ (z) cos ϑ ]dψ

 

 

 

 

 

×

0

 

(6.131)

(6.132)

(6.133)

TEAM LinG