Файл: Ufimtsev P. Fundamentals of the physical theory of diffraction (Wiley 2007)(348s) PEo .pdf
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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 |
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jx(0)disk = 2Y0E0x cos γ e−ikl cos γ eikρ sin γ sin ψ |
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and |
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jy(0)disk = jz(0)disk = 0 |
(13.43) |
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on the left base of the cylinder (Fig. 13.1), and by |
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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
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0 due to the boundary condition on a perfectly conducting surface. The |
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= − |
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noununiform (fringe) currents j concentrate near the left (z |
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edges. The field Ex |
radiated by these currents is calculated in accordance with the |
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theory developed in Section 7.8. The total scattered field is the sum |
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Ex = Ex(0)disk + Ex(0)cyl + Ex(1)left + Ex(1)right . |
(13.45) |
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One can show that |
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Ex(0)disk = us(0)disk , |
Ex(0)cyl = us(0)cyl. |
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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 , |
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Rˆ |
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(13.47) |
rleft |
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y sin ϑ |
z cos ϑ |
(13.48) |
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≈ ˆ |
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+ ˆ |
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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
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θ |
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+ ˆ |
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+ ˆ |
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z, |
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x θ |
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z θ |
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and find their components from the equations |
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Rˆ |
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× ˆ] = |
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sin2 ϑ cos2 ψ , |
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× ˆ |
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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 = − |
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sin ψ |
θy = |
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cos ψ cos2 ϑ |
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1 − sin2 ϑ cos2 ψ |
1 − sin2 ϑ cos2 ψ |
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cos ψ sin ϑ cos ϑ |
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θz = |
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sin2 ϑ cos2 ψ |
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− cos ψ cos ϑ |
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sin ψ cos ϑ |
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φx = − |
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1 − sin2 ϑ cos2 ψ |
1 − sin2 ϑ cos2 |
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sin ψ sin ϑ |
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1 − sin2 ϑ cos2 ψ |
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The angle θ is |
defined by the equation |
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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 |
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z cos ϑ and Q |
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y sin γ |
z cos γ |
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ˆ = −ˆ |
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+ ˆ |
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− ˆ |
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on this plane, one obtains |
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sin φ = − |
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cos ϑ |
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cos φ = |
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sin ϑ sin ψ |
(13.54) |
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1 − sin2 ϑ cos2 ψ |
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1 − sin2 ϑ cos2 ψ |
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sin φ0 = |
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cos γ |
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cos φ0 = |
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sin γ sin ψ |
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1 |
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sin2 ϑ cos2 ψ |
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sin2 ϑ cos2 ψ |
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for the left edge'(z |
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l), and |
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cos ϑ |
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and sin φ = − |
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sin ϑ sin ψ |
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cos φ = − |
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1 − sin2 ϑ cos2 ψ |
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sin φ0 = − |
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sin γ sin ψ |
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cos φ0 = |
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cos γ |
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1 − sin2 ϑ cos2 ψ |
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for the right |
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Now, according to Section 7.8, one obtains |
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a |
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eikR |
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0sin ψ Fθ(1)(ψ , θ , φ) · θx |
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Ex(1)left = E0x |
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ei2kl cos ϑ |
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(1) |
(ψ , θ , φ) |
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G(1)(ψ , θ , φ) |
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ei2ka sin ϑ sin ψ dψ , |
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+ cos |
ϑ cos ψ [Gθ |
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x + |
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Ey(1,z)left = Hx(1)left |
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(13.59) |
TEAM LinG