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

94 Chapter 5 First-Order Diffraction at Strips and Polygonal Cylinders

under the conditions q(1 ± α) 1 and q(1 ± α0) 1. Here, symbol O(qm) is used to show the asymptotic behavior of quantities Qs,h(α, α0, q) when q → ∞. The

Equations (5.57) and (5.58) are the consequences of the fact that the scattered field (5.46) and (5.47) and its normal derivatives at the plane x = 0 are continuous outside the strip. In other words, according to approximation (5.46) and (5.47), no scattering sources exist outside the strip surface.

We have calculated functions (5.46) and (5.47) with the well-established estimations (5.55) and (5.56) to demonstrate the accuracy of the PO and PTD approximations. The results are plotted in Figures 5.5 and 5.6 in the decibel scale as 10 log(σs,h/kl2). It is seen in these figures that the accuracy of the first-order PTD approximation is higher for the soft boundary condition. Only a small discrepancy with the exact TED curve is observed in the vicinity of the direction φ = 90◦, and the PTD and TED curves infact merge in the region 100◦ < φ ≤ 270◦. The reason for the better approximation provided by the PTD for the soft boundary condition is the faster attenuation of the primary edge waves of the scattering sources js(1) compared to the similar waves of jh(1). In this case, the amplitude of these waves (at the location of the opposite edge of the strip) is of the order (2ka)−3/2 for the soft strip, and of the order (2ka)−1/2 for the hard strip.

It is also seen in Figures 5.5 and 5.6 that PO cannot provide a reasonable approximation for the scattered field in the vicinity of the minima of the directivity patterns,

Figure 5.5 Scattering at an acoustically soft strip (Ez -polarization of the electromagnetic field).

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Figure 5.6 Scattering at an acoustically hard strip (Hz -polarization of the electromagnetic field).

where PO predicts wrong pure zeros. The incorrect values of the PTD approximation for the function σh in the vicinity of the directions φ = 90◦ and φ = 270◦ are caused by the fictitious scattering sources jh(1) distributed outside the strip surface, as has already been discussed at the end of Section 5.1.2. This shortcoming of the first-order PTD is eliminated in the next section.

5.1.4 First-Order PTD with Truncated Scattering Sources jh(1)

The relationship uh = Hz exists between the acoustic and electromagnetic fields for this

problem.

The geometry of the problem is shown in Figure 5.1 and the incident wave is given by Equation (5.1). To calculate the scattered field uh, we apply Equation (1.10) and utilize the following observations.

• The symbol r was used in Equation (1.10) for the distance'between the integration and observation points. Now we replace it by ρ = x2 + (y − η)2 + ζ 2 and retain the symbol r for the polar coordinate r = 'x2 + y2 of the observation point, assuming that x = r cos φ and y = r sin φ.

•The surface of integration in Equation (1.10) covers both faces of the strip: S = S− + S+ (Fig. 5.7).

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96 Chapter 5 First-Order Diffraction at Strips and Polygonal Cylinders

•We denote the scattering sources as jh(0) = u0Jh(0) and jh(1) = u0Jh(1).

•Then the scattered field in the far zone can be represented in the following form:

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• According to the definition (1.31), one has Jh(0)(η, x = −0) = 2eikη sin φ0 .

•The nonuniform scattering source jh(1) is found by utilizing the exact solution for the half-plane diffraction problem presented in Section 2.5. According to this solution

–The function jh(1) is anti-symmetrical,

jh(1)(η, x = +0) = −jh(1)(η, x = −0) and consequently

Jh(1)(η, x = +0) = −Jh(1)(η, x = −0);

•In view of above observations, the functions (5.60) and (5.61) can be approximated by

•The asymptotic expression (5.64) was found here by the direct integration of the uniform component of the scattering sources. It is totally identical to the expression (5.11) derived by the summation of two edge waves.

•The field (5.65) is calculated by integration by parts.

•We provide here the final results in terms of the directivity patterns:

ei(kr+π/4)

uh(0,1) = u0 (h0,1)(φ, φ0) √ (5.66)

2π kr

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