f n=l
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L 7jJ~ (Bsl , Bil ; fn)7jJ~* (B s2 , Bi2 ; fn) / VPl P2
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(11.2.15) where N r is the number of frequencies over the frequency range fo - tlf < fn < fo + tlf, and n is the frequency index for fn. Angular averaging: The angular averaging is defined by the small changes of incident and scattering angles around the fixed angles.
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ra(B sl , Bil; BS 2, B ) i2 1
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N a n=l where N a is the number of the angles, and 6n is the small angular difference for index n. Figure 11.2.1 shows the ACF magnitude with the three different ensemble averaging methods. The random rough surface is generated by using the spectrum method with a Gaussian height distribution and a Gaussian correlation function. The tapering parameter 9 chosen to be L/4. We plot the ACF as a function of sin Bi2 and sin Bs2 with fixed angles Bil = 20 and Bsl = -20 . Figure l1.2.1a is the result for one rough surface without any
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L 7jJ~ (Bsl + 6n ; Bil + 6n)7jJ~*(Bs2 + 6n , Bi2 + 6n )/ VHP2 (11.2.16)
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2.4 Scattering by a Buried Object Under a Rough Surface
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averaging. Figure 11.2.1b is that of realization averaging taken over 100 realizations. It is clear that the existence of the memory line becomes apparent if a sufficient number of independent samples are included. Figure 11.2.1c shows the ACF magnitude of a single rough surface based on the frequency averaging method. The single rough surface profile is as shown in Fig. 11.1.1. There are 50 equally spaced frequency samples over the frequency bandwidth of 0.5fo to 1.5fo. Although the ACF magnitude by frequency averaging is noisier than that of realization averaging, a distinct memory line is clearly visible. To suppress fluctuation in Fig. 11.2.1c, more independent samples must be included in the averaging process. It was estimated that only about 10 independent samples can be obtained with the bandwidth of 0.5fo to 1.5fo. For the frequency averaging to be effective, a wide bandwidth may be required. Figure 11.2.1d shows the ACF magnitude by the angular averaging method given in (11.2.16). The results are smoother than those without averaging (Fig. 11.2.1a), but the memory line is not as clearly visible as that of frequency averaging.
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2.4 Scattering by a Buried Object Under a Rough Surface
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In this section we study the scattered field from an object placed below a rough surface. The incident wave is a horizontally polarized (TE) wave, and we assume that the buried object is a perfect conductor as shown in Fig. 11.1.1. Let 1/;0 and 1/;1 be the fields in regions 0 and 1, respectively, and let the a boundary conditions be 1/;0 = 1/;1, = ~ at the rough surface and 1/;1 = 0 on the surface of the object. We make use of the surface integral equations from 4. In region 0 we have the integral equation given by, for r on Sr,
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~1/;o(r) = 1/;i(r) +
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region 0, and
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[1/;o(r') aG~~; r') - Go(r, r') a~~~')] ds'
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(11.2.17)
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where 1/;i(r) is the incident wave field, Go(r, r') is the Green's function for
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In region 1 we have the integral equation given by
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denotes the principal value integral over the rough surface.
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~1/;1(r) =1/;g(r)
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-ir [1/;l(r,)aG~~;r') -G1(r,r,)a~~~')]
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1/;g(r) =
(11.2.18)
where 1/;g(r) is the scattered field from the buried object and is given by
-1 G1(r,r,)a~(~')
(11.2.19)
11 ACF AND DETECTION OF BURlED OBJECT
0.03 002 001
sin(theta_s2)
-1 -1
sin(theta_i2)
:::j
005 00' 0.03 002 001
sJn(theta_s2) -0.5
-1 -1
sin{theta_i2)
Figure 11.2.1 Three-dimensional plots of ACF magnitude by different averaging methods. Reference angles are (l1il = 20 , I1 s l = -20 ). Dielectric constant of region 1 is Er = 3.7 + iO.13. 11, = 0.35A o , 1- 1.0Ao , L = 40A o , 9 = L/4. (a) One realization. (b) Realization averaging over 100 rough surfaces. (c) Frequency averaging over a frequency band of 0.510 to 1.510' (d) Angular averaging over an angular range of 11 _10 to 11 + 10 .
is the surface of the buried object.
For r on the surface of the buried object, we have
1/;: (r) ~
G 1 (r, r') 8i (~') n
ds' = 0
(11.2.20)
where1/;: (7') is the scattered field from the rough surface, given by (11.2.21)
2.4 Scattering by a Buried Object Under a Rough Surface
Thus Eqs. (11.2.18) and (11.2.20) become, respectively,