Let Zl and Z2 be two positions and let El the joint probability density function is in .NET

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Let Zl and Z2 be two positions and let El the joint probability density function is
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= E(Zl) and E2 = E(Z2)' Then
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p(El' E2) =
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21Tcr2~ exp [- 2cr2(11_ r ((EI - Em)2
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- 2r(El - E )(E2 - Em) m
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E )2)] (1.1.2) m
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where r is the correlation coefficient that depends on IZl - z21. If the correlation coefficient is of exponential form, then
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r(zl - Z2) = exp ( _IZI : Z2!)
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where Zz is the correlation length. Let Ef(z) = E(Z) - Em be the fluctuating part of the permittivity. Then the covariance function of the permittivity is
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(Ef(z)Ef(z')) =
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OE~ exp ( _Iz ~ z'l)
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where angular bracket denotes average. Given a realization of permittivity profile, we can discretize the medium into fine layers (say up to 30,000 layers). We note that in Fig. 1.1.2 of the Gaussian random process, the permittivity is a continuous function of depth.
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If we assume that j(z) is a real Gaussian process with normalized correlation function C (z ). Then
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(j(Z)j(Z')) = cr 2 C(z - z')
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Let W(K) be the spectral density
cr 2 C(z) =
dzcr 2 C(z)e- 2Kz 21T -00 For the case that the correlation is exponential C(Z) = exp (
the spectral density is
= -1
(1.1. 7)
(1.1.9) Let the sample be generated for 0 S Z S L. Outside L, we can assume that Ef (z) is periodic. Then using a Fourier series,
j(z) =
The discretization is
L bnen=-oo
tlK = 21T L 21Tn K n = - - = ntlK
(1.1.11) (1.1.12)
1.3 Numerical Results and Applications to Antarctica
The bn's are complex and
(J(zl)f(Z2)) =
dKe iK (ZI-Z2)W(K)
i hn Z eLl -i 2,m Z 2 L
=1 -
L L (bb*)
(1.1.13) (1.1.14)
eiKn (ZI-Z2)W(K n )
is real,
= b":...-n
(1.1.16) (1.1.17)
\Ib~12) = (lb~12) = 1fLW(Kn )
where bn = b~ + ib~. The real and imaginary parts are independent Gaussian random variables. Thus (bnb":.-n) = (bnb n ) = O. Let L be divided with N intervals, L 6.z = (1.1.18) N and N be a power of 2.
f(ZI) =
N /2
( bnexp i~
Then bo and bN / 2 are real. We first obtain N independent Gaussian random numbers with zero mean and unit variance. We next multiply the numbers by a normalization factor to bo, bN / 2 , b~ and b~ with n = 1,2, ... , N/2 - 1, such that (1.1.17) holds. We then use
to get bn with n by (1.1.19).
= -1, -2, ... , -
N /2 + 1. The permittivity is then calculated
1.3 Numerical Results and Applications to Antarctica
In this section we illustrate the numerical results and application to the Antarctic fim. The Antarctic fim has a layering structure. The permittivity of snow is around 1.5E o . Thus if a half-space medium is assumed, the reflectivity at = 0 is 0.01 so that with T = 240 K, the brightness temperature is 237.6 K. However, the measured brightness temperature is significantly
less than that. The difference can be attributed to the reflections by the layering structure. Snow is a mixture of ice and air, so that the density of snow P indicates the fractional volume of ice in snow. Ice has a density of 0.91 g/cm3 . The density of snow is
P = 0.91
where f is the fractional volume of ice in snow. We model p(z) as a random process. First we assume that p(z) is a Gaussian random process with
(p) p(z) (pf(zdpf(Z2))
(1.1.22) (1.1.23)
= Pm + Pf(z)
o-~ exp ( _I Z 1 ~ z21)
In Fig. 1.1.2 we show a simulated density profile of a single realization using 3 lz = 2 mm, o-p = 0.0156 g/cm , and Pm = 0.4 g/cm3 . The continuous profile of each realization is generated down to a depth of 21 m and is discretized with 214 = 32,768 layers. This gives a 6.z discretization thickness of 0.6 mm. The permittivity of each layer is calculated using the following empirical mixing formula for dry snow
- = 1+
1.60 P 1- 0.35p
(0.52p+0.62p 2)
- =..3 f
Eo Eo
where E~~e is the imaginary part of the permittivity of pure ice and is temperature-dependent. If o-p Pm, we can linearize (1.1.25) and (1.1.26), so that E~ and Pm are related by the same relation as in (1.1.25) and E'/n and Pm are related by the same relation as in (1.1.26).
= 1+ =