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A17.5.3 A17.5.3

According to ITU-R Rec. P.526-2, the diffraction loss  L_p (d) can be derived by the received field strength  E referred to the free space E₀ :

     Image Removed                                        

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body $-{{L}_{d}}(p)=20\log \frac{E}{{{E}_{0}}}=F(X)+G({{Y}_{1}})+G({{Y}_{2}})$

   (Eq. 220)

with:

  X

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body $X$

    is the normalized radio path between transmitter and receiver Image Removed   Y1    

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body $X=2.2\beta \ {{f}^{\frac{\ 1}{3}}}\ a_{e}^{\frac{-2}{3}}\ d$

 

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body Y1

  is the normalized antenna height of the transmitter Image Removed    Y2 transmitter  

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body $Y=9.6\ \cdot {{10}^{-3}}\beta \ {{f}^{\frac{2}{3}}}\ a_{e}^{\frac{-1}{3}}\ h{}_{i}$

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body Y2

   is the normalized antenna height of the receiver

where:Image Removed

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body $\beta $

 is a parameter derived from the earth admittance factor K  :Image Removed =1 for > :  

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body $\beta $

 =1 for 

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body $f$

 > 20 MHz.

 f is

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body $f$

 is the frequency in MHz

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body ${{a

_e is

}_{e}}$

 is the equivalent earth radius in km (definition see below)d is

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body $d$

 is the distance in km

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body ${{h

_i is

}_{i}}$

 is the antenna height above ground in m with =1 or 2 for the transmitter or receiver, respectively

The distance-dependent term F

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body $F(X)

 is

$

  is given by the semi-empirical formula:

...

                       

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body $F(X)=11+10\log (X)-17.6X$

 (Eq. 221)

The antenna height gain G

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body $G(Y)

is

$

 is given by the formula set:

            Image Removed                  for Y>2                                                (Eq. 222)

 Image Removedfor  10K < Y < 2                                 

             Image Removed     for    K/10 < Y < 10K                      

                                              Image Removed                for    Y < K/10                                 

...

   

    

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body $G(Y)=17.6{{(Y-1.1)}^{\frac{1}{2}}}-5\log (Y-1.1)-8$

 for 

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body Y>2

 

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body $G(Y)=20\log (Y+0.1{{Y}^{3}})$

 for

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body $10K<Y<2$

 

       

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body $G(Y)=2+20\log K+9\log (Y/K)\left[ \log (Y/K)+1 \right]$

  for  

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body $K/10<Y<10K$

 

                             

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body $G(Y)=2+20\log K$

 for  

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body $Y<K/10$

 

where

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body $K$

 is the normalized earth surface admittance factor (see ITU-R Rec. P.526), default value 10e-5:

...