Exp-4:- Boundary Coverage

Aim
To understand the impact of shadowing and path loss exponent on boundary coverage probability.

Objectives
To calculate boundary coverage probability from measured signal strength.

1 Theory for Experiment 4:-Boundary Coverage

The circular coverage region of a Base Station is the area defined by a certain radius R with the Base
Station at the center where the mean signal level received at the mobile unit remains above a specified
threshold with certain probability. The probability mentioned above is decided based on quality of
service requirement. If there is no shadow fading then the radius R can be calculated for a region
where the signal level crosses the threshold with certainty.
However in real conditions, shadow fading plays an important role in cellular network design. Due to
the random variation of the signal strength received owing to shadow fade, one needs to find the
probability with which received signal strength crosses the predicted threshold
The detail derivation in the context of % boundary coverage and% area coverage are given below:

1.1 % Boundary Coverage:-

The received signal power in log domain at a distance d from the Base Station is given by
d 
0
Pr (d) = Pr (d0 ) + 10np log + xdB
d
Where,
• xdB represents shadow fading
• xdB is a random variable with Gaussian probability density function with mean Pr (d) and
standard deviation σxdB .
The probability that the signal level crosses the certain sensitivity level γ is given by
Z ∞
P rob[Pr (d) > γ] = p(x)dx
γ
Z γ
=1− p(x)dx
−∞
h i
= 1 − Pr (d) < γ
= 1 − FPr (γ)
!
1 γ − Pr (d)
= − erf c √
2 2σxdB
!
γ − Pr (d)
=Q
σxdB

1
1.2 % Area Coverage:-
The % area coverage is determined by the radius Rγ at which the signal level exceeds the sensitivity
level γ with probability Prob Rγ which is the likelihood of coverage at the cell boundary with

d = Rγ .
h i
P robRγ = prob Pr (Rγ ) > γ
Given that
[Pr (d) > γ](probdγ )is the probability that the signal the range 0 < d < Rγ exceeds the sensitivity level
we can associates this with the probabilty that the level exceds γ with in an infinite signal area dA at
the range d.

The % of useful area covered with the boundary of the Rγ with the received signal strength ≥ γ is
Z h
γ 1 i
Fu = P r (d) > γ dA
πRγ2
Z Rγ Z 2π
1 h i
= Pr (d) > γ rdrdθ
πRγ2 0 0

The power received can be referenced to the power received at cell boundary.
d 
0
Pr (d) = Pr (d0 ) + 10np log10
d
d  R 
0 γ
= Pr (d0 ) + 10np log10 + 10np log10
Rγ d

2
Where,
Pr (d0 ) = Pt − P L(d0 )
We shall use the radial distance r instead of d therefore
!
γ − Pr (r)
prob[Pr (r) > γ] = Q
σxdB
!
1 1 γ − Pr (r)
= − erf √
2 2 2σxdB
   !
d0 Rγ
1 1 γ − P r (d 0 ) + 10n p log (
10 Rγ ) + 10n P log r
= − erf √
2 2 2σxdB
r
!
1 1 γ − Pr (Rγ ) 10np log10 (e)ln( Rγ )
= − erf √ + √
2 2 2σxdB 2σxdB
!
1 1 r

prob[Pr (r) > γ] = − erf a + bln
2 2 Rγ
! !
γ − Pr (Rγ ) γ − (Pt − P L(Rγ )) 10np log10 (e)
a= √ = √ ,b = √
2σxdB 2σxdB 2σxdB
Z Rγ !
γ 1 1  r 
Fu = − 2 rerf a + bln dr
2 Rγ 0 Rγ
 
Making variable substitution t = a + bln r/Rγ , it can shown that
" #!
1 1−2ab
 1 − ab 
Fuγ = 1 − erf (a) + e b2 1 − erf
2 b

by choosing the signal level such thatPr (Rγ ) = γ such that a=0,Fuγ can be shown to be
"  !#
1 1
1
Fuγ = 1 + e b2 1 − erf
2 b
! !
γ − Pr (Rγ ) γ − (Pt − P L(Rγ ))
a= √ = √
2σxdB 2σxdB

3
1.3 Examples:-
1. Given,
Pr (d0 ) = 0dBm,
d0 = 100m,
np = 4.5,
Rγ = 3000m,
P robRγ = 0.65,
and σ = 6dB.
find the margin value:-
d 
0
Pr (Rγ ) = Pr (d0 ) + 10np log10
Rγ
 100 
Pr (Rγ ) = 0 + 10 ∗ 4.5 log10
3000
= −66.47dBm
!
1 1  3000 
and, 0.65 = − erf a − bln
2 2 3000
or, erf (a) = −0.3
or, a = −0.2725
!
√
now, γ = aσ 2 − Pr (γ)
!
√  
= − 0.2725 ∗ 6 ∗ 2− − 66.47dBm

4
 = 68.78dBm
So,margin value (-68.78-(-66.47))dB = -2.31 dB.

2. Given,
Pr (d0) = 0dBm,
d0 = 100m,
np = 3,
Rγ = 3000m
P rob(Rγ) = 0.5,
σ = 9dB.
find % Area coverage.
d 
0
P r(Rγ ) = Pr (d0 ) + 10np log10 ,
Rγ
 100 
P r(Rγ ) = 0 + 10 ∗ 3 log10
3000
= −44.3136dBm
!
1 1  3000 
N ow, 0.5 = − erf a − bln ,
2 2 3000
or, erf (a) = 0
or, a = 0

" #
10np 10 log10 (e)
and b= √
σ 2
" #
10 ∗ 3 ∗ 0.4343
=
9 ∗ 1.414
= 1.0238
" #!
1 1−2ab
 1 − ab 
so, Fuγ = 1 − erf (a) + e b2 1 − erf ∗ 100
2 b
" #!
1 1−2∗0∗1.0238
 1 − 0 ∗ 1.0238 
⇒ Fuγ = 1 − erf (0) + e 1.02382 1 − erf ∗ 100
2 1.0238
= 71.71%
so, % area coverage =71.71%

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 Reference

[1]Theodore S. Rappaport,Wireless Communications: Principles and Practice,2nd Edition,Prentice

Hall Communications Engineering and Emerging Technologies Series.