Percentage Coverage Area Suvra Sekhar Das G.S. Sanyal School of Telecommunications, IIT Kharagpur, India e-mail:
[email protected]; Phone: +91 3222 83902
P ERCENTAGE C OVERAGE A REA A. Received Signal The received signal power in log domain at a distance d from the base station is given by Pr (d) = {Pr (d0 ) + 10np log10 (
d0 ) + xdB }, d
(1)
where xdB represents shadow fading and Pr (d0 ) is the received power at the reference point do defined by Pr (d0 ) = Pt − P L(d0 ),
(2)
where P L(d0 ) is the path loss at the reference point d0 . xdB is a random variable with gaussian probability density function with mean as Pr (d) and standard deviation σxdB which is represented by where ∼ N (Pr (d), σxdB ). The probability that the signal level crosses the certain sensitivity level γ is given by Z
∞
Z
γ
p(x)dx = 1 −
P rob[Pr (d) > γ] = γ
p(x)dx = 1 − [Pr (d) < γ] −∞
= 1 − FPr (γ) 1 γ − Pr (d) = erf c{ √ } 2 2σxdB ! γ − Pr (d) =Q σxdB
(3)
similarly, the probability that the received signal level is below the sensitivity threshold level γ is given
by [Pr (Rγ ) < γ] = Q
Pr (d) − γ σxdB
! (4)
Determination of the % coverage area : How the boundary coverage relates to the % coverage within the boundary. Circular coverage area is determined by the radius Rγ at which the signal level exceeds the sensitivity level γ with probability PorbRγ - the likelihood of coverage at cell boundary d = Rγ then ProbRγ ≡ Prob[Pr (Rγ ) > γ]
(5)
Generally given that Prob[Pr (d) > γ] (Probdγ ) is the probability that the signal at the range (0 < d < Rγ ) exceeds the sensitivity level, we can associate this with the probability that the level exceeds γ within an infinitesimal area dA at range d. Then the % of useful area covered within the boundary of Rγ with received signal strength > γ is Fuγ
1 = πRγ2
Z
1 [Pr (d) > γ]dA = πRγ2
Z
Rγ
Z
2π
[Pr (d) > γ]r dr dθ 0
(6)
0
In the above, if γ is infinite then integral evaluates to 0 and if γ is zero then the integral evaluates to one. The power received can be referenced to the power received at cell boundary (if represented in terms of path loss, the same can be done.) d0 ) d d0 Rγ = Pr (d0 ) + 10np log10 ( ) + 10np log10 ( ) Rγ d Pr (d) = Pr (d0 ) + 10np log10 (
(7) (8)
where Pr (d0 ) = Pt − P L(d0 ). In terms of Path Loss Rγ d ) + 10np log10 ( ) d0 Rγ d = P L(Rγ ) + 10np log10 ( ) Rγ
P L(d) = P L(d0 ) + 10np log10 (
(9)
we shall use the radial distance r instead of d. Therefore Prob[Pr (r) > γ] = Q =
γ − Pr (r) σxdB
!
1 1 γ − Pr (r) − erf( √ ) 2 2 2σxdB
γ − (Pr (d0 ) + 10np log10 ( Rd0γ ) + 10np log10 ( Rrγ )) 1 1 √ = − erf( ) 2 2 2σxdB ! γ − (Pr (d0 ) + 10np log10 ( Rd0γ )) 10np log10 ( Rγ ) 1 1 r √ √ = − erf − 2 2 2σxdB 2σxdB ! d0 γ − (Pr (d0 ) + 10np log10 ( Rγ )) 10np log10 (e)ln( Rγ ) 1 1 r √ √ − = − erf 2 2 2σxdB 2σxdB ! r 1 1 γ − Pr (Rγ ) 10np log10 (e)ln( Rγ ) √ √ = − erf + 2 2 2σxdB 2σxdB
(10)
From the above, two parameters are defined a=
(γ − Pr (Rγ )) (γ − (Pt − P L(Rγ ))) 10np log10 (e) √ √ √ = and, b = σxd B 2 σxd B 2 σxd B 2
(11)
Therefore 1 1 r Prob[Pr (r) > γ] = − erf a + bln( ) 2 2 Rγ
(12)
If Fuγ is given, it means that to ensure coverage (probability) in 100.Fuγ % of the area inside Rγ then a normalized margin of a is required at the boundary Rγ . Now, the integral can be written as Fuγ
1 1 = − 2 2 Rγ
Z 0
Rγ
r r erf a + b ln( ) dr Rγ
(13)
Making variable substitution t = a + bln( Rrγ ), it can be shown that Fuγ
1 = 2
1 − erf(a) + e
1−2ab b2
1 − ab 1 − erf b
(14)
By choosing the signal level such that Pr (Rγ ) = γ such that a = 0, Fuγ can be shown to be Fuγ
1 1 1 = 1 + e b2 1 − erf 2 b
(15)
