Appendix for the paper of "Downlink coverage probability in heterogeneous cellular networks on nakagami m fading channel"

1. Appendix of Downlink Coverage Probability in
Heterogeneous Cellular Networks on Nakagami-m
Fading Channel
Chao Li, Abbas Yongacoglu, and Claude D’Amours
School of Electrical Engineering and Computer Science
University of Ottawa
Ottawa, ON, K1N 6N5, CA
Email: {cli026, yongac, cdamours}@uottawa.ca
APPENDIX
A. Proof of Coverage Probability on General Fading
In [1] and [2], the following conclusion has been given. Here
give the more detailed derivation process for the reference.
P(Ir +
σ2
L0
≤ PkGk(rαk
Tk)
−1
)
= P(X ≤ Y )
(a)
=
+∞
−∞
ˆf(s)
¯ˆg(s) − 1
2πis
ds
(b)
=
+∞
−∞
Lx(2πis)
Ly(−2πis) − 1
2πis
ds
(c)
=
+∞
−∞
e−2πis σ2
L0 · LIr
(2πis)·
LG(−2πisPk(rαk
Tk)
−1
) − 1
2πis
ds
The detail explanation for the above equation in each
steps is as follows. In step (a), Refer to [3] (Corollary
12.2.2), in which let X represents the non-negative real
valued random variable Ir + σ2
L0
with a square integrable
density f(x), and Y represents the non-negative real valued
random variable PkGk(rαk
Tk)
−1
with a square integrable
density g(y). In step (b), ˆf(s) =
R
e−2iπxs
· f(x)dx is the
Fourier transform of f(x). Lx(θ) =
R
e−θx
· f(x)dx is
the Laplace transform of f(x). Hence ˆf(s) = Lx(2πis).
¯ˆg(s) is conjugate of ˆg(s) and
¯ˆg(s) = Ly(−2πis). In step
(c), Lx(θ) = LIr+ σ2
L0
(θ) =
R
e−θ(Ir+ σ2
L0
)
· fIr
(x)d(x).
Because σ2
L0
is one constant parameter, Lx(θ) = e−θ σ2
L0 LIr (θ)
and Lx(2πis) = e−2πis σ2
L0 LIr
(2πis). fIr
(x) is probability
density function (PDF) of the random variable of Ir. The same
idea is to LG.
The condition for the above conclusion is from [2], which
is:
+∞
−∞
| e−2πis σ2
L0 | · | LIr (2πis) | ·
|
LG(−2πisPk(rαk
Tk)
−1
) − 1
2πis
| · ds < ∞
Where, | ∗ | represents the abstract value of ∗.
B. Proof of Coverage Probability on Nakagami-m Fading of
m = 1
Pck(r, Tk)
=
+∞
−∞
e−2πis σ2
L0 LIr (2πis) ·
LG(−2πisPk(rαk
Tk)
−1
) − 1
2πis
ds
(a)
=
+∞
−∞
e−2πis σ2
L0 LIr (2πis) ·
Pk(rαk
Tk)
−1
1 − 2πisPk(rαk Tk)−1 ds
(b)
=
+∞
−∞
e−2πis σ2
L0 (
+∞
0
e−2πisx
fIr
(x)dx)
·
Pk(rαk
Tk)
−1
1 − 2πisPk(rαk Tk)−1 ds
(c)
=
+∞
0
(
+∞
−∞
e−2πis( σ2
L0
+x)
·
Pk(rαk
Tk)
−1
1 − 2πisPk(rαk Tk)−1 ds)fIr
(x)dx
=
+∞
0
(
+∞
−∞
e−2πis( σ2
L0
+x)
·
1
(P−1
k rαk Tk) − 2πis
ds)fIr
(x)dx
(d)
=
+∞
0
e−P −1
k rαk Tk( σ2
L0
+x)
fIr
(x)dx
= e−P −1
k rαk Tk
σ2
L0
+∞
0
e−P −1
k rαk Tkx
fIr (x)dx
(e)
= e−P −1
k rαk Tk
σ2
L0 LIr (P−1
k rαk
Tk)

2. The detail explanation for the above equation in each step
is as follows.
In step (a):
LG(s) = 1
1+s is from [4].
In step (b):
Laplace transform is LIr
(2πis) =
+∞
0
e−2πisx
· fIr
(x)dx.
In step (c):
due to Fubini’s theorem.
In step (d):
+∞
−∞
e−2πisf
· 1
|A|−2πis ds is Fourier transform of 1
|A|−2πis ,
it is e−|A|f
u(f), here u(x) is step function, where u(x) =
1 for x > 0, otherwise, u(x) = 0. A = P−1
k rαk
Tk and f =
σ2
L0
+ x.
In step (e):
LIr (θ) =
R
e−θ(x)
· fIr (x)d(x)
C. Proof of Laplace transform LIr of the total interference
LIrj
(s)
(a)
= exp(πλjdj
2
) · exp(−πλjdj
2
Ek(e−skdj
−αj
))·
exp(−πλjs
2
αj Ek(k
2
αj Γ(1 −
2
αj
)))·
exp(πλjs
2
αj Ek(k
2
αj Γ(1 −
2
αj
, skdj
−αj
)))
(b)
= exp(πλjdj
2
) · exp(−πλjdj
2
EG(e−sPj Gj dj
−αj
))·
exp(−πλjs
2
αj EG((PjGj)
2
αj Γ(1 −
2
αj
)))·
exp(πλjs
2
αj EG((PjGj)
2
αj Γ(1 −
2
αj
, sPjGjdj
−αj
)))
(c)
= exp(πλjdj
2
) · exp(−πλjdj
2
LG(sPjd
−αj
j )) · Mj(s)
The detail explanation for the above equation in each step
is as follows.
In Step(a):
[5] give the Laplace transform of the total interference
based on the decay power law impulse response function
f(k, r) = kr−α
, r ≥ 1. Ek(.) is the expectation over k.
Γ(t) =
∞
0
xt−1
e−x
dx is Gamma function.
In Step(b):
Update response function f(k, r) = kr−α
into f(G, r) =
PjGjr−α
.
In Step(c):
the key part of the derivation of LIr
is the calculation of
Mj(s).
Mj(s)
= exp(−πλjs
2
αj EG((PjGj)
2
αj Γ(1 −
2
αj
)))·
exp(πλjs
2
αj EG((PjGj)
2
αj Γ(1 −
2
αj
, sPjGjdj
−αj
)))
(a)
= exp(−πλjs
2
αj EG((PjGj)
2
αj ·
Γ(1 −
2
αj
)(sPjGjdj
−αj
)
1− 2
αj e−sPj Gj dj
−αj
·
∞
n=0
(sPjGjdj
−αj
)n
Γ(2 + n − 2
αj
)
))
= exp(−πλjΓ(1 −
2
αj
)dj
2
·
∞
n=0
(sPjdj
−αj
)n+1
EG(Gn+1
j · e−sPj Gj dj
−αj
)
Γ(2 + n − 2
αj
)
)
(b)
= exp(−πλjΓ(1 −
2
αj
)dj
2
·
∞
n=0
(sPjdj
−αj
)n+1 Γ(n + m + 1)
Γ(2 + n − 2
αj
) · Γ(m)
·
mm
(sPjdj
−αj
+ m)−n−m−1
)
= exp(−πλjΓ(1 −
2
αj
)dj
2
·
mm
Γ(m)
sPjdj
−αj
(sPjdj
−αj
+ m)m+1
·
∞
n=0
Γ(n + m + 1)
Γ(2 + n − 2
αj
)
· (
sPjdj
−αj
sPjdj
−αj
+ m
)n
)
(c)
= exp(−πλjΓ(1 −
2
αj
)dj
2
·
mm
Γ(m)
sPjdj
−αj
(sPjdj
−αj
+ m)m+1
·
Γ(m + 1)
Γ(2 − 2
αj
)
· 2F1(m + 1, 1, 2 −
2
αj
;
sPjdj
−αj
sPjdj
−αj
+ m
))
= exp(−πλjdj
2
mm+1
· (1 −
2
αj
)−1 sPjdj
−αj
(sPjdj
−αj
+ m)m+1
·
2F1(m + 1, 1, 2 −
2
αj
;
sPjdj
−αj
sPjdj
−αj
+ m
))
The detail explanation for the above equation in each step
is as follows.
In Step(a):
[6] and [7] give the gamma function property Γ(a, x) = Γ(a)−
Γ(a)xa
e−x ∞
n=0
xn
Γ(a+n+1) .
In Step(b):
the expectation item of Nakagami-m fading EG =
∞
0
Gn+1
e−sPj Gdj
−αj
fG(G)dG, fG(G) is pdf of channel
power gain G. fG(G) = 1
Γ(m) mm
Gm−1
e−mG
. So EG =
∞
0
Gn+1
e−sPj Gdj
−αj 1
Γ(m) mm
Gm−1
e−mG
dG. After simpli-
fication, EG = 1
Γ(m) mm
(sPjdj
−αj
+m)−n−m−1
Γ(n+m+1)
In Step(c):
according to [6], Γ(a)
Γ(c) ·2F1(a, 1, c : z) =
∞
n=0
Γ(a+n)
Γ(c+n) · zn
.
2F1(.) is the Gauss hypergeometric function.

3. Hence, Laplace transform of total interference in jth tier is
LIrj
(s) = exp(πλjdj
2
) · exp(−πλjdj
2
LG(Rj))·
exp(−πλjdj
2
mm+1
· (1 −
2
αj
)−1 Rj
(Rj + m)m+1
·
2F1(m + 1, 1, 2 −
2
αj
;
Rj
Rj + m
))
where Rj is sPjdj
−αj
. K is the total number of tiers. λj
is jth tier BS density. m is Nakagami-m fading parameter.
dj is (
Pj Cj
PkCk
)
1
αj r
αk
αj and it is the minimum distance from the
closest interfering BS in jth tier. The detailed proof about
this minimum distance please refer to [8]. 2F1(.) is Gauss
hypergeometric function. Cj is biased factor in jth tier. A
bias factor greater than unity enables the cells to have an
incrementally larger coverage area and higher load.
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