Appendix of heterogeneous cellular network user distribution model

Appendix of Heterogeneous Cellular Network
User Distribution Model
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 Lemma 1
The probability that a user associates with the BS in kth
tier (conditioned that the distance between this user to its
associated BS is d) is
P|d(Ak)
(a)
= P|d[PBRPk
> any PBRPi
i=1:K,i=k
]
(b)
=
K
i=1,i=k
P|d[PBRPk
> PBRPi
]
=
K
i=1,i=k
P|d[PkCkd−αk
> PiCid−αi
i ]
=
K
i=1,i=k
P|d[di > (
PiCi
PkCk
)
1
αi · d
αk
αi ]
(c)
=
K
i=1,i=k
exp(
−λiπ(
PiCi
PkCk
)
2
αi ·d
2αk
αi
)
where (a) is for the user connectivity model that is the
maximum average bias received power connectivity model. If
the user associates with its BS in kth tier, PBRPk
=PkCkd−αk
is the maximum average bias received power compared with
PBRPi =PiCid−αi
i in other tiers. (b) is because each tier is
independent and it is the joint probability of all the other tiers
except kth tier. (c) assumes that the R is the distance between
this user to its closest interference in ith tier. For the ith tier
BS, their spatial distribution is Poisson point process. So the
probability P[di > R] above is the null probability, which is
e(−λiπR2
)
in the circle area πR2
.
The above proof is almost similar with [1]. Here only the
conditional associated probability is needed.
B. Proof of Lemma 2
The probability density function (PDF) of the distance d
between the user and its associated kth tier BS includes two
parts. One is the PDF fref (d) because BSs are with PPP
distribution of the density λk. The other is the user density
fdenk
(d) in kth tier we deﬁned. fdenk
(d) can be any non-
uniform user density function such as fdenLin
(d), fdenExp
(d)
and fdenGau
(d) adjusted by the experimentally obtained gain
GT un. If there are no users on kth tiers, the corresponding
fdenk
(d)=0.
fref (d)
(a)
= exp(−λkπd2
) · 2λkπd
fk(d) = fref (d) · fdenk
(d)
= exp(−λkπd2
) · 2λkπd · fdenk
(d)
where (a) is the PDF of the distance d to the associated BS
with PPP of the density λk from [2].
C. Proof of Lemma 3
The proof is almost the same to [1]. Here give the detailed
proof again for reference.
P|d[SINRk(d) > Tk|Ak]
= P|d[
PkGkd
−αk
Ir + σ2
L0
> Tk]
= P|d[Gk > (Ir +
σ2
L0
)TkPk
−1
dαk
]
= EIr
(P|d[Gk > (Ir +
σ2
L0
)TkPk
−1
dαk
])
(a)
= EIr
[exp(−(Ir +
σ2
L0
)TkPk
−1
dαk
)]
= exp(−
σ2
L0
TkPk
−1
dαk
)EIr
[exp(−IrTkPk
−1
dαk
)]
(b)
= exp(−
σ2
L0
TkPk
−1
dαk
)
K
i=1
EIri
[exp(−IriTkPk
−1
dαk
)]
(c)
= exp(−
σ2
L0
TkPk
−1
dαk
)
K
i=1
LIri (TkPk
−1
dαk
)
where (a) is from the assumption of the channel fading
envelope is Rayleigh fading. So Gk is random fading channel
power which is exponential distributed with the unity mean.
(b) comes from Ir =
K
i=1 Iri. (c) gives the deﬁnition of
Laplace transform Lx(s) = Ex[e−xs
]. Ex[∗] is the expectation
of ∗ over x.

D. Proof of Lemma 4
The Laplace transform of the interference in ith tier is
LIri (s)
(a)
= exp(πλiDi
2
) · exp(−πλiDi
2
Ek(e−skDi
−αi
))·
exp(−πλis
2
αi Ek(k
2
αi Γ(1 −
2
αi
)))·
exp(πλis
2
αi Ek(k
2
αi Γ(1 −
2
αi
, skDi
−αi
)))
(b)
= exp(πλiDi
2
) · exp(−πλiDi
2
EG(e−sPiGiDi
−αi
))·
exp(−πλis
2
αi EG((PiGi)
2
αi Γ(1 −
2
αi
)))·
exp(πλis
2
αi EG((PiGi)
2
αi Γ(1 −
2
αi
, sPiGiDi
−αi
)))
(c)
= exp(πλiDi
2
) · exp(−πλiDi
2
LG(sPiD−αi
i )) · Mi(s)
(d)
= exp(πλiDi
2
) · exp(
−πλiDi
2
1 + sPiD−αi
i
) · Mi(s)
The detail explanation of the above equation in each step is
as follows.
(a) gives the Laplace transform of the interference based
on the decay power law impulse response function f(k, r) =
kr−α
, r ≥ 1 [3]. Γ(t) =
∞
0
xt−1
e−x
dx is gamma function.
(b) updates response function f(k, r) = kr−α
into
f(G, r) = PiGir−α
.
(c) shows that the key part of the derivation of LIri
is the
calculation of Mi(s).
(d) comes from LG, which is the Laplace transform of
exponential channel power gain (LG(s) = 1
1+s [4]).
Mi(s)
= exp(−πλis
2
αi EG((PiGi)
2
αi Γ(1 −
2
αi
)))·
exp(πλis
2
αi EG((PiGi)
2
αi Γ(1 −
2
αi
, sPiGiDi
−αi
)))
(a)
= exp(−πλis
2
αi EG((PiGi)
2
αi ·
Γ(1 −
2
αi
)(sPiGiDi
−αi
)
1− 2
αi e−sPiGiDi
−αi
·
∞
n=0
(sPiGiDi
−αi
)n
Γ(2 + n − 2
αi
)
))
= exp(−πλiΓ(1 −
2
αi
)Di
2
·
∞
n=0
(sPiDi
−αi
)n+1 EG(Gn+1
i · e−sPiGiDi
−αi
)
Γ(2 + n − 2
αi
)
)
(b)
2
αi
)Di
2
·
∞
n=0
(sPiDi
−αi
)n+1 Γ(n + 2)
Γ(2 + n − 2
αi
)
·
(sPiDi
−αi
+ 1)−n−2
)
2
αi
)Di
2
·
sPiDi
−αi
(sPiDi
−αi
+ 1)2
·
∞
n=0
Γ(n + 2)
Γ(2 + n − 2
αi
)
· (
sPiDi
−αj
sPiDi
−αi
+ 1
)n
)
(c)
2
αi
)Di
2 sPiDi
−αi
(sPiDi
−αi
+ 1)2
·
Γ(2)
Γ(2 − 2
αi
)
· 2F1(2, 1, 2 −
2
αj
;
sPiDi
−αi
sPiDi
−αi
+ 1
))
= exp(−πλiDi
2
(1 −
2
αi
)−1 sPiDi
−αi
(sPiDi
−αj
+ 1)2
·
2F1(2, 1, 2 −
2
αi
;
sPiDi
−αi
sPiDi
−αi
+ 1
))
The detail explanation of the above equation in each step is
as follows.
(a) comes from [5] and [6] about the gamma function
property Γ(a, x) = Γ(a) − Γ(a)xa
e−x ∞
n=0
xn
Γ(a+n+1) .
(b) gives the expectation item of exponential power gain
EG(Gn+1
i · e−sPiGiDi
−αi
) =
∞
0
Gn+1
e−sPiDi
−αi
fG(G)dG,
fG(G) is the PDF of channel power gain G. fG(G) =
e−G
when G is exponential distributed. So EG =
∞
0
Gn+1
e−sPiDi
−αi
e−G
dG. After simpliﬁcation, EG(Gn+1
i ·
e−sPiGiDi
−αi
) = (sPiDi
−αi
+ 1)−n−2
Γ(n + 2).
(c) is from [5]. Γ(a)
Γ(c) ·2F1(a, 1, c : z) =
∞
n=0
Γ(a+n)
Γ(c+n) · zn
.
2F1(.) is the Gauss hypergeometric function.
Hence, Laplace transform of total interference in the ith tier
is
LIri
(s) = exp(πλiDi
2
) · exp(
−πλiDi
2
1 + Ri
)·
exp(−πλiDi
2
(1 −
2
αi
)−1 Ri
(Ri + 1)2
·
2F1(2, 1, 2 −
2
αj
;
Ri
Ri + 1
))
where Ri is sPiDi
−αi
. λi is ith tier BS density. Di is
( PiCi
PkCk
)
1
αi d
αk
αi and it is the minimum distance from the closest
interfering BS in ith tier. For the detailed proof about this min-
imum distance refer to [1]. 2F1(.) is the Gauss hypergeometric
function. Ci is biased factor in ith tier. A bias factor greater
than unity enables the cells to have an incrementally larger
coverage area and higher load.
E. Proof of Theorem
The coverage probability for non-uniform user model is as
follows.
Pc = Ed[
K
k=1
P|d[SINRk(d) > Tk|Ak] · P|d(Ak)]
=
K
k=1
∞
d=0
P|d[SINRik(d) > Tk|Ak] · P|d(Ak)·
fk(d) dd
=
K
k=1
Pck
where fk(d) is the PDF of the distance d.

The coverage probability associating with the kth tier BS is
Pck =
∞
d=0
P|d[SINRk(d) > Tk|Ak] · P|d(Ak) · fk(d) dd
=
∞
d=0
exp(−
σ2
L0
TkPk
−1
dαk
)
K
i=1
LIri
(TkPk
−1
dαk
)·
P|d(Ak) · fk(d)dd
=
∞
d=0
exp(−
σ2
L0
TkPk
−1
dαk
)
K
i=1
LIri (TkPk
−1
dαk
)·
K
i=1,i=k
exp(−λiπ(
PiCi
PkCk
)
2
αi · d
2αk
αi ) · exp(−λkπd2
)·
2λkπd · fdenk
(d) dd
=
∞
d=0
exp(−
σ2
L0
TkPk
−1
dαk
)
K
i=1
LIri
(TkPk
−1
dαk
)·
K
i=1
exp(−λiπ(
PiCi
PkCk
)
2
αi · d
2αk
αi ) · 2λkπd · fdenk
(d) dd
REFERENCES
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[3] J. Venkataraman, M. Haenggi, and O. Collins, “Shot noise models for
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Communications Conference, 2006. MILCOM 2006. IEEE. IEEE, 2006,
pp. 1–7.
[4] F. Yilmaz and M.-S. Alouini, “A novel uniﬁed expression for the
capacity and bit error probability of wireless communication systems over
generalized fading channels,” IEEE Transactions on Communications,
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[5] M. Abramowitz and I. A. Stegun, Handbook of mathematical functions:
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[6] M. Di Renzo, A. Guidotti, and G. E. Corazza, “Average rate of downlink
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