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Submitted on 16 Jan 2019
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Note technique: Throughput Analysis for different power allocation policies in a random access channel
Asma Selmi, Jean Pierre Cances
To cite this version:
Asma Selmi, Jean Pierre Cances. Note technique: Throughput Analysis for different power allocation policies in a random access channel. [Research Report] Xlim UMR 7252. 2018. �hal-01982964�
Note technique: Throughput Analysis
for different power allocation policies in a random access channel
Asma Selmi, JP Cances, Xlim UMR CNRS 7252
Summary/Introduction:
In this technical note, we give the mathematical derivations necessary to calculate the throughput efficiency of a multiband slotted ALOHA uplink transmission system using NOMA power domain technique. We consider in a first time only the presence of the capture effect at the SIC based receiver and we derive the probability to decode the different packets of the users present at the SIC level. Furthermore, we consider here a uniform power allocation and we study the corresponding decoding probability at the SIC receiver.
I. Capture effect at the SIC receiver, throughput analysis
We consider the case of B orthogonal frequency resources available at the BS station. There are M users in the network who want to access to the BS resources. Denote by ithe index set of active users transmitting signals through the ith sub-channel. Then, the received signal at the BS over the ith subchannel, can be written as:
, , ,
i
i i k i k i k i
k
y h P s n
(1)Where hi k, , Pi k, and si k, represent the channel coefficient, transmit power and signal from user k through the ith subchannel, respectively, and ni (0,N0) is the background noise. Here, N0 is the noise spectral density. We will consider here the capture effect taking into account the main following event: if two packets collide over a given frequency resource, we have the following equation:
i i i j j j
yh P s h P s n (2)
Considering user i as the user of interest, we will be able to recover packet i if the log2(1+SINR) is greater than the spectral efficiency of the used modulation format denoted as in bits/s/Hz.
The SINR is defined as:
2 2 0
SINR i i
j j
P h N P h
(3)
And we have to compare it with the threshold: 2 1. The probability to obtain this capture effect is thus equal to:
2 2 0
( , ) Proba(SINR ) Proba( i i )
cap
j j
P i j P h
N P h
(4)
We assume a classical Rayleigh distribution for hi2and hj2that means:
2( )
i
x
ph x e (5)
We suppose that real part and imaginary part of hi follow a Gaussian distribution with mean zero and variance 0.5. In this case, we have:
2 0
2 0
2 2
2 2 0 2
2 0
( )
( )
Proba( ) = Proba( ( ) )
=
j j
i
j j
i
j j i i
j i j
j j i
N P h x P
N P h P
N P h
P h h h h
N P h P
e dx e
(6)
In fact, the obtained probability is a conditioned probability since we consider a fixed value of
2
hj in equation (6). To obtain the complete probability we have to average over the distribution of hj2, this yields:
0 0
0 0
( )
2 2
0 0
0
( 1)
0
Proba( )
1
j j
i i i
j i
i i
N P x N P
P x P P x x
i i j j
N
N P x P
P P
j i
P h e e dx e e e dx
N P h e e dx e
P P
(7)
Hence, we obtain:
0
( , )
1
i N P cap
j i
P i j e P P
(8)
This result will be helpful to derive the throughput efficiency. In this case, if there are M active users and each active user chooses a sub-channel independently and uniformly at random, the conditional throughput can be written as:
1 2
1
,
1 2
1
,
1 1 1
( , ) [ 1 1 [ ( , ) ( , )].Proba(sélection couple ( , ))]
2.
1 1 1 1
( , ) [ 1 1 [ ( , ) ( , )]]
2 ( 1)
M M
B cap cap
i j
M M
B cap cap
i j
M B M C P i j P j i i j
B B B
M B M C P i j P j i
B B B M M
(9)
With !
!( )!
k n
C n
k n k
which denotes the Newton binomial coefficient. The first term 1 1
1
M
M B
corresponds to the case where there is no collision between packets and the second term is related to the case where two packets collide on a given frequency resource, there are C1BB possible choices for this given resource and the probability that two packets collide on this resource is: 1
B. We obtain eventually:
0 0
1 2
,
1 2
,
1 1 1 1
( , ) [ 1 1 [ ( , ) ( , )]]
2 ( 1)
1 1 1 1
( , ) [ 1 1 [ ]]
2 ( 1)
1 1
i j
M M
cap cap
i j N N
M M M P P
j i
i j i j
M B M P i j P j i
B B M M
e e
M B M
P P
B B M M
P P
(10)We can make some further assumptions taking into account that the signal to noise ratio is sufficiently high to use the following approximation:
0
1 0 i N P
i
e N
P
(11)
We obtain then:
0 0
1 2
1 1, 1 1,
0
1 2
1 1,
1 1
1 1 1 1
( , ) [ 1 1 ]
2 ( 1)
1 1
1 1 1 1
( , ) [ 1 1
( 1)
1
M M M M M M
i i
j j
i j j i j i i j
i i
M M M M
i i j j i j
i
N N
P P
M B M
P P
B B M M
P P
N M B M P
B B M M P
P
(12)
In (12), since M is a random variable, in order to find the average throughput, we need to consider a distribution of M. For convenience, we consider a uniform distribution with a large number of users. To this end, assume that there are K users and each user becomes active with access probability pa. In addition, let N denote the number of active users that choose a sub- channel. Then, [M]Kpaand [ ] Kpa
N B . For a large K, we can use the Poisson approximation for N. That is, N becomes a Poisson random variable as follows:
( ) !
ne N p n
n
(13)
Where p n( ) denotes the probability mass function (p.m.f) of a Poisson random variable with parameter . Here, we have: [M] Kpa
B B
which is assumed constant and is called the traffic intensity as K . Then, the average throughput of multichannel ALOHA can be calculated first, without taking into account the effect capture by:
(M B, )
M[ 1 1 M 1 B (14) We have:
log( (M B, )) log(M) (M 1) log(1 1)
B
With: 1 1
log(1 )
B B
, we have : log( ( , )) log( ) ( 1) / log( ) M
M B M M B M
B , then
using the expectation values, we have: log( [ ( M B, )])log(B), and this yields : [ ( M B, )]B e (15)
Taking into account the capture effect to study the influence of power allocation we have to average over powers Pi and Pj. Moreover, we will consider that M , in these conditions the first task is to characterize the random variable
1 0
1
i j i
N Z P
P P
, this is due to the following
relationship:
0 0
1 1,
1 1
[ 1 ]
( 1)
1 1
M M
i i
j j
i j j i
i i
N N
P P
P P
M M
P P
(16)To go deeper in the derivation, we suppose that the cell has been partionned into circular areas of radius equal to the distance between the current user and the BS, that means that the first power allocation policy is to consider: Pi di2 with an uniform law, i.e. we only compensate for the line of sight losses. In this case we have the maximum accuracy, but of course this is an ideal case. ThusPi is a random variable with a pdf equal to: 2 2
min
( ) 1
f x R d
. To characterize the law of Z, we compute the probability:
0
0
0
0
0
1
Proba( ) Proba( ) Proba(1 1 )
1
Proba(1 ) Proba(1 ( ))
Proba( ( ))
1
N
N V
Z x U x x
V U U
U
V N
x x x xV N
U U U
U xV N
x
(17)
Using the distribution of U we can calculate:
0
2min
( )
1
2
0 2 2 2 2 0 min
min min
Proba( ( ) ) 1 ( )
1 1
x xV N
d
U xV N V du xV N d
x R d R d x
(18)To end the derivation we have to average with the law of V.
2
2 min
0
2
0 min
2 2 2 2
min min
0 2 2 min
2 2
min
Proba( ) Proba( ( ) ) ( )
1
1 ( )
1
1 ( )
1 1
R
d
Z x U xV N V p V dV
x xV N d dV
R d x R d
x N
V d dV
x x
R d
2 2
2 4 min4 0 min2
2 min2
min
2 2 0 2
min min
2 2
min
1 ( )
2(1 ) 1
1 ( )
2(1 ) 1
x N
R d d R d
x x
R d
x N
R d d
R d x x
(19)
We derive (19) and we obtain the p.d.f of Z.
2 2 0
2 2 min 2 2
min
2 2
min 0
2 2 2 2 2 2
min min
2 2
min 0
2 2 2 2 2 2
min min
2 2
min
1 1
( ) ( )
2 1 (1 ) (1 )
( ) 1 1
( ) 2( ) 1 (1 ) (1 )
( ) 1 1
( ) 2( ) (1 ) (1 )
( ) ( )(1
Z
Z
Z
Z
x N
f x R d
R d x x x
R d x N
f x
R d x x R d x
R d N
f x
R d x R d x
f x
R d
2 2
min
2[ 0]
) 2
R d x N
(20)
Then, to compute the mean value of Z, we have to study, the function:
0
0 0
2 2
0 2
( , )
( , ) ( ) ( )
0 ,
( , )
0 ,
x N g x y
x y
g x y x y x N y N
x x y x y x y
g x y x N
y x y x y
It’s straightforward to conclude that 1 0
1
i j i
N P P P
is maximum for Pi R2 and Pj dmin2 , this
maximum is equal to
2 0
2 2 2
min
R N
X R d
and takes its minimum value for Pj R2 and Pi dmin2 with the minimum value equal to
2
min 0
1 2 2
min
d N
X d R
. This yields:
21
2 2
1 1
2 2
min
2 2 2 0
min
2 2 2 2
min min
0 0
2 2 2 2 2 2
min min
( ) [ ]
( )(1 ) 2
[ ] [ ]
1 1
2 2
(1 ) 1 (1 )
X Z
X
X X
X X
x R d
Z x f x dx N dx
R d x
R d R d
N N
x dx dx
R d x R d x x
2
1
2 2
min 0
2 2
min
2 2
min 0
2
2 2
min 1 2 1
[ ]
2 log(1 ) 1
1
[ ]
1 1 1
2 log
1 1 1
X
X
R d N
R d x x
R d
N X
R d X X X
With :
2 0
2 2 2 2 2 2
2 2
min 0 min min
2 min min
2 2 2 2 2 2 2
min 0
1 min 0 min
2 2
min
2 2
2 2 2 2 2 2 2 2
min min
min min min min
2 2
1 1 /
1
1 1 /
1
1 / 1 / 1 / /
R N
d N d R d R
X R d d
d N
X R d R N R d R
d R
d d
d R d R d R d R
R R
2
2 2 2 2
2 min
min min
2 1
2 2
2 min min
2 2
1
log 1 log log 1 / /
1
1 1
log log [ ]
1
X d
d R d R
X R
X d d
X R R
We obtain:
2 2
min 0 2 2 2 2 2 2
min min min min
2 2 2 2 2 2
min min 0 0
2 2
min 0 2 2 2 2 2 2
min min min min
2 2 2 2 2 2
min min
[ ]
2 log [1 ]
( ) ( )
[ ]
2 log [1 ]
R d
N d d R d d R
Z R d R R d N R N
R d
N d d R d d R
Z R d R R d R
(21)
And finally:
2 2
min 0 2 2 2 2 2 2
min min min min
2 2 2 2 2 2
min min
2 2
min 0 2 2 2 2
min min min
2 2 2 2 2 2
min min
( , )
[ ]
2 log [1 ]
[ ]
2 log [1 ]
M B
R d
N d d R d d R
B e B e
R d R R d R
R d
N d d R d
B e B e
R d R R d R
2 2
min 0 2 2 2
min min
2 2 2 2 2
min min
2 2 2 2 2
min min
2 2 2 2
min min min
2 2 min 2
2 min
[ ]
2 log
2( ) 2
2 1
[3 ]
2
R d
N d R d
B e B e
R d R d R
R d R R d
B e B e B e B e
R d d d
B e B e R
d
e R
B d
(22)
If we use a different law i.e. if we consider: Pi dik with an uniform law and k 2, i.e. we only compensate for the line of sight losses. ThusPi is a random variable with a pdf equal to:
min
( ) k 1 k f x R d
. To characterize the law of Z, we compute the probability:
0
0
0
0
0
1
Proba( ) Proba( ) Proba(1 1 )
1
Proba(1 ) Proba(1 ( ))
Proba( ( ))
1
N
N V
Z x U x x
V U U
U
V N
x x x xV N
U U U
U xV N
x
(23)
Using the distribution of U we can calculate:
0
min
( )
1
0 0 min
min min
Proba( ( ) ) 1 ( )
1 k 1
x xV N
k
k k k k
d
U xV N V du xV N d
x R d R d x
(24)To end the derivation we have to average with the law of V.
min0
0 min
min min
0 min 2
min
Proba( ) Proba( ( ) ) ( )
1
1 ( )
1
1 ( )
1 1
k
k
k
k k k k
R
k
k k
d
Z x U xV N V p V dV
x xV N d dV
R d x R d
x N
V d dV
x x
R d
2 2 min2 0 min
min
min
1 ( )
2(1 ) 1
k k k k k
k k
x N
R d d R d
x x
R d
(25)
We derive (25) and we obtain the p.d.f of Z.
2 2 0
min min
2 2 2
min
2 2
min 0
2 2 min 2
min
2 2
min 0 min
2 2
min
1 1
( ) ( ) ( )
2 1 (1 ) (1 )
1 ( )
( ) ( )
2 (1 ) (1 )
( ) 1( ) ( )
(1 ) 2
k k k k
Z k k
k k
k k
Z k k
k k k k
Z k k
x N
f x R d R d
x x x
R d
R d N
f x R d
x x
R d
f x R d N R d
R d x
(26)
The quantity 1 0
1
i j i
N P P P
is maximum for Pi Rk and Pj dmink , this maximum is equal to
0 2
min k
k k
R N
X R d
and takes its minimum value for Pj Rk and Pi dmink with the minimum value equal to 1 min 0
min k
k k
d N
X d R
. This yields:
2
1
2
1 2
1
2 2
min 0 min
2 2
min
2 2
min 0 min
2 2
min
2 2
min 0 min
2 2
min
( ) 1( ) ( )
(1 ) 2
1( ) ( )
2 (1 )
1 1 1
( ) ( )
2 1 (1 )
X
k k k k
Z k k
X
X
k k k k
k k
X X
k k k k
k k
X
Z x f x dx x R d N R d dx
R d x
R d N R d x dx
R d x
R d N R d dx
x x
R d
min
2 2 min2 0 min 21 2 11 1 1 1
( ) ( ) log
2 1 1 1
k k k k
k k
R d N R d X
X X X
R d
(27)
With :
0
2
min 0 min min
2 min min
2
min 0
1 min 0 min
min
min min
min min min min
1 1 /
1
1 1 /
1
1 / 1 / 1 / /
k
k k k k k
k k
k k k k k k
k k
k k
k k k k k k k k
k k
R N
d N d R d R
X R d d
d N
X R d R N R d R
d R
d d
d R d R d R d R
R R
2 min
min min
1
2 min min
1
log 1 log log 1 / /
1
1 1
log log [ ]
1
k
k k k k
k
k k
k k
X d
d R d R
X R
X d d
X R R
We obtain:
2 2 min min min min
min 0 min
2
min 0 0 min
2 2 min min min
min 0 min
2 min min
1 1
( ) ( ) log [ ]
2 ( )
1 1
( ) ( ) [ ]
2
k k k k k k
k k k k
k k k k
k k
k k k
k k k k
k k
k k
Z
d d R d d R
R d N R d
R R d N R N
R d
d R d d
R d N R d
R d
R d
2 2
min min min
2 min min
2 2
min min
2 min min
2 2
min min
2
min min
min min
( ) 1
[ ]
2
( )
2
( ) ( )
2 2
k k
k
k k k k k
k k k
k k
k k k k
k k
k k
k k k k k k
k k k
k k k
R R
R d d R d
R d R
R d
R d d R
R d
R d
R R d R R d
d R d
d R d
And then:
min
minmin
min min
min
( )
( , )
2
[1 ]
2
[3 ]
2
k k k
k k k
k k
k k k
R R d M B B e B e
d R d
R d
B e d
B e R
d
(28)
