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Submitted on 16 Jan 2019
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Note technique: Throughput Analysis for different power allocation policies with uniform power allocation
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 with uniform power allocation in a random access channel. [Research Report] Xlim UMR 7252; XLIM-SRI. 2018. �hal-01982961�
Note technique: Throughput Analysis for different power
allocation policies with uniform power allocation in a random access channel Asma Selmi, JP Cances, Xlim UMR CNRS 7252
Summary/Introduction:
In this technical note we define a power allocation policy for IoT based user equipment which aims at optimizing the capture effect at the SIC receiver. To benefit from this capture effect we estimate that the power allocation policy is to use a uniform law over a given range of available powers. In this case, we are able to calculate accurate lower bounds for the throughput efficiency in the case of a multiband S-ALOHA based random access channel.
I. Uniform power allocation with SIC capture effect
If we consider the reception of l transmit signals on a given subchannel, we have:
1 1 1 2 2 2 ... i i i ... l l l
yh P s h P s h P s h P s n If we consider user 1 as the user of interest, we have the corresponding SINR:
2
1 1
2 0
2
SINR l
j j j
P h
N P h
(1)We will be able to recover packet 1 if the log2(1+SINR) is greater than the spectral efficiency of the used modulation format denoted as in bits/s/Hz. The probability to decode user 1 is thus equal to:
2
1 1
2 0
2
( , ) Proba(SINR ) Proba( )
cap l
j j j
P i j P h
N P h
(2)We assume a classical Rayleigh distribution for hi2 that means:
2( )
i
x
ph x e (3)
In fact, 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 1
2 0
2 1
2
2 0
2 2 2 2 2 2 2 2
1 1
2 3 2 3
2 1
0 2
( )
( )
Proba( , ,..., ) = Proba( ( ) , ,..., )
= (4)
l
j j
j
l
j j
j
l j j j
l i l
l j j j
N P h
P x
N P h
P
N P h
P h h h h h h h h
N P h P
e dx e
The most difficult task now is to evaluate the probability mass function or the probability density function of the random variable:
2 2 l
j j j
R P h
(5)To calculate: f X( )Proba(R X), we begin with the simplest case l = 2, i.e. R1P h2, we have :
2 2
( ) Proba( 1 ) Proba( ) Proba( / )
f X R X P h X h X P
We calculate:
/
2 / /
0 0
Proba( / ) [ ] 1
X P
x x X P X P
h X P P
e dx e e (6) Then, we have to average using the p.d.f of P: 1P( )
p x L, we obtain :
2
( ) Proba( / ) 1 XZ 1/P( )
f X h X P
e p Z dZ (7) We have to determine the p.d.f : f Z( ) p1/P(1 / ) withP Z 1 /P, we have:2 1/P(1 / ) dP2 P( ) 1/P(1 / ) P( )
p P p P dP p P P p P
P
1/ 2 2
1 1
( ) (1 / )
P P
p Z p Z
Z LZ
min
max 1/
2
min max 1/
( ) 1 1
1 / 1 /
P
XZ P
f X e Z dZ
P P
/
/0 0
1 1
( ) 1 1
L L
X u X u
f X e du e du
L L
We have:
/
2 2
0 1/ /
L
X u tX z
L X L
dt dz
e du e X e
t z
(8)And, this yields:
/ 2 1
/ / /
( / )
z z X L
z
X L X L X L
dz e e dz e
e L E X L
z z z X
With: 1
/
( / )
x
X L
E X L e dx
x
We obtain eventually:
/
/ / 10 0
/
1
1 1 1
( ) 1 1 1 [ ( / ) ]
( ) 1 ( / )
L L X L
X u X u
X L
f X e du e du X Le E X L
L L L X
f X e X E X L
L
To obtain the p.d.f, of R we have to calculate the derivative function:
/
' /
1
/ /
1
1
1 1 1
( ) ( ) ( ) ( ) ( / )
/
1 1 1
( / )
1 ( / )
X L X L
R
X L X L
X e
g X f X p x e E X L
L L X L L L
e e E X L
L L L
E X L L
(9)
We check that:
1
0 0
/
1 0
0
/ /
0 0
( ) 1 ( / )
1 1
[ ( / )]
/
1 [ ] 1
X L
X L X L
g X dX E X L dX L
e dX
XE X L X
L L X L L
e dX e
L
(10)
To calculate the p.d.f of: R2 P h2 22P h3 32, we have to evaluate the convolutional product:
1 1
2 0
( ) ( ) ( ) 1 ( ) ( )
X
R R
u X u
f X p u p X u du E E du
L L L
(11)Before developing further (11), we use the following property:
( )
1
1 1
( )
( )
(1 / )
t u a u
a
x a x x
u u n u
a a a n
n n
n n
x x x
e e e
E x a dt du e du
t u a u a
e e a e
E x a e du e du e a du
u a u u u u
(12)
Then, we have:
a u a
a u a
u
a u
u du e a
du e u e u
du e u
e . [ ] 2 . 2 .
2 2 3 2 3
3 3 4 3 4
2 2
3 3
u u u a u
a a a a
u u u a u
a a a a
e e e e e
du du du
u u u a u
e e e e e
du du du
u u u a u
This yields:
2 3 4
1
2
1 1
2 32 ...
( 1) ( 1)!
... ( 1) !
u a a a u
a a
n a u
p
p
n p
n a
e e e e e
du du
u a a a u
n e e
p du
a u
(13)
We eventually obtain:
1
1
( 1) ( 1) ( 1)!
! !
u p u p n p a
p n
a a n
e e n e
du du
u p u p a
(14)1 1
1
0 1
1
0 0 1
( )
( 1) ( 1) ( 1)!
( )
! !
( 1) ( 1)!
( ) ( 1) ( 1)
! !
u
a n
n
n x
n u n n m x
a n
n
n x m
n u n m x
a n a n n
n
n x n m
E x a e a e du u
e m e
E x a e a du
n u n x
a e m e
E x a e du e a
n u n x
( )
1
0 0 1
( )
1 1
0 1
( )
1 1
0
( ) ( 1) ( 1) ( 1) ( 1)!
! !
( ) ( ) ( 1) ( 1) ( 1)!
! ( ) ( ) ( 1)
!
n u n n n
a n x a m
n x n m
n n n
x a m
n m
n n
x a n
n
a e a
E x a e du e m
n u n x
E x a E x e a m
n x
A a E x a E x e
n x
(15)
( )
1 1
0
( ) ( ) ( 1)
!
X u n
n n L
n
X u X A u
E E e
L L n X
If we use (11) we obtain: (by denoting:1
( 1) ( 1)!
n m n
m
A m
)1 1 1
2 2
0 0 0
1 ( 1)
( ) ( ) ( ) ( )
!
X
X L n X u
n n L
n n
X u e A u
f X E E du u e E du
L L L L X n L
(16)We compute at first:
/ /
/
1 1 1 0
0 0 0
/
/
1 1 1
0 0
( ) ( ) [ ( )] ( )
( ) ( ) ( ) (1 )
X X L X L t
X L
X X L
t X L
u e
E du L E t dt L tE t L t dt
L t
u X X
E du XE L e dt XE L e
L L L
(17)
And then:
1 0
( ) ( )
X u
n L n
J X u e E u du
L (18) We have:/
/1 1
0 0 0
( ) ( )
X u X L X L
n L n t n n t n
n n
I X
u e duL
Lt e dtL
t e dtL K X (19) And:/ /
/ 1 /
0 1
0 0
1
/ /
2 1
/ /
2
( ) [ ] ( )
( ) ( 1) ( )
( ) ( 1) ( )
X L X L n
n t t n X L n t X L
n n
n n
X L X L
n n
n n
X L X L
n n
K X t e dt e t n t e dt X e nK X
L
X X
K X e n e n K X
L L
X X
K X e n e n n K X
L L
/ / /
1 0
0 0
/ / 2
2 2 /
2 0 1
0 0
2
/ /
3 3 / 2
3 0
0 0
( ) [ ] ( 1) 1
( ) [ ] 2 2 ( )
2 2 1
( ) [ ] 3
X L X L X X X X
t t X L t L L L L
X L X L X
t t X L t L
X X X
L L L
X L X
t t X L t
X X
K X te dt te e dt e e e e
L L
K X t e dt t e te dt X e K X
L
X X
e e e
L L
K X t e dt t e t e dt
3 23 2
3 2
3 ( )
3 2 2 1
3 6 6 1
L X
L
X X X X
L L L L
X X X X
L L L L
X e K X
L
X X X
e e e e
L L L
X X X
e e e e
L L L
We deduce:
1 2
/ / /
/ 1 /
1
/ 1 /
0
( ) ( 1)
... ( 1) ( 1)...( 1) ... ( 1) !(1 )
( ) ( 1) ( 1)...( 1) ( 1) !(1 )
( ) ( 1)
n n n
X L X L X L
n
n k
k X L n X L
n n k
k X L n X L
n
k
k n
X X X
K X e n e n n e
L L L
n n n k X e n e
L
K X n n n k X e n e
L K X n
1
/ 1 /
0
! ( 1) !(1 )
( )!
n n k
X L n X L
k
X e n e
n k L
(20)
To check (20), we verify:
/ /
1
2
/ / /
2
( ) 1
( ) 2 2 1
X L X L
X L X L X L
K X X e e
L
X X
K X e e e
L L
1
1 1 / 1 /
0 1
/ 1 1 /
0
( ) ( ) ( 1) ! ( 1) !(1 )
( )!
( ) ( 1) ! ( 1) ! [1 ]
( )!
n n k
n n k X L n X L
n n
k n
k n k k X L n n X L
n
k
n X
I X L K X L e n e
n k L
I X L n X L e n L e
n k
(21)
Then, we have to evaluate:
/
1 1 0
0 0
/ 1
0 1 /
/ 1 1 /
1
0 0
( ) ( ) [ ( ) ( )] ( ) 1
( ) ( ) ( ) ( )
( ) ( ) ( ) ( 1) ! ( 1) ! [1 ]
( )!
( ) (
X u X u L
n L X
n n n
X u L
n n n
X n u L
k n k k u L n n u L
n n
k
n n
u u e
J X u e E du I u E I u du
L L u L
L
X e
J X I X E I u du
L u
X n e
J X I X E L u L e n L e du
L n k u
J X I
/
1
1 1 1 1
1
0 0 0
1 1
1 1 1
1
0 0
! 1
) ( ) ( 1) ( 1) !
( )!
( ) ( ) ( ) ( 1) ! ( 1) ! ( 1)
( )( )! !
X X u L
n
k k n k n n
k
X m
n
k k n k n n m
n n m
k
X n e
X E L u du n L du
L n k u
X n u
J X I X E L X n L du
L n k n k m L
1 1
1 1
1
0 1
1 1
1 1
1
0 1
1 1
! ( 1)
( ) ( ) ( ) ( 1) !
( )( )! !
! ( 1)
( ) ( ) ( ) ( 1) !
( )( )! !
( ) ( ) ( ) ( 1) !
!
n m m
n
k k n k n
n n m
k m
n m m
n
k k n k n
n n m
k m
n m n m m
n n
m
X n X
J X I X E L X n L
L n k n k L m m
X n X
J X I X E L X n L
L n k n k L m m
X n
J X I X E L X
L mm
1 1
1 1
1 1
1
1
1 1
1
1
( 1) !
!
( ) ( ) ( ) ( 1) !
!
( ) ( ) ( ) ( 1) ! ( 1) (22)
!
n
n m n m m
m
n m n m m
n n
m n
m
n n m m
n n
m n
L n X
m m
X n
J X I X E L X
L m m
X X
J X I X E L n L
L m m
1
1 1
1
( ) ( ) ( ) ( 1) ! ( 1)
!
m m
n n
n n m
m n
X X
J X I X E L n
L L m m
It is possible to rewrite it in a more readable manner:
1
1 1
1 1
1
0
( ) ( ) ( ) ( 1) ! ( 1)
!
( ) ( ) ( ) ! ( 1) (23)
( 1 ) ( 1 )!
m m
n n
n n m
m n
k k
n
n n k
k
X X
J X I X E L n
L L m m
X X
J X I X E n X
L L n k n k
We obtain then:
1 1 1
2 2
0 0 0
/
2 1 2
0
/ 1
1 1
2 2
0
1 ( 1)
( ) ( ) ( ) ( )
!
1 ( 1)
( ) ( ) (1 ) ( )
!
1 ( 1)
( ) ( ) (1 ) ( ) ( ) !
!
X
X L n X u
n n L
n n
X L n
X L n
n n n X L n
X L n n
n n n
X u e A u
f X E E du u e E du
L L L L X n L
X e A
f X XE L e J X
L L L X n
X e A X
f X XE L e I X E n X
L L L X n L
0
/
1 1
2 2
0
2
0 0
( 1)
( 1 ) ( 1 )!
1 ( 1) ( )
( ) ( ) (1 ) ( ) ...
! ... ( 1) ( 1)
( 1 ) ( 1 )!
k k
k k X
L n
X L n n
n n
X
k k
L
n
n k
n k
X
L n k n k
X X e A I X
f X XE L e E
L L L L X n
e X X
L A L n k n k
We have:
1
/ 1 1 /
0 0 0
1 1
/ 2 /
0 0 0
( 1) ( ) ( 1) !
( 1) ( 1) ! [1 ]
! ! ( )!
( 1) ( 1) ! [1 ]
( )!
n n n
k n k k X L n n X L
n n n
n n
n n k
n k
n k X L n n X L
n k n
n k n
A I X A n
L X L e n L e
X n X n n k
n L A
A e L e
n k X X
And hence:
/ 2 1
1 1
/ 2 /
1 2
0 0 0
2
0 0
( ) 1 ( ) (1 ) ...
... ( ) ( 1) ( 1) ! [1 ] ...
( )!
... ( 1) ( 1)
( 1 ) ( 1 )!
X L
X
n k L
n k X L n n X L
n k n
n k n
X
k k
L
n
n k
n k
f X XE X L e
L L
X e n L A
E A e L e
L L n k X X
e X X
L A L n k n k
/ 2 1
1 1 1
0 0 0
2
0 0
( ) 1 ( ) (1 ) ...
( 1) !
... ( ) ( 1) [ 1] ...
( )!
... ( 1) ( 1)
( 1 ) ( 1 )!
X L
k k X n
n n n L
n k n
n k n
X
k k
L
n
n k
n k
f X XE X L e
L L
X n L A
E A L e
L n k X X
e X X
L A L n k n k
We obtain eventually:
/ 2 1
1 1 1
0 0 0
2
0 0
( ) 1 ( ) (1 ) ...
( 1) !
... ( ) ( 1) [ 1] ...
( )!
... ( 1) ( 1)
( 1 ) ( 1 )!
X L
X
k k
n
n n n L
n k n
n k n
X
k k
L
n
n k
n k
f X XE X L e
L L
X n L A
E A L e
L n k X X
e X X
L A L n k n k
(24)
With:
1
( 1) ( 1)!
n m n
m
A m
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 (25)
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
(26)
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
(27)
We assume a classical Rayleigh distribution for hi2and hj2that means:
2( )
i
x
ph x e (28)
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
2 2 0
0 0
2 0
2
2 2 2 2
0 2
0
1 1 1 ( )
( ) ( )
( ) 2 2
0 1 0
Proba( ) = Proba( ( ) )
= 1 ( / ) ( ) ( ) 1
( ) ( ( ))
j j
j j j j
j j
i i
j j i i j j j j
j j
t
N P h
N P h N P h L
L N P h
L
j j j j
P h P h P h N P h P h
N P h
E u L du E t dt tE t e
L
e N P h E N P h
L L
(29)In fact, the obtained probability is a conditioned probability since we consider a fixed value of
2 j j
P h in equation (29). To obtain the complete probability we have to average over the distribution of P hj j 2, this yields:
0
0 2
2 0
( )
0 1 0 1
0
( )
0
1 2 1 0 1 2 1 0 1
0 0 0
Proba( )
( ) ( ( )) 1 ( / )
1 ( / ) ( ( )) ( / ) ( ( )) ( / )
i i j j N x L
N x L
P h N P h
e N x E N x E x L dx
L L L
e E x L dx N E N x E x L dx x E N x E x L dx
L L L L L
(30) We have:
0
1 0 1 1 0 1 1 0 1
0 0 0
1 1 0 1
0 0
0
1 1
( ( )) ( / ) ( ( )) ( ) ( ) ( )
[ ( ) 1 ] ( ) [ ( ) 1 ]
[ ( ) 1 ] ( )
N u
L
u u
u
U E N x E x L dx L E N Lu E u du L E N u E u du
L L L
L uE u e E N u uE u e e du
L N u
L
L uE u e E u
0
0
0 1
0
1
0 0
[ ( ) 1 ]
( ) 1 (31)
N u L u
N u
u u
L
e uE u e e du
u
Le E u e du e e du
u
