domain: spi info math

text:
J P Cances
email: cances@ensil.unilim.fr

V Meghdadi
email: meghdadi@ensil.unilim.fr

Simple Precoding Algorithms for Multi-User Relay Communications and Optimized Power Allocation

Keywords: Cooperative Diversity, Diversity Gain, Gram-Schmidt Algorithm, Multi-User Interference Mitigation, Expectation Maximization, Power Allocation Optimization

In this paper we propose new simple precoding solutions to be used at the relay station in the context of a multirelay wireless network where the different mobile stations belong to the same network, i.e. we don't consider the case of competing users from different networks as in some recent works based on game theory. Our proposed algorithm is based on the Gram-Schmidt Orthonormalization since our goal is to mitigate the multi-user interference at each mobile station. The strength of this method is that it only requires the knowledge of all channel impulse responses from a given relay to all the mobile stations. In other words, to compute its precoding vectors, each relay does not need to know the channel impulse responses of the other relays. Using our algorithm, we find that, when compared to the centralized reference method where each mobile station benefit from the same diversity gain, some mobile stations will improve their diversity gain whilst the other ones will suffer from a loss. This constitutes a simple solution to supply different quality of service (QoS) in the case of a multi-services network. We propose then an optimized power allocation strategy at each relay and we give complete accurate performance derivations for the different studied contexts.

I. Introduction

Emerging commercial multimedia applications require reliable wireless transmission links operating at high data rates. However, the limitations such as multipath fading and time variations, make this a challenging task. Cooperative relay combined with Multiple Input Multiple Output (MIMO) transmissions are attractive solutions to both extend the coverage of existing cellular networks and to improve the quality of wireless links. Recent relaying techniques, including cooperative relaying, show to be very promising in order to increase the reliability of telecommunication links and consequently to extend the wireless network coverage.

That is why relaying schemes, especially cooperative networks, have become a very hot research area. In general, the relaying protocols are divided into two main categories [1][2]: Amplify-and-Forward (AF) and Decode-and-Forward (DF). In the AF case relay stations (RS's) amplify the received signal (which is the noised version of transmitted signal), and resend it towards its final destination [3][4]. In the DF case, the RS's decode the received signal then retransmit the information via a suitable processing to destination [5][6]. In the case of successful decoding in a relay, the noise effect from the first link (base station to relay stations) is removed. To improve the performance, maximum ratio combining together with distributed beamforming are used to eliminate multiple access interference (MAI) and at the same time to maximize the signal to noise ratio (SNR) at receivers [7].

In this paper we will concentrate on the link between relays and mobile stations (MS). Since base stations (BS) and relays are fixed, we consider that it is easy to build reliable links between them using for example optical fiber. We will study the configuration where the RS's, using multiple transmit antennas, send the information corresponding to all the mobiles at the same time and at the same carrier frequency to all mobile stations using optimized precoding vectors. The objective is to maximize the signal to noise ratio at each MS and to mitigate the MAI. Furthermore, we will consider the case of slowly non-frequency-selective fading channels between RS's and MS's, enabling the MS's to feed back the RS's with their own Channel Impulse Response (CIR) thanks to a low-rate specific channel.

A lot of works have been already published in the open literature concerning the optimization of precoding vectors at relay places [8][9][10][11][12] for a single destination terminal. The optimization criterion is always the maximization of the SNR at the receive terminal. Among them, the paper of Zhihang & al [8] constitutes an outstanding reference since it copes with different realistic contexts including the case where only second order statistics of the channel are available. However, it does not take into account the multi-user case and one can remark that the majority of work is related to the AF context with only one destination terminal [9][10]. The system in [11] uses the precoding vectors to maximize the system performance of a MIMO single user system whilst [12] tries to use the same technique for the case of two distinct users. In fact, the literature concerning the usage of precoding vectors for the case of arbitrary number of MS's is not rich. One can cite the work of Shu & al [13] where the authors try to maximize the system capacity using MSE based precoders together with Particle Swarm Optimization (PSO) algorithm. However the proposed solution in [10] implies a complicated receiver structure at the MS which limits the practical interest of the work by imposing expensive structure to the end line users. At the same time, due to the complicated equivalent channel equations, very little analytic predictions are produced. In a former version of this study [14], we have proposed Zero-Forcing (ZF) equalization based precoders which offer always the same diversity gain at each receive mobile station. This diversity gain is equal to the number of transmit antennas in the network minus the number of interfering users. This system will serve us as a reference to compare our results with the previous ones.

In the new presented work, we concentrate the complexity on the relay station side by computing precoding vectors which aim at cancelling MAI at MS's receiver side. This is done using the well known Gram-Schmidt orthonormalization process. The strength of this method is that it only requires the knowledge of CIR's from a given relay to all the mobile stations. In other words, to compute its precoding vectors, each relay does not need to know the CIR's of the other relays. Using this algorithm, we carefully study the potential diversity gain at the receivers and we show that, unlike in [14], the diversity gain is not constant at the receiver side and depends on the ranking of the mobile station used in the orthogonalization process.

However, the diversity gain for the first MS's can be equal to the total number of transmit antennas in the network, i.e. it may be superior to the value we obtain in the reference system. Obviously this is obtained at the price of some diversity gain loss particularly for the last MS's included in the algorithm.

Furthermore, we study the power allocation policy to the relays when the total power is assumed limited. We obtain a rigorous mathematical analysis on the Symbol Error Rate (SER) at each MS in both cases i.e. when each MS benefits from optimized power and when the allocated power is chosen randomly. This mathematical derivation is obtained thanks to an accurate approximation of the p.d.f of the optimized SNR using an iterative Expectation-Maximization (EM) algorithm. The obtained results clearly demonstrate the advantage of optimizing the allocated power at the cost of a centralized strategy. This means that, in the case of optimized power allocation, each relay knows all the CIR's from all the relays in the network, adding an additional complexity in the signalling feedback channels. Hence, the contributions of this paper are threefold:

-A detailed study of the diversity gain is given with the corresponding theoretical Symbol Error Rate (SER) expressions.

-An accurate optimization of the allocated power to each relay is proposed with the corresponding BER expressions thanks to an iterative E.M based algorithm to find the SNR distribution.

-In the case of randomly distributed power we are able to derive theoretical SER expressions which enable us to compare with the optimized case.

The rest of the paper is organized as follows. The system model is described in part II together with the orthonormalization algorithm. The diversity gain is studied in Section III and the power allocation policy, together with the theoretical BER derivations, are given in part IV. Simulation results illustrating the performances of the proposed system are discussed in part V. Finally the main results are highlighted in the concluding part in Section VI.

II. System Model

Fig. 1: System Model

The complete system model is depicted on Fig. 1. It includes one base station with M antennas which sends N symbols s 1 to s N to N mobile stations MS 1 to MS N via L fixed relays each equipped with R antennas. A two hop communication scheme is considered. In the first hop, the base station sends the signal to the relays. The relays will then decode the received signal and multiply it by some precoding vectors before transmitting them to mobile stations in the second hop. Since the base station and the relays are considered fixed, the communication between the BS and relays is considered to be perfect. In fact a wide variety of low-noise communication media such as optical fibers may be used. Moreover CRC may be employed to detect any possible errors in relays and to ask the base station to resend the missing information. As a result BS to RS link may be considered error-free. In the remainder of this paper we will focus on the second hop of the communication where L relays cooperate in sending each of the N data symbols to their intended mobile ). Since the BS to RS links are considered to be error-free, we will assume that by the end of the first hop s is received correctly by all relays. Each relay will then multiply its received signals by a set of precoding vectors j i w each of size R × 1. The relays then will send these signals to the MS's; the received signal at mobile station j can be written as: , we obtain the following writing form: [ , ,..., ]

i i i i N R N   W w w w and i = 1,…,L (4)
It is then possible to use a global matrix model since we can write:

.

1 1 1,1 2,1 ,1 1,1 2,1 ,1 2 1,2 2,2 ,2 1,2 2,2 ,2 1 1, 2, , 1, 2, , 1 . . ..
.

N R L N jj j T T T T T T LL N T T T T T T jj LL j T T T T T T N N L N N N L N N L jj j s s s                                                           

The reference system:

In the reference system, using equations (3) or ( 5), the goal is to compute matrix W in order to obtain a diagonal matrix for: .

HW. The jth element on the diagonal of HW determines the signal to noise ratio of the jth MS. In this paper we focus at first on the case where all MS's have the same SNR (i.e HW = g.I N with g being a positive real). We have in fact two possibilities depending on the fact that each relay knows completely the CIR's from all relays to the MS's or not.

Case 1: each relay knows all the CIR's from all relays to the MS's.

In this case, the system is mathematically equivalent to a relay with R.L antennas and we find the same behavior as a MIMO system with R.L transmit antennas. As stated before, in order to cancel the MAI while imposing equal received power for all MSs, we must find a precoding vector W that satisfies: .. ..

N L R L R N N g   H W I (6)
For simplicity reason and without loosing any generality we assume g = 1 in the following. It is obvious from (8) that the jth element of y depends only on s j and not on other data symbols. As a result the system can be considered as N parallel channels each one transmitting a symbol s j .

Case 2: each relay knows only its own CIR When each relay knows only the channel coefficients of the links between itself and the MS's, the precoding vector of relay i i W must be calculated only as a function of the CIR between relay station RS i and mobile station MS j . That means that for all i = 1,…,L we have:

..

i i i N g  H W I 1,..., iL  (9)
The equation ( 9) has a solution if H i is full rank (i.e. if the number of antennas of each relay is greater than the number of receivers: R ≥ N). Of course this criterion is much more difficult to obtain than its counterpart in the first case, but, on the other hand, relays do not need to be inter-connected to each other. Precoding vectors of the ith relay are given by: †

† 1 † † 1 ( . ) . ( .
) i i i ii i i i P    H H H W H H H ( 10
)
With P i satisfying the overall constraint power:

1 L i i PP   
. This time, we obtain for the received signal:

1 L i i   yn y (11) With 1 N i  y
is the vector contribution of ith relay in the received signal and is given by: † † 1 .

( . ) . ).

i i i i i P   s H H H y . ( . . ). ( . . ). ( . . ). ( .
T T T T T T T T y s s n y s s n          

h w h w h w h w h w h w h w h w (12)

The goal, in order to compute precoding vectors 1 .

T hw + 2 2,1 1 . T hw , 1 1,2 2 . T hw + 2 2,2 2 .
T hw . This optimization problem is solved using the orthogonal projection theorem and yields to the following solution:

* * * * 1,2 2,1 1,1 2,2 1,1 2,2 1,2 2,1 * * * * 1,2 2,1 1,1 2,2 1,1 2,2 1,2 2,1 * * * * 1 2 1 2 2 1 1 2 * * * * . . . . , , , . . . .                     h h h h h h h h h h h h w w w w h h h h (13) With: 1 .( . ) . HH     X I X X X X . Using definitions (13), it is straightforward to obtain: * 1,2 1,1 ** 1,2 1,2 1,1 1,1 * 1,2 1,1 * * * * 1 * 1,1 1,1 1,1 1,1 1* 1,1 2 2 1,1 2 1,1 1,2 2 ** * * 1 1,1 1,1 1,1 1,1 1,1 1,1 * 1,2 2 * . ( . (

h I h h h h h w h h h hh h h h h h h h h (14)

And similarly, we obtain: It is possible to generalize the former situation to our proposed cooperative relaying system of Fig. 1. We have the following equation:

11 , 1 1, 1 2, 2 , , 1 1 1 1 2 2 2 2 1, 1 1 2 2 2, 1 1 2 2 , 1 1
L T ii k i k k T T T T i i i k i k L i L i T T i i i i N N i i i N N T k k k k i i i yn yn y s s s s s s s s s s s                             hx h x h x h x h x h w w w w h w w w w h w w w , 1 1 2 2 1 2 1 2 1, 1 2, 1 , 1 , 1 1 1, 2 2, 2 ,
k T L L L L N N L i i i N N i T T T j T L T T T j T L i i i j i L i i i j i L i TT i i i i j s s s s s n y s s                           

w h w w w w h w h w h w h w h w h w h w h w h w h w h

12 , , 1, 2, , , . ..
T j T L T T T j T L i i L i i i i N i N j i N L i N N i s s n          w h w h w h w h w h w (15)
Mitigating the MAI implies:

l k l k l k l k l l l k l k l N l k l l l k l k l N       

hh w h h h h h h h h h h h h

Where the notation

l l l k l k l N  h h h h h .
The computation of this solution is easily done when we use the orthonormalization process of Gram-Schmidt.

The obtain the computation of

lk l l l k l k l N   h V h h h h h is R×(N-1) or 1,1 1,2 1, 1 1,
( , . ) / , . lj lj m l m l j mm j l m l m l j l j l j l m l j l j l j m jj l m l j j Nk l N l N l j l j l j l N l j l j l j jj                       e e 1 ,, 11 1 , , , , , , , , , 1 11 ,, 1 11 , , , , , , , , , 11 proj ( ) ( , . ) / , . proj ( )

hh e h e h e h e h e hh e h e h e h e h e

Using this new vector family it is straightforward to express the precoding normalized vectors. We find in this case: ,.

** ,, ** , ,1 ,2 , 1 , 1 , , ,1 ,2 , 1 , 1 ,
l k l k l k l k l l l k l k l N l k l l l k l k l N N l k l k l j l j j j k l k N l k l k l j l j j j k lL          12 1, 2, , , * * * * 1, 1, 1, 1, 2, 2, 2, 2, 1, 1, 1, 2, * * * * 1, 1, 1, 1, 2, 2, 2, 2,
T T T k T L i i i i i k i i L i i i i N N i i j j i i j j j j i j j i T T i i i N N i i j j i i j j j j i j j i y s n y                     

 h w h w h w h w h h e e h h e e h h h h h e e h h e e

,. ,.

N L i L i L j L j j j i T L i i i N L i L i L j L j j j

R l i l i l j l j j

 

h h e e . We obtain then:

* * * * 1, 1, 1, 1, 1, 1, , , , , , , 1 1, 1 1, 1, , * * * * 1, 1, 1, 1, , , , , 1, 1, , . , . , . , . ( .
... . ).

, . , .

R N R N i j j i j j L i L j L j L i L j L j j j j i j j j i TT i i L i i i NN i i j j L i L i L j L j j j i j j i y s n                    

 h e e h e e h e e h e e hh h h e e h h e e

And (assuming that each precoding vector has a normalized power equal to one for simplicity):

* * * * 1, 1 , 1, 1, 1, 1, , , , , , , 11 1,
, * * * * 1, 1 , 1, 1, , , , , 1, 1, 22 1, 1, 1, 1, 1 1, , . , . , . , . ( . ... . ).
, . , .

,, [

RR i i i i j j L i L i L i L i L j L j j N j N TT i i L i i i NN i i j j L i L i L j L j j j i j j i R i i i j jN i i y s n y                   

 h e e h e e h e e h e e hh h h e e h h e e h e h e h

, .

, .

[

[ , . ] ... [ , . ]]. R L i L i L i L j jN ii NN i j j L i L i L j L j j j i j j i NN i i i j j L i L i L j L j i i j j i j j i sn y s n                      

 h e h e h e e h h e e h h e e h h e e (17)

Remark: To derive the precoding vectors we have supposed that the vector family ,1 ,2 , 1 , 1 , ( , ,... , ,..., )

l l l k l k l N  h h h h h is free.
If this condition is not met, we can always extract a free family vector from the subspace: ( , )

lk h V . Denoting ( , ) ( ) ,1 ,2 , 1 , 1 , ( ) vect( , ,... , ,..., ) lk N l l l l k l k l N l   h, V h h h h h with () N l N 
this new free family vector, we can build a new orthonormalized vector family using the Gram-Schmidt algorithm.

III. Diversity gain study

3.1: Diversity gain study:

To obtain an accurate estimate of the potential diversity gain, we have to characterize the random variables:

2 22 2 * * , , , , , , , , , 1, 1 , .
, ,

NR k i k i k i k j k j k i k i k i k j j j i j N         

 h h e e h e h e

. To do this, we come back to the Gram-Schmidt algorithm, we have:

kj kj i k i k i i j k i k i k i k j k i k j i j k i k i j               
kj ki i k i k i j       e hh
. Then, using the orthogonal basis, we have:

, , , , 1 ,
. R k i k j k i k j j   h e h e . Putting
this into (18) this yields to:

1 , , , , , , , , , , , , 11 .( , . , . ) .
, .

R i R k i k i k j k i k j k j k i k j k i k j k i k j j j j i           
 . 2 11 2 2 2 , , , , , , , , , , , , 11 .
, , .

.

[ , , ]

ii

k i k i k i k i k j k i k j k i k i k i k j k i jj         

 h h e h e h h e h (20)

With equation ( 20) we can now obtain the characterization of 2 , ki  by induction.

For I = 1, we have:

,1 ,1 ,1 / k k k  e h h so: 2 2 2 ,1 2 ,1 ,1 ,1 ,1 ,1 ,1 ,, k k k k k k k     h h e h h h . Since
,1 k h is a vector of R complex Gaussian random components each having a null mean value and a variance equal to 0.5, the random variable 2 ,1 k  may be written as:

2 2 ( ) ( ) 2 ( ) 2 ,1 ,1 ,1 ,1 11 [Re( ) Im( ) ] RR i i i k k k k ii h h h     



. It is straightforward to conclude in this case that: 2 ,1 k  is a chi-square variable with 2.R degrees of freedom.

For I = 2, we have :

,2 ,2 ,1 ,2 ,1 ,2 ,1 ,2 ,1 ( , . ) / , . k k k k k k k k k    e h e h
k k k k k k k k k k k k k k k k k k k k k k k kk k k k k k k k k k k               

h e h h e h e h e h e h h e h e h h e h hh h h h h h h h h (21)

To calculate the term:

)] R R I R I k k k k k k i R R R I I R I I R k k k k k k k k i h i j h i h i j h i h i h i h i h i j h i h i h i h i            hh ( 22
)
And this yields to:

22 2 ,1 ,2 ,1 ,2 ,1 ,2 ,1 ,2 ,1 ,2 1 2 22 ,1
,1 ,1 1 [ ( ,:). ( ,:) ( ,:). ( ,:)] [ ( ,:). ( ,:) ( ,:). ( ,:)] ,

[ ( ,:)

( ,:) ] R R R I I R I I R k k k k k k k k kk i R RI k kk i h i h i h i h i h i h i h i h i h i h i          hh h
Taking the mean of this expression and using the approximation [8]: ,:).

R R R I I R I I R k k k k k k k k kk i R RI k kk i R R R I k k k k i E h i h i h i h i h i h i h i h i E E h i h i E h i h i E h i h              hh h 222 ,2 ,1 ,2 ,1 ,2 22 ,1 ,1 1 ( ,:) ] [ ( ,:). ( ,:) ] [ ( ,:). ( ,:) ] [ ( ,:) ] [ ( ,:) ] 1 I R I I R k k k k R RI kk i i E h i h i E h i h i E h i E h i      (23)
This entails that the random variable 2 ,2 k  is a chi-square variable with 2.(R-1) degrees of freedom. Let then suppose that the random variable 2 ,1 ki   is a chi-square random variable with 2.(R-(i-2)) degrees of freedom. In this case, we have:

.[ , , ] .[ , , , ] ii k i k i k i k i k j k i k i k i k i k j k i k i k i jj           

 h h e h h h e h e h (24)

Taking into account the former hypothesis on 2 ,1 ki

  we know that 2 2 , , , , 1 ,, i k i k i k j k i j     h h
e h is a chi-square variable with 2.(R-(i-2)) degrees of freedom, and subtracting the quantity 2 , 1 , , k i k i  ehwe will find exactly in the same way as in equation ( 23) that 2 , ki  is a chi-square random variable with 2.(R-(i-1)) degrees of freedom.

Conclusion:

We have proved by induction that the random variable 2 , ki  is a chi-square random variable with 2.(R-(i-1)) degrees of freedom.

It is possible to check this property using the EM algorithm [15]. We want to identify the pdf of random variables 2 , ki  and we guess that Gamma law will be the best approximation. In this case, the pdf of 2 , ki  will be approximated by: 26) is clearly of prohibitive complexity. The EM algorithm [16][17] is an iterative procedure for solving this ML estimation problem. In the EM algorithm the set of observation data i  is termed as incomplete data. Starting from some initial estimate (0)  , the EM algorithm solves the ML estimation problem (26) by the following iterative procedure:

- Expectation Step (E-step): Compute -   () () ( ) log ( ) i i Q E p  X      (27) - Maximization Step (M-step): Solve ) ( max arg ) ( ) 1 ( i i Q       (28)
We define then the following hidden data  

, 1,..., i in



Zz

where i z is a J-dimensional indicator vector such that:

, 1, s ( , )

0 otherwise i j j ij ig z         (29)
The complete data is hen defined as: ) , ( Z X  

. We have:

, 11 ( , ) . ( ; , )

ij nJ z j i j j ij pg            Z (30)
The log-likelihood of the complete data is then given by: ,, 1 1 1 1 log ( , ) .log .[ .log( ) log( ( )) ...

( 1).log( ) .

n J n J i j j i j j j j i j i j j i j i p z z C                         Z  ( 31
)
Where C is some constant. The E-step can then be calculated as follows:

  ' ( ') log ( , ) Q E p   Z   ,, 1 1 1 1 ˆ( ')
.log .[ .log( ) log( ( )) ...

( 1).log( ) .

n J n J i j j i j j j j i j i j j i j i Q z z C                         (32)
With :

    ' ' ' , , , '' ,, '' ,, 1 ' ' ' ' ' ' 1 ˆ,1 ( 1). ( 1)
( 1). ( 1) ( ; , ). ( ; , ).

i j i j i j i i i j i j J i i l i l l i j j j J i j j l l z E z P z p z P z p z P z                             (33)
For the M-step, we have to compute the following derivative functions:

nn j i j j i i j n ii i j j i i j j j zz p z                   Z 
Setting the above derivatives to zero, we obtain the set of non-linear equations: ( ( ) log( )).

.. log ( , ) 0

nn j i j j i i j ii j n j i i j i j n j ij i p zz z p z                            Z Z   (35)
Using the first equation in (35) we can express  j as :

, 1 , 1 ( ( ) log( )). log( ) n

j i i j i j n ij i z z           , so we obtain:            n i j i n i j i i j j n i j i j i n i i j j z z z z 1 , 1 , 1 , , 1 ˆ. . ˆ)). log( ) ( ( ) ( log        (36)
Substituting the value j  from the first equation into the second one, we obtain the following equality:

                 n i j i n i j i i n i j i j i n i n i i j i j n i j i n i j i i j j z z z z z z z 1 , 1 , 1 , , 1 1 , 1 , 1 , ˆ. ). / ] . ) log( ). ( [ exp( ˆ. .        (37)
We obtain an equation of the equivalent form:

() x f x 
which can be easily solved using the well known Newton-Raphson method.

For example, we test the E.M. based algorithm with J = 1 (Gamma pure law) to test the behavior of variables 2 , ki  . We take R = 10, N = 5 and we found the following result for 2 ,1 k  on Fig. 3 andfor 2 ,2 k  on Fig. 4 (green points denote experimental trials while blue ones represent the EM based approximated pdf). The curve on the right is obtained in log-log scale to compare the slopes of the two curves.  follows a chi-square law with 2.(R-(i-1)) degrees of freedom (since each degrees of freedom contributes to 0.5 in the diversity order D, the diversity order is equal to

( 1) 1 D R i     ). Using this result, we are now ready to compute the potential diversity gain at each MS. We have the relationship:

, 1 [ ]. L i k i i i k y s n     (38)
The SNR is then defined as:

2 2 , , , , 1 1 22 . [] L L k i k i j i ki k k j k k i              . It is clear that the random variable 2 1, 1 L ki k Z     is a chi-square variable with 2.L.(R-(i-1)) degrees of freedom. The variable Z , 1 L ki k   
 is a sum of independent identically distributed (i.i.d) Rayleigh variables and it is difficult to obtain an accurate estimate of its p.d.f [18], so the study of the p.d.f of i  is a complicated task too. In appendix I, we give a proof by induction that i  has a diversity order equal to: L.(R-(i-1))-1. To confirm this, we use the E.M based algorithm described just before with a pure gamma law with p.d.f : When we assume the solution exhibited in (17), we can define the average transmitted power of each relay as:

* 11 ** 1 1 1 [ . ] [( . ) ( . )] [ . . ] [ . . ] [ . ]. [ . ] [ . ] NN H i H i i H i i i i j j j j j j k k j j j k N N N i H i i H i i H i i k k k k k k k k k k k k k P E E s s E s s P E s s E E s s E N                      x x w w w w w w w w w w (39)
Hence, when we normalize the transmitted power of each relay to one:

1 i P    ,
k i k i N  
(see equation ( 20)).

In the case where each relay k has a transmit power k P the received signal is written as:

P P C                  (43)
Consequently, the cost function can be obtained as:

NL k k k i k ik NL k k i k k i ik J P N erfc P P N P P                             (45)
Setting this derivative equal to zero, we obtain the equation:

' ' 2 2 ,, 11 (2. / . . . ). .ex p[ .( . ) / ] [1, ] NL k k i m m i im N P P k L                (46)
This yields to:

' ' 2 2 ,, 11 .exp[ .( . ) / ] NL k k i m m i im PP        (47)
So, we can write:

2 ' ' 2 2 2 ,, 11 .[ .exp[ .( . ) / ]] NL k k i m m i im P A P        (48)
Rewriting equation ( 47) for each relay station and summing all the different equalities, we arrive at:

' '2 ' 2 2 , , , 1 1 1 1 . . .exp[ .( . ) / ] L L N L m m i m i m m i m m i m P A P              (49)
Using equation ( 47) we aim at first at calculating:

' , 1 . L i m m i m XP     . X
i is the solution of the equation:

'2 2 2 , 11 . .exp[ . / ] LN i m i i mi X A X       (50)
The constant A may be expressed as:

1 L kT k PP   .[ .exp[ .( . ) / ]] . [ .exp[ .( . ) / ]] . [ .exp[ . / ]] / [ .exp[ . / ]] L N L k i m m i T k i m L N L k i m m i T k i m LN k i i T ki LN T k i i ki A P P A P P A X P A P X                                   (51)
Finally, we obtain the equation:

'2 2 2 ' 2 2 2 ,, 1 1 1 1 . .exp[ . / ]/ [ .exp[ . / ]] 1,..., L N L N i T m i i k i i m i k i X P X X i N                 
It is possible to solve this set of non-linear equations using the primitive function fsolve. As soon as the quantities i X are found it is possible to compute the optimum power

P k : 2 2 2 , 11 [ .exp[ . / ]] . [ .exp[ . / ]] N k i i i kT LN k i i ki X PP X              (52)
The calculation presented here is complicated to implement. To simplify it we can use an approximation at low SNR's. The solution found for the low SNR regime can be generalized at high SNR's as the simulation results highlight it on Section V. If we consider the low SNR regime we have

' 2 2 , 1 exp[ .( . ) / ] 1 L k k i k P      
and equation ( 48) reduces to:

2 ' 2 , 1 .[ ] N k k i i PA     (53)
And the constraint on the overall transmit power yields to:

2 ' 2 2 , '2 1 1 1 , 11 . [ ] [] L L N T k k i T LN k k i ki ki P P A P A               (54)
And we found:

2 ' 2 ' 2 ,, '2 11 , 11 .[ ] .[ ] [] NN T k k i k i LN ii ki ki P PA        (55)
In this case, the received signal at MS I takes the form:

' , 1 ' 2 ' ,, '2 11 , 11 ( . ). ( .[ ] . ). [] L i k k i i i k LN T i k j k i i i LN kj kj kj y P s n P y s n            '' ,, 11 '2 , 11 [ ].
.

[] LN k j k i kj i T i i LN kj kj y P s n           (56)

Theoretical SER performance derivations:

We consider here two cases: the first one corresponds to the case of equal uniform power allocation whose diversity gain has already been studied at the end of Section III and the second one concerns the case of power optimization just given in the last section.

In the first case, we have demonstrated that a gamma law with p.d.f:

1.

( , , ) . . ()

x x x e          
was a good approximation of the random variable:

'2 , 1 [] L ki k    with: 1 . 1 .( ( 1)) 1 i L q L R i         and 1/ L   . The SNR is given by: '2 , 1 2 [] L ki k i       with /10 2 .10 dB SNR u P    and: '2 , 1 [( ) ] L u k i k PE     . We have: 1 . . 1 0 0 0 . . . . . . . . . . ( ) ( ) ( ).
1 . ( 1) . ( ).

x x u u P x x e dx x e dx u e du L                                        
Its p.d.f can be well approximated by:

22 22 2 2 2 1 . . 1 . . .. ( ) . ( . , , ) .[ . ] . . . ( ) ( ) xx h x x x e x e                             (57)
The average Symbol Error Rate (SER) at a given SNR or 2  value is then given by:

2 2 1 . . 00 2. . . ( ). ( ). . . . (

. ). ()

u ee P P u h u du u e erfc u du                (58)
We have

1 1/ , 3/ 2.( 1) MM      for M-QAM constellations with k M 2  .
                       (
                       and 1 w . . t d t e dt  



. The primitive function of w is defined as:

1 0 ( 1)! w( ) . . ! kt k t t e k        
. We obtain then:

                                                       (60)
To calculate integral form:

2 1/ 2 (1 / . ) 0 .. kt t e dt         , we set: 2 (1 / . ). xt     i.e. 2 (1 / . ). dx dt     : 2 1/ 2 (1 1/ ) 2 1/ 2 2 00 2 1/ 2 1/ 2 0 2 1/ 2 . . [ /(1 / . )] . . /(1 / . ) 1/(1 / . ) . . . 1/(1 / . ) . ( 1/ 2) k t k x k k x k t e dt x e dx x e dx k                                    With:             dt e t z t z . . ) ( 0 1
, we have the following particular values:

2 1 1 3 2( 1) 1 2( 1) 1 (2. )! ( ) ( ).( )...( ). ( ) . 2 2 2 2 2 2 . ! n n n n n n n n n n              
This yields to:

2 1 1/ 2 (1 1/ ) , 22 0 0 1 2 2 2 2 1/ 2 0 1 2 2 2 2 2 1/ 2 0 ( 1)! 1 .[1 . . . . ] [ . ] . .! ( 1)! 1 (2. )! .[1 . . . ] [ . ] . 2 . ! . !.(1 / . ) ( 1)! (2. )! .[1 . ] [ . ] . 2 . ! .(1 / . ) kt k k k k kk k I t e dt k k k k k k                                                         (60)
And, finally we have:

2 , 2 1 2 2 2 2 2 1/ 2 0 1 2 2 2 2 2 0 2. . . . () 2. . . ( 1)! (2. )! . .[1 . ] ( 1)! [ . ] . 2 . ! .(1 / . ) 1 (2. )! 2. .[1 . . ] . 2 . ! .(1 / . ) / . 1 e kk k kk k PI k k k k                                                 (61)
At high SNR's it's possible to simplify the expression of: e P , we have:

22 2 2 / . 1 . 1 / . 1 2. 1 . /                  and 2 2 1. ( / . 1) kk kk       
, this entails the new expression: (2. )!. . 2. .[1 ( 1). ] 2.

2 . . !

kk e kk k k P k              (62)

Conclusion:

In the case where the relay radiates the same power, we have:

1 , 2 2 2 2 2 0 1 (2. )! 2. .[1 . . ] . 2 . ! .(1 / . ) / . 1 e equal kk k k P k                 
In the case of power optimization, we have: We can conclude experimentally that the variable  I is a chi-square random variable with diversity order:

' [ ( 1)]. 1 ii D R i L q      . Its mean value is equal to: '' ' 1 ' ' 00 ! 11 . . .exp( ). . .exp( ). ( 1)! ( 1)! ( 1)! ii qq i ui i i i q P x x x dx x x dx q q q q            



In this case, it is straightforward to obtain the theoretical BER performance of this scheme. We have now:

1 , 2 2 2 2 2 0 1 (2. )! 2. .[1 . . ] 2 . ! .(1 / ) /1 e opti kk k k P k              (63)
It's in fact the same expression as in (61) but with: 1   . Obviously, it's easy to check that: ,, e opti e equal

PP 

.

Remark: There is still an interesting case to study. This situation is related to the case where we consider precoding vectors which are no more normalized. We have then:

** , , , , 1, ** , , , , , , 1 1, ** , , , , , , 1 ,. , . , .
, . , .

N l k l k l k l j l j j j k RN l k l j l j l k l j l j j j j k R l k l k l k l k l j l j jN            w

h h e e h e e h e e h e e h e e

We have: * ,, ,0

l k l j  he for jk  . Hence: * , , ,
,.

l k l k l k l k  w
h e e and the average transmitted power of one relay station is written as:

*2 ,, 11 [ . ] [ | , | ] NN l H l l k k l k l k kk P E E         w w h e . Since each random variable *2 ,, | , | l k l k he is a chi-square random variable with 2.( ( 1)
) Rk  degrees of freedom, we have:

2 1 * ,, 00 ! 11 [ . ] . . .exp( ). . .exp( ). ( 1)! ( 1)! ( 1)! kk qq k l k l k k k k k q E e x x x dx x x dx q q q q                h With: ( 1) k q R k    .
Finally, the transmitted power is equal to:

11 [ ( 1)] ( , ) NN lk kk P q R k R N           (64)
In this case, to have an average transmitted power equal to one, we have to use precoding vectors of the form:

** , , , , , , 1 , . , .
( , ) [

( 1)]

l k l k l k l k l k l k l k N k RN Rk      h e e h e e w ( 65
)
Considering a non-uniform power allocation between the different relays, the received signal is equal to:

2 2 2 1 1, 1 , 2 , , , , 1 [ . , ... .
, ... . , ]]. ( , ) i i i k i k i L L i L i i i y P P P s n RN        h e h e h e ( 66
)
Since the derivations are heavy to obtain, we will simplify at first the studied context and we'll take: L = 2. We have then to characterize the p.d.f of the random variable:

i ii i ii q i ZZ qq i q qq q TF p x g v q j v j v j v j v                  (68)
The decomposition of this fractional function yields to:

i ii i i ii q Z qq q q kk q k q k k gv j v j v ab j v j v                      We found: 1 1 2 1 2 1 2 1 1 1 12 ( 1) . .( 1)...( 1) ( 1)! ( 1) ( 1) .. ( ) ! ( ) ( 1)!. ! ( ) ( 1) .
()

i i i i i i i i q k kk qk i i i i k q k q k q k i q k qk k qk C q q q k q k a k q k C b                                  Remark: if: 21     , we obtain directly: 2 2. 2. 1 ( ) . ( .2. . ) i i q Z q gv jv    and: 2 2. 1 . 2. . ( ) . (2. 1)! i i q x q Z i xe px q    
In the case where:

21

   , we suppose that 21    without any loss of generality. This gives:

2 11 1 11 12 0 2 1 1 1 2 2 1 1 1 2 1 0 2 1 1 1 2 ( 1) . ( 1) . ( ) ( . ) .[ ] ( ) .( .2. . ) ( ) .( .2. . ) 11 ( . ) .[ ( 1) . .( ( ) .( .2. . ) ( ) . ii i ii i i i i i i ii i i i i qq kk q q k q k q Z q k q k q k q k k q qq k qk q k q k q k k CC gv j v j v C jv                                                   2 )] ( .2. . ) i qk jv   
and:

1 2 2 1 2 1 1 1 . . 1 1 2 1 0 2 1 1 2 1 1 . 1 1 2 1 0 21 1 . 1 . ( ) ( . ) .[ ( 1) . .( . . )] ( ) ( 1)! ( ) ( 1)! ( ) ( . ) .[ ( 1) . . ( ( 1)
! ( )

i i i i i i i i i i i i i q q k q k x x qq k Z q k qk qk k i i q q k x qq k Z q k qk k i x e x e p x C q k q k x e e p x C qk                                                 2 . 12 )] () i x qk     
We have to calculate the p.d.f of:

22 2 / Z   , this yields to: 22 2 ( ) ( . ). /(2. . ) ( . ). /(2. ) ZZ f u p u u p u u        2 ( ) ( . ). /(2. ) Z f u p u u    (68) 12 1 ( ) / 2 . . . . 1 1 1 2 1 0 2 1 1 2 . ( ) ( . ) .[ ( 1) . . ( )] 2.( 1)! ( ) ( ) i i i ii i ii q q k q k uu qq k qk q k q k k i u e e f u C qk                            (69)
The computation of the average symbol error (SER) probability yields to:

i i i ii i ii e q q k q k uu qq k e q k q k q k k i P erfc u f u du u e e P erfc u C du qk                                   1 2 1 ( ) / 2 1 . 1 1 2 1 0 21 0 ( ) /
i i i ii i i i i q q k q k u qq k e q k qk k i qk u qk ue P C erfc u du qk ue erfc u du                                       (70) 12 1 ,, 1 1 2 1 0 2 1 1 2 . .( . ) . ( 1) . . [ ] ( 1)! ( ) ( ) ii ii ii i ii q k q k q q k qq k e q k q k q k k i II PC qk                         (71)
To obtain the final result, we have to calculate integrals of the form: ).

( ) / 2 1 . . , 0 . . ( .
q k q k u I u e erfc u du          
We set:

q k q k x q k q k x q k I x e erfc x x dx I x e erfc x dx J                         
We have now to calculate integrals of the form:

                                                With: .( / 2
                                        With: 1 0 [ , ]
. .

x
n mk m n n k inc m m k n k n kn kn mk m n n k inc m n k n kn me J m C k n k n k me J C k n k                                            2 / 4. , ( 1) / 2] kn  And finally: 12 1 . , . , 1 1 2 1 0 2 1 1 2 2. .( . ) . ( 1) . . [ ] ( 1 )! ( ) ( ) ii ii ii i ii q k q k q q k qq k e q k q k q k k i JJ PC qk                            (72) With:   2 / 4. 1 2 , ( ) / 2 00 ( 1)! 1 . .[1 . . .( 1) . . [( 1) / 2] [ / 4. ,( 1) / 2] ! 2 . kn mk m n n k inc m n k n kn me J C k n k n k                   



We need to calculate the average power at the receive side which is equal to:

i i ii i q q k xx qq k qk q k q k k i q q qk q q k x k qk k i x e e E Z x C dx qk C x e dx qk                                                2 1 .
12 0

1 1 1 2 1 22 0 2 1 1 1 2 2 22 1 1 2 2 1 1 2 . . . ) 11 ( . ) . ( 1) .
.

( ).( 1).[ ] (

) . ( ) .

.

( 1) . . ( 1) .

.

( ).( 1).[ (

i i i ii i i i i i i i i i i qk x qk q qq k q k i i q k q k q k q k k q q k q kk q k q k i i x e dx C q k q k C q k q k                                                    1 0 1 ] ) i i q qk k     1 22 1 2 1 2 2 1 21 0 21
.

( 1) .

. [ ]

( 1) .

.

( ).( 1).[ ] ()

i i i i i i i q q k q q kk q k q k i i qk k E Z C q k q k                       (73)
With:

1 1 2 2 1/ , 1/ PP     .
It is possible to generalize this result for a number L of relays. We have in this case the random variable:

2 2 2 1 1, 1, 2 2, 2, , , . , . , ... .
i LL i i i q Z Z L q q q L TF p x g v j v j v j v               (75)
With:

1 1 2 2 1/ ,
i i i i i q L q k k k ZL q k q k q k k L a a a gv j v j v j v                       (76)
To obtain the decomposition of 2 () Z gv into simple elements, we use the following property. If we consider the following function (obviously .2. .

x j v   in our case) 12 1 () ( ) .( ) ...( ) i i i L q q q L Fx x x x        (77)
We have:

1 1 1 1 ( ) . ( ) ( ) ( ) ...

( ) .( ) ...( )

i i i i i q i L L q q q q i i L x F x G x x x x x              (78)
Using the Taylor expansion of ()

L Gx for x around i  , we obtain: 2 ' '' ( ) ( ) ( ) ( ) ( ) ( ). ( ) . ( ) ... . ( ) ... 2! ! n n ii L L i i L i L i L i xx G x G x G G G n                 (79) With: () () ( ) [ ] i n n L L i x n d G x G dx   
; in this case, we obtain:

(

1) ' '' ( ) 12 ( ) ( ) ( ) ( ) ( ) ( ) ... ... ( ) ( ) 2!.( ) !.( ) ( 1)!.( ) i i i i i q n L i L i L i L i L i L q q q q n i i i i i i G G G G G Fx x x x n x q x                           
To calculate these derivative values, we can use the following writing form:

1 1 1 1 1 1 1 2 1 1 1: , 1: , 1 1: , 1/ ( ) ( ) ...( ) .( ) ...( ) [ ( )] [( )...( ).( )...( )] [ ( ). ... ( ). ... ( )] i i i i i i i q q q q q L i i L L q i i L q L L L k k L k i k k L k i L k L k i G x x x x x P x x x x x x x x                                             (80)
With:

1: , 1 1 1 1 ( , ,..., , ,..., ) k L k i i i L            and 12 12 1: , 1 ... , ( ) . ... k kk k k L k i i i i i i i L i i               . Using (80) it is not difficult to obtain the derivative values () () n Li
G  . We have, for example:

1 1 1 1 () [( )...( ).( )...( )] i Li q i i i i i L i G                11 1 2 1 ' ' 1 1 1 1 1 .. ... , ' 1 1 1 1 .[ ( )] () [( )...( ).( )...( )] ( 1) . . [( )...( )...( )] () [( )...( ).( )...( )] i i kL k L k i i L x Li q i i i i i L i n i i i i i i i i i i i L i i Li q i i i i i L i q P x G q G                                                       For the derivation of () () n Li
G  , we have the following relationships:

' '' ' ' '' 1 1 2 '' ' '' 12 (3) (3) 1 . ( ) ( ) ( ) ( ) 1/[ ( )] , ( ) , ( ) .[ ( 1) ] [ ( )] [ ( )] [ ( )] ( ) ( ) ( ) . .( 1). [ ( )] [ ( )] () ( ) .[ ( 1 [ ( )] i i i i ii i q iL LL L L L L i i q q q L L L LL L i i i qq LL L L i i q L q P x P x P x G x P x G x G x q q P x P x P x P x P x G x q q q P x P x Px G x q q Px                     '' '' ' 2 2 3 ' '' (3) (3) 3 2 1 (4) ( ) ( ) ( ) ). ] .( 1).[ ( 2). ] [ ( )] [ ( )] [ ( )] ( ) ( ) ( ) ( )
.( 1).( 2). 2. .( 1). .

[ ( )] [ ( )] [ ( )] ( ) .( 1).( 2
i i i i i i L L L i i i q q q L L L L L L L i i i i i i q q q L L L L i i i
P x P x P x q q q P x P x P x P x P x P x G x q q q q q q P x P x P x

G x q q q                      '' ' (3) '' 3 4 2 3 (4) (3) 12 ' (4) ( ) ( ) ( ) ( ) ).[ ( 3). ] 2. .( 1).[ ( 2). ] [ ( )] [ ( )] [ ( )] [ ( )] ( ) ( ) .[ ( 1). ] [ ( )] [ ( )] () ( )
.( 1).( 2).( 3).

[

i i i i ii L L L L i i i i q q q q L L L L LL ii qq LL L L i i i i
P x P x P x P x q q q q P x P x P x P x

P x P x qq P x P x Px G x q q q q P                   '' (3) (4) 4 3 2 1 ' '' (5) 5 ( ) ( ) ( ) 3. .( 1).( 2) 3. .( 1). . ( )] [ ( )] [ ( )] [ ( )] ( ) ( ) ( )
.( 1).( 2).( 3).( 4).

.( 1).( 2).( 3). [ ( )] [ ( )]

i i i i i L L L i i i i i i q q q q L L L L LL L i i i i i i i i i q LL
P x P x P x q q q q q q x P x P x P x P x P x G x q q q q q q q q q P x P x

                     4 (3) (4) (5) 3 2 1 ... ( ) ( ) ( ) 6. .( 1).( 2). 4. .( 1). . [ ( )] [ ( )] [ ( )] i i i i q L L L i i i i i i q q q L
L L P x P x P x q q q q q q P x P x P x

          
It is not difficult to obtain, by induction, a general expression:

() 1 ( ) 1 1 1 0 () ( ) .( 1) . .( 1)...( ). [ ( )] i nk n n k k L L n i i i qk k L Px G x C q q q k Px            (81)
With:

1 12 () 1 .. ... , ( ) [( )...( )...( )] k L m k L m k m L i i i i i i i L i i P x x x x                 
, taking into account (76), we arrive at:

1 12 ( ) ( ) 1 1 1 1 0 1 1 .. ... , 1 1 0 ( ) ( ) 1 . .( 1) . .( 1)...( ). ! ! [ ( )] [( )...( )...( )] 1 . .( 1) . .( 1)...( ). ! i k L n k k L n k k k n k n i k k L i L i k n i i i qk k Li i i i i i i n i i i i L i i k k k n i i i k GP a C q q q k k k P a C q q q k k                                          1 12 1 1 1 1 1 1 .. ... , 1 1 1 0 1 1 1 [( )...( ).( )...( )] [( )...( )...( )] . . .( 1) . [( )...( ).( )...( )] i k L n k k L n k k i i i i qk i i i i i L i i i i i i i n i i i i L i i q i k k k i n q k qk k i i i i i L i a q C C                                                        (81)
It is now possible to derive the p.d.f ()

L Z px : 1 2 1 1 1 1 . . . 1 2 12 0 . . . ( ) ( . .. ) .[ . . ... . ] ( 1)! ( 1)! ( 1)! i i i i L i L q q k q k q k x x x q L Z L k k k k i i i x e x e x e p x a a a q k q k q k                          (82)
And we eventually obtain:

ii i i i i L q q k q q k q k q k L e L k k k k i P a J a J a J qk                          (83) With:   2 / 4. 1 2 , ( ) / 2 00 ( 1)! 1 . .[1 . . .( 1) . . [( 1) / 2] [ / 4. ,( 1) / 2] ! 2 . kn mk m n n k inc m n k n kn me J C k n k n k                     V. Simulation Results

5.1

The reference System:

The reference system has been described in part 2.1 and in a former paper with a detailed study of the obtained diversity gain. We have the following equation: †

† 1 † † 1 † † 1 . ( . ) . . . . ( . ) . ( . ) P P       

H H H H y H W s n = s n H H H s y = n H H H

When we calculate the signal to noise ratio, using:

) . ( trace

2 H A A A 
, we obtain :

  2 † † 1 † † 1 † † 1 † 1 1 1 1 1 1 1 1 1 2 † † 1 † † 1 † † 1 1 1 1 1 1 1 1 1
.( . ) trace .( . ) .( .( . ) )

.( . ) trace( .( . ) .( . ) . )

       

H H H H H H H H H H H H H H H H H H

2 † † 1 † † 1 † † 1 1 1 1 1 1 1 1 1 2 † † 1 † † † † † 1 1 1 1 1 1 1 1 1 1
.( . ) trace(( . ).( . ) ( . ) )

.( . ) trace(( . ) .( . ) .( . ) )

       

H H H H H H H H H H H H H H H H H H

2 † † 1 † 1 1 1 1 1 1 .( . ) trace(( . ) )   H H H H H ( 84
)
For the signal to noise ratio, this yields to: The normalization condition:

12 () 1 2 2 1 2 1 2 2, 1 2 1 2 1 2 1 ( , ). . . .( . ) . .( ) . . 1 2 n n p d d K λ λ e d d                 , yields to: 2, ( 1)!.( 2)! n K n n    .
We have:

1 12 †1 1 1 1 2 1 2 . 1 1 1 [] trace(( . ) ) Z             HH ; so
Then we calculate the probability that:

12 11    or xy ; this entails that:

1 2 1 1 1 1 1 1 ( , ) ( , ) ( , ) xx          .
The corresponding p.d.f is:

12 1 / .( ) 2 1/ ,1/ 2, 1 1 1 1 ( , ) . . . .( ) 2 .( ) ( ) xx n nn p x x K e x x x x            (89)
To evaluate (92), we expand 1 1

x  for: 1 x  , this yields to:

1/ 2 2. 1 1 (2 2)! 1 (1 1/ ) 1 . 2 .( 1)!. ! nn n n x n n x          for 1/ 1 x  (93)
In this case, we obtain: .( 1).( 2)

                                       .1 1 . ( )] .( 1)!. ! kn k kn G kk        (
                                                                    
.( 1).( 2)...( 1)

.( 1)!

km km m km G e EI m m m m m m k m                         With: 1 1 1 1 0 0 . . ( 1)
                            
This gives the following result:

1 2, 24 1 2, 2. 1 2. 1 2. 1 11 1 ( ) . . 2 1 2 (2 2)! (2 2)! . . .[ ( ) . ( ) . ( )] 2 2 .( 1)!. ! 2 .( 1)!. ! U n n n n n n n k k n n k k k k n p K M kk K F F G k k k k                        With: 2, ( 1)!.( 2)! n K n n    , 1 4/ 1 0 ! ( ) . . ( )! 4 p m m p p m Fe mp           and : 1 4 / 1 1 1 1 4 4 4 ( ) .[ ... ( 1) . ...] ( 1) . . (4 / ) 1
.( 1).( 2)

.( 1).( 2)...( 1) .

(

1)! k m k m m k m G e EI m m m m m m k m                           1 1 1 1 0 0 . . ( 1
                            
However, the target p.d.f is those of 1/U, not those of U; we have immediately:

k k m m k m m G e EI m m m m m m k m                         
Using (95), it is straightforward to deduce the diversity order of the random variable 1/U , this diversity order is due to the term 2. 1 . (1/ )

n n F  
and is equal to: 2. 1 1 2 .

n n n R L N      
 . The corresponding term in the sum giving 1/ () U p  gives the approximate law which can be used to obtain the Symbol Error Rate (SER) performances. We have here: HH.

                             -If N = 3,
1 2 3 3 2 3 1 2 3 1 3, () 32 1 2 3 1 2 1 3 2 3 3, 1 ( , , ) . . . ( ) 3!. 1 .( . . ) . .[( ).( ).( )] 3!. i n i i j i i j n n n pe K e K                                 (96)
The normalization condition yields: 3, ( 3

)!.( 2)!.( 1)! n K n n n     .
We want to calculate the p.d.f of the random variable:

1 1 2 3 1 1 1 1 [] U        .
To do this, we first calculate the p.d.f of the random variables:

1 2 3 1 1 1 ,, X Y Z       .
the Jacobian of this transform is:

111 2 222 1 2 3 2 2 2 2 333 2 1 00 1 1 1 1 1 ( , , ) Det( ) Det( 0 0 ) .. 1 00 x y z x J x y z x y z y x y z z x y z                                              
The joint p.d.f of (X, Y, Z) is thus:   over all possible values of (x,y). This may be written as:

1 2 3 1 3 (1/ 1/ 1/ ) 2 2 2 1/ ,1/ ,1/ 3, 2 2 2 1 1 (1/ 1/ 1/ ) 2 2 2 3, 1 1 1 1 1 1 1 1 1 ( ,
                     1 2 3 1 2 3 1 2 2 2 1 (1/ 1/ 1/ ) 1/ ,1/ ,1/ 3, (1/ 1/ 1/ ) 1 2 2 2 1/ ,1
                        (
[1/ ( ) / .( )] 1 2 2 2 3, 1 1 1 00 1 1 1 1 1 1 1 ( ) . . .
                           (99)
We begin to integrate (99) with variable y; this yields:

[1/ ( ) / .( )] 1 2 2 2 3, 1 1 1 0 ( ) / .( ) 1 1/ 2 2 2 3, 11 0 1 1 1 1 1 1 1 ( , ) . . .( ) .( ) .( ) . 3! . ( )
1 1 1 1 1 1 1 ( , ) . . . .
                                                   ( ) / .( )
                              2 2 2 3 /2 .( ) .( /
                        1 1/ 3,
                                              1 1/ 3, 42 (1)
(2) ,,

. ( , )

.[ ] 3!. . .

([ ] ) . ( / ).([ ] )

. 22

1 1 1/ 1 1 1/ . .( [ ] ) . 1 1/ .( [ ] ) 22 x x nn x n n x n x x x x x x x z n x x x x z n x x x dz zz J z e x z dz zz J z e x z                                          
To further proceed with (1)

, nx J we can expand the quantity: 1 1/ z  for 1 z  ; we have:

1/ 2 2. 1 1 (2 2)! 1 (1 1/ ) 1 . 2 .( 1)!. ! nn n n z n n z          for 1 z  Hence,
                                    2. 5 1 1/ 3, 1 4 2. 3 1 2 4. / 2 2 2 1 2 4. / 2 2 2 1 2 . . ( , ) .[ . 1 1/ . 3!. . . 1 1/ . ] [1 1 1/ ] . .[ ([1 1 1/ ]) . . ] 42 [1 1 1/ ] . .[ ([1 1 1/ ]) . . ] 42 x x nx n n U n x n z x xx z x xx Ke p x z dz x z dz z z e z x z z e z x                                
To obtain the diversity order it is sufficient to study one integral form:

x nx n n U n x x k x k kk kk z x k k k Ke px x a x a dz zz z e b z                          (102)
Looking carefully at equation ( 102) it appears that we have two kinds of integral forms; the first one corresponds to the case of a positive exponent for k z and the second one to the case of a negative exponent for k z . To study the diversity order it is sufficient to take the term corresponding to the lowest exponent degree of x  . Hence, we have only to consider the following integral form:

                   (103)
To conclude it is then necessary to integrate (1, , ) () np U Jx over the set of possible values for x. We have:

2. 2. 1 1 4/ 2 1/ 3, (1, , ) 2 2. 2 0 2. 2. 1 1 2 1/ 4 /( ) 3, (1, , ) 2 2. 2 0 2. .. .. 3! . 2 . . .. .. 3! .( ) x np x n kp x U np x np xx n kp x U np K ee J dx x K ee J dx xx                          
We keep on working with the last integral form:

                                         /3 2 1/ 4 /( ) (1) 2 2. 2 0 2 2 2 2 2 2 2. 2 2. 2 2 2 2 2 2 9/ ( 2
                                               2 2 2 2 2 2 2. 2 2. 2 2 2 2 2 9/ [1 ( . / 3
u u u u u u du u u u u u u u u u                              
                         
Using this expression we can see that the lowest exponent for  is :

2. pn   . Then, using the classical expansion:

9. / 0 ( 1) .9 . !.

n n n x n n x e n        
; it is then straightforward to conclude that (1) ( , )

K n p and () U p  can be written as:

(1) 2.

( , ) 1/ np K n p

  
. However, the target p.d.f is those of 1/U, not those of U; we have immediately: .

n p n n n U p            
(when we take the highest value of p which is equal to n + 1). Hence we have proved that for N = 3 the diversity order is equal to: n -3.

As we showed for N = 3, it is very difficult to find the exact law of the SNR when N > 2. However, we can approximate it by a mixture of Gamma laws following the expression given of Nakagami laws. In appendix 5, we give the maximization step to determine the parameters of the Nakagami law mixture. For example, we found a mixture of six Nakagami laws for R.L = 10, N = 3 with the following results:

We found:

 = [0.0630 0.1084 0.1721 0.3285 0.1149 0.2132] ; = [4.1830 5.0961 5.9405 7.2067 6.5736 7.1543] ;

 = [5.5000 5.9000 6.2800 6.0700 5.9800 3.1872].

Theoretical SER derivation:

We

                                                                            1
                                  
In this case, we can express .) . ( 1).( 2).

x nw n m x m x m k x n x F x w e dw x e m x e m m m k x e m m m m x e m m                                                                      2 . 0 ..( (
                                                           2
                                  With: 2 . 22 0 00 22 0 0 2. 0 1 2. 1 . ( (
                                                              2 0 . ( . ). ] erf u du   
                                                     1/ 2 / 2 5 / 4 0 2. 1 ( 1) . . . ( / 2 5 / 4) ! (2. 1). nn n n dx x n nn                
                              
                                             2 2( ) 2 1 2( ) 2 1 0 2. . ( . ). .[( ). . ]. ] u m k m k mk L erf u e m k u u du                       
This can be separated into two integral forms:

22 2 1/ 2 1/ 2 2( ) 2 1 2( ) 2 1 0 00 1/ 2 2( ) 2 1 1/ 2 0 0 1/ 2 2( 0 2 (
                                                   2 ) 2 1/ 2 0 1/ 2 / 2 1/ 4 0 0 1/ 2 / 2 1/ 4 0 .. 2 ( 1) 1 . . . ( ) . . ! (2. 1) 2. 1 ( 1) ( / 2 1/ 4) . . . ! (2. 1) k n u nn m k n x n nn m k n n e du x dx e nn x m k n nn                                               And: 22 1/ 2 1/ 2 2( ) 2 1 2( ) 2
                         2 2 1/ 2 2( ) 2 1 1/ 2 0 0 1/ 2 2( ) 2 3/ 2 0 0 1/ 2 / 2
                                                    1/ 2 / 2 5 / 4 0 ( / 2 5 / 4) .. ! (2. 1) n m k n n m k n n                  We can now deduce mk L  : 1/ 2 / 2 1/ 4 0 1/ 2 / 2 5 / 4 0 / 2 1/ 4 0 1 ( 1) ( / 2 1/ 4) 2.
.

                                                               ). ( / 2 1/ 4) ( / 2 5 / 4)] k m k n m k n                
Finally, we obtain:

P L n L n n n k L n n n n L n n n                                                                          0 ( 1) ). ] nL   (97) With: / 2 1/ 4 0 ( 1) 2.[ . .[( ). ( / 2 1/ 4) ( / 2 5/ 4)] ! (2. 1). i i nk n k i i L n k n k i n k i ii                                (98)
Using the well known formula: ( 1) . ( )

z z z     , we can simplify (98) into: / 2 1/ 4 0 / 2 1/ 4 0 ( 1) 2.[ . .[( ). ( / 2 1/ 4) ! (2. 1). ( / 2 1/ 4). ( / 2 1/ 4)]] ( 1) . ( / 2 1/ 4) 2.[ . .[( ) ( / 2 1/ 4)]] ! (2. 1). i i nk n k i i ii nk n k i i L n k n k i ii n k i n k i n k i L n k n k i ii L                                                                / 2 1/ 4 0 1 ( 1) . ( / 2 1/ 4) .[ . ] 2! ii nk n k i i n k i i                   (99)

( 1) .[ ( 1)

. . . ( / 2 5/ 4)] ! (2. 1).

.[ ( 1)

. ( ) (2.

u i i i i ii i i i e L x du i ii u Li ii L i                                                              0 . ( / 2 5/ 4)] 1). ! i i i        (
                                                     . ) 0 .. u e
                                                           ) 0 4. 2.( ) 1/ 2 2.( ) 1/ 2 0 0 . ( . )] ... 2.( ) ( 1) . . 1 . . . . . !. 2. .( ) n n n n nt nn n erfc u n t e dt nn                          
We have:

                            



Finally, we obtain the following bound at high SNR's:

4. 4. ,' 2.( ) 1/ 2 0 4.( ) ,, ' ( ) 2.
                                               (102)

Simulation set-up and results:

In this part, we give some simulation results to illustrate the accuracy of the proposed SER bounds and to test the efficacy of the optimization power allocation algorithm given in Section 4.2.

We consider at first the simplest case where we have N = 2 mobile stations in the network since it corresponds to a scenario where we have been able to completely characterize all the p.d.f's. We suppose that the total available power of the relays P T is normalized equal to 1 and that the relays have perfect CSI estimates. We consider the following cases:

1-Uniform power allocation between each transmit relay 2-Optimized power allocation (see Section 4.2) with two sub-cases 2.1-Optimum algorithm (see Section 4, equations (51-52))

2.2-Approximated solution (see Section 4, equations (55-56))

3-Zero forcing with complete CSI knowledge at each relay station (Section 5.1) Each time the theoretical SER bound is available, we compare it with Monte-Carlo simulation results. We study the cases of QPSK and QAM-16. 3Zero forcing with complete CSI knowledge at each relay station (Section 5.1) Appendix 1: Characterization of the square of the sum of i.i.d Rayleigh random variables -We consider at first the case of the sum of two Rayleigh independent distributed random variables U 1 and U 2 . Each of one has a p.d.f: (

( 1)

i q R i    ) , 2. 1 2 2 ( ) . .exp( ) ( 1)! i ki q i g x x x q     (1.1)
We calculate then:

12 ,, 22
2 2 2 2. 1 2. 1 (2) 2 2 2 0 2. 1 2. 1 2.[ . ] 2 0 2. 1 2. 1 2.([ / 2] / 4) 2 0 4 ( ) ( ) * ( ) . .exp( ).( ) .exp( ( ) ). ( 1)! 4 . . .( ) . . ( 1)! 4 . . .( ) . ( 1)! ii k i p i ii ii x qq UU i x qq x u x u i x qq x u x x i p x g x g x u u x u x u du q e u x u e du q e u x u e q                           22 22 22 2. 1 2. 1 / 2 2.[ / 2] 2 0 /2 2. 1 2. 1 / 2 2. 2 /2 /2 2.
                     (1.2)
We use the development:

2 2 2 0 ( / 4 ) ( 1) . .( / 4) m m k k m k k m k x t C t x       ; with 2. 1 i mq  , hence we obtain: 22 12 ,, 22 /2 2. 1 (2) / 2 2 2 2. 2 /2 /2 2. 1 2.(2. 1 ) 1 2 / 2 2. 2. 1 2 0 /2 2. 1 1 2. 1 2 0 4 ( ) ( )* ( ) . . ( / 4 ) . . ( 1)! 4 .
( 1) .( / 4) . .

. . ( 1)! 8 . ( 1) ( 1)! i k i p i i i i i i x q xt UU i x x q qk k k k x t q k i x q kk q k i p x g x g x e x t e dt q C x e t e dt q C q                              22 /2 2.(2.
i i i i i i UU UU x q qk k k k x t q k i x q k qk k k x t q k k i i p x p x x C x e t e dt q x C x e t e dt q q                           2. 1 1/ 2 / 2 2. 1 2. 1 2 0 ( 1) . . . . ( ) 1)! 4 i ii q k k k x q q k k k C x e F x         (1.5) With: 2 2.(2. 1 ) 2. 1 1/ 2 / 2 / 2 2. 1 2. 1 2.
2. 1 2. 1 00

1 1 1 ( ) . . . . . . . ( ) 2 2 2 ii ii i i i xx q k q k ut q k q k q k q k q k F x u e du t e dt G x                    and 1/ 2 / 2 0 ( ) . . x kt k G x t e dt    2 12 2 12 2. 1 2. 1 1/ 2 (2) 1/ 2 / 2 / 2 2. 1 2. 1 2 () 0 0 2. 1 (2) 1/ 2 / 2 2. 1 2. 1 2. 1 2 () 0 4 ( 1) 1 ( ) . . . . . . . ( 1)! 4 2 4 ( 1) ( ) . . . . ( ) ( 1)! 2 i i i i i ii i x q k qk k k x t q qk k UU k i q k k k x q q k qk UU k i p x C x e t e dt q p x C x e G x q                             (1.6)
To obtain the diversity gain, it is necessary to obtain the Taylor expansion of function: () k Gx. We have obviously :

2. 1/ 2 2. 1 1/ 2 2. 3/ 2 /2 2. 1 00 00 ( 1) ( 1) . ( ) . . . 2 . ! 2 .( 2. 1/ 2). ! i ii i xx n q k nn q k n q k t qk nn nn i x G x t e dt t dt n n q k n                        And : 2. 1 1/ 2 / 2 / 2 2. 1 0 ( 1) . . . ( ) . 2 .( 2. 1/ 2). ! i i nq n k x x qk n n i x x e G x e n q k n             
Hence, the diversity order in this case (for the convolution of two Rayleigh random variables) is equal to:

2 2. 1 i  .
We will assume now that we have, at rank n, the following assumptions: the diversity order is equal to:

.1 i nq  and the p.d.f around 0 x  may be well approximated by:

2 .1 ( ) / () ( ) . . i n nq n x n n Z p x K e x    with : 1 n ni i ZU   
We calculate then:



. To obtain the result at rank n +1, we have now to compute the convolution product of () ()

n n Z px with the p.d.f of 1 n U  which is equal to : 2. 1 2 2 . .exp( ) ( 1)! i q i xx q   
. We obtain then:

22 1 22 22 2. . 1 2. 1 ( 1) / 2. . 1 2. 1 / 2. . 1 2. 1 / 1 2 ( ) 2. . . * . . ( 1)! 4. . . . * . ( 1)! . . * . ii n ii ii n q q n x n x Zn i n q q x n x n i n q q x n x n p x K x e x e q K x e x e q K x e x e                (1.7) 22 1 22 22 2. . 1 2. 1 ( 1) / ( ) 1 0 2. . 1 2. 1 (1 1/ ).[ 2. . . /( 1)] 1 0 2. . 1 2. 1 . / 1 [( 1) / ].[ . / ( 1)] 1 0 ( ) . . .( ) . . . . .( ) . . . . . .( ) .
ii

n ii ii x n q q n u n x u Zn x n q q x n u n x u n n x n q q x x n n n n u n x n n p x K u e x u e du K e u x u e du K e e u x u e                             2 22 /( 1) 2. . 1 2. 1 /( 1) [( 1) / ]. 1 . /( 1) . . . [ . /( 1)] .[ /( 1) ] . . ii xn n q q x n n n t n n x n du K e t n x n x n t e dt              (1.8)
We can calculate the polynomial function:

2. . 1 [ . /( 1)] i nq t n x n   . 2. 1 [ /( 1) ] i q x n t  
. We set:

1 1 1 n n     and: 11 1 1 nn n n      
, we obtain in this case:

2. . 1 2. 1 1 2. . 1 2. 1 11 2. . 1 2. 1 2. . 1 2. 1 2. . 1 1 2. 1 1 00 2. . 2. . 1 1 ( , ) [ . /( 1)] .[ /( 1) ] [ . ] .[ . ] [ . .( . ) ].[ .( 1) . .( . ) ] [. ii ii ii ii ii i n q q n n q q nn n q q n q k q k k k k k k n q n q n kk n k n q n P t x t n x n x n t t x x t C t x C t x C                              2. . 1 2. 1 1 2. . 1 2. 1 2. 1 2. 1 1 00 2. . 1 2. 1 2. . 1 2. 1 00 . . ].[ .( 1) . . . ] [ . . ].[ . . ] ii i i i i i ii ii n q q q k n q k q k q k k k k k qn kk n q q n q k q m km km km t x C t x a t x b t x                       With: 2. . 1 2. . 1 1 . i i n q k k k n q n aC    
                         (1.10)
We want now to compute the p.d.f of 2 1 () n Z  . We have the relationship:

2 1 1 ( 1) 2 ( 1) () () ( ).2. . ( ). n n nn Z Z p x x dx p x dx    
; this yields to:

2 1 1 ( 1) ( 1) () () 1 ( ) . ( ) 2. n n nn Z Z p x p x x     (1.11) 2 1 1 2 ( 1) ( 1) () () /( 1) 2.( 1). 1 ( 1). 1 / 2 /( 1) [( 1) / ]. 1 00 . /( 1) 2.( ( 1)
. 1 ( 1) / 2 /( 1) 1 00 1 ( ) . ( ) 2. 1 . . . [ . ]. . . . 2. . . [ . ]. 2 n n i i i nn Z Z xn nq m n q m x n m n n t n k m k mk n x n m n q m xn n k m k mk p x p x x K e a b x t e dt x K e a x                              1). 1 .[ ( /( 1)) ( . /( 1))] i nq F x n F n x n      
x m m n u n x kk m m k k k k k m k m k k n F n x n u e du n nn u du n n k n n x n n k k m                               (1.15)
So, finally, we obtain :

) i i n i m nq k k m k m k n q m n x n nk m k k Z mk m m k m k n q m k m ab K x p x e x n n n k k m n a b x n                                   2.( 1



(1.16)

The power of the current term i x is equal to:

( 1). 1 ( 1) / 2 ( 1) / 2 ( 1).

1 ii n q m m k n q k            (1.17)
In this case it is clear that the diversity order at rank n +1 is equal to: ( 1). 1 i nq  .

Conclusion:

The diversity order of a square sum of n i.i.d Rayleigh random variables, each having a diversity order equal to 2.

1 i q  , is : .1 i nq  .
Appendix 2: Characterization of the p.d.f of: ,, ..

k i l j Z U V  
-We deal at first with the general case: i ≠ j.

At first, we calculate the p.d.f of , ki  . Assuming that each real or imaginary part of each coefficient channel has a power 0.5, the p.d.f of 2 , ki  can be written as:

2 , 1 2 2 1 ( ) . .exp( ) 2. .2 . ( ) i ii ki q qq h hi x f x x q      
(2.1) with : 2 0.5 h   , this yields to : (

( 1) (2.4) with:

i q R i    )
( 1)

j q R j   
We have now to find the p.d.f of the variable: ,, .

k i l j . To do this, we use the cumulative density function (c.d.f) and we set:

XV

V U V U V X U X V V p V dV p y dy p V dV          (2.4) Setting / y z V 
in the first integral, we obtain: / 00 ( ). ( / ). .

X U V U V XX V U V U dz F X U V X p z V p V dV V pV dz p z V p V dV p z V dz dV VV           (2.5)
This can be expressed as:

. 00

() ( )= [ ( / ). ]. . X V U V U p F X p z dz d      
.

We calculate at first:

i i i i i i XX q U i X q i X q q i X q q ii z dz z z dz q u u du q u u u du q x x dx J X qq                              (2.
                                    (2.7)
This yields to:

22 22 22 1 0 1 2 1 2 2 2 2 2 2 / 2 2 12 2 2 2 2 / ( / ). . ( / ) ( 1)! ( / ) ( / ) ( / ) .( 1)!.[1 .[ ... ... / 1]] ( 1)! ( 1)! ( 2)! ( 1 )! ( / ) ( / ) .[1 .[ ... ( 1)! ( 2)! i i i i ii X Uq i q q q k X i i i i i qq X ii p z dz J X q X X X q e X q q q q k XX e qq                                           1 22 22 ( / ) ... / 1]] ( 1 )! i qk i X X qk        (2.
i i i j X V U V U q q q k q X i i i j p F X p z dz d X X X e X d q q q k q                                     2. 2. . ( / ) ( / ) ( / ) [ .exp( ). .[ ... ... / 1]]. ( 1)! ( 1)! ( 1)! ( 2)! ( 1 )! jj i i i qq q q q k X j j i i i e X X X d X d q q q q q k                                  2 2
q q q k X UV k ji q q k q q k X k ji qk qq ji eX F X d q k X ed q q k X I q q k                                        1 .1 0 () i q k k X     (2.9) With: 2 2 2 2.( ) 2. 1 [ / ] 2.( ) 2. 1 0 ( ) . . ji ji q q k X q q k I X e d            
We suppose at first that: ji qq  . The probability density function is the derivative of function:

. ()

ji j i j i ji ji X q q k q q k q q k X q q k X q q k IX Xe e d e d XX X e d X I X                                              
We obtain then: i i i q q q k q q q q q q q k UV i j j j q q q q q q q q q q q i j j j

q j i i i Ix X I X X I X X I x pX X q q q q q q k q qX                                 1 2.( 1
q i j i i i I X X X I X q X I X X X I X q q q q q q q q XI                             2. 1 2.( 2) 1 2. 3 ) 1 2.( ) 1 3 1 ( ) 4. . ( ) 4. . ( ) 4. ( ) ... ( 1)!.( 2)! ( 1)!( 1)! ( 1)!( 3)! ( 1)!( 2)! . j j i i i i i q q q q q q q j i j j j i i i X X I X X I X X I X q q q q q q q q                    2. 1 2.( ) 1 . 4. . ( ) () ( 1)!.( 1)! i ji q qq UV j i X I X pX qq    
(2.10)

We have: q q qq XX p q q z q q p pp qq qq p X

pq p qq qq p IX ee z z X dz z z X dz z X z X e C z z X dz zX C                               2. 2 2 1/ 2 2. . .[ 4. ] . . i ji q q q p pz X z z X e dz       
The final variable change is when we set: q q q q p p X u p q q q q q q p p X u qq p p q q qq pp q q q q p X u u e du

C X u u e du C X K X                         
To conclude, we have to express integral values of the form:

1 2 1/ 2 2. . 1 ( ) .[1 1/ ] . . j i j i q q q q p X u p K X u u e du       
(2.12)

Since we have

2 1/ 1 u  for [1, [ u  
, it is possible to expand the function :

2 1/ 2 [1 1/ ] p uu     over the integration interval: 2 1/ 2 2 4 2. ( 1/ 2) 1/ 2).( 3/ 2) ( 1/ 2).( 3/ 2)...( 1/ 2 1) [1 1/ ] 1 ... ( 1) ... 2!. !. pq q p p p p p p q u u u q u                 
With :

(

k k k q k k k q k k k k k q k q k k k k q k k k k k q k                            2. 1 2. 1 2. 1 2. ) 2 . .( 1).( 2)...( ) (2. ).(2. 1).(2. 2).(2. 3)...(2. 2. 1).(2. 2. ).(2. 2. 1)!.( 1)! 2 . .( 1).( 2)...( ).( 1)!.(2. 2. 1)!. (2. )!.( 1)! 2 . !.(2. 2. 1)! q q q q k k k k q k k k k k q k q k q k q k k k k q k q k q k k q k k q                           
Thus, we obtain: .

2 1/ 2 2. 1 2. 0 (2. )!.( 1)! 1 [1 1/ ] ( 1
.

ji ji q q r Xu q q r L u e du          
. Then, considering whether 1 2.

ji q q r    is positive or not, we have:

-1 if 1 2. 0 ji q q r     : ( ) . . (1 ) . . 1 . . . . . . . . [2. ] . j i j i ji j i j i j i j i j q q r q q r X u X t q q r q q r q q r X k k X t X k k z q q r q q r k kk X q L X
                                           1 2. 1 2. 2. 1 2. 1 2. 1 1 1 00 ( 1) ! . . . [2. ] 2 . j i j i i j i q q r q q r k X k q r q q r k k k kk kk eC XX                     (2.15) -2 if 1 2. 0 ji q q r     : 2. . 2. . 1 2. 2. . 2. 1 2. 2. 1 2. 1 1 1 0 ( ) . . . . . (1 ) ji ji i j i j X u X t q q r X u X q q r r q q r q q ee L X u e du du e dt ut                         (2.16)
Using an induction process, it is possible to express this last integral with the exponential integral, we have:

                                       11 1 1 2. ...... ( 1) . . (2. )] ).( 2)...( 1) ( 1)! nn n X EI X n n k n          With: . . . 1 1 1 1 0 . . ( 1
                           
In conclusion, we obtain: q q r r r rr q q r rr r q q r p p r K X L X r p p r p p r p p r LX r p p r r p p r

p q q qq pp q q q q p IX C X K X        With: 1 2. 2. 1 0 1 2. 2. 1 2. 1 ; 1 2. 0 ( 2
                                1 2. ; 1 2. 0 . ( ) )! ji ji q q r r q q r LX        
We have:

-For n ≥ 0:  . Assuming that each real or imaginary part of each coefficient channel has a power 0.5, the p.d.f of 2 , ki  can be written as:

2. 11 0 ! ( ) . . 2. n Xk nn kk k k L X e C X      -For n < 0: 11 2. 1 1 1 2. 2 . 2 . ( ) .[ ... ( 1) . ...... ( 1) . . (2. )] 1 ( 1).( 2) ( 1).( 2)...( 1) ( 1)! k k n n X k n n X X X L X e EI X n n n n n n k n                   With: . . . 1 1 1 1 0 0 . . ( 1
i k i k i q i g x x f x x x q       So, we have demonstrated that: , 2. 1 2 2 ( ) . .exp( ) ( 1)! i ki q i g x x x q     (2.3)
We have now to find the p.d.f of the variable: ,, .

k i l i , assuming that ,, ,

k i l i
 are two independent identically distributed (i.i.d) random variables. To do this, we use the cumulative density function and we set:

XV

V U V U V X U X V V p V dV p y dy p V dV          (2.4) Setting / y z V 
in the first integral, we obtain: / 00 ( ). ( / ).

X V X

UU dz p y dy p z V V   and (2.4) changes into:

. 00

0 0 0 0 ( ) = Proba( . ) [ ( / ). ]. ( ). () [ ( / ). ]. ( ). [ ( / ). ]. . X U V U V XX V U V U dz F X U V X p z V p V dV V pV dz p z V p V dV p z V dz dV VV           (2.5)
This can be expressed as: . 00

() ( )= [ ( / ). ]. . X V U V U p F X p z dz d      
.

We calculate at first:

i i i i i i XX q U i X q i X q q i X q q ii p z dz z z dz q u u du q u u u du q x x dx I X qq                              (2.
                                    (2.7)
i i i i ii X Uq i q q q k X i i i i i qq X ii p z dz I X q X X X q e X q q q q k XX e qq                                           1 22 22 ( / ) ... / 1]] ( 1 )! i qk i X X qk        (2.8)
And finally, we obtain:

22

. 00

1 2 1 2 2 2 2 2 2 2. 1 / 2 2 2 0 () ( ) [ ( / ). ]. . ( / ) ( / ) ( / ) 2 [1 .[ ... ... / 1]]. . .exp( ). ( 1)! ( 2)! ( 1 )! ( 1)! i i i i X V U V U q q q k q X i i i i p F X p z dz d X X X e X d q q q k q                                     2. 2. . ( / ) ( / ) ( / ) [ .exp( ). .[ ... ... / 1]]. ( 1)! ( 1)! ( 1)! ( 2)! ( 1 )! i i i i i q q q q q k X i i i i i e X X X d X d q q q q q k                                   2 2 2 2 2 2 1 2. 1 1 [ / ] 2 2 . 0 0 1 2.( 1 ) 2. 1 [ / ] 0 0 1 2.( 1 ) 2. 1 0 2. . ( / ) ( ) 1 . . ( 1)! ( 1 )! 2. 1 . . ( 1)!.( 1 )! 2. 1 . ( ) ( 1)!.( 1 )! i i i ii ii q q q k X UV k ii q q k kX k ii q q k k k ii eX F X d q q k X ed q q k X IX q q k                                           (2.9)
The probability density function is the derivative of function: . ()

                                    
We obtain then:

i i i i i i q q k q q k UV i i i i q q i i i i Ix X I X X I x X I X pX X q q q q k q q q X I X X X I X q                               2.( 2) 1 2.( 2) 3 1 2.( 2) 1 2.( 1) 1 2. 1 2. 1 3 1 1 1 2 2 4.( 2). . ( ) ) 2. .2. . ( ) ( 1)!( 2)! ( 1)!( 2)! ( 1)! 4. . ( ) 4. . ( ) 4. . ( ) 4. ( ( 1)!.( 2)! ( 1)!( 3)! ( 1)! ... . i i i i i i q q i i i q q q q i i i i i i i i q X I X X X I X q q q q q X I X X I X X I X X I X q q q q q                            ) ( 1)!( 2)! ... i i qq   2. 1 1 . 2 4. . ( ) () ( 1)! i q UV i X I X pX q    
(2.10)

We have: ; it is easy to show that the diversity order of the product U.V is equal to: 2. 1 i q  ; and that we have for 0 X  :

. This result is confirmed by the E.M fitting algorithm; for example with R = 5 we have the following result for the variable:

1,2 5,2 .

.

We found:

L N L N i k j k i k i k j k i k j k j j i V                 
. Using appendices 1 and 2, we know that each term '2 , ki  has a diversity order equal to: 1 i q  and that each term '' ,, .

k j k i  has a diversity order equal to: 1 ij qq  . However, due to the fact that the random variables in i V are highly correlated, the diversity order study is difficult. It is easier to begin with a simpler case with L = 1 and N = 2; i.e. we have to study the distribution of : X z z q u q q q q k X z z I X z u e du J X z z qq X z z X z z X z z e X z z q q q k . q k q X z z k q k q X z z z z K X q q q k X z z z e dz X ( 2)! ( 1 )! (5.1)

                              (3.
q q q k X z z V X z z X z z X z z F X e X z z p z dz q q q k                         (
q q q k X z z V q qk V k X z z X z z X z z X z z F X e X z z p z dz q q q k F X X z z q q k e                                    1 2 1,
q k V k q q k q X z z k q q k q Xz k z K X q q pX q q k X X z z z e dz q q k X z z z e q q k e                                         2
X U V U V n n n k X V n n n k X F X p z dz p d X X X e X p d n n n k X X X eX n n n k                                                         1 0 1 2 2 3 1 2 1 (
The mixture of Nakagami laws takes the following form:

                             Z  (5.3)
The maximization step yields to : ˆ.(log( ) 1 ( ) 2.log( ) / ) log ( , ) 0 log( )

n jj i j j i i j i j j j n i j j j i i j i j n i j j j i i j i j j p z z z p                                                         Z Z , 1
(5. ; and we solve (5.5) with the classical Newton Raphson method.

  link from the ith relay to the jth mobile station is a flat fading Rayleigh channel , R. since each MS is equipped with one receive antenna. The channel state information is assumed to be known at the transmitter. We define the vector s of transmitted symbols

Fig

  Fig. 3: p.d.f of 2 ,1 k  for R = 10, N = 5 Fig. 4: p.d.f of 2 ,2 k  for R = 10, N = 5

  below which yields to a diversity order of: 43.5. This confirms that the diversity order is nearly equal to: L.(R-(i-1))-1 = 5×(10-1)-1 = 44 in this case. In fact, for each scenario we tested with the E.M algorithm, we found a diversity order very close to: L.(R-(i-1)).

Fig. 5

 5 Fig. 5: p.d.f of i  for R = 10, N = 5, L = 5 and its approximation with E.M based algorithm

..

  The p.d.f of i  is difficult to track; we give in Appendix 2 the characteristics of the p.d.f of the product of two independent Rayleigh variables. Using this result, we give an approximate characterization of the p.d.f of 2 . i  in Appendix 3. However, it is possible, once again, to use the E.M. based fitting algorithm to find, with a good precision, an approximated distribution for the term: For example, using Gamma pure laws at each time, we have obtained the following result. In the case where L = 2, N = 3 and R = 8, we obtained the result just below for 1  :  = 15.81,  = 1; i.e. we obtain a chi-square law with diversity order: 1 .1 D R L  .

Fig. 6 :

 6 Fig. 6: characterization of the p.d.f of 1  with: L = 2, N = 3 and R = 8 (in log-log scale on the right side)

Fig. 7 :

 7 Fig. 7: characterization of the p.d.f of 1  with: L = 2, N = 3 and R = 9 (in log-log scale on the right side)

Fig. 8 :

 8 Fig. 8: characterization of the p.d.f of 1  with: L = 3, N = 4 and R = 8 (in log-log scale on the right side)

Fig. 9 :

 9 Fig. 9: characterization of the p.d.f of 2  with: L = 2, N = 3 and R = 8 (in log-log scale on the right side)

  we have the p.d.f of the non-ordered eigenvalues of the Wishart matrix: † 11 .

  x

  NR  . In this case, we can always extract an orthonormal free vector family

																	Det( , ,..., h h	h	k	1  , k h	,...,	1, h	N	) 0 	. This is equivalent
	to the condition:	1 NR  or :			1
	,1 ( , ,..., ,2	, 1	,	, 1	,...,	,	)
													e		h		h . Then, we take:
						l e	,2			,1 ,1 ll l e ,2 ,2 proj ( ) ,2 ,2  hh l ll e proj ( )  hh		(	l h	,2		,1 ,1 eh , ll , ll ee	,2 ,1	,1 . ) / l e	l h	,2		,1 proj ( ) ,2 l h l e
						l e	,2		(	l h	,2		,1 e h e ,2 l l l , . ) / ,1	l h	,2		,1 e h e ,2 l l l , .	,1
	And naturally, we obtain:

, ,

e h e h e

	2 ,2	,2	,	,2	2	,2	, (	,2	,1	,	,2	,1 . ) /		,2	,1	,	,2	.	,1	2
		22 ,2 ,2 ,2 ,1 ,2 ,1 ,2 . , ( , . )	.[		,2	,	,2		,1	,	,2	2	]
																				2
		2 22 ,2 ,2 ,2 ,1 ,1 ,2 ,2 .[ , / , ]	.[		,2	,	,2		,1	,	,2 2	]
																				,1
																	. Hence, we obtain:

  ,

	Z	L		P	i h e	i		P	i h e	i	 	L P	L i h e	L i	(74)
	The Fourier transform of the p.d.f	() px is equal to:
					Z
					L
												1
	[ ( )]		( ) ( . ... ) .
					12
								(	.2. . ) .(	.2. . ) ...(	.2. . )
									12

  If N = 2, we have the p.d.f of the non-ordered eigenvalues of the Wishart matrix: †

							11 . HH.
	12 () e  .( . ) . 2 1 2 n pK λ λ 1 1 2 2, 1 ( , ) . n 2      1 .(   2	)	2	(86) (	. n R L 	)
				2	† 1 .trace([ . ] ) 1 HH 		(85)
					11
	Since						1. N R L   HH	with:	. N R L 

1

H is a matrix with N rows and R.L columns, we have naturally . The p.d.f of the signal to noise ratio in (

85

) is difficult to track, particularly when N > 2. However, using the well known Wishart distribution, it is possible to accurately obtain it in the case: N = 2.

-

  the first task is to obtain the p.d.f of :

														U	12 11  	.To do
	this, we set:	12 11 , xy   ; the Jacobian of this transform is:
		J	(	12 1 , x  		1 y	) Det( 	11 22 xy xy            	) Det( 	2 1 0 x  2 0 1 y       	)		22 1 . x y	(87)
	The p.d.f of the couple	12 11 ( , ) ( , ) xy  	is then equal to:
				12 1/ ,1/ p 	( , ) x y	1 2, n   2 (1/ 1/ ) 1 1 1 1 . .( ) . .( 2 . x y x y x y 2 22 1 ) . . n x y K e   
				1 2, n 2. 1 1 . .( ) . x y n Ke   (1/ 1/ ) 1 1 .( x y x y  	)	2

  

			11 4/ 11 00 !! . . ! . . !.( )! 4 ( )! 4 pp mm pp pp mm pe p m p m p         
	And:
					1
		1	4	4	4
	4 /			1
	( )	.[	... ( 1) .	...] ( 1) .	. (4 / )
				1	1
		1

  To complete the result, we have to calculate:

	mk L	2 00 2( ) 2 . . . . uu m k ee u du u	u	2(	m k	) 2	. e	u	2	.	( ( . )) d erf u du	. du
	.[[ u	2(	m k	) 2	. e	u	2	0 . ( . )] erf u	...
		erf	( . ). u e	u	2	.[(2.(	m k	) 2. ). u	2(	m k	) 2 1	2. . . u u	2(	m k	) 2	]. ] du
	0

  At high SNR's, due to the high number of terms in the series expansion of , it may appear some divergence cases. To avoid this, we have developed an other bound for the SER working at high SNR's. This bound is obtained as follows:

	Remark: If   , we obtain: 2. . ( ) e Pn / 2 1/ 4 0 ( 1) .[ . ( ) 2. . . ( / 2 ! (2. 1). nn n n nn                       . 4. . 2. 1 00 4. 4. 2.( ) 0 4. . . ( ). ( ). . . ( . ). 1/ 4)] 0 ( ). 4. . . ( 1) . . .[ [ . . ( . )] ( ). !. 2.( ) u ee n n n n e n n P P u h u du u e erfc u du ... (101) u P erfc u nn Remark: 2 4. 2.( 0 ( 1) . . n n n 1 . . . !. 2.( ) n n n u nn	100)

e P

  .2.

	n nn Z Z p x p 	n	x	x	, i.e. :	n ( ) Z p x K x x n ( ) .2. .[ ] 2 n q . 1 ii n 22 2. . 1 / . 2. . e K x e . n q n x n x n /  

2 , 1

Finally, we have: To obtain the diversity order, it is necessary to study the behaviour of 12 , () qq GX around zero. To do this, we set:

We have:

Of course, it is very difficult to calculate this integral but remember that we are only looking for the diversity order of the random variable 1,2 V ; i.e. the first non-null derivative of 1,2 () V pX around zero. To do this, we use the following approximation:

and we use the following ( )] 4 .

( ) 8.

2. 2. .[1 2. / ( )]

(3.12)

It's then straightforward to deduce the diversity order of variable V 1 : 11 2. 1 Dq  . At this stage it is important to compare this result with the case where we consider independent variables. Using appendices 1 and 2, we know that each term '2 , ki  has a diversity order equal to:

 has a diversity order equal to:  were independent, they should exhibit a diversity order:

Compared to this ideal case on can see that we encounter a diversity loss of:

12

It is possible to generalize this result, considering the random variable: [

We have demonstrated in Appendix 1 that around zero we have: .exp( )

We set: 23 ... nn r q q q     ; we obtain thus (similarly as in equation (3.8)):

We can write 1 , () n qr GX :

(3.16)

It is clear that the diversity order remains equal to: 11 2.

1 Dq  .

In this case, it is straightforward to obtain the diversity order of the numerator Although we use here ratio of correlated variables, this may be justified in comparison with the case of the p.d.f of the ratio of two independent Chi-2 variables (see Appendix 4). We proved in Appendix 4 that the diversity order of the ratio of two independent random Chi-2 variables correspond to the diversity order of the numerator random variable and this completes the proof.

Appendix 4 : Characterization of the ratio of two independent Chi 2 random variables.

We want to calculate the p.d.f of U/V with: We calculate at first :

.

. 00 ( ). ( . ). .

V X X UU p y dy p V z V dz  

; we obtain :

We use the well known result :

( Formula (4.5) clearly shows that the diversity order of random variable U/V is equal to: n -1; i.e. the diversity order of U.

Appendix 5 : Characterization of SNR p.d.f for the reference system by means of a mixture of Nakagami laws.

Using the same notations as in Section 3.1 with the use of Gamma laws, we have now: