無線網路編碼系統中的訊號偵測

國

立

交

通

大

學

電信工程研究所

碩

士

論

文

無線網路編碼系統中的訊號偵測

Signal Detection in a Wireless Network-Coded System

研 究 生：鄭 瑩

指導教授：蘇育德 教授

無線網路編碼系統中的訊號偵測

Signal Detection in a Wirework-Coded System

研 究 生：鄭 瑩 Student：Ying Cheng

指導教授：蘇育德 Advisor：Yu-Ted Su

國 立 交 通 大 學

電 信 工 程 研 究 所

碩 士 論 文

Submitted to Department of Communication Engineering College of Electrical and Computer Engineering

National Chiao Tung University

in partial Fulfillment of the Requirements for the Degree of

Master in

Communication Engineering Nov 2010

Hsinchu, Taiwan, Republic of China

無線網路編碼系統中的訊號偵測

學生：鄭 瑩

指導教授：蘇育德 博士

國立交通大學電信工程學系碩士班

摘

要

近來實體層（無線）網路編碼因為可提高了吞吐量而引起不少研

究者的興趣。在無線傳輸環境中，接收器收到的訊號常常是不同來源

的信號和雜訊的疊加。相較於傳統接收器試圖估測其中個別的信號，

作為有網路編碼網路的中繼接收機，則只需要估計兩個同時到達信號

的和或差。

本篇論文主要目的是在研究當中繼接收機所收到的無線信

號為二元相移鍵調變或二元頻率鍵移調變時的訊號檢測。我們首先在

時間同步的情況下推導出訊號檢測的架構及其錯誤率表現。接著討論

在固定和隨機時間偏移時如何檢測兩個信號的編碼和。我們的分析有

助於瞭解實體層網路編碼在什麼樣的條件下方可提升系統吞吐量。

Signal Detection in A Wireless Network-Coded System

Student：Ying Cheng

Advisor：Yu T. Su

Department of Communication Engineering

National Chiao Tung University

ABSTRACT

Physical layer network coding (PNC) has drew much attention due to its potential of bringing about significant throughput enhancement. In a wireless environment, a received signal is often resulted from the superposition of the signals and noises from various sources. Instead of detecting one or more signal, the responsibility of a relay receiver in a wireless network-coded system is to estimate the algebraic sum of the (two) incoming binary waveforms.

The aim of this thesis is to solve the above signal detection problem when binary frequency-shift-keying (BFSK) modulation is used. We first derive the detector structure in a network with perfect timing synchronization and analyze the resulting bit error rate performance. We then consider the scenarios when either a fixed or a random timing offset is present. The performance analysis shows that the network throughput can indeed be improved through PNC if proper conditions are met.

誌

謝

這份論文的完成，首先感謝我的指導老師 蘇育德博士一路以來的包容和提攜。在學

術方面指引我學習的方法和整體的概念，更著重口條訓練以及個人儀態等全方面學

習。此外對於生活體驗的鼓勵與分享，令我受益匪淺。

感謝引領我進入論文研究領域的人仰學長，博學多聞又很有耐心，除了專業知識的

指導，平時遇到挫折時也給我鼓舞和建議。另外由衷的感謝一起努力的同學和學弟妹

們，相處時分享的喜怒哀樂是這段時光的精神寄託。

最後感謝我的家人，默默給我的支持、關心和照顧是我一路以來的力量泉源。謹以

此篇論文代表我最深的敬意。

Contents

Contents i

List of Figures iii

List of Tables iv

1 Introduction 1

2 System model 5

2.1 Case 1: perfect timing synchronization . . . 5

2.2 Case 2: timing alignment error . . . 7

3 Receiver structure over Rayleigh fading channel 11 3.1 Detection schemes in case 1 . . . 12

3.1.1 The optimal detection scheme in case 1 . . . 12

3.1.2 Suboptimal detectors in case 1 . . . 13

3.2 The optimal detection scheme in case 2 . . . 16

4 Bit error rate of receiver over Rayleigh fading channel 21 4.1 Bit error rate of the two suboptimal detectors in case 1 . . . 21

4.1.1 Bit error rate of the first suboptimal detector in case 1 . . . 21

4.1.2 Bit error rate of the second suboptimal detector in case 1 . . . 24

5.1 The performance of the optimal and suboptimal detectors in case 1 . . . 25

5.2 The performance of the optimal and mismatch detector in case 2 . . . 28

6 Conclusion and future works 33

A Notations 34

B Proofs of lemma 3.2.1, 4.1.2 and 4.1.3 36

B.1 Proof of lemma 3.2.1 . . . 36

B.2 Proof of lemma 4.1.2 . . . 37

List of Figures

1.1 Three nodes system which includes one relay node and two end nodes. . . 2

1.2 Traditional system to exchange data by the relay node. . . 2

1.3 TPTC system to exchange data by the relay node. . . 2

1.4 PNC system to exchange data by the relay node. . . 3

2.1 Information exchange between N1 and N2 among R. . . . 5

2.2 The structure of match filter bank. . . 8

5.1 Performances of the optimal detector (3.4) and the first suboptimal detec-tor (3.13). The power loading facdetec-tor is 0.5. The analysis (4.5) is included. 26 5.2 Performances of the first suboptimal detector (3.13) with the analysis (4.5) under various power loading factors. . . 27

5.3 Performances of the second suboptimal detector (3.14) with the analysis (4.5) and (4.8) under various power loading factors. . . 27

5.4 The comparison of throughput of PNC and TPTC under different packet length. . . 29

5.5 The performance of mismatch detector with fixed timing offsets. . . 30

5.6 The number of bb = 1 given b = 0 in the presence of timing offset or not . 31 5.7 Compare the performance of mismatch detector under fixed and random timing offset. . . 32

5.8 Compare the performance of optimal and mismatch detector under fixed and random timing offset (Til = Tih= Ti = 0.005, i = 1, 2). . . . 32

List of Tables

3.1 The distribution of y1 and y2 under various events . . . 13

3.2 The value of t11, t12, t21, and t22 under various events . . . 17

Signal Detection in A Wireless Network-Coded

System

Student : Ying Cheng Advisor : Yu T. Su

Department of Communications Engineering National Chiao Tung University

Abstract

Physical layer network coding (PNC) has drew much attention due to its potential of bringing about significant throughput enhancement. In a wireless environment, a received signal is often resulted from the superposition of the signals and noises from various sources. Instead of detecting individual signals, a relay in a wireless PNC-coded relay network has to estimate the encoded output only. For example, if a relay receives two incoming binary waveforms, it has to estimate the algebraic sum of the (two) binary symbols.

The aim of this thesis is to solve the above signal detection and estimation problem when noncoherent binary frequency-shift-keying (BFSK) modulation is used to avoid the need of phase pre-compensations at the source nodes. We first derive the optimal and suboptimal detector structures in a network with perfect timing synchronization and analyze the resulting bit error rate performance. We then extend our investiga-tion to the scenarios when either a fixed or a random timing timing offset is present. The performance analysis shows that the network throughput can indeed be improved through PNC if proper conditions are met.

Chapter 1

Introduction

Recently, physical-layer Network Coding (PNC) has been appealing lots of attention within wireless communication society due to the significant improvement of throughput. [1]-[4], for example, show the noticeable throughput improvement under various wireless networks. Specifically, a MAC layer protocol (COPE) based on the concept of PNC is proposed in [2] and it is shown that COPE can be applied in an Ad Hoc or Mesh access network. A regular 1 dimension (1D) linear network system is considered in [3] with 100% throughput improvement compared with traditional transmission scheme. Moreover, it is also shown that PNC can achieve the upper-bound capacity. In [4], the authors propose scheduling algorithms for PNC in 2 dimension(2D) regular networks and, in this case, 200% throughput improvement is achieved.

In this thesis, we consider a wireless network composed of two source nodes and a relay node, as shown in Fig. 1.1. It is assumed that the two source nodes want to exchange the information and is accomplished via the relay node. This situation occurs when two source nodes are apart from. Because of the implement issue, a half-duplex system is considered (no transmission and reception occurs simultaneously).

In a traditional system, the relay node receives the signal sent from one source, and send the detected result to the other node at the next time slot. Hence, to complete a exchange for two source nodes, four time slot is required in this system, as illustrated in Fig. 1.2. The throughput in the system is 1

N2

R N1

Figure 1.1: Three nodes system which includes one relay node and two end nodes.

R

N

x

x

y

y

N

Time slot 1 Time slot 2 Time slot 3 Time slot 4

Figure 1.2: Traditional system to exchange data by the relay node.

To improve the throughput, network coding scheme can be applied at relay. As illustrated in Fig. 1.3, two time slots are devoted to the transmission from source nodes to relay. Then, the relay encodes the received signals and broadcast it. In this scheme (called three-phase transmission communication(TPTC)), the throughput of the network is 1

3(Sym/U/TS).

Notice that the property of wireless environment, broadcast, is used in the third time slot in TPTC. In fact, this property can also be used to improve the throughput by

R

N

x

x

y

y

N

Time slot 1 Time slot 2 Time slot 3

R

N

x

x

y

y

N

Time slot 1 Time slot 2

Figure 1.4: PNC system to exchange data by the relay node.

allowing the transmissions of two source nodes at the same time slot (Fig. 1.4). Although the two source nodes interfere with each other, it is clear that the throughput can be enhanced if the relay node can detect the signals without error. Hence, it is important point to study the optimal detection at relay node such that the enhancement can be achieved.

In [5], the optimal mapping function at the relay node that minimizes the bit-error rate (BER) over Rayleigh fading channels and the associated performance analysis were proposed. We point out that the detector considered in this paper is not a maximum likelihood (ML) detector and it can be shown that the former detector can be obtained from ML detector with some approximation. The symbol error rate performance with ML detector and BPSK/QPSK modulation is studied in [6] for the system with PNC over AWGN channels. Different from the previous mentioned work, [7] investigates a relay system with multiple antennas at the source nodes.

As can be seen, synchronization among each node is a key requirement to provide the improvement. [8] shows that the performance loss with BPSK modulation in the presence of timing or phase error (but not both) is about 3dB. However, power control is required such that the received signals from source nodes have the same amplitude. In practical system, it is hard to have exact power control and the performance loss is larger. In order to avoid the synchronization requirement, non-coherent detection is considered in [9] and [10]. [10] shows that the average expected throughput performance

using BFSK has similar behavior as BPSK, and [9] describes a different detection model of BFSK according to various channel information.

In this thesis, we consider the noncoherent detection and the associated performance analysis of PNC under BFSK in Rayleigh fading channel. Different from the work [9], we focus on the detection without any channel information and two suboptimal detectors with negligible performance loss are proposed. The associated performance is also provided. Moreover, we also consider the noncoherent detection with timing misalignment.

The reset of this thesis is organized as follows. In section II, the system model with and without timing error is provided and the optimal and suboptimal detectors are derived based on the system model. Due to the negligible performance loss, performance analysis based on the suboptimal detectors are provided in section IV and we justify our analysis through simulation in the following section. Some discussions of the performance with and without timing misalignment are included in the same section. Finally, we draw some conclusions to end up this thesis.

Chapter 2

System model

In this section, we describe the PNC system models and notations considered in this thesis. For physical-layer network coding (PNC) scheme in a three terminal network, each of the end nodes Ni, i = 1, 2 transmits the information bi ∈ {0, 1} simultaneously to the relay over a flat-fading channel, as shown in Fig. 2.1. Let hi represent the complex value channel coefficient from node Ni to the relay. The coefficient can be represented as

hi = αieθi, where αi is the received amplitude and θi is the phase shift due to the fading, the transmitter’s continuous-phase constraint, and the offset between the transmitter’s oscillator and receiver’s oscillator.

N2

R N1

Figure 2.1: Information exchange between N1 and N2 among R.

2.1

Case 1: perfect timing synchronization

Suppose the timing is synchronized such that the signals from node N1 and N2 are

arrived at the relay at the same time. Then the received waveform at the relay is yp(t) which consists of these components : the combination of signals from two source and

the noise term. The received waveform can be represented as

yp(t) = h1xp1(t) + h2xp2(t) + np(t) (2.1)

where xpi(t) = {ej2π(f0+biMf )t} with bi ∈ {0, 1}, channel spacing M f = _T1_s, and subcarrier

f0. np(t) is zero mean additive white Gaussian noise (AWGN) with two-sided power

spectral density N0

2 W/Hz and the channel coefficient hi is modeled as a complex

Gaus-sian random variable with zero mean and variance Psi. Do notice that the transmission power factor of Ni is incorporated into the variance of hi.

The equivalent baseband model of (2.1) is

y = h1x1+ h2x2+ n (2.2) where y = · y1 y2 ¸ , n = · n1 n2 ¸ ,

y is the equivalent received signal vector at relay node and n is a zero mean circularly

symmetric complex Gaussian noise with covariance matrix σ2

0I2, where I2 is the 2-by-2

identity matrix and σ2

0 = N0. The transmission signal vector xi is represented by

xi = [(1 − bi) bi]T , i = 1, 2 (2.3)

The possible combinations of transmitted bits are represented by four events:

event E1 E2 E3 E4

b1 0 1 0 1

b2 0 1 1 0

Specifically, given the event Ei, the signal received by the relay is y = mi+ n,where

m1 = [(h1+ h2) 0]T m2 = [0 (h1+ h2)]T

2.2

Case 2: timing alignment error

In practical, to achieve the perfect timing alignment, the relay node should estimate the timing offset of signals from N1 and N2, and feedback the information to source nodes

perfectly. It is, however, difficult to estimate the timing offset without error and this information may be outdated due to the delay. Due to the practical consideration, we are interested in asynchronous PNC signal detection scheme and the system model should be provided. In this section, we establish the system model and the signal detection scheme is discussed later.

Under timing alignment error, suppose a timing offset ²iTs for N1 and N2 appears

when the relay node receives signals from end nodes, where ²i is a random variable and

Ts is the symbol duration time. Before deriving the equivalent baseband system model,

we define a duration interval function.

PTs(t) =

½

1, 0 ≤ t < Tc

0, otherwise. (2.5)

Then, the baseband signal xpi(t), i = 0, 1 is

xpi(t) = PTs(t − ²iTs)e

j2π(f0+biMf )t

and the received signal at relay is the same as (2.1). Again, do notice that the power factor is incorporated into the variance of channel coefficient.

In this thesis, we assume that no inter-symbol interference (ISI) appears. This as-sumption is rational if there is a sufficient long guard time between the two successive symbols. Based on this assumption, the match filter output for ith frequency band (Fig. 2.2) is yi = Z _Ts 0 y(t)e−j2π(f0+(i−1)Mf )t_dt = Z _Ts 0 (h1xp1(t) + h2xp2(t))e−j2π(f0+(i−1)Mf )tdt + n 4 = Si+ n i = 1, 2

where Si denotes the signal part and n is the noise. Because the noise n has been discussed in the previous section, to find the equivalent baseband model, we only need to calculate S.

³

t=T

)

(

_t

y

e

e

³

t=T

³

y

y

Figure 2.2: The structure of match filter bank.

Due to the symmetry of the signal part from N1 and N2, it is sufficient to consider

the signal part from N1 only. Suppose ²1 > 0, then we have

Si1 = h1 Z _Tc 0 PTc(t − ²1Tc)(t)e j(2π(b1−i+1)Mf t)_dt = h1 Z _Tc ²1Tc ej(2π(b1−i+1)Mf t)_dt

where Si1 denote the signal part consisting of the signal from N1.

Observe that Z _Tc ²1Tc ej(2π(b1−i+1)Mf t)_{dt =} e jπ(b1−i+1)Mf (1+²1)Tc π(b1− i + 1) M f 1 2j ¡ ejπ(b1−i+1)Mf (1−²1)Tc _{− e}−jπ(b1−i+1)Mf (1−²1)Tc¢ = ejπ(b1−i+1)Mf (1+²1)Tcsin (π(b1− i + 1) M f (1 − ²1)Tc) π(b1− i + 1) M f = ejπ(b1−i+1)Mf (1+²1)Tc_{(1 − ²} 1)Tcsinc ((b1− i + 1) M f (1 − ²1)Tc) where sinc(x)=4 sin(πx) πx

Hence, the signal part Si is

Si1 = h1ejπ(b1−i+1)Mf (1+²1)Tc(1 − ²1)Tcsinc ((b1− i + 1) M f (1 − ²1)Tc) (2.6) Similarly, for ²1 < 0, the signal part Si1 is

Si1 = h1 Z _Tc 0 PTc(t − ²1Tc)(t)e j(2π(b1−i+1)Mf t)_dt = h1 Z _Tc+²1Tc 0 ej(2π(b1−i+1)Mf t)_dt

Again, we focus on the integral and we have Z _(1+²1)Tc 0 ej(2π(b1−i+1)Mf t)_{dt =} e j(2π(b1−i+1)Mf (1+²1)Tc− 1 j2π(b1 − i + 1) M f = e j(π(b1−i+1)Mf (1+²1)Tc π(b1− i + 1) M f ej(π(b1−i+1)Mf (1+²1)Tc _{− e}−j(π(b1−i+1)Mf (1+²1)Tc 2j = ej(π(b1−i+1)Mf (1+²1)Tc π(b1− i + 1) M f sin((π(b1− i + 1) M f (1 + ²1)Tc) = ej(π(b1−i+1)Mf (1+²1)Tc_{(1 + ²} 1)Tcsinc((b1− i + 1) M f (1 + ²1)Tc) and Si1 = h1ejπ(b1−i+1)Mf (1+²1)Tc(1 + ²1)Tcsinc ((b1− i + 1) M f (1 + ²1)Tc) (2.7) From (2.6) and (2.7), the signal part can be represented as

Si1 = h1ejπ(b1−i+1)Mf (1+²1)Tc(1 − |²1|)Tcsinc ((b1− i + 1) M f (1 − |²1|)Tc) (2.8)

To be compact, we assume Tc = 1 without loss of generality and transform it into

the matrix representation (2.2)

y = h1x1+ h2x2+ n (2.9) and y = · y1 y2 ¸ = h1(1 − |²1|) · 1 ejπ(1+²1)_{sinc(1 − |²} 1|) e−jπ(1+²1)_{sinc(1 − |²} 1|) 1 ¸ · 1 − b1 b1 ¸ + h2(1 − |²2|) · 1 ejπ(1+²2)_{sinc(1 − |²} 2|) e−jπ(1+²2)_{sinc(1 − |²} 2|) 1 ¸ · 1 − b2 b2 ¸ + n Similar to the case 1, we have four possibilities

event E1 E2 E3 E4

b1 0 1 0 1

b2 0 1 1 0

and the corresponding received signal for the four event Ei is

y = mi+ n (2.10) where m1 = h1(1 − |²1|) · 1 e−jπ(1+²1)sinc(1 − |² 1|) ¸ + h2(1 − |²2|) · 1 e−jπ(1+²2)sinc(1 − |² 2|) ¸ m2 = h1(1 − |²1|) · ejπ(1+²1)sinc(1 − |² 1|) 1 ¸ + h2(1 − |²2|) · ejπ(1+²2)sinc(1 − |² 2|) 1 ¸ m3 = h1(1 − |²1|) · 1 e−jπ(1+²1)_{sinc(1 − |²} 1|) ¸ + h2(1 − |²2|) · ejπ(1+²2)sinc(1 − |² 2|) 1 ¸ m4 = h1(1 − |²1|) · ejπ(1+²1)_{sinc(1 − |²} 1|) 1 ¸ + h2(1 − |²2|) · 1 e−jπ(1+²2)sinc(1 − |² 2|) ¸ (2.11)

Chapter 3

Receiver structure over Rayleigh

fading channel

For binary modulation, a natural and popular encoding scheme at relay node is the

exclusive or (XOR) scheme and the encoding rule, denoted by L, is

0M0 = 0, 1M1 = 0, (3.1)

1M0 = 1, 0M1 = 1. (3.2)

For PNC, relay node transmits the signal after encoding signals from two end nodes. Hence, it is needless to detect signal separately and is suitable to detect the signal based on the XOR criterion. Let b = b1

L

b2 be the network codeword. Then, the relay detect

b and forward it back to the end nodes. In this case, although we have four possible

for (b1, b2), b have only two possible values and our detection problem becomes a binary

hypothesis testing problem. It has been shown that the optimal detection scheme for a binary hypothesis testing problem is the likelihood ratio test (LRT) [14].

From (16) of [14], an equivalent optimal test is Λ(b)= log∆ ½ Pr(y|b = 1) Pr(y|b = 0) ¾ 1 ≷ 0 0

because Pr(bi = 0) = Pr(bi = 1) = 1₂. Do notice that Λ(b) = log ½ Pr(y|b1⊕ b2 = 1) Pr(y|b1⊕ b2 = 0) ¾ = log ½ Pr(y|E3 S E4) Pr(y|E1 S E2) ¾ = log ½ Pr(y|E3) + Pr(y|E4) Pr(y|E1) + Pr(y|E2) ¾

= log{p(y|E3) + p(y|E4)} − log{p(y|E1) + p(y|E2)} (3.3)

As can be seen, the optimal detector requires the computation of p(y|Ei) and p(y|Ei)

depends on the system model. Hence, the computation of each p(y|Ei) is done in the

following section according to the system model discussed in the last chapter.

3.1

Detection schemes in case 1

3.1.1

The optimal detection scheme in case 1

In this section, we find the conditional probability density function (pdf) of Ei, i = 1, · · · , 4. Recall that. under perfect timing alignment, the equivalent baseband signal is (2.2) and based on a event Ei, the signal part is shown in (2.4).

For E1, we have y1 = h1 + h2 + n1 and y2 = n2. Since the channel coefficients

are assumed to be the zero mean complex Gaussian random variables with variance depending on the transmission power of source nodes, y1 is the sum of three independent

complex Gaussian random variable. The distribution of y1 can be easily obtained by the

following lemma.

Lemma 3.1.1. For two independent zero mean complex Gaussian random variables x and y with variance σx and σy, respectively, the distribution of z = x + y is also zero

mean complex Gaussian random variables with variance σx+ σy.

For convenience, we denote a complex Gaussian random variable with mean m and

variance σ2 _{by CN (m, σ}2_{). Then, according to lemma (3.1.1), the distribution of y}

is CN (0, Ps1 + Ps2 + N0). Since y2 only contain the noise, the distribution of y2 is

CN(0, N0).

Similarly, with the aid of lemma 3.1.1, the distribution of y1 and y2 under various

events are summary in the following table. The joint distribution of y1 and y2 under

Table 3.1: The distribution of y1 and y2 under various events

Event y1 y2

E1 CN (0, Ps1+ Ps2+ N0) CN (0, N0)

E2 CN (0, N0) CN (0, Ps1+ Ps2+ N0)

E3 CN (0, Ps1+ N0) CN (0, Ps2+ N0)

E4 CN (0, Ps2+ N0) CN (0, Ps1+ N0)

event Ei is the product of the two distributions because y1 and y2 are independent.

Hence, the log-likelihood ratio (LLR) (3.3) becomes Λ(b) = log µ N0(N0+ Ps1+ Ps2) (N0+ Ps1)(N0+ Ps2) ¶ − log µ exp ½ − |y1| 2 N0+ Ps1+ Ps2 − |y2| 2 N0 ¾ + exp ½ − |y2| 2 N0+ Ps1+ Ps2 −|y1| 2 N0 ¾¶ + log µ exp ½ − |y1|2 N0+ Ps1 − |y2|2 N0+ Ps2 ¾ + exp ½ − |y1|2 N0+ Ps2 − |y2|2 N0+ Ps1 ¾¶ (3.4)

3.1.2

Suboptimal detectors in case 1

The optimal detector has been shown in (3.3) and (3.4). In practical, it is hard to imple-ment the detector due to the computation of log(exp(−x) + exp(−y)) for some x, y ∈ R, especially large x or y. To overcome this problem, the computation is approximated by the max-log approximation [15]:

Using this approximation, we have − log µ exp ½ − |y1| 2 N0+ Ps1+ Ps2 − |y2| 2 N0 ¾ + exp ½ − |y2| 2 N0 + Ps1+ Ps2 − |y1| 2 N0 ¾¶ = min µ |y1|2 N0+ Ps1+ Ps2 +|y2|2 N0 , |y2|2 N0+ Ps1+ Ps2 +|y1|2 N0 ¶ (3.5) log µ exp ½ − |y1|2 N0 + Ps1 − |y2|2 N0+ Ps2 ¾ + exp ½ − |y1|2 N0+ Ps2 − |y2|2 N0+ Ps1 ¾¶ = − min µ |y1|2 N0+ Ps1 + |y2| 2 N0+ Ps2 , |y1| 2 N0+ Ps2 + |y2| 2 N0+ Ps1 ¶ (3.6) and (3.4) becomes Λ(b) ≈ log µ N0(N0+ Ps1+ Ps2) (N0+ Ps1)(N0+ Ps2) ¶ + min µ |y1|2 N0+ Ps1+ Ps2 + |y2| 2 N0 , |y2| 2 N0+ Ps1+ Ps2 +|y1| 2 N0 ¶ − min µ |y1|2 N0+ Ps1 + |y2|2 N0+ Ps2 , |y1|2 N0 + Ps2 + |y2|2 N0+ Ps1 ¶ (3.7)

To be concise, we define ξ = N0+ Ps1+ Ps2 and observe that min µ |y1|2 N0+ Ps1+ Ps2 +|y2|2 N0 , |y2|2 N0+ Ps1+ Ps2 +|y1|2 N0 ¶ = min µ N0|y1|2+ ξ|y2|2 N0ξ ,ξ|y1| 2_{+ N} 0|y2|2 N0ξ ¶ = ( N0|y1|2+ξ|y2|2 N0ξ if |y1| 2 _{> |y} 2|2 ξ|y1|2+N0|y2|2 N0ξ otherwise (3.8)

Define xM = max(|y1|2, |y2|2) and xm = min(|y1|2, |y2|2). Then, (3.8) becomes

min µ |y1|2 N0+ Ps1+ Ps2 +|y2|2 N0 , |y2|2 N0+ Ps1+ Ps2 +|y1|2 N0 ¶ = ξxm+ N0xM N0ξ = xm N0 +xM ξ (3.9)

Similarly, let ξ1 = N0+ Ps1, ξ2 = N0+ Ps2 and min µ |y1|2 N0+ Ps1 + |y2| 2 N0+ Ps2 , |y1| 2 N0+ Ps2 + |y2| 2 N0 + Ps1 ¶ = min µ ξ2|y1|2+ ξ1|y2|2 ξ1ξ2 ,ξ1|y1|2+ ξ2|y2|2 ξ1ξ2 ¶ = ( ξ2|y1|2+ξ1|y2|2 ξ1ξ2 (ξ2− ξ1)(|y1| 2_{− |y} 2|2) < 0 ξ1|y1|2+ξ2|y2|2 ξ1ξ2 otherwise (3.10)

Suppose ξM = max(ξ1, ξ2) and ξm = min(ξ1, ξ2). (3.10) can be further reduced to be min µ |y1|2 N0+ Ps1 + |y2|2 N0+ Ps2 , |y1|2 N0+ Ps2 + |y2|2 N0 + Ps1 ¶ = ξMxm+ ξmxM ξMξm = xM ξM + xm ξm (3.11)

Substituting (3.9) and (3.11) into (3.7), we have Λ(b) ≈ log µ N0ξ (N0+ ξ1)(N0 + ξ2) ¶ +xm N0 +xM ξ − xM ξM − xm ξm = K + µ 1 N0 − 1 ξm ¶ xm+ µ 1 ξ − 1 ξM ¶ xM = K + kmxm+ kMxM (3.12) where K = log ³ N0ξ (N0+ξ1)(N0+ξ2) ´ , km = ³ 1 N0 − 1 ξm ´ and kM = ³ 1 ξ − ξ1M ´ . Hence, the suboptimal detector is kmxm+ kMxM 1 ≷ 0 −K ∆ = η (3.13) where η = log ³ (N0+ξ1)(N0+ξ2) N0ξ ´ = log ³ N2 0+N0Ps1+N0Ps2+Ps1Ps2 N2 0+N0Ps1+N0Ps2 ´ ≥ 0

Do notice that km > 0 because ξ_Nm₀−N_ξm0 = min(P_N0s1_ξm,Ps2) > 0. Moreover, kM < 0 as ξM−ξ

ξξM =

− min(Ps1,Ps2)

ξξM < 0. These two factors will be used when the error rate formula is

derived.

As can be seen from (3.13), it depends on xM and xm. However, when N0 goes to

zero, the term km is larger compared with kM and is the dominant term. It motives

us to further approximate the LLR by omitting the term kMxM and we get the second

suboptimal detector:

kmxm

1

≷

0 η (3.14)

Later, we will show that these two suboptimal detectors yield negligible performance loss by simulations, hence, the performance analysis is based on these two detectors and is provided in the next section.

3.2

The optimal detection scheme in case 2

From (3.3), to find the optimal detector, the distributions of received signal y under various events are required. In case 2, to find the distributions is more complex than that in case 1. The difficulty comes from the cross terms due to the timing asynchronization. For example, for event E1, we have

m1 = h1(1 − |²1|) · 1 e−jπ(1+²1)_{sinc(1 − |²} 1|) ¸ + h2(1 − |²2|) · 1 e−jπ(1+²2)_{sinc(1 − |²} 2|) ¸

from (2.11). The distribution of y1 and y2 are no longer independent and we should take

y1 and y2 into account together.

The distributions of received signal y under various events can be represented as

p(y|Ei) = Z h2 Z h1 1 πN0 exp ½ − 1 N0 ky − mik2 ¾ f (h1)f (h2)dh1dh2

where mi are described by (2.11).

Expanding the above equation, we have

p(y|Ei) = Z h2 Z h1 1 πN0 exp ½ − 1 N0 £ |y1− (1 − |²1|)t11h1− (1 − |²2|)t12h2|2 + |y2− (1 − |²1|)t21h1− (1 − |²2|)t22h2|2 ¤ª f (h1)f (h2)dh1dh2 = Z h2 Z h1 1 πN0 exp ½ − 1 N0 h |y0 1− (1 − |²1|)t11h1|2+ |y20 − (1 − |²1|)t21h1|2 i¾ f (h1)f (h2)dh1dh2 (3.15) where y0 1 = y1 − (1 − |²2|)t12h2 , y02 = y2− (1 − |²2|)t22h2

and t11, t12, t21, t22 are constants depending on the timing offset and events. The

value of t11, t12, t21, and t22 under various events are summary in Table 3.2, where

t1 = sinc(1 − |²1|), t2 == sinc(1 − |²2|).

In the remaining part of this section, we will find out the distribution case by case and the optimal detector can be found with these results and (3.3).

Table 3.2: The value of t11, t12, t21, and t22 under various events Event t11 t12 t21 t22 E1 1 1 t1exp{−jπ(1 + ²1)} t2exp{−jπ(1 + ²2)} E2 t1exp{jπ(1 + ²1)} t2exp{jπ(1 + ²2)} 1 1 E3 1 t2exp{jπ(1 + ²2)} t1exp{−jπ(1 + ²1)} 1 E4 t1exp{jπ(1 + ²1)} 1 1 t2exp{−jπ(1 + ²2)}

Lemma 3.2.1. Suppose h is a zero mean complex Gaussian random variable with vari-ance σ2 h, then Z h exp ½ −|y − h| 2 N0 ¾ f (h)dh = N0 N0 + σh2 exp ½ − |y| 2 N0+ σ2h ¾ (3.16)

Proof. The proof of this lemma is shown in Appendix. For convenience, (3.15) is repeated below.

p(y|Ei) = Z h2 Z h1 1 πN0 exp ½ − 1 N0 h |y0 1− (1 − |²1|)t11h1|2+ |y20 − (1 − |²1|)t21h1|2 i¾ f (h1)f (h2)dh1dh2 Observe that |y0 1− (1 − |²1|)t11h1|2+ |y20 − (1 − |²1|)t21h1|2 = |h1|2 ¡ (|t11|2+ |t21|2)(1 − |²1|)2 ¢ + 2(1 − |²1|)<{y0∗1t11h1+ y20∗t21h1} + |y10|2+ |y02|2 = (|t11|2+ |t21|2) ¯ ¯ ¯ ¯(1 − ²1)h1 − y0 2t∗21+ y10t∗11 |t11|2+ |t21|2 ¯ ¯ ¯ ¯ 2 − |y 0 1t∗11+ y02t∗21|2 |t11|2+ |t21|2 + |y0₁|2+ |y₂0|2 = (|t11|2+ |t21|2) ¯ ¯ ¯ ¯(1 − ²1)h1 − y0 2t∗21+ y10t∗11 |t11|2+ |t21|2 ¯ ¯ ¯ ¯ 2 + |y 0 2t11− y01t21|2 |t11|2+ |t21|2

Hence, the distribution of y given Ei is

p(y|Ei) = Z h2 1 πN0 Z h1 exp ( − 1 N0 " (|t11|2+ |t21|2) ¯ ¯ ¯ ¯(1 − ²1)h1− y0 2t∗21+ y01t∗11 |t11|2+ |t21|2 ¯ ¯ ¯ ¯ 2 +|y 0 2t11− y10t21|2 |t11|2+ |t21|2 #) f (h1)f (h2)dh1dh2 = Z h2 1 πN0 exp ½ − |y 0 2t11− y01t21|2 N0(|t11|2+ |t21|2) ¾ Z h1 exp ( − 1 N0/(|t11|2+ |t21|2) ¯ ¯ ¯ ¯(1 − |²1|)h1 − y0 2t∗21+ y10t∗11 |t11|2+ |t21|2 ¯ ¯ ¯ ¯ 2) f (h1)f (h2)dh1dh2

Let h0

1 = (1 − |²1|)h1. Then, according to lemma 3.2.1, we have

p(y|Ei) = Z h2 1 πN0 exp ½ − |y02t11− y10t21|2 N0(|t11|2+ |t21|2) ¾ N0/(|t11|2+ |t21|2) N0/(|t11|2+ |t21|2) + (1 − |²1|)2Ps1 exp     − ¯ ¯ ¯y02t∗21+y01t∗11 |t11|2+|t21|2 ¯ ¯ ¯2 N0/(|t11|2+ |t21|2) + (1 − |²1|)2Ps1     f (h2)dh2 = Z h2 1 π 1 N0+ T1K1 exp ½ −|y20t11− y10t21|2 N0T1 − |y20t∗21+ y01t∗11|2 N0T1+ T12K1 ¾ f (h2)dh2 (3.17)

where K1 = (1 − |²1|)2Ps1 and T1 = |t11|2+ |t21|2. Similarly, we want to integrate h2 via

the lemma 3.2.1. Again, we observe that

|y0 2t11− y10t21|2 N0T1 +|y 0 2t∗21+ y01t∗11|2 N0T1 + T12K1 = 1 T1 · |y0 2t11− y10t21|2 N0 + |y 0 2t∗21+ y10t∗11|2 N0+ T1K1 ¸ = 1 T1 T1K1(|y20t11− y10t21|2) + N0(|y20t∗21|2+ |y10t∗11|2+ |y02t11|2+ |y01t21|2) N0(N0+ T1K1) = K1(|y 0 2t11− y10t21|2) + N0(|y02|2+ |y10|2) N0(N0+ T1K1) Recall that y0₁ = y1− (1 − |²2|)t12h2, y20 = y2− (1 − |²2|)t22h2 Then, we have |y0 2t11− y01t21|2 N0T1 +|y02t∗21+ y01t∗11|2 N0T1+ T12K1 = K1(|(y2− (1 − |²2|)t22h2)t11− (y1− (1 − |²2|)t12h2)t21|2) N0(N0+ T1K1) + N0(|y2 − (1 − |²2|)t22h2|2+ |y1− (1 − |²2|)t12h2|2) N0(N0+ T1K1) (3.18)

The numerator can be represented as K1(|(y2− (1 − |²2|)t22h2)t11− (y1 − (1 − |²2|)t12h2)t21|2) +N0(|y2− (1 − |²2|)t22h2|2 + |y1− (1 − |²2|)t12h2|2) = K1|(y2t11− y1t21) − (1 − |²2|)(t11t22− t12t21)h2|2 +N0|y2− (1 − |²2|)t22h2|2+ N0|y1− (1 − |²2|)t12h2|2 = K1|y2t11− y1t21|2+ K1(1 − |²2|)2|t11t22− t12t21|2|h2|2 −2K1(1 − |²2|)<{(y2t11− y1t21)∗(t11t22− t12t21)h2} +N0|y2|2+ N0|y1|2+ N0(1 − |²2|)2(|t12|2+ |t22|2)|h2|2− 2N0(1 − |²2|)<{y2∗t22h2+ y1∗t12h2} = K1|y2t11− y1t21|2+ N0|y2|2+ N0|y1|2+ {K1|t11t22− t12t21|2+ N0(|t12|2+ |t22|2)}(1 − |²2|)2|h2|2 −2(1 − |²2|)<{[K1(y2t11− y1t21)∗(t11t22− t12t21) + N0(y2∗t22+ y1∗t12)]h2} = Mi ¯ ¯ ¯ ¯(1 − |²2|)h2− Ai Mi ¯ ¯ ¯ ¯ 2 − |Ai| 2 Mi + Bi (3.19) where Mi = K1|t11t22− t12t21|2+ N0(|t12|2+ |t22|2) Ai = K1(y2t11− y1t21)(t11t22− t12t21)∗+ N0(y2t∗22+ y1t∗12) Bi = K1|y2t11− y1t21|2+ N0|y2|2 + N0|y1|2

Let G = N0 + T1K1 and K2 = (1 − |²2|)2Ps2. With (3.17)-(3.19) and lemma 3.2.1, we have p(y|Ei) = Z h2 1 πGexp     − Mi ¯ ¯ ¯(1 − |²2|)h2− _MAi_i ¯ ¯ ¯2− |Ai|2 Mi + Bi N0G     f (h2)dh2 = 1 πGexp ½ |Ai|2 MiN0G − Bi N0G ¾ Z h2 exp ( − 1 N0G/Mi ¯ ¯ ¯ ¯(1 − |²2|)h2− Ai Mi ¯ ¯ ¯ ¯ 2) f (h2)dh2 = 1 πG N0G/Mi N0G/Mi+ K2 exp     − ¯ ¯ ¯Ai Mi ¯ ¯ ¯2 N0G/Mi+ K2 + |Ai| 2 MiN0G − Bi N0G      = 1 π N0 N0G + K2Mi exp ½ K2|Ai|2 (N0G + K2Mi)N0G − Bi N0G ¾ (3.20)

Hence, the condition pdf (3.20) for each event can be obtained by substitute each different t11,t12, t21, and t22 in Table 3.2 and the parameters involved in (3.20) are

summary in Table 3.3.

Table 3.3: The parameters involved in (3.20) General term K1 = (1 − |²1|)2Ps1, K2 = (1 − |²2|)2Ps2 t1 = sinc(1 − |²1|), t2 = sinc(1 − |²2|) G = N0+ T1K1, T1 = |t11|2+ |t21|2 Event M E1 K1|t22− t21|2+ N0(|t22|2+ 1) E2 K1|t11− t12|2+ N0(|t12|2+ 1) E3 K1|1 − t12t21|2+ N0(1 + |t12|2) E4 K1|1 − t22t11|2+ N0(1 + |t22|2) Event A E1 K1(y2− y1t21)(t22− t21)∗+ N0(y2t∗22+ y1) E2 K1(y1− y2t11)(t12− t11)∗+ N0(y1t∗12+ y2) E3 K1(y2− y1t21)(1 − t12t21)∗+ N0(y2+ y1t∗12) E4 K1(y1− y2t11)(1 − t22t11)∗+ N0(y1+ y2t∗22) Event B E1, E3 K1|y2− y1t21|2+ N0|y2|2+ N0|y1|2 E2, E4 K1|y1− y2t11|2+ N0|y1|2+ N0|y2|2

Chapter 4

Bit error rate of receiver over

Rayleigh fading channel

4.1

Bit error rate of the two suboptimal detectors

in case 1

In the next section, we will show that these two suboptimal detectors in case 1 have the performance similar to that of the optimal scheme. Therefore, it motives us to consider the performance analysis of these two suboptimal detectors. Do notice that these two detectors not only avoid the numerical sensitivity problem, but the performance analysis of these two detectors are easiler than that of optimal detector because they have a simple representation.

4.1.1

Bit error rate of the first suboptimal detector in case 1

Recall that the first suboptimal detector is

kmxm+ kMxM 1 ≷ 0 ∆ = η (4.1)

and xm = min(|y1|2, |y2|2), xM = max(|y1|2, |y2|2) and its bit error rate (BER) is

Pe = 1

2Pe(kmxm+ kMxM < η|b = 1) + 1

2Pe(kmxm+ kMxM > η|b = 0). (4.2) In fact, the following theorem states the BER formula for the first suboptimal detector.

Theorem 4.1.1. The BER of the first suboptimal detector is Pe = 1 2(1 − Pe(kmxm+ kMxM > η|b = 1) + Pe(kmxm+ kMxM > η|b = 0)) (4.3) where Pe(kmxm+ kMxM > η|b = 1) = kmλ µ N0ξ ξ1ξ2 ¶ 1 kmλ µ 1 |kM|ξ1+ kmλ + 1 |kM|ξ2+ kmλ − 1 |kM|λ + kmλ ¶ , λ = ξ1ξ2 ξ1+ ξ2 (4.4) and Pe(kmxm+ kMxM > η|b = 0) = kmλ µ N0ξ ξ1ξ2 ¶ 1 kmλ µ 1 |kM|ξ + kmλ + 1 |kM|N0+ kmλ − 1 |kM|λ + kmλ ¶ , λ = ξN0 ξ + N0 (4.5)

Proof. To derive the bit error rate, we need to find the distribution of kmxm+ kMxM

under b = 0 or b = 1. As shown in Table 3.1, the distribution of y1 and y2 are zero

mean complex Gaussian random variable with variance depending on the considered event. Hence, |y1|2 and |y2|2 are exponential random variables. Specifically, suppose

x is CN (0, σx), then, |x|2 is a exponential random variable with mean σx (denoted by

exponential(σx)) [16].

Before completing the proof, we need the following two lemmas and their proofs are provided in the appendix.

Lemma 4.1.2. Suppose x1 ∼ exponential(λ1), x2 ∼ exponential(λ2) are independent

random variables and let xM = max{x1, x2}, xm = min{x1, x2}. Then, we have

f (xm) = λ1+ λ2 λ1λ2 exp ½ −λ1+ λ2 λ1λ2 xm ¾ , xm ≥ 0 f (xM) = 1 λ1 exp ½ −xM λ1 ¾ + 1 λ2 exp ½ −xM λ2 ¾ −λ1 + λ2 λ1λ2 exp ½ −λ1+ λ2 λ1λ2 xM ¾ , xM ≥ 0

and

Lemma 4.1.3. Suppose x1 ∼ exponential(λ1), x2 ∼ exponential(λ2). The distributions

of zs = x1+ x2 and zd= x1− x2 are f (zs) = 1 λ2 − λ1 ³ e−λ2zs − e−λ1zs ´ f (zd) = ( 1 λ1+λ2 e −zd_{λ1, z} d≥ 0 1 λ1+λ2 e zd λ2, z_d≤ 0

From the above two lemmas, we are ready to calculate the probability of L0 ∆₌

kMxM + kmxm ≤ η. Do notice that

Pe(kmxm+ kMxM < η|b = 1) = 1 − Pe(kmxm+ kMxM > η|b = 1)

Pe(kmxm+ kMxM < η|b = 0) = 1 − Pe(kmxm+ kMxM > η|b = 0).

Hence, it is sufficient to consider P e(L0 _{< η|b = 1) and P e(L}0 _{< η|b = 0). Moreover, it} can be seen that L0 _{has the same distribution under b = 0 and b = 1, and the difference} is only the moment of L0_{; it is sufficient to consider one of the two cases. Then, according} to the above two lemmas, we have

P r{L0 _{≤ η|b = 1}} = P r{kmxm ≤ η + |kM|xM} = Z _∞ 0 Z _η+|kM|xM 0   e −_{|kM |ξ1}xM |kM|ξ1 +e −_{|kM |ξ2}xM |kM|ξ2 − e − xM |kM |(_ξ1+ξ2ξ1ξ2 ) |kM| ³ ξ1ξ2 ξ1+ξ2 ´   ξ1+ ξ2 kmξ1ξ2 e− xm km(_ξ1+ξ2ξ1ξ2 ) dx mdxM = 1 − kmλe− η kmλ µ 1 kmλ + |kM|ξ1 + 1 kmλ + |kM|ξ2 − 1 kmλ + |kM|λ ¶

where the second line comes from that kM < 0. The third line use the fact that η ≥ 0, lemma 4.1.2 and lemma 4.1.3. λ is defined as ξ1ξ2

ξ1+ξ2. Because η = −K = −log ³ Noξ ξ1ξ2 ´ , we have e−kmλη = ³ Noξ ξ1ξ2 ´ 1 kmλ

. Hence, we can get (4.4) and (4.5) can be easily obtained by replacing ξ1 and ξ2 with ξ and N0, respectively.

4.1.2

Bit error rate of the second suboptimal detector in case

1

Recall that the second suboptimal detector is

kmxm

1

≷

0

η (4.6)

Because km > 0, the detector can be rewritten as

xm

1

≷

0 η/km (4.7)

As stated in lemma 4.1.2, xm is a exponential random variable with parameter λ. λ is

ξ1ξ2

ξ1+ξ2 under b = 1 while λ is

ξN0

ξ+N0 under b = 0. Hence, the error rate under various cases

is Pe{xm > η/km|b = 0} = Z _∞ η/km fxm(x)dx = exp ½ −N0+ ξ N0ξ η/km ¾ Pe{xm < η/km|b = 1} = Z _η/km 0 fxm(x)dx = 1 − exp ½ −ξ1+ ξ2 ξ1ξ2 η/km ¾ = 1 − Pr{xm > C|H1}

and the BER is

Pe = 1 2(Pe{xm > η/km|b = 0} + Pe{xm < η/km|b = 1}) = 1 2 µ exp ½ −N0+ ξ N0ξ η/km ¾ + 1 − exp ½ −ξ1+ ξ2 ξ1ξ2 η/km ¾¶ = 1 2 + 1 2exp ½ −N0+ ξ N0ξ log (ξ1ξ2/N0ξ) 1/N0− 1/ξm ¾ − 1 2exp ½ −ξ1+ ξ2 ξ1ξ2 log (ξ1ξ2/N0ξ) 1/N0− 1/ξm ¾ = 1 2 + 1 2 µ ξ1ξ2 N0ξ ¶₋ N0+ξ N0ξ(1/N0−1/ξm) −1 2 µ ξ1ξ2 N0ξ ¶₋ ξ1+ξ2 ξ1ξ2(1/N0−1/ξm) (4.8)

Chapter 5

Simulation/Numerical results

In the following simulations, signal-to-noise ratio (SNR) is define as the total signal

power divided by noise power P/N0, where the total power is the sum of the power

from two sources Ps1, Ps2. To investigate the power loading problem, the power loading factor is denoted by Pf actor and, without loss of generality, we have Pf actor = P1/P . The

simulation will be stopped if the error occurs more than 1000 times.

5.1

The performance of the optimal and suboptimal

detectors in case 1

For convenience. the optimal detector (3.4) is labelled as “opt. detector” in the figure. The suboptimal detectors (3.13) and (3.14) are denoted by “1st subopt.” and “2nd subopt.”, respectively. Moreover, “analysis 1” and “analysis 2” are abbreviations for the performance analysis of the first and second suboptimal detectors (4.5) and (4.8), respectively.

First, we investigate the performance loss due to the approximation by simulation. As illustrated in Fig. 5.1 with power loading factor Pf actor = 0.5, the BER curves of the optimal detector and the first suboptimal detector are almost identical, especially at high SNR region. Moreover, the simulation result of the first suboptimal detector is identical to the numerical result of (4.5). This justifies our performance analysis.

0 5 10 15 20 25 30 35 40 45 50 10−4 10−3 10−2 10−1 100 SNR (dB) Opt. detector 1st subopt. Analysis 1

Figure 5.1: Performances of the optimal detector (3.4) and the first suboptimal detector (3.13). The power loading factor is 0.5. The analysis (4.5) is included.

be predicted by (4.5) under various power loading factor Pf actor, we consider Pf actor = 0.5, 0.75 and 0.95. Moreover, we can observe that the optimal power loading is 0.5 among the three cases. It is the future work to show that the optimal power loading is 0.5 or near 0.5 for all power loading analytically. Similarly, Fig. 5.3 shows that the the performance of the second detector under various power loading factors and the same conclusion is obtained. (Because the performance of the first suboptimal detector is similar to that of the optimal detector, we use the result of the analysis as the performance of the optimal detector.)

It is interesting to compare the performance of PNC with that of TPTC system to show which system is more suitable for wireless environment. Recall that, in TPTC system, the two source nodes transmit their messages in orthogonal time slots. The relay receiver first detect the individual signals during each time slot based on the decision

0 5 10 15 20 25 30 35 40 45 50 10−4 10−3 10−2 10−1 100 SNR (dB) 1st subopt. P factor=.5 1st subopt. P factor=.75 1st subopt. P factor=.95 Analysis1 P factor=.5 Analysis1 P factor=.75 Analysis1 P factor=.95

Figure 5.2: Performances of the first suboptimal detector (3.13) with the analysis (4.5) under various power loading factors.

0 5 10 15 20 25 30 35 40 45 50 10−4 10−3 10−2 10−1 100 SNR (dB) 2nd subopt. P factor=.5 Anaylsis1 P factor=.5 Analysis1 P factor=.95 Analysis2 P factor=.95 Analysis2 P factor=.5 2nd subopt. P factor=.95

Figure 5.3: Performances of the second suboptimal detector (3.14) with the analysis (4.5) and (4.8) under various power loading factors.

rule

bbi = ½

0 if |y1,i|2− |y2,i|2 > 0

1 otherwise , (5.1)

Then, the detection of bb is obtained by encoding bb1 and bb2 according to the XOR rule.

Through the detection procedure, an error occurs if only one of the two bits b1, b2

is detected incorrectly. Therefore, the error rate of the TPTC system is PT P T C

b =

p1(1 − p2) + p2(1 − p1), where pi = 1/(2 + Psi/N0) is the bit error rate of noncoherent

binary FSK modulation. However, for TPTC system, it requires two time slots to to transmit the signals and an additional time slot is needed compared with PNC system. To compare these two system in terms of throughput, we consider a packet with length

Pl and the normalized throughput is defined as (1 − Pe)Pl/Tp, where Pe is the bit error rate and Tp is the normalized transmission time of a packet. The normalized throughput is equal to (1 − Pe)Pl and (1 − PbT P T C)Pl/2 for PNC and TPTC, respectively.

P

2N0.

Figure 5.4 shows that the normalized throughput of PNC and TPTC with packet

length Pl = 100, 1000 and 10000. The normalized throughput of PNC is higher than

that of TPTC when SNR is higher than a threshold, depending on the packet length. Moreover, when the packet length is small, PNC outperforms TPTC at any SNR.

5.2

The performance of the optimal and mismatch

detector in case 2

In this section, we consider the effect of timing asynchronization. At the beginning, we investigate the performance of the optimal detector in case 1 in the present of timing error and this detector is called as mismatch detector. The timing offset ²i can be either modelled as uniform random variables between [−Til, Tih] or a fixed vale Ti, for i = 1, 2. The former case occurs when the network synchronization is not perfect and relay does not estimate the timing offset. However, relay can estimate the timing offset and, with the information, the performance may be improved further. Hence, we consider an

20 30 40 50 60 70 80

SNR

Figure 5.4: The comparison of throughput of PNC and TPTC under different packet length.

0 5 10 15 20 25 30 35 40 45 50 10−4 10−3 10−2 10−1 100 SNR (dB) T 1=.001, T2=.001 T 1=.001, T2=.003 T 1=.001, T2=.005 T 1=.005, T2=.003 T 1=.005, T2=.005 Analysis 1

Figure 5.5: The performance of mismatch detector with fixed timing offsets.

extreme case that relay can perfectly estimate the timing offset, which is equal to the later case.

As illustrated in Fig. 5.5, the performance of the mismatch detector with fixed delay is similar to that of the optimal detector without timing offset at low SNR region. However, at high SNR region, the performance of the mismatch detector becomes worse. This phenomena can be explained by observing the structure of the second suboptimal detector. Recall that the structure of the second suboptimal detector is

kmxm

1

≷

0 η (5.2)

It only depends on xm = min(|y1|2, |y2|2). Due to the timing offset, the magnitude of y1

and y2 are larger then that of y1 and y2 in case 1 (this can be observed in (2.11)). Hence,

the suboptimal detector will choose 1 with higher probability at high SNR region. To show this effect, we count the number of bb = 1 given b = 0 in the presence of timing offset or not, as illustrated in Fig. 5.6.

0 5 10 15 20 25 30 35 40 45 50 10−5 10−4 10−3 10−2 10−1 100 SNR (dB) Case 1 T 1l=T1h=T2l=T2h=.005

Figure 5.6: The number of bb = 1 given b = 0 in the presence of timing offset or not .

Fig. 5.7. For comparison, the performance of the mismatch detector with fixed timing offsets is also represented. As can be seen, the performance get worse with the increase of timing offsets. Moreover, the performance with rand timing offsets is better that that with fixed timing offsets when Til = Tih= Ti, i = 1, 2.

Because the mismatch detector has great performance loss at high SNR region, it is necessary to study the performance of the optimal detector and to compare it with that of mismatch detector. Recall that the structure of the optimal detector is given by (3.20) and Table 3.3 and its performance is shown in Fig. 5.8. We consider the performance under fixed and random timing offset (Til = Tih = Ti = 0.005, i = 1, 2). As can be seen, the optimal detector outperforms the mismatch detector in high SNR region significantly and the optimal detector will not get worse with the increase of SNR.

0 5 10 15 20 25 30 35 40 45 50 10−4 10−3 10−2 10−1 100 SNR (dB) T 1=T2=.001 fix T 1=T2=.005 fix T 1l=T1h=T2l=T2h=.001 T 1l=T1h=T2l=T2h=.005 Analysis 1

Figure 5.7: Compare the performance of mismatch detector under fixed and random timing offset. 10 20 30 40 50 SNR

Mismatch fixed

Mismatch random

Optimal fixed

Optimal random

Figure 5.8: Compare the performance of optimal and mismatch detector under fixed and random timing offset (Til = Tih= Ti = 0.005, i = 1, 2).

Chapter 6

Conclusion and future works

In this thesis, physical layer network coding is considered with BFSK modulation. The optimal detectors with timing alignment error or not is derived through the likeli-hood ratio test. To make the performance analysis traceable, we propose two suboptimal detectors in the case of perfect timing synchronization. One suboptimal detector is ob-tained through the max-log approximation and the second one is obob-tained by further ignoring a term which is very small when SN R is high. Simulation result shows that these two suboptimal detectors have negligible performance loss. Consequentially, it is sufficient to evaluate the performance of these two suboptimal detectors, which is also part of contributions of this thesis. Moreover, we compare the throughput of PNC with that of TPTC and shows that PNC outperforms TPTC in high SNR region (depending on the packet length) in terms of throughput. Finally, we consider the mismatch detec-tor (the optimal detecdetec-tor in the case of perfect timing synchronization) in the presence of timing error. The performance of mismatch detector gets worse with the increase of

SNR while it is not true in the case of the optimal detector.

In this thesis, we consider BFSK modulation. To increase data rate, MFSK modu-lation could be applied. Consequently, the associated optimal detector and performance should be investigated. Moreover, the Rayleigh fading is assumed. In practice, there may exists a direct path between the transmitter and receiver and this results in Rician fading channel. To investigate the Rician fading effect is also one of the future works.

Appendix A

Notations

To describe the structure of PNC we need the following notations.

bi : the code bit transmitted by node Ni during a particular signaling interval.

bi ∈ {0, 1}, i ∈ {1, 2}.

xi(t) : the baseband signal of node Ni’s modulated signal, which is chosen as the bthi signal of the set of SF.

xpi(t) : bandpass signal which is in connection with xi(t).

xi : signal vector with respect to xi(t), each element shows the bits transmitted by end node Ni.

y(t) : the complex envelope of the signal received by the relay node.

yp(t) : the combined bandpass signal received by the relay during one symbol period.

y_i : the output of the BFSK detector placed into the 2 × 1 complex vector.

hi : the complex-valued channel gain from node Ni to the relay during signaling interval with the energies of the transmitted signals normalized to unity. hi = αiejφi.

αi : the received amplitude due to the fading.

Ei : the power of channel gains corresponding to Ni, where Ei = E[|hi|2] = E[α2].

n(t) : additive white Gaussian noise (AWGN) with two-sided noise spectral density N0.

n : the zero-mean circularly-symmetric complex Gaussian noise with covariance matrix

Appendix B

Proofs of lemma 3.2.1, 4.1.2 and

4.1.3

B.1

Proof of lemma 3.2.1

We prove lemma 3.2.1 in this section. This lemma will be used to derive the optimal detector in case 2. Lemma B.1.1. Suppose h is CN (0, σ2 h). We have Z h exp ½ −|y − h|2 N0 ¾ f (h)dh = N0 N0 + σ_h2 exp ½ − |y|2 N0+ σ2_h ¾ (B.1)

Proof. Recall that the pdf of h with variance σ2

h is f (h) = πσ12 he −|h|2_σ2 h . Then, we have Z h exp ½ −|y − h| 2 N0 ¾ f (h)dh = Z h exp ½ −|y − h| 2 N0 ¾ 1 πσ2 h exp ½ −|h| 2 σ2 h ¾ dh = 1 πσ2 h Z h exp ½ −|y − h|2 N0 − |h|2 σ2 h ¾ dh = 1 πσ2 h exp ½ − |y|2 N0 + σ_h2 ¾ Z h exp ( −N0+ σh2 N0σ_h2 ¯ ¯ ¯ ¯h − σ 2 h N0+ σ2_h y ¯ ¯ ¯ ¯ 2) dh (B.2) Since CN ³ σ2 h N0+σ2_hy, N0σ_h2 N0+σ_h2 ´ is a p.d.f, we have Z h exp ( −N0+ σh2 N0σ2h ¯ ¯ ¯ ¯h − σ 2 h N0+ σ2h y ¯ ¯ ¯ ¯ 2) dh = πN0σh2 N0+ σ2h .

Using this factor, (B.2) becomes ∴ Z h exp ½ −|y − h|2 N0 ¾ f (h)dh = 1 πσ2 h exp ½ − |y|2 N0+ σ_h2 ¾ πN0σh2 N0+ σ2_h = N0 N0+ σ2_h exp ½ − |y|2 N0+ σ_h2 ¾

B.2

Proof of lemma 4.1.2

We prove lemma 4.1.2 in this section and, for convenience, lemma 4.1.2 is repeated below.

Lemma B.2.1. Suppose x1 ∼ exponential(λ1), x2 ∼ exponential(λ2) are independent

random variables and let xM = max{x1, x2}, xm = min{x1, x2}. Then, We have

f (xm) = λ1+ λ2 λ1λ2 exp ½ −λ1+ λ2 λ1λ2 xm ¾ , xm ≥ 0 f (xM) = 1 λ1 exp ½ −xM λ1 ¾ + 1 λ2 exp ½ −xM λ2 ¾ −λ1 + λ2 λ1λ2 exp ½ −λ1+ λ2 λ1λ2 xM ¾ , xM ≥ 0

Proof. To find the distribution of xm, we first calculate the probability of xm > x.

Since x1 and x2 are independent random variables, we have

P r{xm > x} = P r{x1 > x}P r{x2 > x} = Z _∞ x 1 λ1 e−λ1x dx Z _∞ x 1 λ2 eλ2x dx = e−λ1+λ2λ1λ2 x and P r{xm < x} = 1 − P r{xm < x} = 1 − e− λ1+λ2 λ1λ2 x

Hence, the distribution of xm is

f (xm) = d (P r{xm < x}) dxm = λ1+ λ2 λ1λ2 e−λ1+λ2λ1λ2 xm

Similarly, we first calculate the probability of xM > x. Recall that for any two events

A, B, we have the equality [16]:

P r(A ∪ B) = P (A) + P (B) − P (A ∩ B)

Using this equality and the fact that x1 and x2 are independent, we have

P r{xM > x} = P r{x1 > x} + P r{x2 > x} − P r{x1 > x}P r{x2 > x} = Z _∞ x 1 λ1 e−λ1x dx + Z _∞ x 1 λ2 e−λ2x dx − Z _∞ x 1 λ1 e−λ1x dx Z _∞ x 1 λ2 eλ2x dx = e−λ1x + e−λ2x − e−λ1+λ2λ1λ2 x and P r{xM < x} = 1 − e− x λ1 − e−λ2x + e−λ1+λ2λ1λ2 x

Take the derivative of the probability, we have the distribution of xM.

f (xM) = 1 λ1 e−λ11 xM + 1 λ2 e−λ21 xM −λ1+ λ2 λ1λ2 e−λ1+λ2λ1λ2 xM

B.3

Proof of lemma 4.1.3

Lemma B.3.1. Suppose x1 ∼ exponential(λ1), x2 ∼ exponential(λ2). Than the

distri-butions of zs = x1+ x2 and zd = x1− x2 are

f (zs) = 1 λ2 − λ1 ³ e−λ2zs − e−λ1zs ´ f (zd) = ( 1 λ1+λ2 e −zd_{λ1, z} d≥ 0 1 λ1+λ2 e zd λ2, z_d≤ 0