# On the Performance of Power Splitting Energy Harvested Wireless Full-Duplex Relaying Network with Imperfect CSI over Dissimilar Channels

Authors: Tan N. Nguyen; Minh Tran; Phuong T. Tran; Phu Tran Tin; Thanh-Long Nguyen; Duy-Hung Ha; Miroslav Voznak
Journal: Security and Communication Networks
 (2018)

---

## Abstract

The energy harvesting amplify-and-forward full-duplex relaying network over the dissimilar fading environments in imperfect CSI condition is investigated. In this system model, the energy, and information are transferred from the source to the relay nodes by the power splitting protocol with helping of the full-duplex relay node. Firstly, the outage probability, achievable throughput, and the optimal power splitting factor in terms of the analytical mathematical expressions were proposed, analyzed, and demonstrated. Furthermore, the system performance of the proposed model on the connection with all system parameters is rigorously studied. Finally, the numerical results demonstrated and convinced one that the analytical and the simulation results are matched well with each other for all system parameter values using Monte-Carlo simulation. The results show that the system performance degrades significantly but is still in a permissible interval while the channel estimation error increases and the system performance of the mixing scenarios is better in comparison with the Rayleigh-Rayleigh scenario.


---

## Body

## 1. Introduction

In the last few decades, traditional energy-constrained communication network (ECN) systems have such a limited operational lifetime. Periodical battery recharging or replacement is necessary to propose in order to maintain network connectivity and performance, which caused some disadvantages such as high cost and inconvenient and sometimes impossible realization [1, 2]. Nowadays, energy harvesting (EH) from the environment is a prospective direction for enhancing the lifetime of energy-constrained communication networks (ECN). In comparison with other renewable energy sources, the radio frequency (RF) signals could be promised innovative sources for wireless powered communication network (WPCN). It can be easily realized in connection with this that RF signals can carry both information and energy. Hence energy constrained nodes can harvest energy from the received RF signals. This solution seems to be a promising solution for applications where battery-limited devices are not easily accessible, and replacing or recharging their batteries is inconvenient such as devices embedded inside human bodies and sensors placed in dangerous areas [1–6]. However, energy harvesting and information processing from the source node to the destination node may be interrupted because of the performance degradation caused by fading, shadowing, and path loss. In this case, the intermediate helping relay node between the source node and the destination node can become an excellent solution [7]. There are many types of research focused on the WCPN with helping of the intermediate relay perfect and imperfect channel state information (CSI) [8–10]. Moreover, a hybrid EH single relay network with channel and energy state uncertainties in optimizing system throughput over a limited number of transmission intervals is proposed. Fault-tolerant schemes were analyzed in the presence of imperfect CSI [11].Half-duplex (HD) and full-duplex (FD) are considered as the primary architectures, which are popularly used in the relaying network. In the HD relaying model, the relay node cannot receive and transmit data simultaneously in the same frequency band. The HD model is popularly proposed in traditional WPCN due to simple of the system design and implementation. However, the HD led to significant spectrum efficiency loss [12–14]. On another way, an FD scheme, where the reception and the transmission processes occur on the same channel at the same time, can achieve up to double the capacity in comparison with the HD scheme. FD relaying, in which the relay node receives and transmits simultaneously in the same frequency band, is considered as the main direction in the last decades. Hence, FD can reduce the spectral loss, based on innovative antenna technology and signal processing capability [15–17]. However, not many papers have considered the application of FD relaying network in the imperfect CSI condition over the different fading channel [18, 19]. This paper can entirely fill this remaining gap. In some practical applications, the relay should be close to energy station, which is also the source node in our model. Hence, line-of-sight (LOS) should exist between the source and the relay, whereas the channel between the relay and the destination might not necessarily be LOS. The dissimilar channel model has been considered in many previous works, for example, in [17–19].In this research, the system performance analysis (in the term the outage probability and the achievable throughput) of the amplify-and-forward (AF) FD relaying network over different fading environments in the imperfect CSI condition is proposed, analyzed, and demonstrated. In this case, the energy and information are transferred from the source to the relay nodes by the power splitting protocol. Firstly, the outage probability, achievable throughput, and the optimal power splitting factor in terms of the analytical mathematical expressions were proposed, analyzed, and demonstrated. Furthermore, the system performance of the proposed model on the connection with all system parameters is rigorously studied. Finally, the numerical results demonstrated and convinced one that the analytical and the simulation results are matched well with each other for all system parameter values using Monte-Carlo simulation. The main contributions of the paper are summarized as follows:(
1
) The system model of FD relaying network over different fading channels in the imperfect CSI condition with the power splitting protocol is proposed and investigated in two cases: (a) S-R link is Rician Fading Channel, and R-D link is Rayleigh Fading Channel; (b) S-R link is Rayleigh Fading Channel, and R-D link is Rician Fading Channel.(
2
) The closed-form of the throughput and outage probability over different fading environments for FD relay system is derived.(
3
) The influence of the main parameters on the system performance is demonstrated entirely.(
4
) The optimal power splitting factor is investigated and calculated in connection with the main system parameters.The rest of this paper is presented as follows. Section2 proposes the system model of the model system. Section 3 proposes and demonstrates the analytical mathematical expressions of the outage probability, throughput, and the optimal power splitting factor of the system. Section 4 presents the numerical results and some discussions. Finally, Section 5 concludes the paper.


## 2. System Model

In this section, the amplify-and-forward (AF) full-duplex (FD) relaying wireless communications network in an imperfect CSI condition is proposed and illustrated in Figure1. In Figure 1, the information is transferred from the source (S) node to the destination (D) node, through energy constrained intermediate relay (R) [7, 20, 21]. In this model, the following set of assumptions are considered:Figure 1
System model, in which a source, a relay, and destination nodes are denoted by S, R, and D, respectively.(
1
) There is no connection between S and R in the results of elimination transmission information (too poor connection between S and R). In particular, an intermediate FD relay is used for transferring the information from the source to the destination in imperfect CSI condition [16, 17].(
2
) The required power of the data decoding process at the relay is negligible in comparison to the signal transmission energy from the relay to the destination. In this paper, the energy harvesting and information processing at the relay node are proposed with the power splitting protocol at the relay node [7].The energy harvesting and information processing at the relay node by the power splitting protocol are proposed in Figure2. In reality, the power-splitting protocol is more effective because it does not waste the time resource for transmission, so the transmission rate is higher than the time-switching protocol. In Figure 2, T is the block time of the energy harvesting and information process. In this model, half of the time, T/2 is used for the source to relay energy harvesting and in the remaining half, T/2 is used for the source to the relay node to the destination node information transmission. During the all-time interval, in the fraction of the received signal power, ρ
P
S is used for energy harvesting and in the remaining received power, 1
-
ρ
P
S is used for transmitting source information to the destination node via the full-duplex helping relay (where ρ is the power splitting factor, ρ
∈
(
0,1
) and P
s is transmission power from the source) [7, 20, 21]. As introduced before, we consider the case that the channel state information is obtained with an error. The effect of CSI error on the performance of the model of interest is the main concentration of this paper. In this system model, the gain channel factors are denoted as follows:(1)
h
=
h
^
+
Δ
h
g
=
g
^
+
Δ
g
f
=
f
^
+
Δ
fFigure 2
The energy harvesting and information processing by the power splitting protocol.whereh and g denote the channel gains from the source to the relay and the relay to the destination, respectively, and f denotes the loopback interference channel at the relay (Figure 1). Moreover, Δ
h, Δ
g, and Δ
f are the channel estimation error corresponding to h
,
g
, and f, respectively. These estimation errors are all assumed to be Gaussian distributed with zero means.


### 2.1. Energy Harvesting

In the energy harvesting process, the RF signal received from the source S at the FD helping relay R is calculated as(2)
y
r
=
h
x
e
+
n
rwhereh is the channel gain factor from S to R and x
e is the energy symbol with Ε
|
x
e
|
2
=
P
s, where Ε
∙ denotes the expectation operation. n
r is the zero-mean additive white Gaussian noise (AWGN) with variance N
0.Hence, the harvested energy can be calculated as(3)
E
h
=
η
P
s
h
^
2
+
δ
Δ
h
2
ρ
T
2In this case, the relay transmits power can be computed as(4)
P
r
=
η
ρ
P
S
h
^
2
+
δ
Δ
h
2
T
/
2
T
/
2
=
η
ρ
P
S
h
^
2
+
δ
Δ
h
2where0
<
η
<
1 is a constant and denotes the energy conversion efficiency and δ
Δ
h
2 is the standard deviation of the channel estimation error Δ
h, which is a zero-mean Gaussian random variable as mentioned above.


### 2.2. Information Transmission

Now the information transmission phase is investigated in more details. Here, the received signal at the relay is calculated by(5)
y
r
=
1
-
ρ
h
x
s
+
f
x
r
+
n
rwherex
s is the transmitted signal, which satisfies Ε
|
x
s
|
2
=
P
s, x
r is the loopback interference due to full-duplex relaying and satisfies Ε
|
x
r
|
2
=
P
r, f denotes the loopback interference channel, and n
r is the zero-mean AWGN with variance N
0. After applying interference cancellation methods to mitigate the loopback interference [22], we have(6)
y
^
r
=
y
r
-
f
^
x
r
=
1
-
ρ
h
^
+
Δ
h
x
s
+
Δ
f
x
R
+
n
rIn this step, the amplifier factor could be present by(7)
β
=
x
r
y
^
r
=
P
r
1
-
ρ
h
^
2
+
δ
Δ
h
2
P
s
+
δ
Δ
f
2
P
r
+
N
0Finally, the received signal at the destination node is(8)
y
d
=
g
x
r
+
n
d
=
g
^
+
Δ
g
x
r
+
n
d
=
g
^
β
1
-
ρ
h
^
+
Δ
h
x
s
+
Δ
f
x
R
+
n
r
+
Δ
g
β
1
-
ρ
h
^
+
Δ
h
x
s
+
Δ
f
x
R
+
n
r
+
n
dwhereg is the channel gain factor from R to D.Also, the end-to-end signal to noise ratio (SNR) can formulate as(9)
γ
e
2
e
=
Ε
s
i
g
n
a
l
2
Ε
n
o
i
s
e
2
=
A
BwhereA
=
(
1
-
ρ
)
g
^
2
h
^
2
P
S and(10)
B
=
1
-
ρ
P
s
δ
Δ
h
2
g
^
2
+
P
r
g
^
2
δ
Δ
f
2
+
g
^
2
N
0
+
1
-
ρ
P
s
δ
Δ
g
2
h
^
2
+
1
-
ρ
P
s
δ
Δ
g
2
δ
Δ
h
2
+
δ
Δ
g
2
P
r
δ
Δ
f
2
+
δ
Δ
g
2
N
0
+
N
0
β
2Combined with (4), (7) and some algebra helping (9), (9) can be rewritten as(11)
γ
e
2
e
=
C
DIn which, we denoteC
=
(
1
-
ρ
)
φ
1
φ
2
γ
0, and D
=
(
1
-
ρ
)
γ
0
δ
Δ
h
2
φ
2 + η
ρ
γ
0
(
φ
1
+
δ
Δ
h
2
)
φ
2
δ
Δ
f
2 + φ
2
+
(
1
-
ρ
)
γ
0
δ
Δ
g
2
φ
1 + (
1
-
ρ
)
γ
0
δ
Δ
g
2
δ
Δ
h
2 + δ
Δ
g
2
η
ρ
γ
0
(
φ
1
+
δ
Δ
h
2
)
δ
Δ
f
2 + δ
Δ
g
2
+
(
1
-
ρ
)
/
η
ρ
+
δ
Δ
f
2.In the above equation, we denoteφ
1
=
h
^
2, φ
2
=
g
^
2, and γ
0
=
P
s
/
N
0.More details of the analytical mathematical model of the achievable throughput, outage probability, and optimal power splitting factor in FD relaying system in the imperfect CSI condition over different fading channels are proposed and presented in details in the following sections.


## 2.1. Energy Harvesting

In the energy harvesting process, the RF signal received from the source S at the FD helping relay R is calculated as(2)
y
r
=
h
x
e
+
n
rwhereh is the channel gain factor from S to R and x
e is the energy symbol with Ε
|
x
e
|
2
=
P
s, where Ε
∙ denotes the expectation operation. n
r is the zero-mean additive white Gaussian noise (AWGN) with variance N
0.Hence, the harvested energy can be calculated as(3)
E
h
=
η
P
s
h
^
2
+
δ
Δ
h
2
ρ
T
2In this case, the relay transmits power can be computed as(4)
P
r
=
η
ρ
P
S
h
^
2
+
δ
Δ
h
2
T
/
2
T
/
2
=
η
ρ
P
S
h
^
2
+
δ
Δ
h
2where0
<
η
<
1 is a constant and denotes the energy conversion efficiency and δ
Δ
h
2 is the standard deviation of the channel estimation error Δ
h, which is a zero-mean Gaussian random variable as mentioned above.


## 2.2. Information Transmission

Now the information transmission phase is investigated in more details. Here, the received signal at the relay is calculated by(5)
y
r
=
1
-
ρ
h
x
s
+
f
x
r
+
n
rwherex
s is the transmitted signal, which satisfies Ε
|
x
s
|
2
=
P
s, x
r is the loopback interference due to full-duplex relaying and satisfies Ε
|
x
r
|
2
=
P
r, f denotes the loopback interference channel, and n
r is the zero-mean AWGN with variance N
0. After applying interference cancellation methods to mitigate the loopback interference [22], we have(6)
y
^
r
=
y
r
-
f
^
x
r
=
1
-
ρ
h
^
+
Δ
h
x
s
+
Δ
f
x
R
+
n
rIn this step, the amplifier factor could be present by(7)
β
=
x
r
y
^
r
=
P
r
1
-
ρ
h
^
2
+
δ
Δ
h
2
P
s
+
δ
Δ
f
2
P
r
+
N
0Finally, the received signal at the destination node is(8)
y
d
=
g
x
r
+
n
d
=
g
^
+
Δ
g
x
r
+
n
d
=
g
^
β
1
-
ρ
h
^
+
Δ
h
x
s
+
Δ
f
x
R
+
n
r
+
Δ
g
β
1
-
ρ
h
^
+
Δ
h
x
s
+
Δ
f
x
R
+
n
r
+
n
dwhereg is the channel gain factor from R to D.Also, the end-to-end signal to noise ratio (SNR) can formulate as(9)
γ
e
2
e
=
Ε
s
i
g
n
a
l
2
Ε
n
o
i
s
e
2
=
A
BwhereA
=
(
1
-
ρ
)
g
^
2
h
^
2
P
S and(10)
B
=
1
-
ρ
P
s
δ
Δ
h
2
g
^
2
+
P
r
g
^
2
δ
Δ
f
2
+
g
^
2
N
0
+
1
-
ρ
P
s
δ
Δ
g
2
h
^
2
+
1
-
ρ
P
s
δ
Δ
g
2
δ
Δ
h
2
+
δ
Δ
g
2
P
r
δ
Δ
f
2
+
δ
Δ
g
2
N
0
+
N
0
β
2Combined with (4), (7) and some algebra helping (9), (9) can be rewritten as(11)
γ
e
2
e
=
C
DIn which, we denoteC
=
(
1
-
ρ
)
φ
1
φ
2
γ
0, and D
=
(
1
-
ρ
)
γ
0
δ
Δ
h
2
φ
2 + η
ρ
γ
0
(
φ
1
+
δ
Δ
h
2
)
φ
2
δ
Δ
f
2 + φ
2
+
(
1
-
ρ
)
γ
0
δ
Δ
g
2
φ
1 + (
1
-
ρ
)
γ
0
δ
Δ
g
2
δ
Δ
h
2 + δ
Δ
g
2
η
ρ
γ
0
(
φ
1
+
δ
Δ
h
2
)
δ
Δ
f
2 + δ
Δ
g
2
+
(
1
-
ρ
)
/
η
ρ
+
δ
Δ
f
2.In the above equation, we denoteφ
1
=
h
^
2, φ
2
=
g
^
2, and γ
0
=
P
s
/
N
0.More details of the analytical mathematical model of the achievable throughput, outage probability, and optimal power splitting factor in FD relaying system in the imperfect CSI condition over different fading channels are proposed and presented in details in the following sections.


## 3. System Performance

Based on the system model on the above section, the system performance analysis of FD relay network is presented, analyzed, and demonstrated with power splitting protocols in details in this sections. In this analysis, two Scenarios are proposed. In the first Scenario, S-R link is Rician Fading Channel; R-D link is Rayleigh Fading Channel. In another way, S-R link is the Rayleigh Fading Channel; R-D link is the Rician Fading Channel in the second Scenario.


### 3.1. Scenario 1: S-R Link Is Rician Fading Channel and R-D Link Is Rayleigh Fading Channel

#### 3.1.1. The Exact Outage Probability

The probability density function (PDF) of a random variable (RV)φ
1 can be calculated as(12)
f
φ
1
x
=
K
+
1
e
-
K
λ
1
e
-
K
+
1
x
/
λ
i
I
0
2
K
K
+
1
x
λ
1Hereλ
1 is the mean value of RV φ
1. K is the Rician K-factor defined as the ratio of the power of the line-of-sight (LOS) component to the scattered components and I
0
∙ is the zero-th order modified Bessel function of the first kind [23].Equation (12) can be reconstructed as follows:(13)
f
φ
1
x
=
ω
∑
l
=
0
∞
b
K
l
l
!
2
x
l
e
-
θ
xwhere ω
=
(
K
+
1
)
e
-
K
/
λ
1, θ
=
K
+
1
/
λ
1, and I
0
(
x
)
=
∑
l
=
0
∞
x
2
l
/
2
2
l
(
l
!
)
2.Like in [7] the cumulative density function (CDF) of RV φ
1 can be computed as(14)
F
φ
1
ς
=
∫
0
ς
f
φ
1
x
d
x
=
1
-
ω
θ
∑
l
=
0
∞
∑
m
=
0
l
K
l
θ
m
l
!
m
!
ς
m
e
-
θ
ςIn the same way, the probability density function (PDF) of a random variable (RV)φ
2 can be expressed as(15)
f
φ
2
x
=
λ
2
e
-
x
λ
2whereλ
2 is the mean value of RV φ
2.Moreover, the cumulative density function (CDF) of RVφ
2 can be computed as(16)
F
φ
2
ψ
=
∫
0
ω
f
φ
1
x
d
x
=
1
-
e
-
ψ
λ
2The outage probability of the model system is given by(17)
P
o
u
t
=
F
γ
e
2
e
γ
t
h
=
P
r
γ
e
2
e
≤
γ
t
h
=
P
r
φ
2
a
1
φ
1
-
a
2
≤
b
1
φ
1
+
b
2where we denote(18)
a
1
=
1
-
ρ
γ
0
-
γ
t
h
η
ρ
γ
0
δ
Δ
f
2
,
a
2
=
γ
t
h
1
-
ρ
γ
0
δ
Δ
h
2
+
η
ρ
γ
0
δ
Δ
h
2
δ
Δ
f
2
+
1
b
1
=
γ
t
h
1
-
ρ
γ
0
δ
Δ
g
2
+
η
ρ
γ
0
δ
Δ
g
2
δ
Δ
f
2
,
b
2
=
γ
t
h
1
-
ρ
γ
0
δ
Δ
h
2
δ
Δ
g
2
+
η
ρ
γ
0
δ
Δ
h
2
δ
Δ
g
2
δ
Δ
f
2
+
δ
Δ
g
2
+
δ
Δ
f
2
+
1
-
ρ
η
ρIn these equations,γ
t
h
=
2
R
-
1 is the system threshold with the source rate R. Now we have(19)
P
r
φ
2
a
1
φ
1
-
a
2
≤
b
1
φ
1
+
b
2
=
P
r
φ
2
≤
b
1
φ
1
+
b
2
a
1
φ
1
-
a
2
,
i
f
φ
1
>
a
2
a
1
1
,
i
f
φ
1
≤
a
2
a
1
=
∫
0
a
2
/
a
1
f
φ
1
φ
1
d
φ
1
+
∫
a
2
/
a
1
∞
F
φ
2
b
1
φ
1
+
b
2
a
1
φ
1
-
a
2
f
φ
1
φ
1
d
φ
1Using (13) and (14) the outage probability can be rewritten as(20)
P
o
u
t
=
1
-
∫
a
2
/
a
1
∞
e
x
p
-
λ
2
b
1
φ
1
+
b
2
a
1
φ
1
-
a
2
×
ω
∑
l
=
0
∞
θ
K
l
l
!
2
φ
1
l
e
-
θ
φ
1
d
φ
1By changing variable and set:t
=
a
1
φ
1
-
a
2, (20) can be changed into(21)
P
o
u
t
=
1
-
ω
a
1
e
-
θ
a
2
/
a
1
-
λ
2
b
1
/
a
1
×
∫
0
∞
∑
l
=
0
∞
θ
K
l
l
!
2
t
+
a
2
a
1
l
×
e
-
θ
t
/
a
1
×
e
-
ξ
/
t
d
twhere ξ
=
λ
2
(
a
1
b
2
+
a
2
b
1
)
/
a
1.After that, by applying the equation,(
x
+
y
)
m
=
∑
n
=
0
m
m
n
x
m
-
n
y
n, we have(22)
P
o
u
t
=
1
-
ω
a
1
e
-
θ
a
2
/
a
1
-
λ
2
b
1
/
a
1
×
∫
0
∞
∑
l
=
0
∞
θ
K
l
a
2
n
a
1
l
l
!
2
∑
n
=
0
l
l
n
×
t
l
-
n
×
e
-
θ
t
/
a
1
×
e
-
ξ
/
t
d
t
(23)
P
o
u
t
=
1
-
ω
a
1
e
-
θ
a
2
/
a
1
-
λ
2
b
1
/
a
1
×
∑
l
=
0
∞
∑
n
=
0
l
θ
K
l
a
2
n
l
!
n
!
l
-
n
!
a
1
l
×
∫
0
∞
t
l
-
n
×
e
-
θ
t
/
a
1
×
e
-
ξ
/
t
d
tUsing Table of Integral Equation[
3.471,9
] [24], the final exact-form of the outage probability can be expressed as(24)
P
o
u
t
=
1
-
2
ω
a
1
e
-
θ
a
2
/
a
1
-
λ
2
b
1
/
a
1
×
∑
l
=
0
∞
∑
n
=
0
l
θ
K
l
a
2
n
l
!
n
!
l
-
n
!
a
1
l
×
ξ
a
1
θ
l
-
n
+
1
/
2
×
K
l
-
n
+
1
2
ξ
θ
a
1
P
o
u
t
=
1
-
2
ω
×
e
-
θ
a
2
/
a
1
-
λ
2
b
1
/
a
1
×
∑
l
=
0
∞
∑
n
=
0
l
K
l
θ
l
+
n
-
1
/
2
a
2
n
ξ
l
-
n
+
1
/
2
l
!
n
!
l
-
n
!
a
1
l
+
n
+
1
/
2
×
K
l
-
n
+
1
2
ξ
θ
a
1whereK
v
(
∙
) is the modified Bessel function of the second kind and v
t
h order.


#### 3.1.2. Average Throughput

From the above analysis, the average throughput can be calculated by(25)
τ
=
1
-
P
o
u
t
R
2
=
2
ω
×
e
-
θ
a
2
/
a
1
-
λ
2
b
1
/
a
1
×
∑
l
=
0
∞
∑
n
=
0
l
K
l
θ
l
+
n
-
1
/
2
a
2
n
ξ
l
-
n
+
1
/
2
l
!
n
!
l
-
n
!
a
1
l
+
n
+
1
/
2
×
K
l
-
n
+
1
2
ξ
θ
a
1
×
R
2


#### 3.1.3. Asymptotic Outage Probability and Average Throughput

In high SNR regime, (11) can be recomputed as(26)
γ
e
2
e
∞
=
E
FwhereE
=
(
1
-
ρ
)
φ
1
φ
2 and(27)
F
=
1
-
ρ
δ
Δ
h
2
φ
2
+
η
ρ
φ
1
+
δ
Δ
h
2
φ
2
δ
Δ
f
2
+
1
-
ρ
δ
Δ
g
2
φ
1
+
1
-
ρ
δ
Δ
g
2
δ
Δ
h
2
+
δ
Δ
g
2
η
ρ
φ
1
+
δ
Δ
h
2
δ
Δ
f
2In this case, the outage probability at the high SNR can be formulated as(28)
P
o
u
t
∞
=
F
γ
e
2
e
∞
γ
t
h
=
P
r
γ
e
2
e
∞
≤
γ
t
h
=
P
r
φ
2
a
3
φ
1
-
a
4
≤
b
3
φ
1
+
b
4In which, we denoted a
3
=
(
1
-
ρ
)
-
γ
t
h
η
ρ
δ
Δ
f
2, a
4
=
γ
t
h
(
1
-
ρ
)
δ
Δ
h
2
+
η
ρ
δ
Δ
h
2
δ
Δ
f
2, and(29)
b
3
=
γ
t
h
1
-
ρ
δ
Δ
g
2
+
η
ρ
δ
Δ
g
2
δ
Δ
f
2
,
b
4
=
γ
t
h
1
-
ρ
δ
Δ
h
2
δ
Δ
g
2
+
η
ρ
δ
Δ
h
2
δ
Δ
g
2
δ
Δ
f
2
.Similar to the exact case, we have:(30)
P
o
u
t
∞
=
1
-
2
ω
×
e
-
θ
a
4
/
a
3
-
λ
2
b
3
/
a
3
×
∑
l
=
0
∞
∑
n
=
0
l
K
l
θ
l
+
n
-
1
/
2
a
4
n
μ
l
-
n
+
1
/
2
l
!
n
!
l
-
n
!
a
3
l
+
n
+
1
/
2
×
K
l
-
n
+
1
2
μ
θ
a
3where μ
=
λ
2
(
a
3
b
4
+
a
4
b
3
)
/
a
3.Moreover, the average throughput is(31)
τ
∞
=
1
-
P
o
u
t
R
2
=
2
ω
×
e
-
θ
a
4
/
a
3
-
λ
2
b
3
/
a
3
×
∑
l
=
0
∞
∑
n
=
0
l
K
l
θ
l
+
n
-
1
/
2
a
4
n
μ
l
-
n
+
1
/
2
l
!
n
!
l
-
n
!
a
3
l
+
n
+
1
/
2
×
K
l
-
n
+
1
2
μ
θ
a
3
×
R
2


### 3.2. Scenario 2: S-R Link Is Rayleigh Fading Channel and R-D Link Is Rician Fading Channel

#### 3.2.1. Exact Probability and Average Throughput

Similar to scenario 1, the cumulative density function (CDF) of RVφ
2 can be computed as(32)
F
φ
2
ς
=
∫
0
ς
f
φ
2
x
d
x
=
1
-
a
b
∑
l
=
0
∞
∑
m
=
0
l
K
l
b
m
l
!
m
!
ς
m
e
-
b
ςwhere a
=
(
K
+
1
)
e
-
K
/
λ
2, b
=
K
+
1
/
λ
2.The probability density function (PDF) of a random variable (RV)φ
1 is(33)
f
φ
1
x
=
λ
1
e
-
x
λ
1Also, the outage probability can be formulated as(34)
P
o
u
t
=
∫
0
a
2
/
a
1
f
φ
1
φ
1
d
φ
1
+
∫
a
2
/
a
1
∞
F
φ
2
b
1
φ
1
+
b
2
a
1
φ
1
-
a
2
f
φ
1
φ
1
d
φ
1Using (32) and (33) we can reformulate (21) as(35)
P
o
u
t
=
1
-
λ
1
×
∫
a
2
/
a
1
∞
a
b
∑
l
=
0
∞
∑
m
=
0
l
K
l
b
m
l
!
m
!
b
1
φ
1
+
b
2
a
1
φ
1
-
a
2
m
e
-
b
b
1
φ
1
+
b
2
/
a
1
φ
1
-
a
2
×
e
-
λ
1
φ
1
d
φ
1By changing the variable, t
=
1
/
a
1
φ
1
-
a
2 (35) can be rewritten as(36)
P
o
u
t
=
1
-
a
a
1
b
λ
1
∫
0
∞
∑
l
=
0
∞
∑
m
=
0
l
K
l
b
m
l
!
m
!
b
1
a
1
+
κ
t
m
t
-
2
e
-
b
b
1
/
a
1
+
κ
t
×
e
-
λ
1
1
/
a
1
t
+
a
2
/
a
1
d
t
P
o
u
t
=
1
-
a
λ
1
a
1
b
×
e
-
b
b
1
/
a
1
-
λ
1
a
2
/
a
1
∫
0
∞
∑
l
=
0
∞
∑
m
=
0
l
K
l
b
m
l
!
m
!
b
1
a
1
+
κ
t
m
t
-
2
e
-
b
κ
t
×
e
-
λ
1
/
a
1
t
d
t where κ
=
b
1
a
2
/
a
1
+
b
2.Equation (36) becomes as follows:(37)
P
o
u
t
=
1
-
a
λ
1
a
1
b
×
e
-
b
b
1
/
a
1
-
λ
1
a
2
/
a
1
∫
0
∞
∑
l
=
0
∞
∑
m
=
0
l
K
l
b
m
κ
n
l
!
m
!
∑
n
=
0
m
m
n
b
1
a
1
m
-
n
×
t
n
-
2
e
-
b
κ
t
×
e
-
λ
1
/
a
1
t
d
t
P
o
u
t
=
1
-
a
λ
1
a
1
×
e
-
b
b
1
/
a
1
-
λ
1
a
2
/
a
1
∑
l
=
0
∞
∑
m
=
0
l
∑
n
=
0
m
K
l
b
m
-
1
κ
n
l
!
n
!
m
-
n
!
b
1
a
1
m
-
n
∫
0
∞
t
n
-
2
×
e
-
b
κ
t
×
e
-
λ
1
/
a
1
t
d
tUsing Table of Integral Equation[
3.471,9
] in [24], (37) can be changed into(38)
P
o
u
t
=
1
-
2
a
λ
1
a
1
×
e
-
b
b
1
/
a
1
-
λ
1
a
2
/
a
1
∑
l
=
0
∞
∑
m
=
0
l
∑
n
=
0
m
K
l
b
m
-
1
κ
n
l
!
n
!
m
-
n
!
b
1
a
1
m
-
n
λ
1
a
1
b
κ
n
-
1
/
2
×
K
n
-
1
2
λ
1
b
κ
a
1
P
o
u
t
=
1
-
2
a
×
e
-
b
b
1
/
a
1
-
λ
1
a
2
/
a
1
∑
l
=
0
∞
∑
m
=
0
l
∑
n
=
0
m
K
l
b
2
m
-
n
-
1
/
2
κ
n
+
1
/
2
b
1
m
-
n
λ
1
n
+
1
/
2
l
!
n
!
m
-
n
!
a
1
2
m
-
n
+
1
/
2
×
K
n
-
1
2
λ
1
b
κ
a
1Then the average throughput can be calculated as(39)
τ
=
1
-
P
o
u
t
R
2
=
2
a
×
e
-
b
b
1
/
a
1
-
λ
1
a
2
/
a
1
∑
l
=
0
∞
∑
m
=
0
l
∑
n
=
0
m
K
l
b
2
m
-
n
-
1
/
2
κ
n
+
1
/
2
b
1
m
-
n
λ
1
n
+
1
/
2
l
!
n
!
m
-
n
!
a
1
2
m
-
n
+
1
/
2
×
K
n
-
1
2
λ
1
b
κ
a
1
×
R
2


#### 3.2.2. Asymptotic Outage Probability and Average Throughput

Similar proof as scenario 1 we have(40)
P
o
u
t
∞
=
1
-
2
a
×
e
-
b
b
3
/
a
3
-
λ
1
a
4
/
a
3
∑
l
=
0
∞
∑
m
=
0
l
∑
n
=
0
m
K
l
b
2
m
-
n
-
1
/
2
ϑ
n
+
1
/
2
b
3
m
-
n
λ
1
n
+
1
/
2
l
!
n
!
m
-
n
!
a
3
2
m
-
n
+
1
/
2
×
K
n
-
1
2
λ
1
b
ϑ
a
3where ϑ
=
b
3
a
4
/
a
3
+
b
4.Moreover, the average throughput can formulate as(41)
τ
∞
=
1
-
P
o
u
t
R
2
=
2
a
×
e
-
b
b
3
/
a
3
-
λ
1
a
4
/
a
3
∑
l
=
0
∞
∑
m
=
0
l
∑
n
=
0
m
K
l
b
2
m
-
n
-
1
/
2
ϑ
n
+
1
/
2
b
3
m
-
n
λ
1
n
+
1
/
2
l
!
n
!
m
-
n
!
a
3
2
m
-
n
+
1
/
2
×
K
n
-
1
2
λ
1
b
ϑ
a
3
×
R
2


### 3.3. Optimal Power Splitting Factor

The optimal valueρ
∗ can be obtained by solving the equation d
τ
(
ρ
)
/
d
ρ
=
0. Given the average throughput expression in (25) and (39), this optimization problem does not admit a closed-form solution. However, the optimal ρ
∗ is efficiently solved via numerical calculation, as illustrated below [25–27].Here, we can use the Golden section search algorithm to find the optimal factorρ
∗. This algorithm has been used in many global optimization problems in communications, for example, in [25]. The detailed algorithm, as well as the related theory, is described in [25–27].


## 3.1. Scenario 1: S-R Link Is Rician Fading Channel and R-D Link Is Rayleigh Fading Channel

### 3.1.1. The Exact Outage Probability

The probability density function (PDF) of a random variable (RV)φ
1 can be calculated as(12)
f
φ
1
x
=
K
+
1
e
-
K
λ
1
e
-
K
+
1
x
/
λ
i
I
0
2
K
K
+
1
x
λ
1Hereλ
1 is the mean value of RV φ
1. K is the Rician K-factor defined as the ratio of the power of the line-of-sight (LOS) component to the scattered components and I
0
∙ is the zero-th order modified Bessel function of the first kind [23].Equation (12) can be reconstructed as follows:(13)
f
φ
1
x
=
ω
∑
l
=
0
∞
b
K
l
l
!
2
x
l
e
-
θ
xwhere ω
=
(
K
+
1
)
e
-
K
/
λ
1, θ
=
K
+
1
/
λ
1, and I
0
(
x
)
=
∑
l
=
0
∞
x
2
l
/
2
2
l
(
l
!
)
2.Like in [7] the cumulative density function (CDF) of RV φ
1 can be computed as(14)
F
φ
1
ς
=
∫
0
ς
f
φ
1
x
d
x
=
1
-
ω
θ
∑
l
=
0
∞
∑
m
=
0
l
K
l
θ
m
l
!
m
!
ς
m
e
-
θ
ςIn the same way, the probability density function (PDF) of a random variable (RV)φ
2 can be expressed as(15)
f
φ
2
x
=
λ
2
e
-
x
λ
2whereλ
2 is the mean value of RV φ
2.Moreover, the cumulative density function (CDF) of RVφ
2 can be computed as(16)
F
φ
2
ψ
=
∫
0
ω
f
φ
1
x
d
x
=
1
-
e
-
ψ
λ
2The outage probability of the model system is given by(17)
P
o
u
t
=
F
γ
e
2
e
γ
t
h
=
P
r
γ
e
2
e
≤
γ
t
h
=
P
r
φ
2
a
1
φ
1
-
a
2
≤
b
1
φ
1
+
b
2where we denote(18)
a
1
=
1
-
ρ
γ
0
-
γ
t
h
η
ρ
γ
0
δ
Δ
f
2
,
a
2
=
γ
t
h
1
-
ρ
γ
0
δ
Δ
h
2
+
η
ρ
γ
0
δ
Δ
h
2
δ
Δ
f
2
+
1
b
1
=
γ
t
h
1
-
ρ
γ
0
δ
Δ
g
2
+
η
ρ
γ
0
δ
Δ
g
2
δ
Δ
f
2
,
b
2
=
γ
t
h
1
-
ρ
γ
0
δ
Δ
h
2
δ
Δ
g
2
+
η
ρ
γ
0
δ
Δ
h
2
δ
Δ
g
2
δ
Δ
f
2
+
δ
Δ
g
2
+
δ
Δ
f
2
+
1
-
ρ
η
ρIn these equations,γ
t
h
=
2
R
-
1 is the system threshold with the source rate R. Now we have(19)
P
r
φ
2
a
1
φ
1
-
a
2
≤
b
1
φ
1
+
b
2
=
P
r
φ
2
≤
b
1
φ
1
+
b
2
a
1
φ
1
-
a
2
,
i
f
φ
1
>
a
2
a
1
1
,
i
f
φ
1
≤
a
2
a
1
=
∫
0
a
2
/
a
1
f
φ
1
φ
1
d
φ
1
+
∫
a
2
/
a
1
∞
F
φ
2
b
1
φ
1
+
b
2
a
1
φ
1
-
a
2
f
φ
1
φ
1
d
φ
1Using (13) and (14) the outage probability can be rewritten as(20)
P
o
u
t
=
1
-
∫
a
2
/
a
1
∞
e
x
p
-
λ
2
b
1
φ
1
+
b
2
a
1
φ
1
-
a
2
×
ω
∑
l
=
0
∞
θ
K
l
l
!
2
φ
1
l
e
-
θ
φ
1
d
φ
1By changing variable and set:t
=
a
1
φ
1
-
a
2, (20) can be changed into(21)
P
o
u
t
=
1
-
ω
a
1
e
-
θ
a
2
/
a
1
-
λ
2
b
1
/
a
1
×
∫
0
∞
∑
l
=
0
∞
θ
K
l
l
!
2
t
+
a
2
a
1
l
×
e
-
θ
t
/
a
1
×
e
-
ξ
/
t
d
twhere ξ
=
λ
2
(
a
1
b
2
+
a
2
b
1
)
/
a
1.After that, by applying the equation,(
x
+
y
)
m
=
∑
n
=
0
m
m
n
x
m
-
n
y
n, we have(22)
P
o
u
t
=
1
-
ω
a
1
e
-
θ
a
2
/
a
1
-
λ
2
b
1
/
a
1
×
∫
0
∞
∑
l
=
0
∞
θ
K
l
a
2
n
a
1
l
l
!
2
∑
n
=
0
l
l
n
×
t
l
-
n
×
e
-
θ
t
/
a
1
×
e
-
ξ
/
t
d
t
(23)
P
o
u
t
=
1
-
ω
a
1
e
-
θ
a
2
/
a
1
-
λ
2
b
1
/
a
1
×
∑
l
=
0
∞
∑
n
=
0
l
θ
K
l
a
2
n
l
!
n
!
l
-
n
!
a
1
l
×
∫
0
∞
t
l
-
n
×
e
-
θ
t
/
a
1
×
e
-
ξ
/
t
d
tUsing Table of Integral Equation[
3.471,9
] [24], the final exact-form of the outage probability can be expressed as(24)
P
o
u
t
=
1
-
2
ω
a
1
e
-
θ
a
2
/
a
1
-
λ
2
b
1
/
a
1
×
∑
l
=
0
∞
∑
n
=
0
l
θ
K
l
a
2
n
l
!
n
!
l
-
n
!
a
1
l
×
ξ
a
1
θ
l
-
n
+
1
/
2
×
K
l
-
n
+
1
2
ξ
θ
a
1
P
o
u
t
=
1
-
2
ω
×
e
-
θ
a
2
/
a
1
-
λ
2
b
1
/
a
1
×
∑
l
=
0
∞
∑
n
=
0
l
K
l
θ
l
+
n
-
1
/
2
a
2
n
ξ
l
-
n
+
1
/
2
l
!
n
!
l
-
n
!
a
1
l
+
n
+
1
/
2
×
K
l
-
n
+
1
2
ξ
θ
a
1whereK
v
(
∙
) is the modified Bessel function of the second kind and v
t
h order.


### 3.1.2. Average Throughput

From the above analysis, the average throughput can be calculated by(25)
τ
=
1
-
P
o
u
t
R
2
=
2
ω
×
e
-
θ
a
2
/
a
1
-
λ
2
b
1
/
a
1
×
∑
l
=
0
∞
∑
n
=
0
l
K
l
θ
l
+
n
-
1
/
2
a
2
n
ξ
l
-
n
+
1
/
2
l
!
n
!
l
-
n
!
a
1
l
+
n
+
1
/
2
×
K
l
-
n
+
1
2
ξ
θ
a
1
×
R
2


### 3.1.3. Asymptotic Outage Probability and Average Throughput

In high SNR regime, (11) can be recomputed as(26)
γ
e
2
e
∞
=
E
FwhereE
=
(
1
-
ρ
)
φ
1
φ
2 and(27)
F
=
1
-
ρ
δ
Δ
h
2
φ
2
+
η
ρ
φ
1
+
δ
Δ
h
2
φ
2
δ
Δ
f
2
+
1
-
ρ
δ
Δ
g
2
φ
1
+
1
-
ρ
δ
Δ
g
2
δ
Δ
h
2
+
δ
Δ
g
2
η
ρ
φ
1
+
δ
Δ
h
2
δ
Δ
f
2In this case, the outage probability at the high SNR can be formulated as(28)
P
o
u
t
∞
=
F
γ
e
2
e
∞
γ
t
h
=
P
r
γ
e
2
e
∞
≤
γ
t
h
=
P
r
φ
2
a
3
φ
1
-
a
4
≤
b
3
φ
1
+
b
4In which, we denoted a
3
=
(
1
-
ρ
)
-
γ
t
h
η
ρ
δ
Δ
f
2, a
4
=
γ
t
h
(
1
-
ρ
)
δ
Δ
h
2
+
η
ρ
δ
Δ
h
2
δ
Δ
f
2, and(29)
b
3
=
γ
t
h
1
-
ρ
δ
Δ
g
2
+
η
ρ
δ
Δ
g
2
δ
Δ
f
2
,
b
4
=
γ
t
h
1
-
ρ
δ
Δ
h
2
δ
Δ
g
2
+
η
ρ
δ
Δ
h
2
δ
Δ
g
2
δ
Δ
f
2
.Similar to the exact case, we have:(30)
P
o
u
t
∞
=
1
-
2
ω
×
e
-
θ
a
4
/
a
3
-
λ
2
b
3
/
a
3
×
∑
l
=
0
∞
∑
n
=
0
l
K
l
θ
l
+
n
-
1
/
2
a
4
n
μ
l
-
n
+
1
/
2
l
!
n
!
l
-
n
!
a
3
l
+
n
+
1
/
2
×
K
l
-
n
+
1
2
μ
θ
a
3where μ
=
λ
2
(
a
3
b
4
+
a
4
b
3
)
/
a
3.Moreover, the average throughput is(31)
τ
∞
=
1
-
P
o
u
t
R
2
=
2
ω
×
e
-
θ
a
4
/
a
3
-
λ
2
b
3
/
a
3
×
∑
l
=
0
∞
∑
n
=
0
l
K
l
θ
l
+
n
-
1
/
2
a
4
n
μ
l
-
n
+
1
/
2
l
!
n
!
l
-
n
!
a
3
l
+
n
+
1
/
2
×
K
l
-
n
+
1
2
μ
θ
a
3
×
R
2


## 3.1.1. The Exact Outage Probability

The probability density function (PDF) of a random variable (RV)φ
1 can be calculated as(12)
f
φ
1
x
=
K
+
1
e
-
K
λ
1
e
-
K
+
1
x
/
λ
i
I
0
2
K
K
+
1
x
λ
1Hereλ
1 is the mean value of RV φ
1. K is the Rician K-factor defined as the ratio of the power of the line-of-sight (LOS) component to the scattered components and I
0
∙ is the zero-th order modified Bessel function of the first kind [23].Equation (12) can be reconstructed as follows:(13)
f
φ
1
x
=
ω
∑
l
=
0
∞
b
K
l
l
!
2
x
l
e
-
θ
xwhere ω
=
(
K
+
1
)
e
-
K
/
λ
1, θ
=
K
+
1
/
λ
1, and I
0
(
x
)
=
∑
l
=
0
∞
x
2
l
/
2
2
l
(
l
!
)
2.Like in [7] the cumulative density function (CDF) of RV φ
1 can be computed as(14)
F
φ
1
ς
=
∫
0
ς
f
φ
1
x
d
x
=
1
-
ω
θ
∑
l
=
0
∞
∑
m
=
0
l
K
l
θ
m
l
!
m
!
ς
m
e
-
θ
ςIn the same way, the probability density function (PDF) of a random variable (RV)φ
2 can be expressed as(15)
f
φ
2
x
=
λ
2
e
-
x
λ
2whereλ
2 is the mean value of RV φ
2.Moreover, the cumulative density function (CDF) of RVφ
2 can be computed as(16)
F
φ
2
ψ
=
∫
0
ω
f
φ
1
x
d
x
=
1
-
e
-
ψ
λ
2The outage probability of the model system is given by(17)
P
o
u
t
=
F
γ
e
2
e
γ
t
h
=
P
r
γ
e
2
e
≤
γ
t
h
=
P
r
φ
2
a
1
φ
1
-
a
2
≤
b
1
φ
1
+
b
2where we denote(18)
a
1
=
1
-
ρ
γ
0
-
γ
t
h
η
ρ
γ
0
δ
Δ
f
2
,
a
2
=
γ
t
h
1
-
ρ
γ
0
δ
Δ
h
2
+
η
ρ
γ
0
δ
Δ
h
2
δ
Δ
f
2
+
1
b
1
=
γ
t
h
1
-
ρ
γ
0
δ
Δ
g
2
+
η
ρ
γ
0
δ
Δ
g
2
δ
Δ
f
2
,
b
2
=
γ
t
h
1
-
ρ
γ
0
δ
Δ
h
2
δ
Δ
g
2
+
η
ρ
γ
0
δ
Δ
h
2
δ
Δ
g
2
δ
Δ
f
2
+
δ
Δ
g
2
+
δ
Δ
f
2
+
1
-
ρ
η
ρIn these equations,γ
t
h
=
2
R
-
1 is the system threshold with the source rate R. Now we have(19)
P
r
φ
2
a
1
φ
1
-
a
2
≤
b
1
φ
1
+
b
2
=
P
r
φ
2
≤
b
1
φ
1
+
b
2
a
1
φ
1
-
a
2
,
i
f
φ
1
>
a
2
a
1
1
,
i
f
φ
1
≤
a
2
a
1
=
∫
0
a
2
/
a
1
f
φ
1
φ
1
d
φ
1
+
∫
a
2
/
a
1
∞
F
φ
2
b
1
φ
1
+
b
2
a
1
φ
1
-
a
2
f
φ
1
φ
1
d
φ
1Using (13) and (14) the outage probability can be rewritten as(20)
P
o
u
t
=
1
-
∫
a
2
/
a
1
∞
e
x
p
-
λ
2
b
1
φ
1
+
b
2
a
1
φ
1
-
a
2
×
ω
∑
l
=
0
∞
θ
K
l
l
!
2
φ
1
l
e
-
θ
φ
1
d
φ
1By changing variable and set:t
=
a
1
φ
1
-
a
2, (20) can be changed into(21)
P
o
u
t
=
1
-
ω
a
1
e
-
θ
a
2
/
a
1
-
λ
2
b
1
/
a
1
×
∫
0
∞
∑
l
=
0
∞
θ
K
l
l
!
2
t
+
a
2
a
1
l
×
e
-
θ
t
/
a
1
×
e
-
ξ
/
t
d
twhere ξ
=
λ
2
(
a
1
b
2
+
a
2
b
1
)
/
a
1.After that, by applying the equation,(
x
+
y
)
m
=
∑
n
=
0
m
m
n
x
m
-
n
y
n, we have(22)
P
o
u
t
=
1
-
ω
a
1
e
-
θ
a
2
/
a
1
-
λ
2
b
1
/
a
1
×
∫
0
∞
∑
l
=
0
∞
θ
K
l
a
2
n
a
1
l
l
!
2
∑
n
=
0
l
l
n
×
t
l
-
n
×
e
-
θ
t
/
a
1
×
e
-
ξ
/
t
d
t
(23)
P
o
u
t
=
1
-
ω
a
1
e
-
θ
a
2
/
a
1
-
λ
2
b
1
/
a
1
×
∑
l
=
0
∞
∑
n
=
0
l
θ
K
l
a
2
n
l
!
n
!
l
-
n
!
a
1
l
×
∫
0
∞
t
l
-
n
×
e
-
θ
t
/
a
1
×
e
-
ξ
/
t
d
tUsing Table of Integral Equation[
3.471,9
] [24], the final exact-form of the outage probability can be expressed as(24)
P
o
u
t
=
1
-
2
ω
a
1
e
-
θ
a
2
/
a
1
-
λ
2
b
1
/
a
1
×
∑
l
=
0
∞
∑
n
=
0
l
θ
K
l
a
2
n
l
!
n
!
l
-
n
!
a
1
l
×
ξ
a
1
θ
l
-
n
+
1
/
2
×
K
l
-
n
+
1
2
ξ
θ
a
1
P
o
u
t
=
1
-
2
ω
×
e
-
θ
a
2
/
a
1
-
λ
2
b
1
/
a
1
×
∑
l
=
0
∞
∑
n
=
0
l
K
l
θ
l
+
n
-
1
/
2
a
2
n
ξ
l
-
n
+
1
/
2
l
!
n
!
l
-
n
!
a
1
l
+
n
+
1
/
2
×
K
l
-
n
+
1
2
ξ
θ
a
1whereK
v
(
∙
) is the modified Bessel function of the second kind and v
t
h order.


## 3.1.2. Average Throughput

From the above analysis, the average throughput can be calculated by(25)
τ
=
1
-
P
o
u
t
R
2
=
2
ω
×
e
-
θ
a
2
/
a
1
-
λ
2
b
1
/
a
1
×
∑
l
=
0
∞
∑
n
=
0
l
K
l
θ
l
+
n
-
1
/
2
a
2
n
ξ
l
-
n
+
1
/
2
l
!
n
!
l
-
n
!
a
1
l
+
n
+
1
/
2
×
K
l
-
n
+
1
2
ξ
θ
a
1
×
R
2


## 3.1.3. Asymptotic Outage Probability and Average Throughput

In high SNR regime, (11) can be recomputed as(26)
γ
e
2
e
∞
=
E
FwhereE
=
(
1
-
ρ
)
φ
1
φ
2 and(27)
F
=
1
-
ρ
δ
Δ
h
2
φ
2
+
η
ρ
φ
1
+
δ
Δ
h
2
φ
2
δ
Δ
f
2
+
1
-
ρ
δ
Δ
g
2
φ
1
+
1
-
ρ
δ
Δ
g
2
δ
Δ
h
2
+
δ
Δ
g
2
η
ρ
φ
1
+
δ
Δ
h
2
δ
Δ
f
2In this case, the outage probability at the high SNR can be formulated as(28)
P
o
u
t
∞
=
F
γ
e
2
e
∞
γ
t
h
=
P
r
γ
e
2
e
∞
≤
γ
t
h
=
P
r
φ
2
a
3
φ
1
-
a
4
≤
b
3
φ
1
+
b
4In which, we denoted a
3
=
(
1
-
ρ
)
-
γ
t
h
η
ρ
δ
Δ
f
2, a
4
=
γ
t
h
(
1
-
ρ
)
δ
Δ
h
2
+
η
ρ
δ
Δ
h
2
δ
Δ
f
2, and(29)
b
3
=
γ
t
h
1
-
ρ
δ
Δ
g
2
+
η
ρ
δ
Δ
g
2
δ
Δ
f
2
,
b
4
=
γ
t
h
1
-
ρ
δ
Δ
h
2
δ
Δ
g
2
+
η
ρ
δ
Δ
h
2
δ
Δ
g
2
δ
Δ
f
2
.Similar to the exact case, we have:(30)
P
o
u
t
∞
=
1
-
2
ω
×
e
-
θ
a
4
/
a
3
-
λ
2
b
3
/
a
3
×
∑
l
=
0
∞
∑
n
=
0
l
K
l
θ
l
+
n
-
1
/
2
a
4
n
μ
l
-
n
+
1
/
2
l
!
n
!
l
-
n
!
a
3
l
+
n
+
1
/
2
×
K
l
-
n
+
1
2
μ
θ
a
3where μ
=
λ
2
(
a
3
b
4
+
a
4
b
3
)
/
a
3.Moreover, the average throughput is(31)
τ
∞
=
1
-
P
o
u
t
R
2
=
2
ω
×
e
-
θ
a
4
/
a
3
-
λ
2
b
3
/
a
3
×
∑
l
=
0
∞
∑
n
=
0
l
K
l
θ
l
+
n
-
1
/
2
a
4
n
μ
l
-
n
+
1
/
2
l
!
n
!
l
-
n
!
a
3
l
+
n
+
1
/
2
×
K
l
-
n
+
1
2
μ
θ
a
3
×
R
2


## 3.2. Scenario 2: S-R Link Is Rayleigh Fading Channel and R-D Link Is Rician Fading Channel

### 3.2.1. Exact Probability and Average Throughput

Similar to scenario 1, the cumulative density function (CDF) of RVφ
2 can be computed as(32)
F
φ
2
ς
=
∫
0
ς
f
φ
2
x
d
x
=
1
-
a
b
∑
l
=
0
∞
∑
m
=
0
l
K
l
b
m
l
!
m
!
ς
m
e
-
b
ςwhere a
=
(
K
+
1
)
e
-
K
/
λ
2, b
=
K
+
1
/
λ
2.The probability density function (PDF) of a random variable (RV)φ
1 is(33)
f
φ
1
x
=
λ
1
e
-
x
λ
1Also, the outage probability can be formulated as(34)
P
o
u
t
=
∫
0
a
2
/
a
1
f
φ
1
φ
1
d
φ
1
+
∫
a
2
/
a
1
∞
F
φ
2
b
1
φ
1
+
b
2
a
1
φ
1
-
a
2
f
φ
1
φ
1
d
φ
1Using (32) and (33) we can reformulate (21) as(35)
P
o
u
t
=
1
-
λ
1
×
∫
a
2
/
a
1
∞
a
b
∑
l
=
0
∞
∑
m
=
0
l
K
l
b
m
l
!
m
!
b
1
φ
1
+
b
2
a
1
φ
1
-
a
2
m
e
-
b
b
1
φ
1
+
b
2
/
a
1
φ
1
-
a
2
×
e
-
λ
1
φ
1
d
φ
1By changing the variable, t
=
1
/
a
1
φ
1
-
a
2 (35) can be rewritten as(36)
P
o
u
t
=
1
-
a
a
1
b
λ
1
∫
0
∞
∑
l
=
0
∞
∑
m
=
0
l
K
l
b
m
l
!
m
!
b
1
a
1
+
κ
t
m
t
-
2
e
-
b
b
1
/
a
1
+
κ
t
×
e
-
λ
1
1
/
a
1
t
+
a
2
/
a
1
d
t
P
o
u
t
=
1
-
a
λ
1
a
1
b
×
e
-
b
b
1
/
a
1
-
λ
1
a
2
/
a
1
∫
0
∞
∑
l
=
0
∞
∑
m
=
0
l
K
l
b
m
l
!
m
!
b
1
a
1
+
κ
t
m
t
-
2
e
-
b
κ
t
×
e
-
λ
1
/
a
1
t
d
t where κ
=
b
1
a
2
/
a
1
+
b
2.Equation (36) becomes as follows:(37)
P
o
u
t
=
1
-
a
λ
1
a
1
b
×
e
-
b
b
1
/
a
1
-
λ
1
a
2
/
a
1
∫
0
∞
∑
l
=
0
∞
∑
m
=
0
l
K
l
b
m
κ
n
l
!
m
!
∑
n
=
0
m
m
n
b
1
a
1
m
-
n
×
t
n
-
2
e
-
b
κ
t
×
e
-
λ
1
/
a
1
t
d
t
P
o
u
t
=
1
-
a
λ
1
a
1
×
e
-
b
b
1
/
a
1
-
λ
1
a
2
/
a
1
∑
l
=
0
∞
∑
m
=
0
l
∑
n
=
0
m
K
l
b
m
-
1
κ
n
l
!
n
!
m
-
n
!
b
1
a
1
m
-
n
∫
0
∞
t
n
-
2
×
e
-
b
κ
t
×
e
-
λ
1
/
a
1
t
d
tUsing Table of Integral Equation[
3.471,9
] in [24], (37) can be changed into(38)
P
o
u
t
=
1
-
2
a
λ
1
a
1
×
e
-
b
b
1
/
a
1
-
λ
1
a
2
/
a
1
∑
l
=
0
∞
∑
m
=
0
l
∑
n
=
0
m
K
l
b
m
-
1
κ
n
l
!
n
!
m
-
n
!
b
1
a
1
m
-
n
λ
1
a
1
b
κ
n
-
1
/
2
×
K
n
-
1
2
λ
1
b
κ
a
1
P
o
u
t
=
1
-
2
a
×
e
-
b
b
1
/
a
1
-
λ
1
a
2
/
a
1
∑
l
=
0
∞
∑
m
=
0
l
∑
n
=
0
m
K
l
b
2
m
-
n
-
1
/
2
κ
n
+
1
/
2
b
1
m
-
n
λ
1
n
+
1
/
2
l
!
n
!
m
-
n
!
a
1
2
m
-
n
+
1
/
2
×
K
n
-
1
2
λ
1
b
κ
a
1Then the average throughput can be calculated as(39)
τ
=
1
-
P
o
u
t
R
2
=
2
a
×
e
-
b
b
1
/
a
1
-
λ
1
a
2
/
a
1
∑
l
=
0
∞
∑
m
=
0
l
∑
n
=
0
m
K
l
b
2
m
-
n
-
1
/
2
κ
n
+
1
/
2
b
1
m
-
n
λ
1
n
+
1
/
2
l
!
n
!
m
-
n
!
a
1
2
m
-
n
+
1
/
2
×
K
n
-
1
2
λ
1
b
κ
a
1
×
R
2


### 3.2.2. Asymptotic Outage Probability and Average Throughput

Similar proof as scenario 1 we have(40)
P
o
u
t
∞
=
1
-
2
a
×
e
-
b
b
3
/
a
3
-
λ
1
a
4
/
a
3
∑
l
=
0
∞
∑
m
=
0
l
∑
n
=
0
m
K
l
b
2
m
-
n
-
1
/
2
ϑ
n
+
1
/
2
b
3
m
-
n
λ
1
n
+
1
/
2
l
!
n
!
m
-
n
!
a
3
2
m
-
n
+
1
/
2
×
K
n
-
1
2
λ
1
b
ϑ
a
3where ϑ
=
b
3
a
4
/
a
3
+
b
4.Moreover, the average throughput can formulate as(41)
τ
∞
=
1
-
P
o
u
t
R
2
=
2
a
×
e
-
b
b
3
/
a
3
-
λ
1
a
4
/
a
3
∑
l
=
0
∞
∑
m
=
0
l
∑
n
=
0
m
K
l
b
2
m
-
n
-
1
/
2
ϑ
n
+
1
/
2
b
3
m
-
n
λ
1
n
+
1
/
2
l
!
n
!
m
-
n
!
a
3
2
m
-
n
+
1
/
2
×
K
n
-
1
2
λ
1
b
ϑ
a
3
×
R
2


## 3.2.1. Exact Probability and Average Throughput

Similar to scenario 1, the cumulative density function (CDF) of RVφ
2 can be computed as(32)
F
φ
2
ς
=
∫
0
ς
f
φ
2
x
d
x
=
1
-
a
b
∑
l
=
0
∞
∑
m
=
0
l
K
l
b
m
l
!
m
!
ς
m
e
-
b
ςwhere a
=
(
K
+
1
)
e
-
K
/
λ
2, b
=
K
+
1
/
λ
2.The probability density function (PDF) of a random variable (RV)φ
1 is(33)
f
φ
1
x
=
λ
1
e
-
x
λ
1Also, the outage probability can be formulated as(34)
P
o
u
t
=
∫
0
a
2
/
a
1
f
φ
1
φ
1
d
φ
1
+
∫
a
2
/
a
1
∞
F
φ
2
b
1
φ
1
+
b
2
a
1
φ
1
-
a
2
f
φ
1
φ
1
d
φ
1Using (32) and (33) we can reformulate (21) as(35)
P
o
u
t
=
1
-
λ
1
×
∫
a
2
/
a
1
∞
a
b
∑
l
=
0
∞
∑
m
=
0
l
K
l
b
m
l
!
m
!
b
1
φ
1
+
b
2
a
1
φ
1
-
a
2
m
e
-
b
b
1
φ
1
+
b
2
/
a
1
φ
1
-
a
2
×
e
-
λ
1
φ
1
d
φ
1By changing the variable, t
=
1
/
a
1
φ
1
-
a
2 (35) can be rewritten as(36)
P
o
u
t
=
1
-
a
a
1
b
λ
1
∫
0
∞
∑
l
=
0
∞
∑
m
=
0
l
K
l
b
m
l
!
m
!
b
1
a
1
+
κ
t
m
t
-
2
e
-
b
b
1
/
a
1
+
κ
t
×
e
-
λ
1
1
/
a
1
t
+
a
2
/
a
1
d
t
P
o
u
t
=
1
-
a
λ
1
a
1
b
×
e
-
b
b
1
/
a
1
-
λ
1
a
2
/
a
1
∫
0
∞
∑
l
=
0
∞
∑
m
=
0
l
K
l
b
m
l
!
m
!
b
1
a
1
+
κ
t
m
t
-
2
e
-
b
κ
t
×
e
-
λ
1
/
a
1
t
d
t where κ
=
b
1
a
2
/
a
1
+
b
2.Equation (36) becomes as follows:(37)
P
o
u
t
=
1
-
a
λ
1
a
1
b
×
e
-
b
b
1
/
a
1
-
λ
1
a
2
/
a
1
∫
0
∞
∑
l
=
0
∞
∑
m
=
0
l
K
l
b
m
κ
n
l
!
m
!
∑
n
=
0
m
m
n
b
1
a
1
m
-
n
×
t
n
-
2
e
-
b
κ
t
×
e
-
λ
1
/
a
1
t
d
t
P
o
u
t
=
1
-
a
λ
1
a
1
×
e
-
b
b
1
/
a
1
-
λ
1
a
2
/
a
1
∑
l
=
0
∞
∑
m
=
0
l
∑
n
=
0
m
K
l
b
m
-
1
κ
n
l
!
n
!
m
-
n
!
b
1
a
1
m
-
n
∫
0
∞
t
n
-
2
×
e
-
b
κ
t
×
e
-
λ
1
/
a
1
t
d
tUsing Table of Integral Equation[
3.471,9
] in [24], (37) can be changed into(38)
P
o
u
t
=
1
-
2
a
λ
1
a
1
×
e
-
b
b
1
/
a
1
-
λ
1
a
2
/
a
1
∑
l
=
0
∞
∑
m
=
0
l
∑
n
=
0
m
K
l
b
m
-
1
κ
n
l
!
n
!
m
-
n
!
b
1
a
1
m
-
n
λ
1
a
1
b
κ
n
-
1
/
2
×
K
n
-
1
2
λ
1
b
κ
a
1
P
o
u
t
=
1
-
2
a
×
e
-
b
b
1
/
a
1
-
λ
1
a
2
/
a
1
∑
l
=
0
∞
∑
m
=
0
l
∑
n
=
0
m
K
l
b
2
m
-
n
-
1
/
2
κ
n
+
1
/
2
b
1
m
-
n
λ
1
n
+
1
/
2
l
!
n
!
m
-
n
!
a
1
2
m
-
n
+
1
/
2
×
K
n
-
1
2
λ
1
b
κ
a
1Then the average throughput can be calculated as(39)
τ
=
1
-
P
o
u
t
R
2
=
2
a
×
e
-
b
b
1
/
a
1
-
λ
1
a
2
/
a
1
∑
l
=
0
∞
∑
m
=
0
l
∑
n
=
0
m
K
l
b
2
m
-
n
-
1
/
2
κ
n
+
1
/
2
b
1
m
-
n
λ
1
n
+
1
/
2
l
!
n
!
m
-
n
!
a
1
2
m
-
n
+
1
/
2
×
K
n
-
1
2
λ
1
b
κ
a
1
×
R
2


## 3.2.2. Asymptotic Outage Probability and Average Throughput

Similar proof as scenario 1 we have(40)
P
o
u
t
∞
=
1
-
2
a
×
e
-
b
b
3
/
a
3
-
λ
1
a
4
/
a
3
∑
l
=
0
∞
∑
m
=
0
l
∑
n
=
0
m
K
l
b
2
m
-
n
-
1
/
2
ϑ
n
+
1
/
2
b
3
m
-
n
λ
1
n
+
1
/
2
l
!
n
!
m
-
n
!
a
3
2
m
-
n
+
1
/
2
×
K
n
-
1
2
λ
1
b
ϑ
a
3where ϑ
=
b
3
a
4
/
a
3
+
b
4.Moreover, the average throughput can formulate as(41)
τ
∞
=
1
-
P
o
u
t
R
2
=
2
a
×
e
-
b
b
3
/
a
3
-
λ
1
a
4
/
a
3
∑
l
=
0
∞
∑
m
=
0
l
∑
n
=
0
m
K
l
b
2
m
-
n
-
1
/
2
ϑ
n
+
1
/
2
b
3
m
-
n
λ
1
n
+
1
/
2
l
!
n
!
m
-
n
!
a
3
2
m
-
n
+
1
/
2
×
K
n
-
1
2
λ
1
b
ϑ
a
3
×
R
2


## 3.3. Optimal Power Splitting Factor

The optimal valueρ
∗ can be obtained by solving the equation d
τ
(
ρ
)
/
d
ρ
=
0. Given the average throughput expression in (25) and (39), this optimization problem does not admit a closed-form solution. However, the optimal ρ
∗ is efficiently solved via numerical calculation, as illustrated below [25–27].Here, we can use the Golden section search algorithm to find the optimal factorρ
∗. This algorithm has been used in many global optimization problems in communications, for example, in [25]. The detailed algorithm, as well as the related theory, is described in [25–27].


## 4. Numerical Results and Discussion

In this section, the system performance (the throughput and the outage probability) of FD relaying network are validated in details by using the Monte Carlo simulation. The system performance is analyzed in connection withα, ρ, P
s
/
N
0, and R. The system network is considered with one source, one relay, and one destination nodes, where source-relay and relay-destination distances are both normalized to unit value. The channel gains are generated from Rayleigh/Rician distributed random variables with parameters as given in Table 1. The number of generated samples for each simulation is 106. Other simulation parameters are also listed in Table 1. These parameters are chosen specifically to illustrate the correctness of the derived formulas. In fact, these parameters can be varied depending on practical systems, but the simulation methods presented here are still applicable.Table 1
Simulation parameters.
Symbol
Name
Values
η
Energy harvesting efficiency
0.7
λ
1
Mean ofh
^
2
0.5
λ
2
Mean ofg
^
2
0.5
K
Rician K-factor
3
γ
t
h
SNR threshold
7
Ps/N0
Source power to noise ratio
0-50dB
δ
0
=
δ
Δ
h
=
δ
Δ
g
=
δ
Δ
f
Channel estimation error
0; 0,1
R
Source rate
3 bit/s/HzIn this analysis, Figures3(a) and 3(b) show the influence of the standard deviation of the channel estimation error on the outage probability and throughput of the model system for two scenarios, respectively. In this simulation process, the standard deviations of the channel estimation error are set the same for all links δ
Δ
h
=
δ
Δ
g
=
δ
Δ
f and ρ
=
0.5 and 0.7. The results show that the outage probability increased and the throughput decreased crucially while the standard deviation of channel estimation error δ
Δ
h
=
δ
Δ
g
=
δ
Δ
f increases from 0 to 0.25. From that point of view, the system performance degrades significantly after the error exceeds specific threshold value. Furthermore, the analytical results agree well with the Monte Carlo simulation results, validating the theoretical derivations.Figure 3
Outage probability (a) and achievable throughput (b) of the system model versusδ.
(a)
(b)On another hand, Figures4(a) and 4(b) illustrate the influence of the power splitting ratio ρ on the achievable throughput and the outage probability at the destination node. Here, δ
0 is set at 0 and 0.1 at rate 3 bps and the ratio P
s
/
N
0 at 20 dB. From the simulation, it is clear fond that the achievable throughput increases and the outage probability decrease significantly at the destination node while the ratio ρ varies from 0 to the optimal value (0.5-0.6). After the optimal value, both decreased with ratio ρ from 0.6 to 1.0. The optimal value ρ is about 0.5-0.6 as shown in Figure 4 and the corresponding maximum throughput and minimum outage probability are shown in Table 2. Moreover, the analytical and the Monte Carlo simulation results are the same in these cases.Table 2
The maximum throughput and minimum outage probability.
Scenario 1-
Scenario 2-
Scenario 1-
Scenario 2-
δ
0
=
0
δ
0
=
0
δ
0
=
0.1
δ
0
=
0.1
Optimalρ
∗
0.55
0.65
0.45
0.6
Minimum outage
0.4110
0.2957
0.5412
0.4284
Maximum throughput
0.8835
1.0565
0.6882
0.8574Figure 4
Outage probability (a) and achievable throughput (b) of the system model versusρ.
(a)
(b)Moreover, Figures5(a) and 5(b) present the influence of the ratio P
s
/
N
0 on the outage probability and the achievable throughput with δ
0 at 0 and 0.1, at rate 3 bps and ρ
=
0.5 for the two scenarios. From the results, it is shown that the achievable throughput increases and the outage probability decreases significantly at the destination node while the ratio P
s
/
N
0 increased from 0 to 50 dB. Furthermore, the exact and the asymptotical throughput and outage probability get close to each other at the end of the interval of the ratio P
s
/
N
0. In two scenarios, all the analytical and the simulation results agreed well with each other. Furthermore, it is confirmed that, in the low SNR regime, the channel estimation error has little effect on system performance (the outage probability and the throughput). However, when the transmit power increases to the significant value, the channel estimation error has an increasing impact on both achievable throughput and outage probability.Figure 5
Outage probability (a) and achievable throughput (b) of the system mode versus ratio PS/N0.
(a)
(b)In the same way, the influence of the system rate R on the outage probability and the achievable throughput withP
s
/
N
0
=
20 dB and ρ
=
0.5 are illustrated in Figures 6(a) and 6(b). Figure 6(a) showed that the outage probability increase crucially with increasing the rate R. However, the system throughput only increased in the interval R from 0 to the optimal value of around 3. After that, it significantly fell to 0 at the rate around 7. In particular, the simulation lines wholly matched with the analytical lines in Figures 1–6. Moreover, the comparison between our dissimilar channel model (Scenarios 1 and 2) and the similar channel model that means both source-relay and relay-destination links is assumed to be Rayleigh fading channels, in Figure 7(a) for throughput analysis and Figure 7(b) for outage probability analysis, respectively. From the simulation results, we show that the system performance of the scenarios 1 and 2 is better in comparison with the Rayleigh-Rayleigh scenario. Finally, the optimal power splitting factor versus the ratio P
s
/
N
0 in the two scenarios is proposed in Figures 8 and 9, respectively.Figure 6
Outage probability (a) and achievable throughput (b) of the system mode versus R.
(a)
(b)Figure 7
Scenarios 1 and 2 versus the Rayleigh-Rayleigh scenario.
(a)
(b)Figure 8
The optimal splitting factor versus Ps/N0 for the first case.Figure 9
The optimal splitting factor versus Ps/N0 for the second case.


## 5. Conclusions

In this paper, the wireless communication network with FD relaying in imperfect condition with the power splitting protocol is proposed and demonstrated. In this system model, the two scenarios with the dissimilar fading channels are presented and analyzed. To analyze the system performance at the destination node, analytical expressions for the outage probability, the throughput, and optimal power splitting factor are proposed and investigated. The numerical results show that the analytical mathematical expression and the simulation results using Monte Carlo method totally matched each other. Moreover, this paper has provided practical insights into the effect of various system parameters on the system performance. The results show that the system performance degrades significantly but is still in a permissible interval while the channel estimation error increases. Finally, the system performance of the mixing scenarios is better in comparison with the Rayleigh-Rayleigh scenario. The results could be providing the prospective solution for the communication network.



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*Source: 6036087-2018-12-16.xml*