# Joint Time and Power Allocation Algorithm in NOMA Relaying Network

Authors: Su Zhao; Chuan Mei; Qi Zhu
Journal: Mobile Information Systems
 (2019)

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## Abstract

Nonorthogonal multiple access (NOMA) is one of the promising access techniques in 5G network. The application of relay in NOMA system is a hotspot in recent research. NOMA-based cooperative relay network can achieve a higher spectral efficiency and a lower outage probability. In this paper, we analyse the performance of the two-hop DF relay NOMA network scenario, where the number of cell edge users is more than the cell center user, and obtained the closed-form expression of the user's ergodic rates and outage probabilities under the high signal-to-noise (SNR) ratio. Then, we establish an optimization model to maximize the system rates, and a joint optimal time and power allocation algorithm based on the exhaustive search and the binary algorithm is proposed. Simulation results show that the proposed scheme can outperform exiting scheme in terms of achieving a higher ergodic sum rate, a lower outage probability under the premise of fairness.


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## Body

## 1. Introduction

As the demand for smart terminals and new mobile services continues to grow, wireless transmission rates will increase exponentially and the 4G system will be difficult to meet the communications requirements of high-speed, low-latency. Besides, the unexpected excessive energy consumption of the fourth-generation (4G) and pre-4G wireless networks causes serious carbon dioxide emissions. To achieve green wireless networks, the fifth-generation (5G) wireless networks are expected to significantly increase the network energy efficiency while guaranteeing the quality of service (QoS) for time-sensitive multimedia wireless traffics [1]. NOMA has been recognized as one of the key technologies of the fifth-generation mobile communication (5G) for higher spectral efficiency [2–4]. The key idea of NOMA is accommodating multiple users in the same frequency band but each user has a different power. Compared with the traditional OMA system, NOMA can achieve a higher spectral efficiency [5, 6]. Since cooperative relay can improve system capacity and expand network coverage [7], NOMA-based cooperative relay network has become a hotspot in wireless field research.A preliminary study has been conducted on the resource allocation of the NOMA relay network. In [8], the strong users work as the relay, and the proposed cooperative scheme is strong user decode and transmits the weak user’s signals. The ergodic sum rate and outage probability of this cooperative NOMA scheme are analysed. In [9], a dedicated relay node is used to provide services for users equipped with multiple antennas, and the literature obtains the lower bound of the outage probability. In [10], the author studied the outage performance of the NOMA system that relay operates in amplify-and-forward (AF) strategy and derived the exact approximation of the outage probability. In [11], the author has derived the system outage probability and the ergodic sum rate of the scenario where users are directly communication with the base station (BS) and the relay. The relay operates in decode-and-forward (DF) strategy. In [12], the time and power allocation of two-hop relay using DF protocol is studied. The closed-form expression of the outage probability is derived, and the optimal time allocation for minimizing the outage probability is obtained but without considering the fairness of the system and the user communicating directly with the base station. In [13], the author analyses the scenario that the user communicates either directly with the base station or through relay with the base station and derivate the system outage probability and the expression of user ergodic rate. The theoretical and simulation results show that the ergodic sum rate of cooperative NOMA system can be significantly improved compared with noncooperation. However, the paper only analyses the case where the number of users in the cell center is equal to the number of users at the cell edge, and the time shared by the two hops is not optimized. In [14], the author analyses the performance of the scenario, where the number of cell edge users are more than the cell center users by introducing time sharing technology. However, the user is considered to communicate with the BS through the relay, and the strategy of equal time slot division and fixed power allocation factor is adopted, and the fairness of the system is not considered.In this paper, for the NOMA relay scenario where the cell edge users are more than the cell center users and the channel of each user is significantly different, we use the method of dividing the time slot to transmit the information of the user and derive the expressions of the system's ergodic sum rate and the outage probability. Under the condition of system fairness index factor, the optimization problem of maximizing system rates is constructed. In order to get the solution to this problem, we proposed an optimal time and power allocation algorithm based on exhaustive search and binary algorithm. The algorithm can obtain the optimal time and power factor allocation strategy under different fairness index factors. In this allocation policy, we obtained the maximum rates of the system. The main contributions of this paper are as follows:(1)
Modelled the scenario that the number of cell center users is lower than users at the edge(2)
Derived the ergodic rate and outage probability of the system under the new scenario(3)
Proposed the joint and power allocation algorithm to maximize the system sum rateThe rest of the paper is organized as follows: section2 gives the system model of this paper, section 3 deduces the system performance, including the derivation of the system’s ergodic rates and the outage probability, section 4 proposes an optimal allocation algorithm for time and power factor, section 5 performs simulation analysis to analyse the effect of fairness factors on the overall system rates, and section 6 summarizes the full text.


## 2. System Model

Figure1 is the proposed system model for this paper, which contains one base station (BS), one relay (R), and four users (UE1, UE2, UE3, and UE4). We clarify that UE1 is the cell center user directly connected to the base station and has better channel conditions. UE2, UE3, and UE4 are cell edge users that need to forward information through the relay which operates in the half-duplex mode using DF strategy. The hb1 and hbr indicate the channel coefficients from the BS to the UE1, from the BS to the relay. hr2, hr3, and hr4 denote the channel coefficients from the relay to each cell edge user. Channels are independent of each other and are subject to Rayleigh fading; we model these channels as hb1∼CN0,λb1, hbr∼CN0,λbr, hr2∼CN0,λr2, hr3∼CN0,λr3, and hr4∼CN0,λr4. We assume that UE2 and UE4 have similar channel conditions and the same variance, then λr2=λr4. The channel condition of UE3 is significantly worse than other users. In the transmission process, relay forwards signals to UE2, UE3, and UE4 in NOMA strategy. We divide a transmission time slot into four subslots, and cell center users are paired with different cell edge users in different subslots for information transmission.Figure 1
System model.It is assumed that one time slot is divided into four subslots which are denoted ast1, t2, t3, and t4, respectively, and the channel state in one subslot does not change.(1)
During thet1 subslot, the BS transmits xNt1=a1t1Psx1t1+a2t1Psx2t1+a3t1Psx3t1 to the UE1 and the relay, where x1t1, x2t1, and x3t1 are data symbols for UE1, UE2, and UE3 with Ex1t12=Ex2t12=Ex3t12=1. Ps is the transmission power of BS, and a1t1, a2t1, and a3t1 are the power allocation coefficients, where a1t1+a2t1+a3t1=1 and a1t1<a2t1<a3t1. The received signals at UE1 and relay are given by(1)y1t1=hb1xNt1+n1,yRt1=hbrxNt1+nR,where n• is the additive white Gaussian noise (AWGN) at each node.When receiving the superimposed signal,UE1 needs to apply the SIC to obtain its own signal after decoding the signals of UE2 and UE3. It can be seen from [13] that the optimal decode order is the user with the worst channel condition decode first. In this method, UE1 first decodes the signal of UE3, and the decoded signal to interference and noise ratio (SINR) is given by(2)γ3⟶1t1=a3t1ρshb12a1t1+a2t1ρshb12+1,where ρs is the transmit signal-to-noise ratio (SNR) of the base station, ρs=Ps/σ2, Ps is the transmission power of the BS, and σ2 is the variance of Gaussian additive white noise.After decoding the signal of theUE3, the SIC is applied to remove the signal of UE3 and then the UE2 is decoded, and the decoded SINR is(3)γ2⟶1t1=a2t1ρshb12a1t1ρshb12+1.Assuming thatUE2 signal can successfully decode, after applying SIC, the SNR of UE1 is given by(4)γBR1t1=a1t1ρshb12.At relayR, UE3 is decoded first and then UE2 is decoded. The SINR of decoding UE2 and UE3 are, respectively, given by(5)γBR2t1=a2t1ρshbr2a1t1ρshbr2+1,γBR3t1=a3t1ρshbr2a1t1ρshbr2+a2t1ρshbr2+1.(2)
In thet2 subslot, the relay first regenerates the superimposed signal xNt2=a2t2Prx2t1+a3t2Prx3t1 of UE2 and UE3. Then, the relay transmits it to users. Pr is the transmission power of relay and a2t2 and a3t2 are the power allocation coefficients where a2t2+a3t2=1 and a2t2<a3t2. The received signals at UE2 and UE3 are given by(6)y2t2=hr2xNt2+n2,y3t2=hr3xNt2+n3.After receiving the superimposed signal, theUE2 first decodes the UE3 signal and the decoded SINR is given by(7)γ3⟶2t2=a3t2ρrhr22a2t2ρrhr22+1.When SIC is applied atUE2, the signal of UE3 has been removed and the received SNR of the UE2 is(8)γR2t2=a2t2ρrhr22,where ρr is the transmit SNR of the relay, ρr=Pr/σ2, and Pr is the transmission power of the relay.UE
3 has the worst channel condition and directly decodes its own signal, and the decoded SINR is given by(9)γR3t2=a3t2ρrhr32a2t2ρrhr32+1.(3)
In thet3 subslot, the BS transmits the superimposed signals of the users 1, 3, and 4. The signal is xNt3=a1t3Psx1t3+a3t3Psx3t3+a4t3Psx4t3, where a1t3, a3t3, and a4t3 are the power allocation coefficients for users 1, 3, and 4, where a1t3+a3t3+a4t3=1 and a1t3<a4t3<a3t3. The received signals at UE1 and relay are given by(10)y1t3=hb1xNt3+n1,yRt3=hbrxNt3+n1.After receiving the superimposed signal,UE1 first decodes the signal of UE3 and then decodes the signal of UE4. SIC is applied to remove the two signals, and finally UE1 decodes its own signal. The relay decodes the signals of UE3 and UE4 in the same manner, and similar to the t1 subslot analysis, the SINR of the UE1 decodes signals of UE3 and UE4 can be obtained as follows:(11)γ3⟶1t3=a3t3ρshb12a1t3+a4t3ρshb12+1,γ4⟶1t3=a4t3ρshb12a1t3ρshb12+1.Assuming that the relay successfully decodes the two-user signal, the SINR ofUE1 after applying the SIC is given by(12)γBR1t3=a1t3ρshb12.The SINR of decodingUE3 and UE4 at the relay is given by(13)γBR3t3=a3t3ρshbr2a1t3ρshbr2+a4t3ρshbr2+1,γBR4t3=a4t3ρshbr2a1t3ρshbr2+1.(4)
In thet4 subslot, the situation is similar to the t2 subslot, we can obtain the results as follows:(14)γ3⟶4t4=a3t4ρrhr42a4t4ρrhr42+1,γR4t4=a4t4ρrhr42,γR3t4=a3t4ρrhr32a4t4ρrhr32+1,where a3t4 and a4t4 are the power allocation factors of UE3 and UE4 in the t4 sub-time slot, respectively.Since the user channel conditions are unchanged in one time slot and the channel conditions of user 2 and user 4 are similar, then(15)PA1=a1t1=a1t3,a2t1=a4t3,a2t2=a4t4,a3t1=a3t3,a3t2=a3t4,(16)PA2=a1t1+a2t1+a3t1=1a2t2+a3t2=1t1=t3,t2=t4.After applying SIC technique, the achievable data rates ofUE1 in the t1 subslot is given by(17)C1t1=t1log21+γBR1t1.BecauseUE2 must be decoded at UE1 for SIC and the capacity of DF relaying is dominated by the weakest link, the achievable data rates of UE2 is given by [15](18)C2t2=t2log21+minγBR2t1,γ2⟶1t1,γR2t2.The achievable data rates ofUE3 is given by(19)C3t2=t2log21+minγBR3t1,γ3⟶1t1,γR3t2.In thet3∼t4 subslot, the system status is similar to t1∼t2, then(20)C1t3=t3log21+γBR1t3,C3t4=t4log21+minγBR3t3,γ3⟶1t3,γR3t4,C4t4=t4log21+minγBR4t3,γ4⟶1t3,γR4t4.


## 3. Performance Analysis

In this section, we analyse the system’s ergodic rates and the outage probabilities. The closed-form expression of user’s ergodic rates in high SNR is derived, and the outage probability of each user and system is obtained.


### 3.1. Ergodic Rates

Ergodic rates refer to the time average of the maximum information rates of a random channel in a fast fading state. The system ergodic sum rate is given by(21)C¯sum≜EC1+EC2+EC3+EC4=C¯1+C¯2+C¯3+C¯4,where C¯1, C¯2, C¯3, and C¯4 indicate the ergodic rates of users 1, 2, 3, and 4, respectively.The ergodic rates of theUEi in the subslot t is given by(22)C¯i=t∫0∞log21+ωfWωdω=tln2∫0∞1−FWω1+ωdω,where ω is the SINR of UEi and FWω and fWω are the distribution function (CDF) and the probability density function (PDF) of ω.


#### 3.1.1. Ergodic Rates ofUE1

In thet1 subslot, assuming that UE1 successfully decodes the UE2 and UE3 signals, using (22), the C¯1t1 is given by(23)C¯1t1=t1ln2∫0∞1−FγBR1t1ω1+ωdω,where FγBR1t1ω is the CDF of the γBR1t1. Using the definition of the distribution function, there is(24)FγBR1t1ω=PγBR1t1≤ω=Pa1t1ρshb12≤ω=Phb12≤ωa1t1ρs.Since channelhb1∼CN0,λb1 obeys the complex Gaussian distribution, according to the literature [16], hb12 obeys the exponential distribution of λ=1/λb1. The distribution function of γBR1t1 is given by(25)FγBR1t1ω=1−e−ω/a1t1ρsλb1.Taking (25) into equation (23) and the ergodic rates of UE1 in t1 subslot is given by(26)C¯1t1=−t1ln2Ei−1λb1ρsa1t1e1/λb1ρsa1t1,where Eix=∫−∞xet/tdt.From equation (26), we know that the ergodic rates of UE1 is decided by t1, λb1, ρs, and a1t1. Since t1 = t3, we can obtain that C¯1t3=C¯1t1, so the ergodic rates of UE1 is given by(27)C¯1=C¯1t1+C¯1t3=−2t1ln2Ei−1λb1ρsΨB1t1e1/λb1ρsΨB1t1.


#### 3.1.2. Ergodic Rates ofUE2

Assuming that the SINR ofUE2 is γR2, UE2 transmits its signal via the relay in the subslot t1 and subslot t2, because the SINR is decided by the weakest link of the relay, we can obtain that γR2=minγBR2t1,γ2⟶1t1,γR2t2. The ergodic rates of UE2 is given by(28)C¯2=t2ln2∫0∞1−FγR2ω1+ωdω.To get theC¯2, we need to obtain the CDF of FγR2ω. Since the channels hbr, hb1, and hr2 are independent of each other, FγR2 can be derived as follows:(29)FγR2=PγR2≤ω=PminγBR2t1,γ2⟶1t1,γR2t2≤ω=1−PminγBR2t1,γ2⟶1t1,γR2t2>ω=1−PγBR2t1>ωPγ2⟶1t1>ωPγR2t2>ω=1−1−PγBR2t1≤ω1−Pγ2⟶1t1≤ωPγR2t2≤ω=1−FγBR2t1ω1−Fγ2⟶1t1ω1−FγR2t2ω,where FγBR2t1ω, Fγ2⟶1t1ω, and FγR2t2ω are the CDF of the γBR2t1, γ2⟶1t1, and γR2t2. The CDFs are derived separately as follows:(30)FγBR2t1ω=PγBR2t1≤ω=Pa2t1ρshbr2ΨB1t1ρshbr2+1≤ω=Phbr2≤ωa2t1−ΨB1t1ωρs=1−e−ω/λbrρsa2t1−ωΨB1t1Fγ2⟶1t1ω=1−exp−ωλb1ρsa2t1−ωΨB1t1,FγR2t2ω=1−exp−ωa2t2ρrλr2.The expression ofFγR2 and C¯2 are given by(31)FγR2=1−e−ω/λbrρsa2t1−ωΨB1t1+ω/λb1ρsa2t1−ωΨB1t1+ω/a2t2ρrλr2,ω>0,1,ω≤0,(32)C¯2=t2ln2∫0∞e−ω/λbrρsa2t1−ωΨB1t1+ω/λb1ρsa2t1−ωΨB1t1+ω/a2t2ρrλr21+ωdω.In order to obtain the closed-form expression of equation (32), a high SNR approximation is used. When ρs⟶∞ and ρr⟶∞, equation (29) can be rewrite as follows:(33)FγR2=1−Pa2t1ΨB1t1>ωPγR2t2>ω=1−e−ω/a2t2ρrλr2,0<ω<a2t1ΨB1t1,1,otherwise.The closed-form expression ofC¯2 is given by(34)C¯2=t2ln2∫0∞1−FγR21+ωdω=t2ln2∫0a2t1/ΨB1t111+ωe−ω/a2t2ρrλr2dω=t2e1/λr2ρra2t2ln2Ei−1λr2ρra2t21+a2t1ΨB1t1−Ei−1λr2ρra2t2.


#### 3.1.3. Ergodic Rates ofUE3

Assuming that the SINR ofUE3 is γR3, it decided by the weakest link of the relay, that is γR3=minγBR3t1,γ3⟶1t1,γR3t2. We can obtain the ergodic rates of UE3 in the subslots t1 and t2 as follows:(35)C¯3t2=t2ln2∫0∞1−FγR3ω1+ωdω.Since the channelshbr, hb1, and hr2 are independent of each other, the FγR3 is given by(36)FγR3ω=PγR3≤ω=PminγBR3t1,γ3⟶1t1,γR3t2≤ω=1−PminγBR3t1,γ3⟶1t1,γR3t2>ω=1−PγBR3t1>ωPγ3⟶1t1>ωPγR3t2>ω=1−1−PγBR3t1≤ω1−Pγ3⟶1t1≤ωPγR3t2≤ω=1−FγBR3t1ω1−Fγ3⟶1t1ω1−FγR3t2ω,where FγBR3t1ω, Fγ3⟶1t1ω, and FγR3t2ω represent the CDF of γBR3t1, γ3⟶1t1, and γR3t2, respectively, and we can obtain(37)FγBR3t1ω=PγBR3t1≤ω=Pa3t1ρshbr2ΨB1t1ρshbr2+a2t1ρshbr2+1≤ω=Phbr2≤ωa3t1−ΨB1t1+a2t1ωρs=1−e−ω/λbrρsa3t1−ΨB1t1ω+a2t1ω,(38)Fγ3⟶1t1ω=Pγ3⟶1t1≤ω=Pa3t1ρshb12ΨB1t1+a2t1ρshb12+1≤ω=Pa3t1ρshb121−a3t1ρshb12+1≤ω=Phb12≤ωa3t1−1−a3t1ωρs=1−e−ω/λb1a3t1−1−a3t1ωρs,(39)FγR3t2ω=PγR3t2≤ω=Pa3t2ρrhr32a2t2ρrhr32+1≤ω=Phr32≤ωa3t2−a2t2ωρr=1−e−ω/λr3ρra3t2−a2t2ω.Taking equations (37)–(39) into (36), the FγR3ω and C¯3t2 are given by(40)FγR3ω=1−e−ω/λbrρsa3t1−ΨB1t1ω+a2t1ω+ω/λb1a3t1−1−a3t1ωρs+ω/λr3ρra3t2−a2t2ω,ω>0,1,ω≤0,C¯3t2=t2ln2∫0∞1−FW1ω1+ωdω=t2ln2∫0∞e−ω/λbrρsa3t1−ΨB1t1ω+a2t1ω+ω/λb1a3t1−1−a3t1ωρs+ω/λr3ρra3t2−a2t2ω1+ωdω.Whenρs⟶∞ and ρr⟶∞, equation (36) can be rewritten as follows:(41)FγR3ω=1−Pa3t11−a3t1>ωPa3t21−a3t2>ω.The closed-form expression ofC¯3t2 is given by(42)C¯3t2=t2ln2∫0∞Pa3t1/1−a3t1>ωPa3t2/1−a3t2>ω1+ωdω,where C¯3t2 is determined by a3t1/1−a3t1 and a3t2/1−a3t2, since fx=x/1−x is an monotonically increasing function about x, we only need to discuss a3t1 and a3t2.(a)
Ifa3t2≤a3t1, we can obtain(43)C¯3t2=t2ln2∫0a3t2/1−a3t2Pa3t1/1−a3t1>ωPa3t2/1−a3t2>ω1+ω=t2ln2ln1+a3t22−a3t2.(b)
Ifa3t1<a3t2, we can obtain(44)C¯3t2=t2ln2∫0a3t1/1−a3t1Pa3t1/1−a3t1>ωPa3t2/1−a3t2>ω1+ω=t2ln2ln1+a3t12−a3t1.Combine a andb, we can get(45)C¯3t2=t2ln2ln1+a3t22−a3t2,a3t2≤a3t1,t2ln2ln1+a3t12−a3t1,a3t2>a3t1.Sincet2 = t4, we can obtain that C¯3t4=C¯3t2 combined with (16). Therefore, the ergodic rates of UE3 is given by(46)C¯3=C¯3t2+C¯3t4=2t2ln2ln1+a3t22−a3t2,a3t2≤a3t1,2t2ln2ln1+a3t12−a3t1,a3t2>a3t1.


#### 3.1.4. Ergodic Rates ofUE4

It can be seen from (34) that C¯2 is related to t2, λr2, ρr, a1t1, a2t1, and a2t2. Combined with (16), the C¯4 is given by(47)C¯4=C¯2=t2e1/λr2ρra2t2ln2Ei−1λr2ρra2t21+a2t1ΨB1t1−Ei−1λr2ρra2t2.TakingC¯1,C¯2,C¯3,andC¯4 into (21), the system ergodic sum rate is given by(48)C¯sum≜EC1+EC2+EC3+EC4=C¯1+C¯2+C¯3+C¯4=−2t1ln2Ei−1λb1ρsΨB1t1e1/λb1ρsΨB1t1+2t2ln2e1/λr2ρra2t2Ei−1+a2t1/ΨB1t1λr2ρra2t2−Ei−1λr2ρra2t2+2t2ln2ln1+a3t22−a3t2,a3t2≤a3t1,2t2ln2ln1+a3t12−a3t1,a3t2>a3t1.


### 3.2. Outage Probability

The outage probability is the probability of the outage event happened in the communication. When the user’s reachable rates are lower than target rates, the outage event happened. In this paper, the outage probability of the cell center user is determined by its channel state. The outage probability of the cell edge users are determined by the two-hop system, and the state of the previous hop will affect the state of the next hop [17].(1)
Outage probability ofUE1Assumed thatE1t1O and E1t3O, respectively, represent the outage event happened of UE1 in t1 and t3 subslots, the outage probability in t1 time slot is PDFO1,t1=PrE1t1O and in t3 time slot is PDFO11,t3=PrE1t3O. For UE1, if it cannot detect the signal of UE2 and UE3 or the throughput does not reach the target rates RQ in the t1 time slot, the communication will interrupted. Denote these three cases as ER2⟶1t1O, ER3⟶1t1O, and EU1t1O, respectively, we can obtain the outage probability PrE1t1O as follows:(49)PDFO1,t1=PrE1t1O=PrEU1t1O∪ER2⟶1t1O∪ER3⟶1t1O=PrRBR1t1<RQ∪RR2⟶1t1<RQ∪RR3⟶1t1<RQ=PrminRBR1t1,RR2⟶1t1,RR3⟶1t1<RQ=1−PrminRBR1t1,RR2⟶1t1,RR3⟶1t1≥RQ=1−PrRBR1t1≥RQPrRR2⟶1t1≥RQPrRR3⟶1t1≥RQ=1−1−PrRBR1t1<RQ1−PrRR2⟶1t1<RQ1−PrRR3⟶1t1<RQ=1−1−FRBR1t1RQ1−FRR2⟶1t1RQ1−FRR3⟶1t1RQ,where RBR1t1=t1log21+γBR1t1, R2⟶1t1=t1log21+γ2⟶1t1, and R3⟶1t1=t1log21+γ3⟶1t1.The expressionsFRBR1t1RQ, FRR2⟶1t1RQ, and FRR3⟶1t1RQ are given by(50)FRBR1t1RQ=PrRBR1t1<RQ=Prt1log21+γBR1t1<RQ=PrγBR1t1<β1=Prhb12≤β1a1t1ρs=1−e−β1/λb1a1t1ρs,0<β1,1,otherwise,FRR2⟶1t1RQ=PrRR2⟶1t1<RQ=Prt1log21+γ2⟶1t1<RQ=Prγ2⟶1t1<β1=Prhb12≤β1a2t1−a1t1β1ρs=1−e−β1/λb1a2t1−a1t1β1ρs,0<β1<a2t1a1t1,1,otherwise,FRR3⟶1t1RQ=PrRR3⟶1<RQ=Prt1log21+γ3⟶1t1<RQ=Prγ3⟶1t1<β1=Pra3t1ρshb121−a3t1ρshb12+1<β1=Prhb12<β1a3t1−1−a3t1β1ρs=1−e−β1/λb1a3t1−1−a3t1β1ρs,0<β1<a3t11−a3t1,1,otherwise,where β1=2RQ/t1−1. The PDFO11,t1 is given by(51)PDFO1,t1=1−e−β1/λb1a1t1ρs−β1/λb1a2t1−a1t1β1ρs−β1/λb1a3t1−1−a3t1β1ρs,β1<a2t1a1t1,β1<a3t11−a3t1,1,otherwise.Similarly,PrE1t3O is given by(52)PrE1t3O=1−1−FRBR1t3RQ1−FRR2⟶1t3RQ1−FRR3⟶1t3RQ.Since the channel state does not change in the time slotst1 and t3, the relationship PrE1t3O=PrE1t1O is established. Therefore, the outage probability of UE1 is given by(53)PDFO1,t1=PDFO1,t3=1−e−β1/λb1a1t1ρs−β1/λb1a2t1−a1t1β1ρs−β1/λb1a3t1−1−a3t1β1ρs,β1<a2t1a1t1,β1<a3t11−a3t1,1,otherwise.(2)
Outage probability ofUE2TheUE2 communicates with the BS through the relay in the t1 and t2 sub-time slots, and its outage probability is related to each hop of the two-hop system. If UE2 cannot be successfully decoded in the relay or cannot decode the signal of UE3, communication will be interrupted. If UE2 can be decoded in the relay, but in the second hop transmission rate does not reach the target rate, the outage event will also occur. Denote the events that UE2 cannot successfully decode at relay, UE2 cannot decode the signal of the UE3, and in the second hop, the communication rates of UE2 does not reach the target rate RQ, respectively, as ER2O, ER3⟶2O, and EU2O. The outage probability PDFO2 of UE2 can be obtained as follows:(54)PDFO2=PrER2O+PrER2PrER3⟶2OER2+PrER2PrER3⟶2ER2PrEU2OER2ER3⟶2=1−PrER2ER3⟶2EU2=1−PrminRBR2,R3⟶2,RR2≥RQ=1−1−PrRBR2<RQ1−PrR3⟶2<RQ1−PrRR2<RQ=1−1−FRBR2RQ1−FR3⟶2RQ1−FRR2RQ,where RBR2=t1log21+γBR2t1, R3⟶2=t2log21+γ3⟶2t2, and RR2=t2log21+γR2t2.The expressionsFRBR2RQ, FR3⟶2RQ, and FRR2RQ are given by(55)FRBR2RQ=PrRBR2<RQ=Prt1log21+γBR2t1<RQ=PrγBR2t1<β1=Prhbr2≤β1a2t1−a1t1β1ρs=1−e−β1/λbra2t1−a1t1β1ρs,0<β1<a2t1a1t1,1,otherwise,FR3⟶2RQ=PrR3⟶2<RQ=Prt2log21+γ3⟶2t2<RQ=Prγ3⟶2t2<β2=Prhr22≤β2a2t2−a3t2β2ρr=1−e−β2/λr2a3t2−a2t2β2ρr,0<β2<a3t2a2t2,1,otherwise,FRR2RQ=PrRR2<RQ=Prt2log21+γR2t2<RQ=PrγR2t2<β2=Prhr22≤β2a2t2ρr=1−e−β2/λr2a2t2ρr,0<β2,1,otherwise,where β2=2RQ/t2−1, and PDFO2 is given by(56)PDFO2=1−e−β1/λbra2t1−a1t1β1ρs−β2/λr2a3t2−a2t2β2ρr−β2/λr2a2t2ρr,β1<a2t1a1t1,β2<a3t2a2t2,1,otherwise.(3)
Outage probability ofUE3Assumed thatE3t12O and E3t34O, respectively, indicate the event that UE3 interrupts in t1∼t2 and t3∼t4 subtime slots. The corresponding outage probability is PDFO3,t12=PrE3t12O and PDFO3,t34=PrE3t34O; obviously, the equation PDFO3,t12=PDFO3,t34 is established. Taking PDFO3,t12 for example, here we denote the event that UE3 cannot be decoded at the relay as ER3t1O, and the rates at the second hop does not reach the target rate RQ as EU3t2O. The PrE3t12O is given by(57)PDFO3,t12=PrER3t1O+PrEU3t2OER3t1PrER3t1=1−PrEU3t2,ER3t1=1−PrminRBR3t1,RR3t2≥RQ=1−PrRBR3t1≥RQPrRR3t2≥RQ=1−1−FRBR3t1RQ1−FRR3t2RQ,where RBR3t1=t1log21+γBR3t1 and RR3t2=t2log21+γR3t2.From the previous analysis, the outage probability ofUE3 is given by(58)PDFO3,t12=PDFO3,t34=1−e−β1/λbra3t1−1−a3t1β1ρs−β2/λr3a3t2−a2t2β2ρr,β1<a3t11−a3t1,β2<a3t2a2t2,1,otherwise.(4)
Outage probability ofUE4Since the power allocation factor unchanged in one time slot, obviously the power allocation factor of theUE4 in the t4 subslot is the same as the UE2 in the t2 subslot. Therefore, the outage probability is the same for both, we can obtain(59)PDFO4=PDFO2=1−e−β1/λbra2t1−a1t1β1ρs−β2/λr2a3t2−a2t2β2ρr−β2/λr2a2t2ρr,β1<a2t1a1t1,β2<a3t2a2t2,1,otherwise.The overall system outage event is defined as the event that any user in the system cannot achieve reliable detection, which means the overall outage probability is defined as follows [16]:(60)PDFO≜1−∏k=131−PDFOk.Substituting equations (53), (56), (58), and (59) into (60), we can obtain the system outage probability.


## 3.1. Ergodic Rates

Ergodic rates refer to the time average of the maximum information rates of a random channel in a fast fading state. The system ergodic sum rate is given by(21)C¯sum≜EC1+EC2+EC3+EC4=C¯1+C¯2+C¯3+C¯4,where C¯1, C¯2, C¯3, and C¯4 indicate the ergodic rates of users 1, 2, 3, and 4, respectively.The ergodic rates of theUEi in the subslot t is given by(22)C¯i=t∫0∞log21+ωfWωdω=tln2∫0∞1−FWω1+ωdω,where ω is the SINR of UEi and FWω and fWω are the distribution function (CDF) and the probability density function (PDF) of ω.


### 3.1.1. Ergodic Rates ofUE1

In thet1 subslot, assuming that UE1 successfully decodes the UE2 and UE3 signals, using (22), the C¯1t1 is given by(23)C¯1t1=t1ln2∫0∞1−FγBR1t1ω1+ωdω,where FγBR1t1ω is the CDF of the γBR1t1. Using the definition of the distribution function, there is(24)FγBR1t1ω=PγBR1t1≤ω=Pa1t1ρshb12≤ω=Phb12≤ωa1t1ρs.Since channelhb1∼CN0,λb1 obeys the complex Gaussian distribution, according to the literature [16], hb12 obeys the exponential distribution of λ=1/λb1. The distribution function of γBR1t1 is given by(25)FγBR1t1ω=1−e−ω/a1t1ρsλb1.Taking (25) into equation (23) and the ergodic rates of UE1 in t1 subslot is given by(26)C¯1t1=−t1ln2Ei−1λb1ρsa1t1e1/λb1ρsa1t1,where Eix=∫−∞xet/tdt.From equation (26), we know that the ergodic rates of UE1 is decided by t1, λb1, ρs, and a1t1. Since t1 = t3, we can obtain that C¯1t3=C¯1t1, so the ergodic rates of UE1 is given by(27)C¯1=C¯1t1+C¯1t3=−2t1ln2Ei−1λb1ρsΨB1t1e1/λb1ρsΨB1t1.


### 3.1.2. Ergodic Rates ofUE2

Assuming that the SINR ofUE2 is γR2, UE2 transmits its signal via the relay in the subslot t1 and subslot t2, because the SINR is decided by the weakest link of the relay, we can obtain that γR2=minγBR2t1,γ2⟶1t1,γR2t2. The ergodic rates of UE2 is given by(28)C¯2=t2ln2∫0∞1−FγR2ω1+ωdω.To get theC¯2, we need to obtain the CDF of FγR2ω. Since the channels hbr, hb1, and hr2 are independent of each other, FγR2 can be derived as follows:(29)FγR2=PγR2≤ω=PminγBR2t1,γ2⟶1t1,γR2t2≤ω=1−PminγBR2t1,γ2⟶1t1,γR2t2>ω=1−PγBR2t1>ωPγ2⟶1t1>ωPγR2t2>ω=1−1−PγBR2t1≤ω1−Pγ2⟶1t1≤ωPγR2t2≤ω=1−FγBR2t1ω1−Fγ2⟶1t1ω1−FγR2t2ω,where FγBR2t1ω, Fγ2⟶1t1ω, and FγR2t2ω are the CDF of the γBR2t1, γ2⟶1t1, and γR2t2. The CDFs are derived separately as follows:(30)FγBR2t1ω=PγBR2t1≤ω=Pa2t1ρshbr2ΨB1t1ρshbr2+1≤ω=Phbr2≤ωa2t1−ΨB1t1ωρs=1−e−ω/λbrρsa2t1−ωΨB1t1Fγ2⟶1t1ω=1−exp−ωλb1ρsa2t1−ωΨB1t1,FγR2t2ω=1−exp−ωa2t2ρrλr2.The expression ofFγR2 and C¯2 are given by(31)FγR2=1−e−ω/λbrρsa2t1−ωΨB1t1+ω/λb1ρsa2t1−ωΨB1t1+ω/a2t2ρrλr2,ω>0,1,ω≤0,(32)C¯2=t2ln2∫0∞e−ω/λbrρsa2t1−ωΨB1t1+ω/λb1ρsa2t1−ωΨB1t1+ω/a2t2ρrλr21+ωdω.In order to obtain the closed-form expression of equation (32), a high SNR approximation is used. When ρs⟶∞ and ρr⟶∞, equation (29) can be rewrite as follows:(33)FγR2=1−Pa2t1ΨB1t1>ωPγR2t2>ω=1−e−ω/a2t2ρrλr2,0<ω<a2t1ΨB1t1,1,otherwise.The closed-form expression ofC¯2 is given by(34)C¯2=t2ln2∫0∞1−FγR21+ωdω=t2ln2∫0a2t1/ΨB1t111+ωe−ω/a2t2ρrλr2dω=t2e1/λr2ρra2t2ln2Ei−1λr2ρra2t21+a2t1ΨB1t1−Ei−1λr2ρra2t2.


### 3.1.3. Ergodic Rates ofUE3

Assuming that the SINR ofUE3 is γR3, it decided by the weakest link of the relay, that is γR3=minγBR3t1,γ3⟶1t1,γR3t2. We can obtain the ergodic rates of UE3 in the subslots t1 and t2 as follows:(35)C¯3t2=t2ln2∫0∞1−FγR3ω1+ωdω.Since the channelshbr, hb1, and hr2 are independent of each other, the FγR3 is given by(36)FγR3ω=PγR3≤ω=PminγBR3t1,γ3⟶1t1,γR3t2≤ω=1−PminγBR3t1,γ3⟶1t1,γR3t2>ω=1−PγBR3t1>ωPγ3⟶1t1>ωPγR3t2>ω=1−1−PγBR3t1≤ω1−Pγ3⟶1t1≤ωPγR3t2≤ω=1−FγBR3t1ω1−Fγ3⟶1t1ω1−FγR3t2ω,where FγBR3t1ω, Fγ3⟶1t1ω, and FγR3t2ω represent the CDF of γBR3t1, γ3⟶1t1, and γR3t2, respectively, and we can obtain(37)FγBR3t1ω=PγBR3t1≤ω=Pa3t1ρshbr2ΨB1t1ρshbr2+a2t1ρshbr2+1≤ω=Phbr2≤ωa3t1−ΨB1t1+a2t1ωρs=1−e−ω/λbrρsa3t1−ΨB1t1ω+a2t1ω,(38)Fγ3⟶1t1ω=Pγ3⟶1t1≤ω=Pa3t1ρshb12ΨB1t1+a2t1ρshb12+1≤ω=Pa3t1ρshb121−a3t1ρshb12+1≤ω=Phb12≤ωa3t1−1−a3t1ωρs=1−e−ω/λb1a3t1−1−a3t1ωρs,(39)FγR3t2ω=PγR3t2≤ω=Pa3t2ρrhr32a2t2ρrhr32+1≤ω=Phr32≤ωa3t2−a2t2ωρr=1−e−ω/λr3ρra3t2−a2t2ω.Taking equations (37)–(39) into (36), the FγR3ω and C¯3t2 are given by(40)FγR3ω=1−e−ω/λbrρsa3t1−ΨB1t1ω+a2t1ω+ω/λb1a3t1−1−a3t1ωρs+ω/λr3ρra3t2−a2t2ω,ω>0,1,ω≤0,C¯3t2=t2ln2∫0∞1−FW1ω1+ωdω=t2ln2∫0∞e−ω/λbrρsa3t1−ΨB1t1ω+a2t1ω+ω/λb1a3t1−1−a3t1ωρs+ω/λr3ρra3t2−a2t2ω1+ωdω.Whenρs⟶∞ and ρr⟶∞, equation (36) can be rewritten as follows:(41)FγR3ω=1−Pa3t11−a3t1>ωPa3t21−a3t2>ω.The closed-form expression ofC¯3t2 is given by(42)C¯3t2=t2ln2∫0∞Pa3t1/1−a3t1>ωPa3t2/1−a3t2>ω1+ωdω,where C¯3t2 is determined by a3t1/1−a3t1 and a3t2/1−a3t2, since fx=x/1−x is an monotonically increasing function about x, we only need to discuss a3t1 and a3t2.(a)
Ifa3t2≤a3t1, we can obtain(43)C¯3t2=t2ln2∫0a3t2/1−a3t2Pa3t1/1−a3t1>ωPa3t2/1−a3t2>ω1+ω=t2ln2ln1+a3t22−a3t2.(b)
Ifa3t1<a3t2, we can obtain(44)C¯3t2=t2ln2∫0a3t1/1−a3t1Pa3t1/1−a3t1>ωPa3t2/1−a3t2>ω1+ω=t2ln2ln1+a3t12−a3t1.Combine a andb, we can get(45)C¯3t2=t2ln2ln1+a3t22−a3t2,a3t2≤a3t1,t2ln2ln1+a3t12−a3t1,a3t2>a3t1.Sincet2 = t4, we can obtain that C¯3t4=C¯3t2 combined with (16). Therefore, the ergodic rates of UE3 is given by(46)C¯3=C¯3t2+C¯3t4=2t2ln2ln1+a3t22−a3t2,a3t2≤a3t1,2t2ln2ln1+a3t12−a3t1,a3t2>a3t1.


### 3.1.4. Ergodic Rates ofUE4

It can be seen from (34) that C¯2 is related to t2, λr2, ρr, a1t1, a2t1, and a2t2. Combined with (16), the C¯4 is given by(47)C¯4=C¯2=t2e1/λr2ρra2t2ln2Ei−1λr2ρra2t21+a2t1ΨB1t1−Ei−1λr2ρra2t2.TakingC¯1,C¯2,C¯3,andC¯4 into (21), the system ergodic sum rate is given by(48)C¯sum≜EC1+EC2+EC3+EC4=C¯1+C¯2+C¯3+C¯4=−2t1ln2Ei−1λb1ρsΨB1t1e1/λb1ρsΨB1t1+2t2ln2e1/λr2ρra2t2Ei−1+a2t1/ΨB1t1λr2ρra2t2−Ei−1λr2ρra2t2+2t2ln2ln1+a3t22−a3t2,a3t2≤a3t1,2t2ln2ln1+a3t12−a3t1,a3t2>a3t1.


## 3.1.1. Ergodic Rates ofUE1

In thet1 subslot, assuming that UE1 successfully decodes the UE2 and UE3 signals, using (22), the C¯1t1 is given by(23)C¯1t1=t1ln2∫0∞1−FγBR1t1ω1+ωdω,where FγBR1t1ω is the CDF of the γBR1t1. Using the definition of the distribution function, there is(24)FγBR1t1ω=PγBR1t1≤ω=Pa1t1ρshb12≤ω=Phb12≤ωa1t1ρs.Since channelhb1∼CN0,λb1 obeys the complex Gaussian distribution, according to the literature [16], hb12 obeys the exponential distribution of λ=1/λb1. The distribution function of γBR1t1 is given by(25)FγBR1t1ω=1−e−ω/a1t1ρsλb1.Taking (25) into equation (23) and the ergodic rates of UE1 in t1 subslot is given by(26)C¯1t1=−t1ln2Ei−1λb1ρsa1t1e1/λb1ρsa1t1,where Eix=∫−∞xet/tdt.From equation (26), we know that the ergodic rates of UE1 is decided by t1, λb1, ρs, and a1t1. Since t1 = t3, we can obtain that C¯1t3=C¯1t1, so the ergodic rates of UE1 is given by(27)C¯1=C¯1t1+C¯1t3=−2t1ln2Ei−1λb1ρsΨB1t1e1/λb1ρsΨB1t1.


## 3.1.2. Ergodic Rates ofUE2

Assuming that the SINR ofUE2 is γR2, UE2 transmits its signal via the relay in the subslot t1 and subslot t2, because the SINR is decided by the weakest link of the relay, we can obtain that γR2=minγBR2t1,γ2⟶1t1,γR2t2. The ergodic rates of UE2 is given by(28)C¯2=t2ln2∫0∞1−FγR2ω1+ωdω.To get theC¯2, we need to obtain the CDF of FγR2ω. Since the channels hbr, hb1, and hr2 are independent of each other, FγR2 can be derived as follows:(29)FγR2=PγR2≤ω=PminγBR2t1,γ2⟶1t1,γR2t2≤ω=1−PminγBR2t1,γ2⟶1t1,γR2t2>ω=1−PγBR2t1>ωPγ2⟶1t1>ωPγR2t2>ω=1−1−PγBR2t1≤ω1−Pγ2⟶1t1≤ωPγR2t2≤ω=1−FγBR2t1ω1−Fγ2⟶1t1ω1−FγR2t2ω,where FγBR2t1ω, Fγ2⟶1t1ω, and FγR2t2ω are the CDF of the γBR2t1, γ2⟶1t1, and γR2t2. The CDFs are derived separately as follows:(30)FγBR2t1ω=PγBR2t1≤ω=Pa2t1ρshbr2ΨB1t1ρshbr2+1≤ω=Phbr2≤ωa2t1−ΨB1t1ωρs=1−e−ω/λbrρsa2t1−ωΨB1t1Fγ2⟶1t1ω=1−exp−ωλb1ρsa2t1−ωΨB1t1,FγR2t2ω=1−exp−ωa2t2ρrλr2.The expression ofFγR2 and C¯2 are given by(31)FγR2=1−e−ω/λbrρsa2t1−ωΨB1t1+ω/λb1ρsa2t1−ωΨB1t1+ω/a2t2ρrλr2,ω>0,1,ω≤0,(32)C¯2=t2ln2∫0∞e−ω/λbrρsa2t1−ωΨB1t1+ω/λb1ρsa2t1−ωΨB1t1+ω/a2t2ρrλr21+ωdω.In order to obtain the closed-form expression of equation (32), a high SNR approximation is used. When ρs⟶∞ and ρr⟶∞, equation (29) can be rewrite as follows:(33)FγR2=1−Pa2t1ΨB1t1>ωPγR2t2>ω=1−e−ω/a2t2ρrλr2,0<ω<a2t1ΨB1t1,1,otherwise.The closed-form expression ofC¯2 is given by(34)C¯2=t2ln2∫0∞1−FγR21+ωdω=t2ln2∫0a2t1/ΨB1t111+ωe−ω/a2t2ρrλr2dω=t2e1/λr2ρra2t2ln2Ei−1λr2ρra2t21+a2t1ΨB1t1−Ei−1λr2ρra2t2.


## 3.1.3. Ergodic Rates ofUE3

Assuming that the SINR ofUE3 is γR3, it decided by the weakest link of the relay, that is γR3=minγBR3t1,γ3⟶1t1,γR3t2. We can obtain the ergodic rates of UE3 in the subslots t1 and t2 as follows:(35)C¯3t2=t2ln2∫0∞1−FγR3ω1+ωdω.Since the channelshbr, hb1, and hr2 are independent of each other, the FγR3 is given by(36)FγR3ω=PγR3≤ω=PminγBR3t1,γ3⟶1t1,γR3t2≤ω=1−PminγBR3t1,γ3⟶1t1,γR3t2>ω=1−PγBR3t1>ωPγ3⟶1t1>ωPγR3t2>ω=1−1−PγBR3t1≤ω1−Pγ3⟶1t1≤ωPγR3t2≤ω=1−FγBR3t1ω1−Fγ3⟶1t1ω1−FγR3t2ω,where FγBR3t1ω, Fγ3⟶1t1ω, and FγR3t2ω represent the CDF of γBR3t1, γ3⟶1t1, and γR3t2, respectively, and we can obtain(37)FγBR3t1ω=PγBR3t1≤ω=Pa3t1ρshbr2ΨB1t1ρshbr2+a2t1ρshbr2+1≤ω=Phbr2≤ωa3t1−ΨB1t1+a2t1ωρs=1−e−ω/λbrρsa3t1−ΨB1t1ω+a2t1ω,(38)Fγ3⟶1t1ω=Pγ3⟶1t1≤ω=Pa3t1ρshb12ΨB1t1+a2t1ρshb12+1≤ω=Pa3t1ρshb121−a3t1ρshb12+1≤ω=Phb12≤ωa3t1−1−a3t1ωρs=1−e−ω/λb1a3t1−1−a3t1ωρs,(39)FγR3t2ω=PγR3t2≤ω=Pa3t2ρrhr32a2t2ρrhr32+1≤ω=Phr32≤ωa3t2−a2t2ωρr=1−e−ω/λr3ρra3t2−a2t2ω.Taking equations (37)–(39) into (36), the FγR3ω and C¯3t2 are given by(40)FγR3ω=1−e−ω/λbrρsa3t1−ΨB1t1ω+a2t1ω+ω/λb1a3t1−1−a3t1ωρs+ω/λr3ρra3t2−a2t2ω,ω>0,1,ω≤0,C¯3t2=t2ln2∫0∞1−FW1ω1+ωdω=t2ln2∫0∞e−ω/λbrρsa3t1−ΨB1t1ω+a2t1ω+ω/λb1a3t1−1−a3t1ωρs+ω/λr3ρra3t2−a2t2ω1+ωdω.Whenρs⟶∞ and ρr⟶∞, equation (36) can be rewritten as follows:(41)FγR3ω=1−Pa3t11−a3t1>ωPa3t21−a3t2>ω.The closed-form expression ofC¯3t2 is given by(42)C¯3t2=t2ln2∫0∞Pa3t1/1−a3t1>ωPa3t2/1−a3t2>ω1+ωdω,where C¯3t2 is determined by a3t1/1−a3t1 and a3t2/1−a3t2, since fx=x/1−x is an monotonically increasing function about x, we only need to discuss a3t1 and a3t2.(a)
Ifa3t2≤a3t1, we can obtain(43)C¯3t2=t2ln2∫0a3t2/1−a3t2Pa3t1/1−a3t1>ωPa3t2/1−a3t2>ω1+ω=t2ln2ln1+a3t22−a3t2.(b)
Ifa3t1<a3t2, we can obtain(44)C¯3t2=t2ln2∫0a3t1/1−a3t1Pa3t1/1−a3t1>ωPa3t2/1−a3t2>ω1+ω=t2ln2ln1+a3t12−a3t1.Combine a andb, we can get(45)C¯3t2=t2ln2ln1+a3t22−a3t2,a3t2≤a3t1,t2ln2ln1+a3t12−a3t1,a3t2>a3t1.Sincet2 = t4, we can obtain that C¯3t4=C¯3t2 combined with (16). Therefore, the ergodic rates of UE3 is given by(46)C¯3=C¯3t2+C¯3t4=2t2ln2ln1+a3t22−a3t2,a3t2≤a3t1,2t2ln2ln1+a3t12−a3t1,a3t2>a3t1.


## 3.1.4. Ergodic Rates ofUE4

It can be seen from (34) that C¯2 is related to t2, λr2, ρr, a1t1, a2t1, and a2t2. Combined with (16), the C¯4 is given by(47)C¯4=C¯2=t2e1/λr2ρra2t2ln2Ei−1λr2ρra2t21+a2t1ΨB1t1−Ei−1λr2ρra2t2.TakingC¯1,C¯2,C¯3,andC¯4 into (21), the system ergodic sum rate is given by(48)C¯sum≜EC1+EC2+EC3+EC4=C¯1+C¯2+C¯3+C¯4=−2t1ln2Ei−1λb1ρsΨB1t1e1/λb1ρsΨB1t1+2t2ln2e1/λr2ρra2t2Ei−1+a2t1/ΨB1t1λr2ρra2t2−Ei−1λr2ρra2t2+2t2ln2ln1+a3t22−a3t2,a3t2≤a3t1,2t2ln2ln1+a3t12−a3t1,a3t2>a3t1.


## 3.2. Outage Probability

The outage probability is the probability of the outage event happened in the communication. When the user’s reachable rates are lower than target rates, the outage event happened. In this paper, the outage probability of the cell center user is determined by its channel state. The outage probability of the cell edge users are determined by the two-hop system, and the state of the previous hop will affect the state of the next hop [17].(1)
Outage probability ofUE1Assumed thatE1t1O and E1t3O, respectively, represent the outage event happened of UE1 in t1 and t3 subslots, the outage probability in t1 time slot is PDFO1,t1=PrE1t1O and in t3 time slot is PDFO11,t3=PrE1t3O. For UE1, if it cannot detect the signal of UE2 and UE3 or the throughput does not reach the target rates RQ in the t1 time slot, the communication will interrupted. Denote these three cases as ER2⟶1t1O, ER3⟶1t1O, and EU1t1O, respectively, we can obtain the outage probability PrE1t1O as follows:(49)PDFO1,t1=PrE1t1O=PrEU1t1O∪ER2⟶1t1O∪ER3⟶1t1O=PrRBR1t1<RQ∪RR2⟶1t1<RQ∪RR3⟶1t1<RQ=PrminRBR1t1,RR2⟶1t1,RR3⟶1t1<RQ=1−PrminRBR1t1,RR2⟶1t1,RR3⟶1t1≥RQ=1−PrRBR1t1≥RQPrRR2⟶1t1≥RQPrRR3⟶1t1≥RQ=1−1−PrRBR1t1<RQ1−PrRR2⟶1t1<RQ1−PrRR3⟶1t1<RQ=1−1−FRBR1t1RQ1−FRR2⟶1t1RQ1−FRR3⟶1t1RQ,where RBR1t1=t1log21+γBR1t1, R2⟶1t1=t1log21+γ2⟶1t1, and R3⟶1t1=t1log21+γ3⟶1t1.The expressionsFRBR1t1RQ, FRR2⟶1t1RQ, and FRR3⟶1t1RQ are given by(50)FRBR1t1RQ=PrRBR1t1<RQ=Prt1log21+γBR1t1<RQ=PrγBR1t1<β1=Prhb12≤β1a1t1ρs=1−e−β1/λb1a1t1ρs,0<β1,1,otherwise,FRR2⟶1t1RQ=PrRR2⟶1t1<RQ=Prt1log21+γ2⟶1t1<RQ=Prγ2⟶1t1<β1=Prhb12≤β1a2t1−a1t1β1ρs=1−e−β1/λb1a2t1−a1t1β1ρs,0<β1<a2t1a1t1,1,otherwise,FRR3⟶1t1RQ=PrRR3⟶1<RQ=Prt1log21+γ3⟶1t1<RQ=Prγ3⟶1t1<β1=Pra3t1ρshb121−a3t1ρshb12+1<β1=Prhb12<β1a3t1−1−a3t1β1ρs=1−e−β1/λb1a3t1−1−a3t1β1ρs,0<β1<a3t11−a3t1,1,otherwise,where β1=2RQ/t1−1. The PDFO11,t1 is given by(51)PDFO1,t1=1−e−β1/λb1a1t1ρs−β1/λb1a2t1−a1t1β1ρs−β1/λb1a3t1−1−a3t1β1ρs,β1<a2t1a1t1,β1<a3t11−a3t1,1,otherwise.Similarly,PrE1t3O is given by(52)PrE1t3O=1−1−FRBR1t3RQ1−FRR2⟶1t3RQ1−FRR3⟶1t3RQ.Since the channel state does not change in the time slotst1 and t3, the relationship PrE1t3O=PrE1t1O is established. Therefore, the outage probability of UE1 is given by(53)PDFO1,t1=PDFO1,t3=1−e−β1/λb1a1t1ρs−β1/λb1a2t1−a1t1β1ρs−β1/λb1a3t1−1−a3t1β1ρs,β1<a2t1a1t1,β1<a3t11−a3t1,1,otherwise.(2)
Outage probability ofUE2TheUE2 communicates with the BS through the relay in the t1 and t2 sub-time slots, and its outage probability is related to each hop of the two-hop system. If UE2 cannot be successfully decoded in the relay or cannot decode the signal of UE3, communication will be interrupted. If UE2 can be decoded in the relay, but in the second hop transmission rate does not reach the target rate, the outage event will also occur. Denote the events that UE2 cannot successfully decode at relay, UE2 cannot decode the signal of the UE3, and in the second hop, the communication rates of UE2 does not reach the target rate RQ, respectively, as ER2O, ER3⟶2O, and EU2O. The outage probability PDFO2 of UE2 can be obtained as follows:(54)PDFO2=PrER2O+PrER2PrER3⟶2OER2+PrER2PrER3⟶2ER2PrEU2OER2ER3⟶2=1−PrER2ER3⟶2EU2=1−PrminRBR2,R3⟶2,RR2≥RQ=1−1−PrRBR2<RQ1−PrR3⟶2<RQ1−PrRR2<RQ=1−1−FRBR2RQ1−FR3⟶2RQ1−FRR2RQ,where RBR2=t1log21+γBR2t1, R3⟶2=t2log21+γ3⟶2t2, and RR2=t2log21+γR2t2.The expressionsFRBR2RQ, FR3⟶2RQ, and FRR2RQ are given by(55)FRBR2RQ=PrRBR2<RQ=Prt1log21+γBR2t1<RQ=PrγBR2t1<β1=Prhbr2≤β1a2t1−a1t1β1ρs=1−e−β1/λbra2t1−a1t1β1ρs,0<β1<a2t1a1t1,1,otherwise,FR3⟶2RQ=PrR3⟶2<RQ=Prt2log21+γ3⟶2t2<RQ=Prγ3⟶2t2<β2=Prhr22≤β2a2t2−a3t2β2ρr=1−e−β2/λr2a3t2−a2t2β2ρr,0<β2<a3t2a2t2,1,otherwise,FRR2RQ=PrRR2<RQ=Prt2log21+γR2t2<RQ=PrγR2t2<β2=Prhr22≤β2a2t2ρr=1−e−β2/λr2a2t2ρr,0<β2,1,otherwise,where β2=2RQ/t2−1, and PDFO2 is given by(56)PDFO2=1−e−β1/λbra2t1−a1t1β1ρs−β2/λr2a3t2−a2t2β2ρr−β2/λr2a2t2ρr,β1<a2t1a1t1,β2<a3t2a2t2,1,otherwise.(3)
Outage probability ofUE3Assumed thatE3t12O and E3t34O, respectively, indicate the event that UE3 interrupts in t1∼t2 and t3∼t4 subtime slots. The corresponding outage probability is PDFO3,t12=PrE3t12O and PDFO3,t34=PrE3t34O; obviously, the equation PDFO3,t12=PDFO3,t34 is established. Taking PDFO3,t12 for example, here we denote the event that UE3 cannot be decoded at the relay as ER3t1O, and the rates at the second hop does not reach the target rate RQ as EU3t2O. The PrE3t12O is given by(57)PDFO3,t12=PrER3t1O+PrEU3t2OER3t1PrER3t1=1−PrEU3t2,ER3t1=1−PrminRBR3t1,RR3t2≥RQ=1−PrRBR3t1≥RQPrRR3t2≥RQ=1−1−FRBR3t1RQ1−FRR3t2RQ,where RBR3t1=t1log21+γBR3t1 and RR3t2=t2log21+γR3t2.From the previous analysis, the outage probability ofUE3 is given by(58)PDFO3,t12=PDFO3,t34=1−e−β1/λbra3t1−1−a3t1β1ρs−β2/λr3a3t2−a2t2β2ρr,β1<a3t11−a3t1,β2<a3t2a2t2,1,otherwise.(4)
Outage probability ofUE4Since the power allocation factor unchanged in one time slot, obviously the power allocation factor of theUE4 in the t4 subslot is the same as the UE2 in the t2 subslot. Therefore, the outage probability is the same for both, we can obtain(59)PDFO4=PDFO2=1−e−β1/λbra2t1−a1t1β1ρs−β2/λr2a3t2−a2t2β2ρr−β2/λr2a2t2ρr,β1<a2t1a1t1,β2<a3t2a2t2,1,otherwise.The overall system outage event is defined as the event that any user in the system cannot achieve reliable detection, which means the overall outage probability is defined as follows [16]:(60)PDFO≜1−∏k=131−PDFOk.Substituting equations (53), (56), (58), and (59) into (60), we can obtain the system outage probability.


## 4. Time and Power Optimization Allocation Algorithm

In this paper, we consider maximizing the system sum rate under the premise of the system fairness. Since the system sum rate in NOMA system is mainly determined by the cell center user [18, 19], it is necessary to allocate time and power resources to the cell center user as much as possible. However, due to the system fairness constrain, sufficient resources should allocate to the cell edge users to ensure users communication does not interrupt and meets the corresponding communication service quality. Therefore, the system sum rate is affected by the fairness factor F′: When the transmitting power of the base station is fixed, the more time and power resources the edge users are allocated, the higher the fairness index of the system will be, but the overall rate of the system will decrease. Therefore, the sum rates maximum is contradictory to the system fairness, and we need to compromise on the fairness and rates of the system.The system sum rate is maximized under the condition of the cell edge user’s outage probability and system fairness limit. The optimization problem can be expressed as follows:(61)maxC1+C2+C3+C4,s.t.t1+t2+t3+t4=Tslot,PDFOi≤k,i=1,2,3,4,F=F′,PA1,PA2,where C1=C1t1+C1t3, C2=C2t2, C3=C3t2+C3t4, C4=C4t4. Tslot, k, and F′, respectively, represent the time-slot length, the user outage probability target, and the given system target fairness factor.The system fairness index is measured by introducing the fairness index factor. In literature [20], the fairness index factor expression is given by(62)F=∑i=14Ci24∑i=14Ci2.From the above analysis, the system sum rate maximization is negatively correlated with the system fairness. Therefore, it is necessary to consider how to schedule time and power to maximize the system sum rate while ensuring the given fairness factor. Since the system sum rate is a piecewise function and the objective function contains multiple parameters such as time and power allocation factor, the KKT method [21] cannot be used to find the optimal value in our proposed scenario.In this paper, we propose the joint optimal time power allocation algorithm based on exhaustive search and binary algorithm to obtain the optimal solution: firstly, the user's power allocation factor is obtained by the following method,UE1 is the cell center user, and the allocated power is small. Since the user power allocation scheme of the t1∼t2 time slot is the same as the t3∼t4 time slot, the time slot t1∼t2 is taken as an example. According to equation (16), if a1t1=k,k<0.5, then a2t1+a3t1=1−k, a2t1<a3t1, a2t2+a3t2=1, and a2t2<a3t2. Let a1t1 take values in a certain step in the interval 0.1,0.5. When a1t1=k, a2t1 searches in 0.1,1−k/2. a2t2 and a3t2 are also obtained in this way, so that all the power allocation factor list PA_MATRIX satisfying the conditions can be obtained; secondly, according to each value set of power allocation factors in PA_MATRIX, the corresponding user outage probability can be obtained. If the user outage probability satisfies the given threshold value, the corresponding time set Li,i=1,2,3,4 of each user when the outage probability is satisfied can be further calculated. The feasible time allocation list L can be obtained by taking the intersection of L1∼L4. Binary algorithm is then employed to find the optimal transmission time at a certain fairness factor index F′ premise; finally, the sorting algorithm is used to find the power allocation factor corresponding to the maximum system rates. This process shown in Algorithm 1.Algorithm 1:Time and power optimization allocation algorithm.
(1)
InitializationPower_list=zerosm,n, Time_list=zerosm,1
(2)
fori=1:m
(3)
Calculate user outage probabilityPDFOj,j=1,2,3,4 based on PA_MATRIXi,:
(4)
LetPDFOj≤k then find the time set Tj meets the criteria and the intersection L
(5)
Lettmax=Lend calculate system sum rates capacityi, fairness factor fairnessIndex
(6)
Low=1,High=end, set calculation accuracy ε.
(7)
whilefairnessIndex≠F′
(8)
ifLow≠High
(9)
mid=High+Low/2, • indicates rounding down
(10)
tmax=Lmid, calculate capacityi and fairness factor fairnessIndex
(11)
ifabsfairnessIndex−F′>ε
(12)
iffairnessIndex>F′
(13)
Low=mid
(14)
elseiffairnessIndex<F′
(15)
High=mid
(16)
elseifabsfairnessIndex−F′≤ε
(17)
Power_listi,:=PA_MATRIXi,:,Time_listi,:=tmax
(18)
break
(19)
elseiflow==High
(20)
break
(21)
Time_Pa=Time_list,Power_list, optimal time and power allocation factor matrix
(22)
maxCapacity=maxIndex=Optimal_Time_PA=0, initial system maximum rate and index
(23)
forj=1:lengthTime_Pa
(24)
Calculate the system sum ratescapacityj according to Time_Paj,:
(25)
ifcapacityj≥maxCapacity
(26)
maxCapacity=capacityj,maxIndex=j
(27)
Optimal_PA_Time = Time_PamaxIndex,:When acquiringPA_MATRIX, the complexity of this algorithm to calculate a1t1, a2t1, and a3t1 is On2, and the complexity to calculate a2t2 and a3t2 is On. Therefore, the algorithm complexity of obtaining the power allocation factor is On2. The calculation of time list L and system capacity can be completed in constant time, and the complexity is O1. The complexity of the time allocation list under the feasible power set is Olog2n.After the feasible power and time solution set is obtained, the sorting algorithm is used to obtain the optimal time and power allocation scheme. The complexity of this step isOn. The system complexity is T1n=On2log2n+n=On2log2n. When the exhaustive search method is used, the corresponding system complexity is T2n=On3, and it can be seen that T1n<T2n, so the proposed algorithm can effectively reduce the computational time complexity.


## 5. Numerical and Simulation Results

The simulation scenario is shown in Figure1. It includes a base station (BS), a relay (R), and four users (UEi,i=1,2,3,4). The relay operates in the half-duplex mode using DF strategy. UE1 is the cell center user, and hb1 denotes the channel coefficients from the BS to UE1. UE2, UE3, and UE4 are cell edge users, and hr2, hr3, and hr4 denote the channel coefficients from the relay to each cell edge user. We model these channels as hb1∼CN0,λb1, hbr∼CN0,λbr, hr2∼CN0,λr2, hr3∼CN0,λr3, and hr4∼CN0,λr4. The cell center user is more concerned with the ergodic rates, and the cell edge users are more concerned with the outage probability; we set different target outage probability for different users, and the simulation parameters are shown in Table 1.Table 1
System performance parameters.
Parameter
Value
Notes
ρ
s, dB
25∼50
Transmits SNR of BS
ρ
r
=
0.5
ρ
s, dB
12.5∼25
Transmits SNR of relay
λ
b
1
1
−
4
Channel variance ofBS⟶UE1
λ
b
r
0.5
−
4
Channel variance ofBS⟶R
λ
r
2, λr4
0.3
−
4
Channel variance ofBS⟶UE2,4
λ
r
3
0.15
−
4
Channel variance ofBS⟶UE3
k
2,3,4
10
−
3
User outage probability target
R
Q, bps/Hz
0.2
Target rates
F
′
0.5
System target fair factor indexFigure2 shows the theoretical and simulated rates of the user when F′=0.5. It can be seen from the figure that the simulation value agrees with the theoretical value. When ρs increases, the rates of each user gradually increase. UE1 rates increase rapidly at low SNR, and the growth become slower due to the constrain of system fairness index factor at high SNR. Users 2, 3, and 4 are the cell edge users, and the growth rate is slow.Figure 2
Relationship between user rate and SNR (ρs). The users’ rates are positively correlated with ρs.Figure3 shows the ergodic sum rate under different fairness index factors. We have compared the system throughput optimized by Algorithm 1 with the equal time slot allocation scheme. It can be seen that the smaller the fairness index factor F′ is, the larger the system ergodic sum rate is. The larger the F′ is, the smaller the system ergodic sum rate is. When F′=0.4, the ergodic sum rate optimized by the proposed algorithm is greater than the equal-time transmission [11] at any SNR. When F′=0.5 and ρs<45dB, the ergodic sum rate is greater than equal-time transmission, and when ρs>45dB, the fairness factor of equal-time transmission is low, and the ergodic sum rate is higher than that in this paper. When F = 0.6, the value of the intersection of the ergodic sum rate by the proposed algorithm and the equal-time transmission becomes smaller.Figure 3
Relationship between ergodic sum rate andρs under different F′.Figure4 shows the relationship between user outage probability and ρs when F′=0.5. It can be seen that the user outage probability decreases as ρs increases. Since different power allocation methods are adopted for different ρs, the logarithm of the outage probability and ρs are nonlinear.Figure 4
Relationship between user outage probability andρs.Figure5 shows a comparison of the system outage probability of this paper and the equal-time transmission strategy. The algorithm proposed in this paper has a lower system outage probability when ρs is low. When ρs is high, the probability of two systems is close.Figure 5
System outage probability.Figure6 shows the relationship between the fairness index factor and ρs. It shows that in the case of equal-time transmission, the fairness index factor will gradually decrease as the ρs increases. However, the proposed algorithm enables the system fairness index factor maintained near the given fairness factor F′.Figure 6
Relationship between fairness index factor andρs.Figure7 shows the relationship between user rates and F′ when ρs=50dB. It can be seen that the UE1 rates and the system sum rate are negatively correlated with F′, and users 2, 3, and 4 are positively correlated with F′. The rates of UE1 is always the largest, as F increases, and these rates of increase in users 2, 3, and 4 slow down. Because when F′ increases, the power and time resources allocated to the cell edge users rise, and the cell center user decreases accordingly. Besides, the throughput difference between users decreases. Since the cell center user has the greatest impact on system sum rate, although the cell edge users rates increase, the system sum rate still drops.Figure 7
Relationship between rates andF′.


## 6. Conclusion

In this paper, we study the time and power optimization allocation algorithm in the NOMA relay network. Firstly, the scenario of two-hop DF relay is established. Based on this model, the expressions of user outage probability and ergodic rates are derived. Secondly, an optimization model for maximizing the system rates is constructed. Thirdly, the joint time and power factor allocation algorithm is proposed to obtain the solution of the model. The simulation result shows, under the premise of considering the fairness of the system, the system rates optimized by the proposed algorithm is significantly improved compared with the equal-time allocation.



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*Source: 7842987-2019-04-22.xml*