Performance Analysis of a New Non-Orthogonal Multiple Access Design for Mitigating Information Loss

Abstract

This paper proposes a scheme that adds XOR bit operations into the encoding and decoding process of the conventional non-orthogonal multiple access (NOMA) system to alleviate performance degradation caused by the power distribution of the original signal. Because the conventional NOMA combines and sends multiple data within limited resources, it has a higher data rate than orthogonal multiple access (OMA), at the expense of error performance. However, by using the proposed scheme, both error performance and sum rate can be improved. In the proposed scheme, the transmitter sends the original data and the redundancy data in which the exclusive OR (XOR) values of the data are compressed using the superposition coding (SC) technique. After this process, the data rate of users decreases due to redundancy data, but since the original data are sent without power allocation, the data rate of users with poor channel conditions increases compared to the conventional NOMA. As a result, the error performance and sum rate of the proposed scheme are better than those of the conventional NOMA. Additionally, we derive an exact closed-form bit error rate (BER) expression for the proposed downlink NOMA design over Rayleigh fading channels.

1. Introduction

Non-orthogonal multiple access (NOMA) is a promising technology to increase the spectral efficiency of next-generation communication systems [1,2]. Unlike orthogonal multiple access (OMA), which transmits only one data point in one resource block (RB), NOMA can transmit multiple data in one resource block at once. Therefore, NOMA is a promising technology that can enhance the overall system data rate [3]. Among the various NOMA techniques, power-domain NOMA (PD-NOMA), which utilizes power as a resource to transmit multiple data streams simultaneously, is currently receiving significant research attention [4]. The key techniques of PD-NOMA are superposition coding (SC) at the transmitter and successive interference cancellation (SIC) at the receiver [5]. SC is a technique that combines two or more types of data with power allocation, and SIC is a technique that decodes the stronger signal first and removes the decoded signal from the combined signal to decode the weaker signal. In a NOMA system with multiple users, users with better channel conditions perform more SIC to decode their signals [6].

Since NOMA, which has been studied so far, transmits data to multiple users in one cluster by dividing power resources, there is a problem that it is easily modified by noise power during the SIC process at the receiver [7]. Unlike OMA, in which multiple users’ data are sent by dividing the power resource, NOMA compresses multiple data into one resource, leading to enhanced data rates but potentially decreased error performance. In a system with two users, when combining the data of both users through SC, if the bits of the two data are the same, they would be indistinguishable from sending data of one user in an OMA system. The decrease in error performance arises from differences in the bits of the two data. Taking this into account, we have researched a direction to reduce the data length combined through SC while modifying the bits similar to OMA, aiming to improve bit error rate (BER) performance. In this paper, we propose a coding method that transforms the data of two users through exclusive OR (XOR) bitwise operations and adds redundancy data. The advantage of this approach is that it can improve the error performance of both users. Furthermore, since power allocation in SC does not apply to the original signal, this method leads to an enhancement in sum rate and better user fairness.

Notations: ${(·)}^{\mathrm{OMA}}$ and ${(·)}^{\mathrm{NOMA}}$ represent information in OMA and NOMA systems, respectively; $\widetilde{\left(·\right)}$ denotes information in the proposed design; ⊕ is the XOR bitwise operator; ${(·)}_{\mathrm{I}}$ and ${(·)}_{\mathrm{Q}}$ represent the in-phase and quadrature components, respectively; $\mathrm{P}(·)$ denotes the probability of an event; and ${\mathrm{P}}_k\left(\mathit{e}\right)$ denotes the bit error probability of user k.

2. System Model

Although previous studies have typically been conducted on systems with many clusters containing numerous users, in this paper, we consider only one cluster with two users because it explains how a new encoding and decoding method is better than the existing methods. Figure 1 shows the NOMA wireless communication system in a downlink scenario with one base station (BS) and two users. Each node has a single antenna. The channels between the BS and the users experience independent and identically distributed Rayleigh fading. It is assumed that the channel state of user 1 is better than that of user 2, i.e., user 1 is considered as the cell-center user, and user 2 is the cell-edge user. The received signal of the users is as follows:

$$\mathbf{Y}=\sqrt{P_{\mathrm{t}}}\mathbf{H}x+\mathbf{n},$$

(1)

where $P_{\mathrm{t}}$ denotes the BS transmit power, $\mathbf{H}\in {\mathbb{C}}^{2\times 1}$ denotes the channel between the BS and users, and $\mathbf{n}$ denotes additive white Gaussian noise (AWGN) with a zero mean and ${\sigma }^2$ variance. x is a signal sent from the BS, which is described later in Section 2.1, Section 2.2 and Section 3.1.

2.1. OMA System

In the OMA system, it is assumed that two users receive their data by allocating resources to each user in one resource block. When user k’s signal is defined as $s_k$, the received signal at user i is as follows:

$$y_k^{\mathrm{OMA}}=\sqrt{P_{\mathrm{t}}}{h}_k{s}_k+n_k,k=1,2$$

(2)

where $h_k$ and $n_k$ are row vectors of $\mathbf{H}$ and $\mathbf{n}$, respectively. $h_k$ represents the channel between the BS and user k, while $n_k$ represents the AWGN noise at user k. Additionally, the data rate for each user can be expressed as follows:

$$R_k^{\mathrm{OMA}}={\lambda }_k^{\mathrm{OMA}}{log}_2\left(1+\rho {\left|h_k\right|}^2\right),k=1,2$$

(3)

where ${\lambda }_k$ denotes the ratio of user k’s information in one resource block, ${\lambda }_1^{\mathrm{OMA}}$ + ${\lambda }_2^{\mathrm{OMA}}$ = 1, and $\rho$ denotes the transmit signal-to-noise ratio (SNR), $\rho =P_{\mathrm{t}}/{\sigma }^2$.

2.2. Conventional NOMA System

In the conventional NOMA system, the signals of each user are combined at the BS through SC, and the combined signal is sent to users within a single resource block. The signal x sent from the BS can be expressed as follows:

$$x=\sqrt{{\alpha }_1}{s}_1+\sqrt{{\alpha }_2}{s}_2,$$

(4)

where ${\alpha }_k$ denotes the power allocation coefficient (${\alpha }_1+{\alpha }_2=1$ and ${\alpha }_2>{\alpha }_1$, for $k=1,2$). Since user 2 is far away from the BS, in general, more power is allocated to $s_2$ to achieve better fairness. According to (1) and (4), the received signal at user k for the conventional NOMA is represented as follows:

$$y_k^{\mathrm{NOMA}}=\sqrt{{\alpha }_1{P}_{\mathrm{t}}}{h}_k{s}_1+\sqrt{{\alpha }_2{P}_t}{h}_k{s}_2+n_k.$$

(5)

Since user 2 decodes $s_2$ without SIC, the term corresponding to $s_1$ is considered as interference. Therefore, the data rate for $s_2$ at user 2 is given as follows:

$$R_2^{\mathrm{NOMA}}={log}_2\left(1+{\displaystyle \frac{{\alpha }_2{\left|h_2\right|}^2}{{\alpha }_1{\left|h_2\right|}^2+\frac{1}{\rho }}}\right).$$

(6)

After performing SIC, the data rate of $s_1$ at user 1 is calculated as follows:

$$R_1^{\mathrm{NOMA}}={log}_2\left(1+\rho {\alpha }_1{\left|h_1\right|}^2\right).$$

(7)

3. Proposed Scheme

3.1. Proposed Transmitter Design

The characteristic of NOMA is a high transmission rate, but it is susceptible to noise because two or more users’ data are transmitted on one frequency or time resource block. To address this, our approach can reduce the impact of noise by not dividing the power of the user’s actual data and adding redundancy data to help decode the data. Redundancy data can be obtained by performing an XOR bit operation on the data of two users.

Figure 2 shows the block diagram of the proposed transmitter design. When $b_k$ denotes the bit stream of user k’s data, the XOR logic bit operation value of $b_1$ and $b_2$ is defined as $b_{\mathrm{XOR}}$:

$$b_{\mathrm{XOR}}=b_1\oplus b_2.$$

(8)

Next, the partial bits obtained by splitting $b_{\mathrm{XOR}}$ into N are defined as $b_{\mathrm{XOR},n}$, $n\in \left\{1,2,\cdots ,N\right\}$. Then, the parts are combined using the SC technique. The redundancy data are as follows:

$$s_{\mathrm{c}}={\displaystyle \sum _{n=1}^N}{\alpha }_n{s}_{\mathrm{c},n},$$

(9)

where $\alpha$ denotes the power allocation coefficient, ${\sum }_{n=1}^N{\alpha }_n=1$, and $s_{\mathrm{c},n}$ denotes the modulated signal of $b_{\mathrm{XOR},k}$. Finally, the transmitter transmits $s_2$ and $s_{\mathrm{c}}$ serially to the receiver.

The reason for sending the signal $s_2$ instead of $s_1$ is to increase the system’s user fairness. If $s_1$ is sent, user 2 should obtain $b_{\mathrm{XOR}}$ to decode $b_2$. Since $s_{\mathrm{c}}$ is a signal combined with SC, the required SNR to decode it is high. Therefore, assigning the task of converting $s_{\mathrm{c}}$ to $b_{\mathrm{XOR}}$ to user 1, who has a better channel condition, helps increase the overall system’s bit error probability.

3.2. Proposed Receiver Design

Figure 3 shows the block diagram of the proposed receiver design. User 2 can obtain $s_2$ by decoding the received signal, and $b_2$ is given by demodulating $s_2$. In contrast, user 1 can obtain both $s_2$ and $s_{\mathrm{c},2}$ by decoding the received signal. Similar to user 2, user 1 can obtain $b_2$ by demodulating $s_2$. User 1 can then perform SIC to remove $s_{\mathrm{c},2}$ from the received signal and obtain $s_{\mathrm{c},1}$. $s_{\mathrm{c},1}$ and $s_{\mathrm{c},2}$ are demodulated to $b_{\mathrm{XOR},1}$ and $b_{\mathrm{XOR},2}$, respectively, which are then serially combined to $b_{\mathrm{XOR}}$. Finally, user 1 can obtain $b_1$, which is the XOR logical operation value of $b_2$ and $b_{\mathrm{XOR}}$($b_2\oplus b_{\mathrm{XOR}}$). Since user 1 and user 2 both carry information in $s_2$, the data rate of user k in the proposed system is given as follows:

$${\tilde{R}}_k={\tilde{\lambda }}_k{log}_2\left(1+\rho {\left|h_k\right|}^2\right).$$

(10)

As user 1 and user 2 share common information, the amount of information in a single resource block is the same for both. Furthermore, as N increases, the proportion of redundancy decreases. Therefore, $\tilde{\lambda }$ is given as follows:

$${\tilde{\lambda }}_1={\tilde{\lambda }}_2={\displaystyle \frac{N}{1+N}}.$$

(11)

4. Performance Analysis

In this section, we analyze the exact BER expression for both the conventional NOMA and the proposed NOMA. In this paper, it is assumed that the data for user 1 and user 2 are modulated by QPSK. Each user’s symbol is detected as one of $\left\{00,01,10,11\right\}$, assuming that the priori probabilities of each symbol are equal, and each symbol is modulated according to QPSK as {$\sqrt{1/2}+j\sqrt{1/2},-\sqrt{1/2}+j\sqrt{1/2},\sqrt{1/2}-j\sqrt{1/2},-\sqrt{1/2}-j\sqrt{1/2}$}.

4.1. Exact BER of Conventional NOMA

In the conventional NOMA system, both user 1 and user 2 decode the received signal to detect $s_2$. Additionally, user 1 detects and removes $s_2$ to decode its own signal, $s_1$. Therefore, the error probability for $s_2$ at user k is first calculated. In QPSK modulation, since each symbol consists of 2 bits, the probability that user k fails to decode $s_2$ can be expressed as the average of the error probabilities for the first and second bits of $s_2$, as follows:

$$\mathrm{P}\left({\mathit{X}}^k\right)={\displaystyle \frac{1}{2}}\left[\mathrm{P}\left({\mathit{X}}_1^k\right)+\mathrm{P}\left({\mathit{X}}_2^k\right)\right],$$

(12)

where ${\mathit{X}}_j$ denotes the event that user k incorrectly detects $s_2$ and ${\mathit{X}}_j^k$ denotes the event that user k incorrectly detects the j-th bit of $s_2$ ($j=1,2$) [8]. Figure 4 shows the constellation diagram of the signals received by each user in a conventional NOMA system. Each point represents the combination of signals $s_1$ and $s_2$ under SC. It is denoted as $\left\{s_1,s_2\right\}$, and the received signals at each point are expressed by (5) as $(\pm \sqrt{{\alpha }_1{P}_{\mathrm{t}}/2}\pm j\sqrt{{\alpha }_1{P}_{\mathrm{t}}/2})h_k+(\pm \sqrt{{\alpha }_2{P}_{\mathrm{t}}/2}\pm j\sqrt{{\alpha }_2{P}_{\mathrm{t}}/2})h_k+n_k$.

In Figure 4, an error occurs if any point on the constellation diagram deviates from its appropriate decision region. It is evident that the first bit of the QPSK symbol is determined by the quadrature part, and the second bit is determined by the in-phase part. For instance, if the actual first bit of $s_2$ is 1, in order to detect the first bit correctly, points ${\mathrm{A}}_k$ and ${\mathrm{B}}_k$ should be located below the boundary of the decision region. The second bit can also be approached in the same way as before. That is, since user k mistakenly decides bit 1 as 0 when point ${\mathrm{A}}_k$ or ${\mathrm{B}}_k$ is greater than 0, $\mathrm{P}\left({\mathit{X}}_1^k\right)$ and $\mathrm{P}\left({\mathit{X}}_2^k\right)$ can be expressed as follows:

$$\begin{array}{cc}\hfill \mathrm{P}\left({\mathit{X}}_1^k\right)& ={\displaystyle \frac{1}{2}}\left[\mathrm{P}\left({\mathit{X}}_{1,{\mathrm{A}}_k}^k\right)+\mathrm{P}\left({\mathit{X}}_{1,{\mathrm{B}}_k}^k\right)\right]\hfill \\ \hfill \mathrm{P}\left({\mathit{X}}_2^k\right)& ={\displaystyle \frac{1}{2}}\left[\mathrm{P}\left({\mathit{X}}_{2,{\mathrm{A}}_k}^k\right)+\mathrm{P}\left({\mathit{X}}_{2,{\mathrm{B}}_k}^k\right)\right],\hfill \end{array}$$

(13)

where ${\mathit{X}}_{j,i}^k$ denotes the event that user k incorrectly detects the j-th bit of $s_2$ due to the position of point i ($i={\mathrm{A}}_k,{\mathrm{B}}_k$). By substituting (5) and the QPSK symbols, (13) can be expressed as follows:

$$\begin{array}{cc}\hfill \mathrm{P}\left({\mathit{X}}_1^k\right)=& {\displaystyle \frac{1}{2}}\left[\mathrm{P}\left(n_{\mathrm{Q}}\ge \sqrt{{\displaystyle \frac{{ϵ}_2}{2}}}{h}_k+\sqrt{{\displaystyle \frac{{ϵ}_1}{2}}}{h}_k\right)+\mathrm{P}\left(n_{\mathrm{Q}}\ge \sqrt{{\displaystyle \frac{{ϵ}_2}{2}}}{h}_k-\sqrt{{\displaystyle \frac{{ϵ}_1}{2}}}{h}_k\right)\right],\hfill \\ \hfill \mathrm{P}\left({\mathit{X}}_2^k\right)=& {\displaystyle \frac{1}{2}}\left[\mathrm{P}\left(n_{\mathrm{I}}\ge \sqrt{{\displaystyle \frac{{ϵ}_2}{2}}}{h}_k+\sqrt{{\displaystyle \frac{{ϵ}_1}{2}}}{h}_k\right)+\mathrm{P}\left(n_{\mathrm{I}}\ge \sqrt{{\displaystyle \frac{{ϵ}_2}{2}}}{h}_k-\sqrt{{\displaystyle \frac{{ϵ}_1}{2}}}{h}_k\right)\right],\hfill \end{array}$$

(14)

where $n_{\mathrm{I}}$ and $n_{\mathrm{Q}}$ denote the in-phase and quadrature components of the noise, respectively, and ${ϵ}_k$ is the symbol energy, ${\alpha }_k{P}_{\mathrm{t}}$ [9]. By substituting (14) into (12), the error probability of user 2 is defined as follows:

$$\begin{array}{cc}\hfill {\mathrm{P}}_2\left(\mathit{e}\right)=& \mathrm{P}\left({\mathit{X}}^2\right)\hfill \\ \hfill =& {\displaystyle \frac{1}{2}}\left[\mathit{Q}\left({\displaystyle \frac{(\sqrt{{ϵ}_2/2}+\sqrt{{ϵ}_1/2})h_2}{\sqrt{{\sigma }^2/2}}}\right)+\mathit{Q}\left({\displaystyle \frac{(\sqrt{{ϵ}_2/2}-\sqrt{{ϵ}_1/2})h_2}{\sqrt{{\sigma }^2/2}}}\right)\right],\hfill \end{array}$$

(15)

where Q(·) denotes the Gaussian Q function. The simplified expression of (15) is as follows:

$${\mathrm{P}}_2\left(\mathit{e}\right)={\displaystyle \frac{1}{2}}\left[\mathit{Q}\left(\sqrt{{\gamma }_{{\mathrm{A}}_2}}\right)+\mathit{Q}\left(\sqrt{{\gamma }_{{\mathrm{B}}_2}}\right)\right],$$

(16)

where ${\gamma }_i$ is the SNR for the i signal points given in Figure 4. For $i={\mathrm{A}}_k,{\mathrm{B}}_k$, ${\gamma }_i$ can be expressed as follows:

$$\begin{array}{c}\hfill {\gamma }_{{\mathrm{A}}_k}={\displaystyle \frac{{(\sqrt{{ϵ}_2}+\sqrt{{ϵ}_1})}^2{\left|h_k\right|}^2}{{\sigma }^2}},\\ \hfill {\gamma }_{{\mathrm{B}}_k}={\displaystyle \frac{{(\sqrt{{ϵ}_2}-\sqrt{{ϵ}_1})}^2{\left|h_k\right|}^2}{{\sigma }^2}}.\end{array}$$

(17)

Then, the average BER at user 2 can be obtained as follows:

$$\begin{array}{cc}\hfill \overline{{\mathrm{P}}_2\left(\mathit{e}\right)}=& {\displaystyle \frac{1}{2}}\left[{\int }_0^{\infty }\mathit{Q}\left(\sqrt{{\gamma }_{{\mathrm{A}}_2}}\right)f_{{\gamma }_{{\mathrm{A}}_2}}\left({\gamma }_{{\mathrm{A}}_2}\right)\mathrm{d}{\gamma }_{{\mathrm{A}}_2}+{\int }_0^{\infty }\mathit{Q}\left(\sqrt{{\gamma }_{{\mathrm{B}}_2}}\right)f_{{\gamma }_{{\mathrm{B}}_2}}\left({\gamma }_{{\mathrm{B}}_2}\right)\mathrm{d}{\gamma }_{{\mathrm{B}}_2}\right],\hfill \end{array}$$

(18)

where $f_{{\gamma }_i}$ is the probability density function (PDF) of SNR. In this paper, assuming a Rayleigh fading channel, the PDF of the SNR is defined as follows:

$$f_{{\gamma }_i}\left(\gamma \right)={\displaystyle \frac{1}{\overline{{\gamma }_i}}}{e}^{-\gamma /\overline{{\gamma }_i}},\gamma \ge 0,$$

(19)

where $\overline{{\gamma }_i}$ denotes the average of ${\gamma }_i$.

Using ([10], Equations (13.3)–(7)) and (19), (18) is given as follows:

$$\overline{{\mathrm{P}}_2\left(\mathit{e}\right)}={\displaystyle \frac{1}{4}}\left(2-\sqrt{{\displaystyle \frac{\overline{{\gamma }_{{\mathrm{A}}_2}}}{2+\overline{{\gamma }_{{\mathrm{A}}_2}}}}}-\sqrt{{\displaystyle \frac{\overline{{\gamma }_{{\mathrm{B}}_2}}}{2+\overline{{\gamma }_{{\mathrm{B}}_2}}}}}\right).$$

(20)

Next, user 1 decodes $s_1$ after performing SIC to remove $s_2$. The error probability of user 1 depends on whether $s_2$ is detected correctly or incorrectly. Therefore, the error probability of user 1 is defined as follows:

$${\mathrm{P}}_1\left(\mathit{e}\right)={\mathrm{P}}_1\left(\mathit{e}\right|{\left({\mathit{X}}^1\right)}^{\prime })+{\mathrm{P}}_1\left(\mathit{e}\right|{\mathit{X}}^1),$$

(21)

where ${(·)}^{\prime }$ denotes the complement of $(·)$. Figure 5a shows the constellation of $s_1$ after user 1 correctly detects $s_2$ and performs SIC. The expressions for each signal are derived from the signal expressions in Figure 4 with the terms containing ${ϵ}_2$ removed. Therefore, the signal at each point is expressed as $(\pm \sqrt{{ϵ}_1/2}\pm j\sqrt{{ϵ}_1/2})h_1+n_1$. For the correct detection of the bits in $s_1$, the points C or C′ on the in-phase and quadrature axes should not fall outside their respective decision regions.

User 1 incorrectly detects the first bit of $s_1$ if either $\sqrt{{ϵ}_1/2}{h}_1+n_{\mathrm{Q}}\le 0$ or $-\sqrt{{ϵ}_1/2}{h}_1+n_{\mathrm{Q}}\ge 0$, and incorrectly detects the second bit of $s_1$ if either $\sqrt{{ϵ}_1/2}{h}_1+n_{\mathrm{I}}\le 0$ or $-\sqrt{{ϵ}_1/2}{h}_1+n_{\mathrm{I}}\ge 0$. Hence, the bit error probability of user 1, under the condition that no error occurs when detecting $s_2$ symbols, is given as follows:

$$\begin{array}{cc}\hfill {\mathrm{P}}_1\left(\mathit{e}\right|{\left({\mathit{X}}^1\right)}^{\prime })=& {\displaystyle \frac{1}{2}}\left[{\displaystyle \frac{1}{4}}\mathrm{P}\left({\left({\mathit{X}}_{1,{\mathrm{A}}_1}^1\right)}^{\prime }\right)\right.\hfill \\ \hfill & \times \left\{\mathrm{P}\left(n_{\mathrm{Q}}\ge \sqrt{{\displaystyle \frac{{ϵ}_1}{2}}}{h}_1\left|{\left({\mathit{X}}_{1,{\mathrm{A}}_1}^1\right)}^{\prime }\right.\right)+\mathrm{P}\left(n_{\mathrm{I}}\ge \sqrt{{\displaystyle \frac{{ϵ}_1}{2}}}{h}_1\left|{\left({\mathit{X}}_{1,{\mathrm{A}}_1}^1\right)}^{\prime }\right.\right)\right\}\hfill \\ \hfill & +{\displaystyle \frac{1}{4}}\mathrm{P}\left({\left({\mathit{X}}_{1,{\mathrm{B}}_1}^1\right)}^{\prime }\right)\hfill \\ \hfill & \times \left\{\mathrm{P}\left(n_{\mathrm{Q}}\le -\sqrt{{\displaystyle \frac{{ϵ}_1}{2}}}{h}_1\left|{\left({\mathit{X}}_{1,{\mathrm{B}}_1}^1\right)}^{\prime }\right.\right)+\mathrm{P}\left(n_{\mathrm{I}}\ge \sqrt{{\displaystyle \frac{{ϵ}_1}{2}}}{h}_1\left|{\left({\mathit{X}}_{1,{\mathrm{B}}_1}^1\right)}^{\prime }\right.\right)\right\}\hfill \\ \hfill & +{\displaystyle \frac{1}{4}}\mathrm{P}\left({\left({\mathit{X}}_{2,{\mathrm{A}}_1}^1\right)}^{\prime }\right)\hfill \\ \hfill & \times \left\{\mathrm{P}\left(n_{\mathrm{Q}}\ge \sqrt{{\displaystyle \frac{{ϵ}_1}{2}}}{h}_1\left|{\left({\mathit{X}}_{2,{\mathrm{A}}_1}^1\right)}^{\prime }\right.\right)+\mathrm{P}\left(n_{\mathrm{I}}\ge \sqrt{{\displaystyle \frac{{ϵ}_1}{2}}}{h}_1\left|{\left({\mathit{X}}_{2,{\mathrm{A}}_1}^1\right)}^{\prime }\right.\right)\right\}\hfill \\ \hfill & +{\displaystyle \frac{1}{4}}\mathrm{P}\left({\left({\mathit{X}}_{2,{\mathrm{B}}_1}^1\right)}^{\prime }\right)\hfill \\ \hfill & \left.\times \left\{\mathrm{P}\left(n_{\mathrm{Q}}\le -\sqrt{{\displaystyle \frac{{ϵ}_1}{2}}}{h}_1\left|{\left({\mathit{X}}_{2,{\mathrm{B}}_1}^1\right)}^{\prime }\right.\right)+\mathrm{P}\left(n_{\mathrm{I}}\ge \sqrt{{\displaystyle \frac{{ϵ}_1}{2}}}{h}_1\left|{\left({\mathit{X}}_{2,{\mathrm{B}}_1}^1\right)}^{\prime }\right.\right)\right\}\right],\hfill \end{array}$$

(22)

where ${\displaystyle \frac{1}{4}}\mathrm{P}\left({\left({\mathit{X}}_{j,{\mathrm{A}}_1}^1\right)}^{\prime }\right)$ and ${\displaystyle \frac{1}{4}}\mathrm{P}\left({\left({\mathit{X}}_{j,{\mathrm{B}}_1}^1\right)}^{\prime }\right)$ are the probabilities that user 1 correctly detects the j-th bit of $s_2$ as given in (13). If the fundamental theorem of conditional probability, $\mathrm{P}\left(\mathrm{A}\right|\mathrm{B})=\mathrm{P}(\mathrm{A}\cap \mathrm{B})/\mathrm{P}(\mathrm{B})$, is applied, (22) can be redefined as follows:

$$\begin{array}{cc}\hfill {\mathrm{P}}_1\left(\mathit{e}\right|{\left({\mathit{X}}^1\right)}^{\prime })=& {\displaystyle \frac{1}{8}}\left[\mathrm{P}\left(\sqrt{{\displaystyle \frac{{ϵ}_1}{2}}}{h}_1\le n_{\mathrm{Q}}\le \sqrt{{\displaystyle \frac{{ϵ}_2}{2}}}{h}_1+\sqrt{{\displaystyle \frac{{ϵ}_1}{2}}}{h}_1\right)+\mathrm{P}\left(n_{\mathrm{Q}}\le -\sqrt{{\displaystyle \frac{{ϵ}_1}{2}}}{h}_1\right)\right.\hfill \\ \hfill & +\mathrm{P}\left(n_{\mathrm{I}}\ge \sqrt{{\displaystyle \frac{{ϵ}_1}{2}}}{h}_1\right)\hfill \\ \hfill & \times \left\{\mathrm{P}\left(n_{\mathrm{Q}}\le \sqrt{{\displaystyle \frac{{ϵ}_2}{2}}}{h}_1+\sqrt{{\displaystyle \frac{{ϵ}_1}{2}}}{h}_1\right)+\mathrm{P}\left(n_{\mathrm{Q}}\le \sqrt{{\displaystyle \frac{{ϵ}_2}{2}}}{h}_1-\sqrt{{\displaystyle \frac{{ϵ}_1}{2}}}{h}_1\right)\right\}\hfill \\ \hfill & +\mathrm{P}\left(\sqrt{{\displaystyle \frac{{ϵ}_1}{2}}}{h}_1\le n_{\mathrm{I}}\le \sqrt{{\displaystyle \frac{{ϵ}_2}{2}}}{h}_1+\sqrt{{\displaystyle \frac{{ϵ}_1}{2}}}{h}_1\right)+\mathrm{P}\left(n_{\mathrm{I}}\le -\sqrt{{\displaystyle \frac{{ϵ}_1}{2}}}{h}_1\right)\hfill \\ \hfill & +\mathrm{P}\left(n_{\mathrm{Q}}\ge \sqrt{{\displaystyle \frac{{ϵ}_1}{2}}}{h}_1\right)\hfill \\ \hfill & \times \left\{\mathrm{P}\left(n_{\mathrm{I}}\le \sqrt{{\displaystyle \frac{{ϵ}_2}{2}}}{h}_1+\sqrt{{\displaystyle \frac{{ϵ}_1}{2}}}{h}_1\right)+\left.\mathrm{P}\left(n_{\mathrm{I}}\le \sqrt{{\displaystyle \frac{{ϵ}_2}{2}}}{h}_1-\sqrt{{\displaystyle \frac{{ϵ}_1}{2}}}{h}_1\right)\right\}\right].\hfill \end{array}$$

(23)

For the purpose of simplifying (23), we define the SNR for the C or C′ signal point given in Figure 5a as follows:

$$\begin{array}{c}\hfill {\gamma }_{\mathrm{C}}={\displaystyle \frac{{ϵ}_1{\left|h_1\right|}^2}{{\sigma }^2}}.\end{array}$$

(24)

Then, the error probability of $s_1$, given that user 1 correctly detects $s_2$, is expressed as follows:

$$\begin{array}{cc}\hfill {\mathrm{P}}_1\left(\mathit{e}\right|{\left({\mathit{X}}^1\right)}^{\prime })=& {\displaystyle \frac{1}{4}}\left[\mathit{Q}\left(\sqrt{{\gamma }_{\mathrm{C}}}\right)\times \left\{4-\mathit{Q}\left(\sqrt{{\gamma }_{{\mathrm{A}}_1}}\right)-\mathit{Q}\left(\sqrt{{\gamma }_{{\mathrm{B}}_1}}\right)\right\}-\mathit{Q}\left(\sqrt{{\gamma }_{{\mathrm{A}}_1}}\right)\right].\hfill \end{array}$$

(25)

Figure 5b,c show the constellations of $s_1$ after user 1 incorrectly detects $s_2$ and performs SIC. These represent the cases where the first bit of $s_2$ is in error and where the second bit is in error, respectively. The expressions at points D and E, contrary to the case of ${\mathit{X}}^1$, represent the in-phase and quadrature components of the signal, respectively, where terms containing ${ϵ}_1$ or ${ϵ}_2$ are not removed, but instead added to the signals in Figure 4: $(\pm \sqrt{{ϵ}_1/2}-2\sqrt{{ϵ}_1/2})h_1+n_1$. In Figure 5b, user 1 incorrectly detects the first bit of $s_1$ if either $(-\sqrt{{ϵ}_1/2}-2\sqrt{{ϵ}_2/2})h_1+n_{\mathrm{Q}}\ge 0$ or $(\sqrt{{ϵ}_1/2}-2\sqrt{{ϵ}_2/2})+n_{\mathrm{Q}}\le 0$, and incorrectly detects the second bit of $s_1$ if either $\sqrt{{ϵ}_1/2}{h}_1+n_{\mathrm{I}}\le 0$ or $-\sqrt{{ϵ}_1/2}{h}_1+n_{\mathrm{I}}\ge 0$. Similarly, in Figure 5c, user 1 incorrectly detects the first bit of $s_1$ if either $\sqrt{{ϵ}_1/2}{h}_1+n_{\mathrm{Q}}\le 0$ or $-\sqrt{{ϵ}_1/2}{h}_1+n_{\mathrm{Q}}\ge 0$, and incorrectly detects the second bit of $s_1$ if either $(-\sqrt{{ϵ}_1/2}-2\sqrt{{ϵ}_2/2})h_1+n_{\mathrm{I}}\ge 0$ or $(\sqrt{{ϵ}_1/2}-2\sqrt{{ϵ}_2/2})+n_{\mathrm{I}}\le 0$. Hence, the bit error probability of user 1, under the condition that error occurs when detecting $s_2$ symbols, is given as follows:

$$\begin{array}{cc}\hfill {\mathrm{P}}_1\left(\mathit{e}\right|{\mathit{X}}^1)& ={\displaystyle \frac{1}{2}}\left[{\displaystyle \frac{1}{4}}\mathrm{P}\left({\mathit{X}}_{1,{\mathrm{A}}_1}^1\right)\right.\hfill \\ \hfill & \times \left\{\mathrm{P}\left(n_{\mathrm{Q}}\ge 2\sqrt{{\displaystyle \frac{{ϵ}_2}{2}}}{h}_1+\sqrt{{\displaystyle \frac{{ϵ}_1}{2}}}{h}_1\left|{\mathit{X}}_{1,{\mathrm{A}}_1}^1\right.\right)+\mathrm{P}\left(n_{\mathrm{I}}\ge \sqrt{{\displaystyle \frac{{ϵ}_1}{2}}}{h}_1\left|{\mathit{X}}_{1,{\mathrm{A}}_1}^1\right.\right)\right\}\hfill \\ \hfill & +{\displaystyle \frac{1}{4}}\mathrm{P}\left({\mathit{X}}_{1,{\mathrm{B}}_1}^1\right)\hfill \\ \hfill & \times \left\{\mathrm{P}\left(n_{\mathrm{Q}}\le 2\sqrt{{\displaystyle \frac{{ϵ}_2}{2}}}{h}_1-\sqrt{{\displaystyle \frac{{ϵ}_1}{2}}}{h}_1\left|{\mathit{X}}_{1,{\mathrm{B}}_1}^1\right.\right)+\mathrm{P}\left(n_{\mathrm{I}}\ge \sqrt{{\displaystyle \frac{{ϵ}_1}{2}}}{h}_1\left|{\mathit{X}}_{1,{\mathrm{B}}_1}^1\right.\right)\right\}\hfill \\ \hfill & +{\displaystyle \frac{1}{4}}\mathrm{P}\left({\mathit{X}}_{2,{\mathrm{A}}_1}^1\right)\hfill \\ \hfill & \times \left\{\mathrm{P}\left(n_{\mathrm{Q}}\ge \sqrt{{\displaystyle \frac{{ϵ}_1}{2}}}{h}_1\left|{\mathit{X}}_{2,{\mathrm{A}}_1}^1\right.\right)+\mathrm{P}\left(n_{\mathrm{I}}\ge 2\sqrt{{\displaystyle \frac{{ϵ}_2}{2}}}{h}_1+\sqrt{{\displaystyle \frac{{ϵ}_1}{2}}}{h}_1\left|{\mathit{X}}_{2,{\mathrm{A}}_1}^1\right.\right)\right\}\hfill \\ \hfill & +{\displaystyle \frac{1}{4}}\mathrm{P}\left({\mathit{X}}_{2,{\mathrm{B}}_1}^1\right)\hfill \\ \hfill & \left.\times \left\{\mathrm{P}\left(n_{\mathrm{Q}}\ge \sqrt{{\displaystyle \frac{{ϵ}_1}{2}}}{h}_1\left|{\mathit{X}}_{2,{\mathrm{B}}_1}^1\right.\right)+\mathrm{P}\left(n_{\mathrm{I}}\le 2\sqrt{{\displaystyle \frac{{ϵ}_2}{2}}}{h}_1-\sqrt{{\displaystyle \frac{{ϵ}_1}{2}}}{h}_1\left|{\mathit{X}}_{2,{\mathrm{B}}_1}^1\right.\right)\right\}\right],\hfill \end{array}$$

(26)

where ${\displaystyle \frac{1}{4}}\mathrm{P}\left({\mathit{X}}_{j,{\mathrm{A}}_1}^1\right)$ and ${\displaystyle \frac{1}{4}}\mathrm{P}\left({\mathit{X}}_{j,{\mathrm{B}}_1}^1\right)$ are the probabilities that user 1 incorrectly detects the j-th bit of $s_2$ as given in (13). Similar to (23), (26) can be redefined as follows:

$$\begin{array}{cc}\hfill {\mathrm{P}}_1\left(\mathit{e}\right|{\mathit{X}}^1)=& {\displaystyle \frac{1}{8}}\left[\mathrm{P}\left(\sqrt{{\displaystyle \frac{{ϵ}_2}{2}}}{h}_1-\sqrt{{\displaystyle \frac{{ϵ}_1}{2}}}{h}_1\le n_{\mathrm{Q}}\le 2\sqrt{{\displaystyle \frac{{ϵ}_2}{2}}}{h}_1-\sqrt{{\displaystyle \frac{{ϵ}_1}{2}}}{h}_1\right)\right.\hfill \\ \hfill & +\mathrm{P}\left(n_{\mathrm{Q}}\ge 2\sqrt{{\displaystyle \frac{{ϵ}_2}{2}}}{h}_1+\sqrt{{\displaystyle \frac{{ϵ}_1}{2}}}{h}_1\right)+\mathrm{P}\left(n_{\mathrm{I}}\ge \sqrt{{\displaystyle \frac{{ϵ}_1}{2}}}{h}_1\right)\hfill \\ \hfill & \times \left\{\mathrm{P}\left(n_{\mathrm{Q}}\ge \sqrt{{\displaystyle \frac{{ϵ}_2}{2}}}{h}_1+\sqrt{{\displaystyle \frac{{ϵ}_1}{2}}}{h}_1\right)+\mathrm{P}\left(n_{\mathrm{Q}}\ge \sqrt{{\displaystyle \frac{{ϵ}_2}{2}}}{h}_1-\sqrt{{\displaystyle \frac{{ϵ}_1}{2}}}{h}_1\right)\right\}\hfill \\ \hfill & +\mathrm{P}\left(\sqrt{{\displaystyle \frac{{ϵ}_2}{2}}}{h}_1-\sqrt{{\displaystyle \frac{{ϵ}_1}{2}}}{h}_1\le n_{\mathrm{I}}\le 2\sqrt{{\displaystyle \frac{{ϵ}_2}{2}}}{h}_1-\sqrt{{\displaystyle \frac{{ϵ}_1}{2}}}{h}_1\right)\hfill \\ \hfill & +\mathrm{P}\left(n_{\mathrm{I}}\ge 2\sqrt{{\displaystyle \frac{{ϵ}_2}{2}}}{h}_1+\sqrt{{\displaystyle \frac{{ϵ}_1}{2}}}{h}_1\right)+\mathrm{P}\left(n_{\mathrm{Q}}\ge \sqrt{{\displaystyle \frac{{ϵ}_1}{2}}}{h}_1\right)\hfill \\ \hfill & \left.\times \left\{\mathrm{P}\left(n_{\mathrm{I}}\ge \sqrt{{\displaystyle \frac{{ϵ}_2}{2}}}{h}_1+\sqrt{{\displaystyle \frac{{ϵ}_1}{2}}}{h}_1\right)+\mathrm{P}\left(n_{\mathrm{I}}\ge \sqrt{{\displaystyle \frac{{ϵ}_2}{2}}}{h}_1-\sqrt{{\displaystyle \frac{{ϵ}_1}{2}}}{h}_1\right)\right\}\right]\hfill \end{array}$$

(27)

For the purpose of simplifying (27), we define the SNR for the D and E signal point given in Figure 5b,c as follows:

$$\begin{array}{cc}\hfill & {\gamma }_{\mathrm{D}}={\displaystyle \frac{{(2\sqrt{{ϵ}_2}+\sqrt{{ϵ}_1})}^2{\left|h_1\right|}^2}{{\sigma }^2}},\hfill \\ \hfill & {\gamma }_{\mathrm{E}}={\displaystyle \frac{{(2\sqrt{{ϵ}_2}-\sqrt{{ϵ}_1})}^2{\left|h_1\right|}^2}{{\sigma }^2}}.\hfill \end{array}$$

(28)

Then, the error probability of $s_1$, given that user 1 incorrectly detects $s_2$, is expressed as follows:

$$\begin{array}{cc}\hfill {\mathrm{P}}_1\left(\mathit{e}\right|{\mathit{X}}^1)=& {\displaystyle \frac{1}{4}}\left[\mathit{Q}\left(\sqrt{{\gamma }_{\mathrm{C}}}\right)\times \left\{\mathit{Q}\left(\sqrt{{\gamma }_{{\mathrm{A}}_1}}\right)+\mathit{Q}\left(\sqrt{{\gamma }_{{\mathrm{B}}_1}}\right)\right\}\right.\hfill \\ \hfill & \left.+\mathit{Q}\left(\sqrt{{\gamma }_{{\mathrm{B}}_1}}\right)+\mathit{Q}\left(\sqrt{{\gamma }_{\mathrm{D}}}\right)-\mathit{Q}\left(\sqrt{{\gamma }_{\mathrm{E}}}\right)\right].\hfill \end{array}$$

(29)

By substituting (25) and (29) into (21), the error probability of user 1 is obtained as follows:

$$\begin{array}{cc}\hfill {\mathrm{P}}_1\left(\mathit{e}\right)=& \mathit{Q}\left(\sqrt{{\gamma }_{\mathrm{C}}}\right)+{\displaystyle \frac{1}{4}}\left[-\mathit{Q}\left(\sqrt{{\gamma }_{{\mathrm{A}}_1}}\right)+\mathit{Q}\left(\sqrt{{\gamma }_{{\mathrm{B}}_1}}\right)+\mathit{Q}\left(\sqrt{{\gamma }_{\mathrm{D}}}\right)-\mathit{Q}\left(\sqrt{{\gamma }_{\mathrm{E}}}\right)\right].\hfill \end{array}$$

(30)

Then, the average BER at user 1 can be obtained as follows:

$$\begin{array}{cc}\hfill \overline{{\mathrm{P}}_1\left(\mathit{e}\right)}=& {\int }_0^{\infty }\mathit{Q}\left(\sqrt{{\gamma }_{\mathrm{C}}}\right)f_{{\gamma }_{\mathrm{C}}}\left({\gamma }_{\mathrm{C}}\right)\mathrm{d}{\gamma }_{\mathrm{C}}\hfill \\ \hfill & +{\displaystyle \frac{1}{4}}\left[-{\int }_0^{\infty }\mathit{Q}\left(\sqrt{{\gamma }_{{\mathrm{A}}_1}}\right)f_{{\gamma }_{{\mathrm{A}}_1}}\left({\gamma }_{{\mathrm{A}}_1}\right)\mathrm{d}{\gamma }_{{\mathrm{A}}_1}+{\int }_0^{\infty }\mathit{Q}\left(\sqrt{{\gamma }_{{\mathrm{B}}_1}}\right)f_{{\gamma }_{{\mathrm{B}}_1}}\left({\gamma }_{{\mathrm{B}}_1}\right)\mathrm{d}{\gamma }_{{\mathrm{B}}_1}\right.\hfill \\ \hfill & \left.+{\int }_0^{\infty }\mathit{Q}\left(\sqrt{{\gamma }_{\mathrm{D}}}\right)f_{{\gamma }_{\mathrm{D}}}\left({\gamma }_{\mathrm{D}}\right)\mathrm{d}{\gamma }_{\mathrm{D}}-{\int }_0^{\infty }\mathit{Q}\left(\sqrt{{\gamma }_{\mathrm{E}}}\right)f_{{\gamma }_{\mathrm{E}}}\left({\gamma }_{\mathrm{E}}\right)\mathrm{d}{\gamma }_{\mathrm{E}}\right].\hfill \end{array}$$

(31)

Using ([10], Equations (13.3)–(7)) and (19), (31) is given as follows:

$$\begin{array}{cc}\hfill \overline{{\mathrm{P}}_1\left(\mathit{e}\right)}=& {\displaystyle \frac{1}{2}}\left(1-\sqrt{{\displaystyle \frac{\overline{{\gamma }_{\mathrm{C}}}}{2+\overline{{\gamma }_{\mathrm{C}}}}}}\right)+{\displaystyle \frac{1}{8}}\left(\sqrt{{\displaystyle \frac{\overline{{\gamma }_{{\mathrm{A}}_1}}}{2+\overline{{\gamma }_{{\mathrm{A}}_1}}}}}\right.\hfill \\ \hfill & -\left.\sqrt{{\displaystyle \frac{\overline{{\gamma }_{{\mathrm{B}}_1}}}{2+\overline{{\gamma }_{{\mathrm{B}}_1}}}}}+\sqrt{{\displaystyle \frac{\overline{{\gamma }_{\mathrm{D}}}}{2+\overline{{\gamma }_{\mathrm{D}}}}}}-\sqrt{{\displaystyle \frac{\overline{{\gamma }_{\mathrm{E}}}}{2+\overline{{\gamma }_{\mathrm{E}}}}}}\right).\hfill \end{array}$$

(32)

4.2. Exact BER of Proposed Transmitter and Receiver

In the proposed method, the transmitter sends two types of data: one is the information for user 2, and the other are the redundancy data used to transform the former data into the information for user 1. In this method, user 2 only requires the former data, while user 1 requires both the former and the latter data. Therefore, we proceed by separating the former and latter data to calculate the error probability. In this section, we assume $N=2$. First, the constellation diagram of the former data is represented as shown in Figure 6.

The received signals at each point are expressed as $(\pm \sqrt{P_{\mathrm{t}}/2}\pm j\sqrt{P_{\mathrm{t}}/2})h_k+n_k$. User k incorrectly detects the first bit of $s_2$ if $-\sqrt{P_{\mathrm{t}}/2}{h}_k+n_{\mathrm{Q}}\ge 0$, and incorrectly detects the second bit of $s_2$ if $-\sqrt{P_{\mathrm{t}}/2}{h}_k+n_{\mathrm{I}}\ge 0$. Thus, the error probability for $s_2$ at user k is expressed as follows:

$$\begin{array}{c}\hfill \tilde{\mathrm{P}}\left({\mathit{X}}^k\right)={\displaystyle \frac{1}{2}}\left[\mathrm{P}\left(n_{\mathrm{Q}}\ge \sqrt{{\displaystyle \frac{P_t}{2}}}{h}_k\right)+\mathrm{P}\left(n_{\mathrm{I}}\ge \sqrt{{\displaystyle \frac{P_t}{2}}}{h}_k\right)\right].\end{array}$$

(33)

To simplify (33), SNR at points ${\mathrm{F}}_k$ can be expressed as follows:

$${\gamma }_{{\mathrm{F}}_k}={\displaystyle \frac{P_t{\left|h_k\right|}^2}{{\sigma }^2}}.$$

(34)

By substituting (34) into (33), the error probability of user 2 is defined as follows:

$${\tilde{\mathrm{P}}}_2\left(\mathit{e}\right)=\tilde{\mathrm{P}}\left({\mathit{X}}^2\right)=\mathit{Q}\left(\sqrt{{\gamma }_{{\mathrm{F}}_2}}\right).$$

(35)

Then, the average BER at user 2 for the proposed receiver can be obtained as follows:

$$\begin{array}{c}\hfill \overline{{\tilde{\mathrm{P}}}_2\left(\mathit{e}\right)}={\int }_0^{\infty }\mathit{Q}\left(\sqrt{{\gamma }_{{\mathrm{F}}_2}}\right)f_{{\gamma }_{{\mathrm{F}}_2}}\left({\gamma }_{{\mathrm{F}}_2}\right)\mathrm{d}{\gamma }_{{\mathrm{F}}_2}.\end{array}$$

(36)

Using ([10], Equations (13.3)–(7)) and (19), (36) is given as follows:

$$\begin{array}{cc}\hfill \overline{{\tilde{\mathrm{P}}}_2\left(\mathit{e}\right)}=& {\displaystyle \frac{1}{2}}\left(1-\sqrt{{\displaystyle \frac{\overline{{\gamma }_{{\mathrm{F}}_2}}}{2+\overline{{\gamma }_{{\mathrm{F}}_2}}}}}\right).\hfill \end{array}$$

(37)

Next, for user 1, the error probabilities of both $s_2$ and $s_{\mathrm{c}}$ should be considered. The error probability of $s_2$ is the probability of incorrectly detecting $b_2$, and the error probability of $s_{\mathrm{c}}$ is the probability of incorrectly detecting $b_{\mathrm{XOR}}$. Examining the error probability of user 1, the XOR bit operation results in 1 when the two bits are different, and 0 when they are the same. Due to this characteristic, if both $b_2$ and $b_{\mathrm{XOR}}$ are incorrectly detected, the final $b_1$ obtained in Section 3.2 would be detected without error. Therefore, the error for user 1 occurs only when either $s_2$ or $s_{\mathrm{c}}$ has an error. The average error probability of user 1 in the proposed receiver is expressed as follows:

$$\begin{array}{c}\hfill \overline{{\tilde{\mathrm{P}}}_1\left(\mathit{e}\right)}=\left(1-\overline{\tilde{\mathrm{P}}\left({\mathit{X}}^1\right)}\right)\overline{\tilde{\mathrm{P}}\left({\mathit{Y}}^1\right)}+\left(1-\overline{\tilde{\mathrm{P}}\left({\mathit{Y}}^1\right)}\right)\overline{\tilde{\mathrm{P}}\left({\mathit{X}}^1\right)},\end{array}$$

(38)

where ${\mathit{Y}}^1$ denotes the event that user 1 incorrectly detects $s_{\mathrm{c}}$. Due to (33) and (34), the error probability of $s_2$ for user 1 is given as follows:

$$\tilde{\mathrm{P}}\left({\mathit{X}}^1\right)=\mathit{Q}\left(\sqrt{{\gamma }_{{\mathrm{F}}_1}}\right).$$

(39)

Then, the average error probability of $s_2$ at user 1 is given as follows:

$$\begin{array}{cc}\hfill \overline{\tilde{\mathrm{P}}\left({\mathit{X}}^1\right)}& ={\int }_0^{\infty }\mathit{Q}\left(\sqrt{{\gamma }_{{\mathrm{F}}_1}}\right)f_{{\gamma }_{{\mathrm{F}}_1}}\left({\gamma }_{{\mathrm{F}}_1}\right)\mathrm{d}{\gamma }_{{\mathrm{F}}_1}\hfill \\ \hfill & ={\displaystyle \frac{1}{2}}\left(1-\sqrt{{\displaystyle \frac{\overline{{\gamma }_{{\mathrm{F}}_1}}}{2+\overline{{\gamma }_{{\mathrm{F}}_1}}}}}\right).\hfill \end{array}$$

(40)

Since $s_{\mathrm{c}}$ is a combination of $s_{\mathrm{c},1}$ and $s_{\mathrm{c},2}$ combined in series, its average error probability for user 1 can be represented as the average value of the error probabilities for the $s_1$ and $s_2$ of user 1 in conventional NOMA, as follows:

$$\overline{\tilde{\mathrm{P}}\left({\mathit{Y}}^1\right)}={\displaystyle \frac{1}{2}}\left[\overline{{\mathrm{P}}_1\left(\mathit{e}\right)}+\overline{\mathrm{P}\left({\mathit{X}}^1\right)}\right].$$

(41)

By substituting (12) and (32), (41) is obtained as follows:

$$\begin{array}{cc}\hfill \overline{\tilde{\mathrm{P}}\left({\mathit{Y}}^1\right)}=& {\displaystyle \frac{1}{4}}\left(2-\sqrt{{\displaystyle \frac{\overline{{\gamma }_{\mathrm{C}}}}{2+\overline{{\gamma }_{\mathrm{C}}}}}}\right)+{\displaystyle \frac{1}{16}}\left(-\sqrt{{\displaystyle \frac{\overline{{\gamma }_{{\mathrm{A}}_1}}}{2+\overline{{\gamma }_{{\mathrm{A}}_1}}}}}\right.\hfill \\ \hfill & -3\left.\sqrt{{\displaystyle \frac{\overline{{\gamma }_{{\mathrm{B}}_1}}}{2+\overline{{\gamma }_{{\mathrm{B}}_1}}}}}+\sqrt{{\displaystyle \frac{\overline{{\gamma }_{\mathrm{D}}}}{2+\overline{{\gamma }_{\mathrm{D}}}}}}-\sqrt{{\displaystyle \frac{\overline{{\gamma }_{\mathrm{E}}}}{2+\overline{{\gamma }_{\mathrm{E}}}}}}\right).\hfill \end{array}$$

(42)

By substituting (40) and (42) into (38), the average error probability of user 2 is given as follows:

$$\begin{array}{cc}\hfill \overline{{\tilde{\mathrm{P}}}_1\left(\mathit{e}\right)}=& {\displaystyle \frac{1}{2}}\left(1-\sqrt{{\displaystyle \frac{\overline{{\gamma }_{{\mathrm{F}}_1}}}{2+\overline{{\gamma }_{{\mathrm{F}}_1}}}}}\right)+\sqrt{{\displaystyle \frac{\overline{{\gamma }_{{\mathrm{F}}_1}}}{2+\overline{{\gamma }_{{\mathrm{F}}_1}}}}}\times \left\{{\displaystyle \frac{1}{4}}\left(2-\sqrt{{\displaystyle \frac{\overline{{\gamma }_{\mathrm{C}}}}{2+\overline{{\gamma }_{\mathrm{C}}}}}}\right)\right.\hfill \\ \hfill & \left.+{\displaystyle \frac{1}{16}}\left(-\sqrt{{\displaystyle \frac{\overline{{\gamma }_{{\mathrm{A}}_1}}}{2+\overline{{\gamma }_{{\mathrm{A}}_1}}}}}-3\sqrt{{\displaystyle \frac{\overline{{\gamma }_{{\mathrm{B}}_1}}}{2+\overline{{\gamma }_{{\mathrm{B}}_1}}}}}+\sqrt{{\displaystyle \frac{\overline{{\gamma }_{\mathrm{D}}}}{2+\overline{{\gamma }_{\mathrm{D}}}}}}-\sqrt{{\displaystyle \frac{\overline{{\gamma }_{\mathrm{E}}}}{2+\overline{{\gamma }_{\mathrm{E}}}}}}\right)\right\}.\hfill \end{array}$$

(43)

5. Numerical Results

In this section, we present a performance comparison of simulation results for the proposed scheme (

Table 1. Simulation settings.

Parameters	Value
$d_1$	100 [m]
$d_1$	200 [m]
$\beta$	3
Noise Power	−80 [dBm]
${\alpha }_1$	1/5
N	2
${\lambda }_1^{OMA}$, ${\lambda }_2^{OMA}$	1/2

). The channel experiences a distance-dependent attenuation of large-scale fading, and the pathloss factor is expressed as $L_k\left(d\right)=L_0{(d_k/d_0)}^{-\beta }$, where $L_0$ represents the pathloss at the reference distance $d_0$ = 1(m), $d_k$ is the distance between the BS and user k, and $\beta$ is the pathloss exponent.

Figure 7 shows the sum rate performance versus transmit power. The data rate of user 1 is reduced due to redundancy data. On the other hand, the data rate of user 2 is further improved because the signal is sent without power allocation. Therefore, the overall sum rate of users in the proposed scheme shows greater improvement than that of the other schemes. At a transmission power of 15 dBm, there is approximately 15% improvement in sum rate performance, and at 30 dBm, there is about 20% improvement. Figure 8 shows the BER performance versus transmit power, comparing the proposed scheme with other conventional schemes. In the proposed scheme, it can be seen that the BER performance is better than in the conventional NOMA because users experience no power loss in their original information. Significantly, the proposed scheme reduces the required transmit power by 5 dB compared to existing methods to achieve a BER performance of 10⁻³. The BER for user 2 with poor channel conditions has the same performance as OMA, but the BER for user 1 with good channel conditions is reduced compared to OMA due to errors that occur in demodulating the redundancy data. If user 1 achieves a sufficient SNR to decode the redundancy data without errors, the BER performance of the proposed scheme becomes similar to that of OMA. Figure 9 shows the BER performance of user 1 and user 2 in the proposed scheme. The analytical results show the calculations using (43) for user 1 and (37) for user 2 for the exact and approximate results. In the proposed design, the BER performance of user 1 and user 2 using the same modulation scheme demonstrates that the overall system maintains user fairness compared to the conventional NOMA.

6. Conclusions

In this paper, we propose a scheme that adds XOR bit operations into the transmitter and receiver designs of conventional NOMA to alleviate performance degradation caused by the power distribution of the original signal. The addition of the XOR bit operation can improve the overall sum rate of users while mitigating the error performance degradation caused by power allocation. Additionally, by improving the BER and data rate of users with poor channel conditions, the user fairness of the entire system can be improved. Therefore, in systems where the required SNR is high for users with poor channel conditions, high communication performance can be expected by applying the proposed scheme. In this paper, as we have demonstrated through simulations that the addition of this XOR operation reduced the required power by 5 dB, the proposed transceiver model can be expected to achieve higher energy efficiency and increased network capacity.