Non Orthogonal Multiple Access with Energy Harvesting Using Reconfigurable Intelligent Surfaces for Rayleigh Channels

Abstract

In this paper, we derive the throughput of non orthogonal multiple access (NOMA) through reconfigurable intelligent surfaces (RIS) with energy harvesting. The transmitter harvests energy using the radio frequency (RF) signal received from another node N. Node N can be any node transmitting RF signals. The harvested energy is used to transmit data to N NOMA users using RIS. RIS is placed between the transmitter and users. Different sets of reflectors are dedicated to different users. Each user receives all reflections over RIS with the same phase. We also optimize harvesting duration and power allocation coefficients to maximize the total throughput. We suggest the use of two RIS: the first RIS improves the energy harvesting process and is placed between node N and transmitter T. The second RIS is placed between T and users so that reflections have a null phase at all NOMA users. The derived results are valid for Rayleigh fading channels where the transmitter harvests energy using the received RF signals. We derive the throughput at weak and strong users as well as the total throughput of NOMA systems using RIS with energy harvesting. In this paper, we derive the packet error probability and total throughput for NOMA using RIS with energy harvesting using RF signals. When there are two users and 16QAM modulation is used and for a throughput of 3.5 bit/s/Hz, the use of $R=512,256,128,64,32,16,8$ reflectors offers 69 dB, 62.8 dB, 57 dB, 50.9 dB, 45 dB, 38.5 dB, 31.8 dB gain with respect to the absence of RIS (Mondal et al. in Wirel Pers Commun, 2021; Le and Kong in Wirel Pers Commun 116:3577–3597, 2021).

Appendices

Appendix A: PDF and CDF of X

Let $Z=[\sum _{p\in I_i}a_pc_p]^2|f|^2$ be the product of two r.v. $|f|^2$ that follows a central chisquare distribution of two degrees of freedom and $[\sum _{p\in I_i}a_pc_p]^2$ a non-central chisquare distribution with one degree of freedom. We have $X=\frac{\mu \alpha E_N}{1-\alpha }Z$. We will compute the PDF and CDF of Z and deduce that of X.

We use the Mellin transform of PDF p(x) defined as

$$\begin{aligned} M(p(x))=\int _0^{+\infty }x^{s-1}p(x)dx=M(s). \end{aligned}$$

(29)

The inverse Mellin transformation is defined as

$$\begin{aligned} p(x)=M^{-1}(M(s))=\frac{1}{2\pi j}\int _{e-j\infty }^{e+j\infty }x^{-s}g(s)ds. \end{aligned}$$

(30)

We assume that $|f|^2$ and $[\sum _{p\in I_i}a_pc_p]^2$ are independent r.v.. Therefore, the Mellin transform of PDF of Z is the product of Mellin transforms of PDF of $|f|^2$ and $[\sum _{p\in I_i}a_pc_p]^2$.

Let $p_1(x)$ be the PDF of $|f|^2$

$$\begin{aligned} p_1(x)=\frac{1}{\zeta }e^{-\frac{x}{\zeta }} \end{aligned}$$

(31)

where $\zeta =E(|f|^2)=\frac{1}{d_1^{\alpha }}$.

The Mellin transform of $p_1(x)$ is

$$\begin{aligned} M(p_1(x))=M_1(s)=\zeta ^{s-1}\Gamma (s), \end{aligned}$$

(32)

where $\Gamma (s)$ is the Gamma function.

The PDF of $[\sum _{p\in I_i}a_pc_p]^2$ is written as [37]

$$\begin{aligned} p_2(y)=0.5(\frac{y}{\delta ^2})^{-0.25}e^{-0.5(y+\delta ^2)}I_{-0.5}(\delta \sqrt{y}) \end{aligned}$$

(33)

where $\delta ^2=\frac{m^2}{\sigma ^2}$ is the non centrality parameter and $I_n(y)$ is the first kind modified Bessel function with n-th order.

We deduce

$$\begin{aligned} M(p_2(y))=0.5e^{-\frac{\delta ^2}{2}}\sqrt{\delta }\int _0^{+\infty }y^{s-\frac{5}{4}}e^{-\frac{y}{2}}I_{-0.5}(\delta \sqrt{y})dy. \end{aligned}$$

(34)

We use the power series expansion of $I_n(y)$:

$$\begin{aligned} I_n(y)=\sum _{q=0}^{+\infty }\frac{y^{2q+n}}{2^{2q+n}q!\Gamma (q+n+1)} \end{aligned}$$

(35)

We deduce

$$\begin{aligned}&M(p_2(y))=0.5e^{-\frac{\delta ^2}{2}}\sqrt{\delta }\sum _{q=0}^{+\infty }\frac{\delta ^{2q-0.5}}{2^{2q-0.5}q!\Gamma (q+0.5)} \nonumber \\&\quad \int _0^{+\infty }y^{s+q-\frac{3}{2}}e^{-\frac{y}{2}}dy. \end{aligned}$$

(36)

We have

$$\begin{aligned} \int _0^{+\infty }e^{-0.5y}y^{s+q-1.5}dy=\Gamma (s+q-0.5)2^{s+q-0.5}. \end{aligned}$$

(37)

Therefore, we obtain

$$\begin{aligned} M_2(s)=M(p_2(y))=e^{-\frac{\delta ^2}{2}}\sum _{q=0}^{+\infty }\frac{\delta ^{2q}}{q!\Gamma (q+0.5)}\Gamma (s+q-0.5)2^{s-q-1}. \end{aligned}$$

(38)

The PDF of Z can be obtained by inverse Mellin transformation

$$\begin{aligned} p_Z(x)=M^{-1}(M_1(s)M_2(s))=\frac{1}{2\pi j}\int _{e-j\infty }^{e+j\infty }x^{-s}M_1(s)M_2(s)ds. \end{aligned}$$

(39)

Using (32) and (38), we have

$$\begin{aligned}&p_Z(x)=\frac{1}{2\pi j}e^{-\frac{\delta ^2}{2}}\sum _{q=0}^{+\infty }\frac{\delta ^{2q}}{q!\Gamma (q+0.5)} \nonumber \\&\quad \int _{e-j\infty }^{e+j\infty }x^{-s}\Gamma (s+q-0.5)2^{s-m-1}\zeta ^{s-1}\Gamma (s)ds. \end{aligned}$$

(40)

Let $0.5s'=s+\frac{q}{2}-0.25$, we deduce

$$\begin{aligned}&p_Z(x)=\frac{1}{2\pi j}e^{-\frac{\delta ^2}{2}}\sum _{q=0}^{+\infty }\frac{\delta ^{2q}}{q!\Gamma (q+0.5)} \nonumber \\&\quad \int _{e-j\infty }^{e+j\infty }x^{-0.5s'+0.5m-0.25}\Gamma (0.5s'+0.5q-0.25)2^{0.5s'-\frac{3}{2}q-\frac{7}{4}} \nonumber \\&\quad \times \zeta ^{0.5s'-0.5m-0.75}\Gamma (0.5s'-0.5q+0.25)ds'. \end{aligned}$$

(41)

We use

$$\begin{aligned} M^{-1}(\Gamma (0.5s-0.5n)\Gamma (0.5s+0.5n)2^{s-2}a^{-s})=K_n(ax), \end{aligned}$$

(42)

where $K_n(x)$ is the n-th order modified Bessel function of second kind.

Using (41) and (42), we obtain the PDF of X

$$\begin{aligned} p_Z(x)=\frac{e^{-0.5\delta ^2}}{\zeta }\sum _{q=0}^{+\infty }\frac{\delta ^{2q}2^{-1.5q+0.25}}{q!\Gamma (q+0.5)}K_{q-0.5}(\sqrt{\frac{2x}{\zeta }})(\frac{x}{\zeta })^{0.5q-0.25} \end{aligned}$$

(43)

Using (10), we obtain the CDF of Z:

$$\begin{aligned} P_Z(x)=e^{-0.5\delta ^2}\sum _{q=0}^{+\infty }\frac{\delta ^{2q} 2^{-q}}{q!\Gamma (q+0.5)}G_{1,3}^{2,1}\left( \frac{x}{2\zeta }|\begin{array}{ccc} 1 &{} &{} \\ q+0.5, &{} 1, &{} 0 \end{array} \right) \end{aligned}$$

(44)

The CDF of X is given in (11) and deduced from that of Z using $P_X(x)=P_Z(\frac{x(1-\alpha )}{\mu \alpha E_N})$. The PDF of X is given in (8) and obtained from the PDF of Z $p_X(x)=\frac{1-\alpha }{\mu \alpha E_N}p_Z(\frac{x(1-\alpha )}{\mu \alpha E_N})$.

Appendix B: PDF and CDF of Y

It is assumed that $[\sum _{p\in I_i}a_pc_p]^2$ and $[\sum _{p=1}^{R^1}e_pz_p]^2$ are independent r.v. so that the Mellin transform of PDF of $W=[\sum _{p\in I_i}a_pc_p]^2[\sum _{p=1}^{R^1}e_pz_p]^2$ is the product of the Mellin transforms of PDF of $[\sum _{p=1}^{R^1}e_pz_p]^2$ and $[\sum _{p\in I_i}a_pc_p]^2$. The PDF of $[\sum _{p\in I_i}a_pc_p]^2$ is given in (33). The PDF of $[\sum _{p=1}^{R^1}e_pz_p]^2$ is written similarly

$$\begin{aligned} p_3(y)=0.5(\frac{y}{\delta _2^2})^{-0.25}e^{-0.5(y+\delta _2^2)}I_{-0.5}(\delta _2 \sqrt{y}) \end{aligned}$$

(45)

where $\delta _2^2=\frac{m_2^2}{\sigma _2^2}$ is the non centrality parameter.

The Mellin transform of $p_3(y)$ is written similarly to (38)

$$\begin{aligned} M_3(s)=M(p_3(y))=e^{-\frac{\delta _2^2}{2}}\sum _{q=0}^{+\infty }\frac{\delta _2^{2q}}{q!\Gamma (q+0.5)}\Gamma (s+q-0.5)2^{s-q-1}. \end{aligned}$$

(46)

We deduce

$$\begin{aligned}&M_2(s)M_3(s)=e^{-\frac{\delta _2^2+\delta ^2}{2}}\sum _{q=0}^{+\infty }\sum _{p=0}^{+\infty }\frac{\delta _2^{2p}\delta ^{2q}}{q!p!\Gamma (q+0.5)\Gamma (p+0.5)} \nonumber \\&\qquad \qquad \times \Gamma (s+p-0.5)\Gamma (s+q-0.5)2^{2s-q-p-2}. \end{aligned}$$

(47)

The PDF of W is obtained by inverse Mellin transform of $M_2(s)M_3(s)$ as

$$\begin{aligned}&p_W(z)=\frac{1}{2\pi j}e^{\frac{-\delta ^2-\delta _2^2}{2}}\sum _{q=0}^{+\infty }\sum _{p=0}^{+\infty }\frac{\delta _2^{2p}\delta ^{2q}}{q!p!\Gamma (q+0.5)\Gamma (p+0.5)} \nonumber \\&\qquad \qquad \times \int _{e-j\infty }^{e+j\infty }z^{-s}\Gamma (s+q-0.5)\Gamma (s+p-0.5)2^{2s-q-p-2}ds. \end{aligned}$$

(48)

Let $s'=2s+p+q-1$, we deduce

$$\begin{aligned}&p_W(z)=\frac{1}{2\pi j}e^{\frac{-\delta ^2-\delta _2^2}{2}}\sum _{q=0}^{+\infty }\sum _{p=0}^{+\infty }\frac{\delta _2^{2p}\delta ^{2q}}{q!p!\Gamma (q+0.5)\Gamma (p+0.5)} \nonumber \\&\qquad \times \int _{e'-j\infty }^{e'+j\infty }z^{-0.5s'+0.5p+0.5q-0.5}\Gamma (0.5s'+0.5q-0.5p) \nonumber \\&\qquad \times \Gamma (0.5s'+0.5p-0.5q)2^{s'-2q-2p-2}ds'. \end{aligned}$$

(49)

Using (42), we obtain the PDF of W as

$$\begin{aligned} p_W(z)=e^{\frac{-\delta ^2-\delta _2^2}{2}}\sum _{q=0}^{+\infty }\sum _{p=0}^{+\infty }\frac{\delta _2^{2p}\delta ^{2q}K_{p-q}(\sqrt{z})2^{-2q-2p}z^{\frac{p+q-1}{2}}}{q!p!\Gamma (q+0.5)\Gamma (p+0.5)} \end{aligned}$$

(50)

We use (10) to write the CDF of Z as

$$\begin{aligned}&P_W(z)=e^{\frac{-\delta ^2-\delta _2^2}{2}}\sum _{q=0}^{+\infty }\sum _{p=0}^{+\infty }\frac{2^{-q-p}\delta _2^{2p}\delta ^{2q}}{q!p!\Gamma (q+0.5)\Gamma (p+0.5)} \nonumber \\&\qquad \quad \times G_{1,3}^{2,1}\left( \frac{z}{4}|\begin{array}{ccc} 1 &{} &{} \\ p+0.5, &{} q+0.5, &{} 0 \end{array} \right) \end{aligned}$$

(51)

The CDF of Y is given in (27) and deduced from that of W using $P_Y(x)=P_W(\frac{x(1-\alpha )}{\mu \alpha E_N})$. The PDF of Y is given in (26) and obtained from the PDF of W $p_Y(x)=\frac{1-\alpha }{\mu \alpha E_N}p_W(\frac{x(1-\alpha )}{\mu \alpha E_N})$.