On performance of a full duplex SWIPT enabled cooperative NOMA network

Abstract

This paper investigates a novel full-duplex (FD) simultaneous wireless information and power transfer (SWIPT) based cooperative non-orthogonal multiple access (FDS-NOMA) system, where far distant NOMA IoT users are helped by the NOMA strong user with the capability of energy harvesting (EH) and FD communications. The downlink (DL-NOMA) principle is used to transmit the superimposed signals from source i.e a base station (BS) to an energy-limited FD relay where relay harvests energy from the transmit power of the BS. The relay after detecting the superimposed signals using successive interference cancellation (SIC), retransmits the decoded signals using superposition coding to the IoT users. We derive the analytical expressions of the outage probability and ergodic rates experienced by the NOMA strong user as well as NOMA IoT users. The results shows that proposed cooperative NOMA scheme can be used to serve IoT users using EH relay. Extensive simulations are performed to verify the accuracy of the derived approximate closed-from expressions.

Appendices

Appendix A

Proof of Proposition 2

Substituting (8), (9), (13), and (14) in to (19), we get

$$\begin{aligned}{} & {} \begin{aligned} P_{out}^{U2}&= 1-Pr\{|h_{sr}|^2>b_2,|h_{sr}|^2>b_1,|h_{sr}|^2|h_{rd_{2}}|^2>b_5 \\&,|h_{sr}|^2|h_{rd_{2}}|^2>b_6\} \end{aligned} \end{aligned}$$

(36)

$$\begin{aligned}{} & {} P_{out}^{U2}=1-Pr\{|h_{sr}|^2>b_7,|h_{sr}|^2|h_{rd_{2}}|^2>b_8\} \end{aligned}$$

(37)

where $b_7=max.(b_2,b_1)$, $b_8=max.(b_5,b_6)$,

$b_6=\frac{\gamma _{th_{3}}d_{sr}^md_{rd_{2}}^mN_o}{\nu P_{s}(a_3'-a_2'\gamma _{th_{3}})}$, $b_5=\frac{\gamma _{th_{2}}d_{sr}^md_{rd_{2}}^mN_o}{\nu P_{s}a_2'}$.

Let $Z=XY$, $X=|h_{sr}|^2$, $Y=|h_{rd_{2}}|^2$ Using properties of Random process and PDF of random variable (see equation no. 2), the above equation furthger reduces to

$$\begin{aligned} P_{out}^{U2}=1-\int _{b_7}^{\infty } \frac{1}{\lambda _{sr}} e^{-(\frac{x}{\lambda _{sr}}+\frac{b_8}{x\lambda _{rd_{2}}})}dx \end{aligned}$$

(38)

Proof of Proposition 3

Substituting (8), and (13) in to (22), we get

$$\begin{aligned} P_{out}^{U3}=1-Pr\{(|h_{sr}|^2>b_1,|h_{sr}|^2|h_{rd_{2}}|^2>b_6\} \end{aligned}$$

(39)

Following the procedure as shown in Proposition 2, we get the expression of $P_{out}^{U3}$ in terms of Bessel function.

Appendix B

Proof of Proposition 4

With the instantaneous rate given in (25), The ergodic rate can be written as:-

$$\begin{aligned} R_{U1}=(1-\theta )E\{log_2 (1+\gamma _{r,x_1})\} \end{aligned}$$

(40)

Let $U=\gamma _{r,x_1}$ The CDF of U is expressed as

$$\begin{aligned} F_U(u)=1-e^{-c_1u} \end{aligned}$$

where, $c_1=\frac{(I_s+N_o)d_{sr}^m}{a_1P_s\lambda _{sr}}$ Therefore, Ergodic rate is written as

$$\begin{aligned} R_{U1}=(1-\theta )E\{log_2 (1+U\} \end{aligned}$$

(41)

$$\begin{aligned} R_{U1}=\frac{(1-\theta )}{\ln 2}\int _{0}^{\infty }\frac{1-F_{U}(u)}{1+u} \end{aligned}$$

(42)

Substituting $F_U(u)$ in (42), and using relation: $\int _{0}^{\infty }e^\frac{-ax}{1+x}=-e^{-c_1}Ei(-c_1)$ Ergodic rates of $U_1$ is obtained as

$$\begin{aligned} R_{U1}=\frac{(\theta -1)}{\ln 2}e^{c_{1}}Ei(-c_1) \end{aligned}$$

(43)

PROOF OF PROPOSITION 5: With the instantaneous rate of U2 given by (27), The ergodic rate can be expressed as:-

$$\begin{aligned} R_{U2}=(1-\theta )E\{log_2 \left( 1+\min (\gamma _{r,x_2},\gamma _{d,x_2})\right) \} \end{aligned}$$

(44)

Let $X=\gamma _{r,x_2}$, $Y=\gamma _{d,x_2}$, and $Z=\min (\gamma _{r,x_2}, \gamma _{d,x_2})$

$$\begin{aligned} R_{U2}=(1-\theta )E\{log_2 \left( 1+Z\right) \} \end{aligned}$$

(45)

To find the ergodic rates, first we need to find the CDF of X, Y, and Z. The CDF of X and Y are derived as

$$\begin{aligned} F_X(x)=1-e^{\frac{-m_2x}{(a_2-a_1x)}} \end{aligned}$$

(46)

The CDF $F_X(x)$ is obtained under the condition $(a_2-a_1x)>0$, so $x<\left( \frac{a_2}{a_1}\right)$

$$\begin{aligned} F_Y(y)=1-2\sqrt{m_4y}K_1({2\sqrt{m_4y}}) \end{aligned}$$

(47)

Thus the CDF of Z can be written as

$$\begin{aligned} F_Z(z)=Pr\left[ Z<z\right] \end{aligned}$$

(48)

$$\begin{aligned} F_Z(z)=1-\left[ 1-F_X(z)\right] \left[ 1-F_Y(z)\right] \end{aligned}$$

(49)

where $m_2$=$\frac{d_{sr}^md_{rd_2}^mN_o}{P_{s}\lambda _{sr}}$, and $m_4$=$\frac{d_{sr}^md_{rd_2}^mN_o}{a_2'\nu P_{s}\lambda _{sr}\lambda _{rd_{2}}}$

Therefore the ergodic capacity of U2 is written as

$$\begin{aligned} R_{U2}=(1-\theta )E\{log_2 (1+Z)\} \end{aligned}$$

(50)

$$\begin{aligned} R_{U2}=\frac{(1-\theta )}{\ln 2}\int _{0}^{k}\frac{1-F_Z(z)}{1+z}dz \end{aligned}$$

(51)

where $t=\frac{a_2}{a_1}$ The above integration is difficult to solve. Therefore using Gaussian-Chebyshev approximation, we derive the ergodic rate. Gaussian Chebyshev approximation is given by:-

$$\begin{aligned} \int _{-1}^{1}\frac{f(x)}{\sqrt{(1-x^2)}}dx=\sum _{i=1}^{N_1} w_if(x_{i}) \end{aligned}$$

(52)

where, $w_i$=$\frac{pi}{N_1}$ are weights, $x_i$=$cos\left[ \frac{(2i-1)\pi }{2N_1}\right]$, and $N_1$ is a parameter to achieve an accuracy-complexity trade-off. So, the Ergodic rate is given by:-

$$\begin{aligned} R_{U2}=\frac{(1-\theta )m_5}{\ln 2}\sum _{i=1}^{N_1}w_ig(x_{i}) \end{aligned}$$

(53)

where $g(x)=\frac{2\sqrt{m_4m_5(1-x^2)(1+x)}}{1+m_5(1+x)}K_1(2\sqrt{m_4m_5(1+x)})e^\frac{m_2(1+x)}{2(x-1)}$

Proof of Proposition 6

With the instantaneous rate given by (31), the ergodic rate of U3 is expressed as:-

$$\begin{aligned} R_{U3}=(1-\theta )E\{log_2 \left( 1+\min (\gamma _{r,x_3},\gamma _{d,x_3})\right) \} \end{aligned}$$

(54)

Let $U=\gamma _{r,x_3}$, $V=\gamma _{d,x_3}$, and $W=\min (\gamma _{r,x_3},\gamma _{d,x_3})$

$$\begin{aligned} R_{U3}=(1-\theta )E\{log_2 \left( 1+W\right) \} \end{aligned}$$

(55)

Similar to the solution obtained in finding ergodic capacity of U2, we follow the same procedure. CDF of U, V, can be obtained as

$$\begin{aligned} F_U(u)=1-e^{\frac{-l_2x}{(a_3-a_2x-a_1x)}} \end{aligned}$$

(56)

The CDF of U must satisfy $(a_3-a_2x-a_1x)>0$ such that $x<\frac{a_3}{a_2+a_1}$

$$\begin{aligned} F_V(v)=1-2\sqrt{l_4y}K_1({2\sqrt{l_4y}}) \end{aligned}$$

(57)

$l_4=\frac{N_od_{sr}^md_{rd_{2}^m}}{(a_3'\nu P_s-a_2'\nu P_sy)\lambda _{sr}\lambda _{rd_3}}$ Therefore, CDF of W is expressed as

$$\begin{aligned} F_W(w)=Pr\left[ W<w\right] \end{aligned}$$

$$\begin{aligned} F_W(w)=1-\left[ 1-F_U(w)\right] \left[ 1-F_V(w)\right] \end{aligned}$$

(58)

Therefore the ergodic rate of U3 is written as

$$\begin{aligned}{} & {} R_{U3}=(1-\theta )E\{log_2 (1+W)\} \end{aligned}$$

(59)

$$\begin{aligned}{} & {} R_{U3}=\frac{(1-\theta )}{\ln 2}\int _{0}^{m}\frac{1-F_W(w)}{1+w}dw \end{aligned}$$

(60)

where $m=\min (u,v)$, $u=\frac{a_3}{a_2+a_1}$, and $v=\frac{a_3'}{a_2'}$.

Using Gaussian-Chebyshev approximation, we obtain the ergodic rate of U3 as

$$\begin{aligned} R_{U3}=\frac{(1-\theta )l_7}{\ln 2}\sum _{i=1}^{N_2}w_ih(x_{i}) \end{aligned}$$

(61)

where $h(x)=2\sqrt{ \frac{l_6l_7(1+x)(1-x^2)}{a_3'-a_2'l_7(1+x)}}K_1(2\sqrt{\frac{l_6l_7(1+x)(1-x^2)}{a_3'-a_2'l_7(1+x)}})\times e^(\frac{-l_2l_7(1+x)}{a_3-(a_2+a_1)l_7(1+x)})$ $l_7=\frac{a_3}{2(a_2+a_1)}$ $N_2$ is a parameter to achieve an accuracy-complexity trade-off.