Non-orthogonal Multiple Access in Coordinated LEO Satellite Networks

Abstract

Non-orthogonal multiple access (NOMA) has been widely considered to improve the spectral efficiency in terrestrial wireless networks. In this paper, we extend the idea to satellite networks and propose a NOMA-based scheme for coordinated low Earth orbit satellite systems, where the beam-edge user is supported by two satellites. By exploiting the difference of the equivalent downlink channel gains, users located both at the beam-center and beam-edge can be served simultaneously using NOMA. It is shown that the NOMA-based cooperative method is capable of providing a higher system capacity while guarantee the rate quality of the beam-edge user.

Appendices

Appendix

A Proof of Theorem 1

Since $|h_{i,i}|^2\sim \exp (1/L)$ and $|h_{i,e}|^2\sim \exp (1/(\eta \bar{L}))$, we have

$$\begin{aligned} \mathbb {E}[R_1] = \mathbb {E}[R_2] = \beta \frac{1}{\ln 2} \int _{0}^{\infty } \frac{1-(1-e^{-\frac{\beta }{L\gamma (1-\alpha )}x})}{1+x}dx. \end{aligned}$$

(23)

Based on the result derived in [19] (Page 341, line 2), (23) can be derived as

$$\begin{aligned} \mathbb {E}[R_1] = \mathbb {E}[R_2]&= \beta \frac{1}{\ln 2} \int _{0}^{\infty } \frac{e^{-\frac{\beta }{L\gamma (1-\alpha )}x}}{1+x}dx\nonumber \\&= -\beta \frac{1}{\ln 2} e^{\frac{\beta }{L\gamma (1-\alpha )}} \mathrm {Ei}\big (-\frac{\beta }{L\gamma (1-\alpha )}\big ), \end{aligned}$$

(24)

where $\mathrm {Ei}(x) = -\int _{-x}^{\infty } e^{-r}/rdr$.

From (14), the ergodic rate of User e can be calculated as

$$\begin{aligned}&\mathbb {E}[R_e] = \mathbb {E}[(1-\beta ) \log _2(1+(|h_{1,e}|^2 + |h_{2,e}|^2) \frac{\alpha }{1-\beta }\gamma )]\nonumber \\&= (1-\beta ) \frac{1}{\ln 2} \int _{0}^{\infty } \frac{1-\mathrm {Pr}((|h_{1,e}|^2 + |h_{2,e}|^2) \frac{\alpha }{1-\beta }\gamma \le x)}{1+x}dx. \end{aligned}$$

(25)

Let $Y\triangleq |h_{1,e}|^2 + |h_{2,e}|^2$. Then, the probability density function of Y can be expressed as $f_Y(y) = (\frac{1}{\eta \bar{L}})^2 y e^{-y/(\eta \bar{L})}$. From this, we have

$$\begin{aligned}&\mathrm {Pr}((|h_{1,e}|^2 + |h_{2,e}|^2) \frac{\alpha }{1-\beta }\gamma \le x)\nonumber \\&= \int _{0}^{(1-\beta )x/(\alpha \gamma )} \big (\frac{1}{\eta \bar{L}}\big )^2 ye^{-y/(\eta \bar{L})}dy\nonumber \\&= 1-\big (1 + \frac{(1-\beta )x}{\eta \bar{L} \alpha \gamma }\big ) e^{-\frac{(1-\beta )x}{\eta \bar{L} \alpha \gamma }}. \end{aligned}$$

(26)

Substituting (26) into (25), it can be derived that

$$\begin{aligned}&\mathbb {E}[R_e] = (1-\beta ) \frac{1}{\ln 2} \int _{0}^{\infty } \frac{(1 + \frac{(1-\beta )x}{\eta \bar{L} \alpha \gamma }) e^{-\frac{(1-\beta )x}{\eta \bar{L} \alpha \gamma }}}{1+x}dx\nonumber \\&= (1-\beta ) \frac{1}{\ln 2} \bigg ( \int _{0}^{\infty } \frac{e^{-\frac{(1-\beta )x}{\eta \bar{L} \alpha \gamma }}}{1+x}dx + \int _{0}^{\infty } \frac{\frac{(1-\beta )x}{\eta \bar{L}\alpha \gamma }e^{-\frac{(1-\beta )x}{\eta \bar{L} \alpha \gamma }}}{1+x}dx \bigg )\nonumber \\&= (1-\beta ) \frac{1}{\ln 2} \big [ -e^{\frac{1-\beta }{\eta \bar{L}\gamma \alpha }} \mathrm {Ei}\big (-\frac{1-\beta }{\eta \bar{L}\gamma \alpha }\big )\nonumber \\&~~~~~~~~+ \frac{1-\beta }{\eta \bar{L}\gamma \alpha } e^{\frac{1-\beta }{\eta \bar{L}\gamma \alpha }}\mathrm {Ei}\big (-\frac{1-\beta }{\eta \bar{L}\gamma \alpha }\big ) + 1 \big ]. \end{aligned}$$

(27)

Then, we can derive the result in Theorem 1.

The proof is completed.

B Proof of Theorem 2

According to Lemma 1, the ergodic rate for User e is given by

$$\begin{aligned} \mathbb {E}[R_{e}^{'}] = \mathbb {E} \big [ \log _2\big ( 1 + X\big ) \big ] = \frac{1}{\ln 2} \int _{0}^{\infty } \frac{1-F_X(x)}{1+x}dx, \end{aligned}$$

(28)

where $X=\frac{(|h_{1,e}|^2 + |h_{2,e}|^2) \alpha \gamma }{(|h_{1,e}|^2 + |h_{2,e}|^2) (1-\alpha )\gamma + 1}$.

Since $Y\triangleq |h_{1,e}|^2 + |h_{2,e}|^2$, which has been defined in above, $F_X(x)$ can be further expressed as

$$\begin{aligned} F_X(x)&= \mathrm {Pr} (X\le x) = \mathrm {Pr} \big (\frac{Y\alpha \gamma }{Y(1-\alpha )\gamma +1}\le x\big )\nonumber \\&= \mathrm {Pr} (Y(\alpha \gamma -x(1-\alpha )\gamma )\le x). \end{aligned}$$

(29)

In (29), when $\alpha \gamma -x(1-\alpha )\gamma >0$, i.e., $x<\frac{\alpha \gamma }{(1-\alpha )\gamma }$, we have

$$\begin{aligned}&\mathrm {Pr} (Y(\alpha \gamma -x(1-\alpha )\gamma )\le x) = \mathrm {Pr} \big (Y\le \frac{x}{\alpha \gamma -x(1-\alpha )\gamma }\big )\nonumber \\&= \int _0^{\frac{x}{(\alpha \gamma -x(1-\alpha )\gamma )}} \big (\frac{1}{\eta \bar{L}}\big )^2 y e^{-y/(\eta \bar{L})}dy\nonumber \\&= \big ( -\frac{x}{\eta \bar{L} (\alpha \gamma -x(1-\alpha )\gamma )}-1 \big ) e^{-\frac{x}{\eta \bar{L} (\alpha \gamma -x(1-\alpha )\gamma )}} + 1. \end{aligned}$$

(30)

For the case of $\alpha \gamma -x(1-\alpha )\gamma <0$, it can be found that $\mathrm {Pr} (Y(\alpha \gamma -x(1-\alpha )\gamma )\le x)=\mathrm {Pr} (Y\ge \frac{x}{\alpha \gamma -x(1-\alpha )\gamma })=1$ as Y is always a positive random variable. Thus, (29) becomes

$$\begin{aligned} F_X(x) = \left\{ \begin{array}{ll} \big ( -\frac{x}{\eta \bar{L} (\alpha \gamma -x(1-\alpha )\gamma )}-1 \big ) e^{-\frac{x}{\eta \bar{L} (\alpha \gamma -x(1-\alpha )\gamma )}} + 1,\\ ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ \qquad \qquad x<\frac{\alpha \gamma }{(1-\alpha )\gamma }\\ 1, ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~\qquad \qquad x>\frac{\alpha \gamma }{(1-\alpha )\gamma }. \end{array} \right. \end{aligned}$$

(31)

Substituting (31) into (28), we have

$$\begin{aligned} \mathbb {E}[R_e^{'}] = \frac{1}{\ln 2} \int _{0}^{\frac{\alpha }{1-\alpha }} \frac{(\frac{x}{\eta \bar{L} (\alpha \gamma -x(1-\alpha )\gamma )} + 1) e^{-\frac{x}{\eta \bar{L} (\alpha \gamma -x(1-\alpha )\gamma )}}}{1+x}dx. \end{aligned}$$

(32)

From (16) and (32), we can derive Theorem 2.

The proof is completed.