Throughput optimization of cooperative non orthogonal multiple access

Abstract

This paper derives the outage and packet error probabilities of Non Orthogonal Multiple Access (NOMA) systems. In the first time slot, the Base Station transmits a combination of two symbols $s_w$ and $s_s$ dedicated for weak and strong users. This signal is received by the two users and a relay. In the second time slot, the relay amplifies the received signal to the two users. Both users use the signal with the highest Signal to Interference plus Noise Ratio among direct and relayed signals. The weak user detects only its signal. Strong user first detects symbol $s_w$ of weak user. After removing the contribution of weak user, strong user detects its own symbol $s_s$. In this article, expressions for outage probability, Packet Error Probability and the throughput of cooperative NOMA are derived. We also optimize the power allocated to weak and strong users to maximize the system throughput.

Appendices

Appendix A

The channel gain of weak user is lower than that of strong user i.e.

$$\begin{aligned} |h_{BSu_w}|^2<|h_{BSu_s}|^2 \end{aligned}$$

(50)

Therefore, we have

$$\begin{aligned} |h_{BSu_w}|^2=min(|h_{BSu_1}|^2,|h_{BSu_2}|^2) \end{aligned}$$

(51)

We deduce

$$\begin{aligned}&P(|h_{BSu_w}|^2\le x)=P(min(|h_{BSu_1}|^2,|h_{BSu_2}|^2)\le x)\nonumber \\&\quad =1-P(min(|h_{BSu_1}|^2,|h_{BSu_2}|^2)> x) \end{aligned}$$

(52)

If $h_{BSu_1}$ and $h_{BSu_2}$ are independent random variables (r.v.), we deduce

$$\begin{aligned}&P(|h_{BSu_w}|^2\le x)\nonumber \\&\quad =1-P(|h_{BSu_1}|^2> x)P(|h_{BSu_2}|^2> x) \end{aligned}$$

(53)

For Rayleigh fading channels $|h_{BSu_1}|^2$ and $|h_{BSu_2}|^2$ are exponentially distributed with mean

$$\begin{aligned} E(|h_{BSu_i}|^2)=\frac{1}{\lambda _{BSu_i}} \end{aligned}$$

(54)

for $i=1,2$.

Therefore, we can write

$$\begin{aligned} P(|h_{BSu_w}|^2\le x)=1-e^{-x(\lambda _{BSu_1}+\lambda _{BSu_2})} \end{aligned}$$

(55)

We have

$$\begin{aligned} |h_{BSu_s}|^2=max(|h_{BSu_1}|^2,|h_{BSu_2}|^2) \end{aligned}$$

(56)

We deduce

$$\begin{aligned} P(|h_{BSu_s}|^2\le x)=P(max(|h_{BSu_1}|^2,|h_{BSu_2}|^2)\le x) \end{aligned}$$

(57)

If $h_{BSu_1}$ and $h_{BSu_2}$ are independent random variables (r.v.), we deduce

$$\begin{aligned}&P(|h_{BSu_s}|^2\le x)=P(|h_{BSu_1}|^2\le x)P(|h_{BSu_2}|^2\le x)\nonumber \\&\quad =(1-e^{-x\lambda _{BSu_1}})(1-e^{-x\lambda _{BSu_2}}) \end{aligned}$$

(58)

Appendix B

Using (6), we have

$$\begin{aligned}&P(\Gamma _{BSu_w}\le x)=P\left( \frac{|h_{BSu_w}|^2\mu b_w}{1+|h_{BSu_w}|^2\mu b_s}\le x\right) \nonumber \\&\quad =P\left( |h_{BSu_w}|^2 \le \frac{x}{\mu (b_w-b_sx)}\right) \end{aligned}$$

(59)

Using the results of “Appendix A”, we deduce

$$\begin{aligned} P(\Gamma _{BSu_w}\le x)=1-e^{-\frac{x}{\mu (b_w-b_sx)}(\lambda _{BSu_1}+\lambda _{BSu_2})} \end{aligned}$$

(60)

Using (13), the outage probability of relayed link of weak user is expressed as

$$\begin{aligned}&P(\Gamma _{ru_w}\le x)\nonumber \\&\quad =P\left( \frac{|h_{BSr}|^2|h_{ru_w}|^2b_w\mu }{|h_{BSr}|^2|h_{ru_w}|^2b_s\mu +|h_{ru_w}|^2+C}\le x \right) \end{aligned}$$

(61)

We deduce

$$\begin{aligned}&P(\Gamma _{ru_w}\le x)=P(|h_{BSr}|^2|h_{ru_w}|^2\mu (b_w-b_sx)\nonumber \\&\quad \le x|h_{ru_w}|^2+xC) \end{aligned}$$

(62)

Let

$$\begin{aligned} \epsilon =\frac{x}{\mu (bf-b_sx)} \end{aligned}$$

(63)

Therefore, we can write

$$\begin{aligned}&P(\Gamma _{ru_w}\le x)=P(|h_{BSr}|^2|h_{ru_w}|^2\le \epsilon |h_{ru_w}|^2+\epsilon C)\nonumber \\&\quad =P(|h_{ru_w}|^2(|h_{BSr}|^2-\epsilon )\le \epsilon C) \end{aligned}$$

(64)

We deduce

$$\begin{aligned}&P(\Gamma _{ru_w}\le x)=P(|h_{ru_w}|^2(|h_{BSr}|^2-\epsilon )\nonumber \\&\quad \le \epsilon C,|h_{BSr}|^2<\epsilon )+P(|h_{ru_w}|^2(|h_{BSr}|^2-\epsilon )\nonumber \\&\quad \le \epsilon C,|h_{BSr}|^2\ge \epsilon ) \end{aligned}$$

(65)

The first term of (33) is equal to

$$\begin{aligned}&P(|h_{ru_w}|^2(|h_{BSr}|^2-\epsilon )\le \epsilon C,|h_{BSr}|^2<\epsilon )\nonumber \\&\quad =P(|h_{ru_w}|^2(|h_{BSr}|^2-\epsilon )\nonumber \\&\quad \le \epsilon C| |h_{BSr}|^2<\epsilon )P(|h_{BSr}|^2<\epsilon ) \nonumber \\&\quad =P(|h_{BSr}|^2<\epsilon )=1-e^{-\epsilon \lambda _{BSr}} \end{aligned}$$

(66)

The second term of (33) is equal to

$$\begin{aligned}&P(|h_{ru_w}|^2(|h_{BSr}|^2-\epsilon )\le \epsilon C,|h_{BSr}|^2\ge \epsilon )\nonumber \\&\quad =\int _{\epsilon }^{+\infty }\lambda _{Bsr}e^{-\lambda _{Bsr}y}\int _0^{\frac{\epsilon C}{y-\epsilon }}\lambda _{ru_w}e^{-\lambda _{ru_w}z}dzdy \nonumber \\&\quad =\int _{\epsilon }^{+\infty }\lambda _{Bsr}e^{-\lambda _{Bsr}y}[1-e^{-\lambda _{ru_w}\frac{\epsilon C}{y-\epsilon }}]dy \end{aligned}$$

(67)

Let $u=y-\epsilon$, we have

$$\begin{aligned}&P(|h_{ru_w}|^2(|h_{BSr}|^2-\epsilon )\le \epsilon C,|h_{BSr}|^2\ge \epsilon )\nonumber \\&\quad =\int _{0}^{+\infty }\lambda _{Bsr}e^{-\lambda _{Bsr}(u+\epsilon )}[1-e^{-\lambda _{ru_w}\frac{\epsilon C}{u}}]dy \end{aligned}$$

(68)

We have [23]

$$\begin{aligned} \int _0^{+\infty }e^{-\frac{b}{4x}-ax}dx=\sqrt{\frac{b}{a}}K_1(\sqrt{ba}) \end{aligned}$$

(69)

where $K_1(.)$ is the modified Bessel function of second kind and first order.

Therefore, we obtain

$$\begin{aligned}&P(|h_{ru_w}|^2(|h_{BSr}|^2-\epsilon )\le \epsilon C,|h_{BSr}|^2\ge \epsilon )\nonumber \\&\quad =e^{-\lambda _{BSr}\epsilon }\left[ 1-2\sqrt{\frac{\lambda _{ru_w}\epsilon C}{\lambda _{BSr}}}K_1(2\sqrt{\lambda _{ru_w}\epsilon C\lambda _{BSr}})\right] \end{aligned}$$

(70)

Finally, (33), (34) and (38) give the outage probability of weak user is written as

$$\begin{aligned}&P(\Gamma _{ru_w}\le x)=1-e^{-\epsilon \lambda _{BSr}}\nonumber \\&\quad +e^{-\lambda _{BSr}\epsilon }\left[ 1-2\sqrt{\frac{\lambda _{ru_w}\epsilon C}{\lambda _{BSr}}}K_1(2\sqrt{\lambda _{ru_w}\epsilon C\lambda _{BSr}})\right] \end{aligned}$$

(71)

Appendix C

The first term of (17) is expressed as

$$\begin{aligned}&1-P(\Gamma _{BSu_s}^{(1)}> x,\Gamma _{BSu_s}^{(2)}> x)\nonumber \\&\quad =1-P\left( \frac{|h_{BSu_s}|^2b_w\mu }{1+|h_{BSu_s}|^2b_s\mu }>,|h_{BSu_s}|^2b_s\mu>x\right) \nonumber \\&\quad =1-P(|h_{BSu_s}|^2>\alpha ) \end{aligned}$$

(72)

where

$$\begin{aligned} \alpha =max\left( \frac{x}{b_s\mu },\frac{x}{b_w\mu -b_s\mu x}\right) \end{aligned}$$

(73)

Using the results of “Appendix A” (26), we deduce

$$\begin{aligned}&1-P(\Gamma _{BSu_s}^{(1)}> x,\Gamma _{BSu_s}^{(2)}> x)\nonumber \\&\quad =(1-e^{-\alpha \lambda _{BSu_1}})(1-e^{-\alpha \lambda _{BSu_2}}). \end{aligned}$$

(74)

The second term of (17) is expressed as

$$\begin{aligned}&1-P(\Gamma _{ru_s}^{(1)}> x,\Gamma _{ru_s}^{(2)}> x) \nonumber \\&\quad =1-P\left( \frac{|h_{BSr}|^2|h_{ru_s}|^2b_w\mu }{C+|h_{BSr}|^2|h_{ru_s}|^2b_s\mu +|h_{ru_s}|^2} \right. \nonumber \\&\left. \quad>x,\frac{|h_{BSr}|^2|h_{ru_s}|^2b_s\mu }{C+|h_{ru_s}|^2}>x\right) \nonumber \\&\quad =1-P(|h_{ru_s}|^2>\frac{Cb}{|h_{BSr}|^2-b},|h_{BSr}|^2>b) \end{aligned}$$

(75)

where

$$\begin{aligned} b=max(\epsilon ,a) \end{aligned}$$

(76)

$$\begin{aligned} a=\frac{x}{b_s\mu } \end{aligned}$$

(77)

$$\begin{aligned} \epsilon =\frac{x}{\mu (bf-b_sx)} \end{aligned}$$

(78)

Therefore, last equation gives

$$\begin{aligned}&1-P(\Gamma _{ru_s}^{(1)}> x,\Gamma _{ru_s}^{(2)}> x)=1\nonumber \\&\quad -\int _b^{+\infty }\lambda _{BSr}e^{-\lambda _{BSr}y}P\left( |h_{ru_s}|^2>\frac{Cb}{y-b}\right) dy \nonumber \\&\quad =1-\int _b^{+\infty }\lambda _{BSr}e^{-\lambda _{BSr}y}e^{-\lambda _{ru_s}\frac{Cb}{y-b}}dy \end{aligned}$$

(79)

Let $z=y-b$, we deduce

$$\begin{aligned}&1-P(\Gamma _{ru_s}^{(1)}> x,\Gamma _{ru_s}^{(2)}> x)\nonumber \\&\quad =1-e^{-\lambda _{BSr}b}\lambda _{BSr}\int _0^{+\infty }e^{-\lambda _{BSr}z}e^{-\lambda _{ru_s}\frac{Cb}{z}}dz \nonumber \\&\quad =1-2e^{-\lambda _{BSr}b}\lambda _{BSr}\sqrt{\frac{\lambda _{ru_s}Cb}{\lambda _{BSr}}}K_1(2\sqrt{\lambda _{ru_s}Cb\lambda _{BSr}}) \end{aligned}$$

(80)

Using (42) and (48), the outage probability of strong user is written as

$$\begin{aligned}&P_{outage,s}(x)=(1-e^{-\alpha \lambda _{BSu_1}})(1-e^{-\alpha \lambda _{BSu_2}})\nonumber \\&\quad \times \left[ 1-2e^{-\lambda _{BSr}b}\lambda _{BSr}\sqrt{\frac{\lambda _{ru_s}Cb}{\lambda _{BSr}}}K_1(2\sqrt{\lambda _{ru_s}Cb\lambda _{BSr}})\right] \nonumber \\ \end{aligned}$$

(81)

Appendix D

User $u_i$ has the i-th largest channel gain. Its PDF is given by [24]

$$\begin{aligned}&p_{|h_{BSu_i}|^2}(x)\nonumber \\&\quad =\sum _{q=1}^N \lambda _{BSu_q} e^{-x\lambda _{BSu_q}}\sum _{p_1,p_2,\ldots ,p_{N-1}} \prod _{j=1}^{N-i}[1-e^{-x\lambda _{BSu_{p_j}}}]\nonumber \\&\qquad \prod _{l=N-i+1}^{N-1}e^{-x\lambda _{BSu_{p_l}}} \end{aligned}$$

(82)

where $p_1$, $p_2$, ..., $p_N$ are relay indexes different from q, $p_1\ne p_2 \ne p_{N-1}$, $p_1<p_2<\cdots <p_{N-i}$ and $p_{N-i+1}<p_{N-i+2}<\cdots <p_{N-1}$.

We have

$$\begin{aligned} \prod _{j=1}^{N-i}\![1-e^{-x\lambda _{BSu_{p_j}}}]\!=\!\sum _{n=0}^{2^{N-i}-1}(-1)^{d(n)}e^{-x\sum _{m=1}^{N-i}\!\lambda _{BSu_{p_m}}b_n(m)}.\nonumber \\ \end{aligned}$$

(83)

Therefore, (82) and (83) give

$$\begin{aligned} p_{|h_{BSu_i}|^2}(x)=\sum _{q=1}^N \!\sum _{p_1,p_2,\ldots ,p_N}\lambda _{BSu_q} \!\sum _{n=0}^{2^{N-i}-1}(-1)^{d(n)}e^{-xf_{q,n,i}}\nonumber \\ \end{aligned}$$

(84)

where

$$\begin{aligned} f_{q,n,i}=\lambda _{BSu_q}+\sum _{l=N-i+1}^{N-1}\lambda _{Su_{p_l}}+\sum _{m=1}^{N-i}\lambda _{BSu_{p_m}}b_n(m) \end{aligned}$$

(85)

The CDF of the channel gain of i-th user is deduced by a primitive of the PDF

$$\begin{aligned}&P_{|h_{BSu_i}|^2}(x)\nonumber \\&\quad =\sum _{q=1}^N \sum _{p_1,p_2,\ldots ,p_N}\lambda _{BSu_q} \sum _{n=0}^{2^{N-i}-1}(-1)^{d(n)}[1-\frac{e^{-xf_{q,n,i}}}{f_{q,n,i}}].\nonumber \\ \end{aligned}$$

(86)