#  SCIENCE CHINA Information Sciences, Volume 64 , Issue 8 : 182310(2021) https://doi.org/10.1007/s11432-020-2935-6

## Full-duplex two-way AF relaying systems with imperfect interference cancellation in Nakagami-m fading channels • AcceptedMay 27, 2020
• PublishedJun 2, 2021
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### Abstract ### Acknowledgment

This work was supported by National Key RD Program of China (Grant No. 2019YFB1803400) and National Natural Science Foundation of China (Grant No. 61922071).

### Supplement

Appendix

Proof of Theorem 1

Without loss of generality, we focus on the outage probability at node A. The c.d.f. of $\gamma_{\rm~A}^{\rm~up}$ can be written as \begin{align}F_{\gamma_{\rm A}^{\rm up} }(x)& = \mbox{Pr}\left(\rho_{\rm A}\min\left(\frac{\gamma_2}{\gamma_3+\rho_{\rm A}\gamma_4},\frac{\gamma_1}{\gamma_3}\right)\leq x\right) \tag{49} \\ &=1-\mbox{Pr}\left(\rho_{\rm A}\min\left(\frac{\gamma_2}{\gamma_3+\rho_{\rm A}\gamma_4},\frac{\gamma_1}{\gamma_3}\right)\geq x\right). \tag{50} \end{align} Since the two items inside the min function are correlated, it is difficult to handle directly. To circumvent this, we adopt the conditional approach; i.e., we first compute the conditional c.d.f. of $\gamma_{\rm~A}^{\rm~up}$ by conditioning on $\gamma_3$ and $\gamma_4$, and then take expectation over $\gamma_3$ and $\gamma_4$.

Now, conditioned on $\gamma_3$ and $\gamma_4$, we have \begin{align}F_{\gamma_{\rm A}^{\rm up} }(x|\gamma_3, \gamma_4)&=1-\mbox{Pr}\left(\left.\frac{\gamma_2}{\gamma_3+\rho_{\rm A}\gamma_4}\geq \frac{x}{\rho_{\rm A}}\right|\gamma_3, \gamma_4\right)\mbox{Pr}\left(\left.\frac{\gamma_1}{\gamma_3}\geq \frac{x}{\rho_{\rm A}}\right|\gamma_3, \gamma_4\right). \tag{51} \end{align} Since both $\gamma_1$ and $\gamma_2$ are Gamma distributed, the conditional c.d.f. can be computed as \begin{align}F_{\gamma_{\rm A}^{\rm up} }(x|\gamma_3, \gamma_4)&=1-\left(1-\gamma\left(m_2,\frac{\beta_2 x(\gamma_3+\rho_{\rm A}\gamma_4)}{\rho_{\rm A}}\right)\right)\left(1-\gamma\left(m_1,\frac{\beta_1\gamma_3x}{\rho_{\rm A}}\right)\right) \tag{52} \\ &=1-\left(\sum_{j=0}^{m_2-1}\frac{1}{j!}\left(\frac{\beta_2 x(\gamma_3+\rho_{\rm A}\gamma_4)}{\rho_{\rm A}}\right)^j{\rm e}^{-\frac{\beta_2x(\gamma_3+\rho_{\rm A}\gamma_4)}{\rho_{\rm A}}}\right)\left(\sum_{i=0}^{m_1-1}\frac{1}{i!}\left(\frac{\beta_1\gamma_3x}{\rho_{\rm A}}\right)^i{\rm e}^{-\frac{\beta_1\gamma_3x}{\rho_{\rm A}}}\right), \tag{53} \end{align} where $\gamma(n,x)$ is the incomplete Gamma function.

Applying the binomial expansion, and taking expectations over $\gamma_3$ and $\gamma_4$, we have \begin{align}F_{\gamma_{\rm A}^{\rm up} }(x)&=1-\sum_{i=0}^{m_1-1}\frac{1}{i!}\left(\frac{\beta_1x}{\rho_{\rm A}}\right)^i\sum_{j=0}^{m_2-1}\frac{1}{j!}\left(\frac{\beta_2 x}{\rho_{\rm A}}\right)^j\sum_{k=0}^j{j\choose k}{\rm E}_{\gamma_3}\left\{\gamma_3^{k+i}{\rm e}^{-\frac{\beta_1x+\beta_2x}{\rho_{\rm A}}\gamma_3}\right\}{\rm E}_{\gamma_4}\left\{\left(\rho_{\rm A}\gamma_4\right)^{j-k}{\rm e}^{-\beta_2x\gamma_4}\right\}. \tag{54} \end{align} To compute ${\rm~E}_{\gamma_3}\{\gamma_3^{k+i}{\rm~e}^{-\frac{\beta_1x+\beta_2x}{\rho_{\rm~A}}\gamma_3}\}$, the p.d.f. of random variable $\gamma_3$ is required. Utilizing a result 4), after some algebraic manipulations, it can be shown that the p.d.f. of $\gamma_3$ can be obtained as \begin{align}f_{\gamma_3}(x)= & \beta_3^{m_3}\beta_4^{m_4}\sum_{s=1}^{m_3}\frac{\prod_{j=1}^{s-1}(1-m_4-j)\left(\beta_4-\beta_3\right)^{1-m_4-s}}{(m_3-s)!(s-1)!}(x-1)^{m_3-s}{\rm e}^{-\beta_3(x-1)} \\ &+\beta_3^{m_3}\beta_4^{m_4}\sum_{s=1}^{m_4}\frac{\prod_{j=1}^{s-1}(1-m_3-j)\left(\beta_3-\beta_4\right)^{1-m_3-s}}{(m_4-s)!(s-1)!}(x-1)^{m_4-s}{\rm e}^{-\beta_4(x-1)}. \tag{55} \end{align} Hence, we have \begin{align}&{\rm E}_{\gamma_3}\left\{\gamma_3^{k+i}{\rm e}^{-\frac{\beta_1x+\beta_2x}{\rho_{\rm A}}\gamma_3}\right\} = \beta_3^{m_3}\beta_4^{m_4}{\rm e}^{-\frac{\beta_1x+\beta_2x}{\rho_{\rm A}}} \\ &\times\left(\sum_{s=1}^{m_3}\frac{\prod_{j=1}^{s-1}(1-m_4-j)\left(\beta_4-\beta_3\right)^{1-m_4-s}}{(m_3-s)!(s-1)!}\sum_{t=0}^{k+i}{k+i\choose t}\left(\beta_3+\frac{\beta_1x+\beta_2x}{\rho_{\rm A}}\right)^{-(m_3+t-s+1)}\Gamma(m_3+t-s+1)\right. \\ &+\left.\sum_{s=1}^{m_4}\frac{\prod_{j=1}^{s-1}(1-m_3-j)\left(\beta_3-\beta_4\right)^{1-m_3-s}}{(m_4-s)!(s-1)!}\sum_{t=0}^{k+i}{k+i\choose t}\left(\beta_4+\frac{\beta_1x+\beta_2x}{\rho_{\rm A}}\right)^{-(m_4+t-s+1)}\Gamma(m_4+t-s+1)\right). \tag{56} \end{align} Finally, the term ${\rm~E}_{\gamma_4}\{\left(\rho_{\rm~A}\gamma_4\right)^{j-k}{\rm~e}^{-\beta_2x\gamma_4}\}$ can be computed as \begin{align}{\rm E}_{\gamma_4}\left\{\left(\rho_{\rm A}\gamma_4\right)^{j-k}{\rm e}^{-\beta_2x\gamma_4}\right\} &=\frac{\beta_5^{m_5}}{\Gamma(m_5)}\int_1^{\infty}(\gamma_4-1)^{m_5-1}{\rm e}^{-\beta_5(\gamma_4-1)}(\rho_{\rm A}\gamma_4)^{j-k}{\rm e}^{-\beta_2x\gamma_4}{\rm d}\gamma_4 \tag{57} \\ &=\frac{\beta_5^{m_5}\rho_{\rm A}^{j-k}{\rm e}^{-\beta_2x}}{\Gamma(m_5)}\sum_{t=0}^{j-k}{j-k\choose t}\int_0^{\infty}y^{m_5+t-1}{\rm e}^{-(\beta_5+\beta_2x)y}{\rm d}y \tag{58} \\ &=\frac{\beta_5^{m_5}\rho_{\rm A}^{j-k}{\rm e}^{-\beta_2x}}{\Gamma(m_5)}\sum_{t=0}^{j-k}{j-k\choose t}(\beta_5+\beta_2x)^{-(m_5+t)}\Gamma(m_5+t). \tag{59} \end{align} To this end, pulling everything together yields the desired result.

Abu-Dayya A, Beaulieu N. Outage probability of cellular mobile radio systems with multiple Nakagami interferers. IEEE Trans Veh Tech, 1991, 41: 757–768.

Proof of Theorem 2

Following the similar lines as the full-duplex relaying case, the SINRs at nodes A and B can be upper bounded by \begin{align}v_{\rm A} \leq v_{\rm A}^{\rm up}= \rho_{\rm A}\min\left(\frac{\gamma_2}{\gamma_6+\rho_{\rm A}},\frac{\gamma_1}{\gamma_6}\right) \tag{60} \end{align} and \begin{align}v_{\rm B} \leq v_{\rm B}^{\rm up}= \rho_{\rm B}\min\left(\frac{\gamma_1}{\gamma_7+\rho_{\rm B}},\frac{\gamma_2}{\gamma_7}\right). \tag{61} \end{align} Without loss of generality, we focus on $P_{\rm~OA}^{\rm~HL}$. Conditioned on $\gamma_6$ and invoking the c.d.f. of $\gamma_1$ and $\gamma_2$, the outage probability lower bound can be expressed as \begin{align}P_{\rm OA}^{\rm HL}=1- {\rm e}^{-\beta_2\eta_{\rm A}}\sum_{i=0}^{m_1-1}\frac{1}{i!}\left(\frac{\beta_1 \eta_{\rm A}}{\rho_{\rm A}}\right)^i\sum_{j=0}^{m_2-1}\frac{1}{j!}\left(\frac{\beta_2 \eta_{\rm A}}{\rho_{\rm A}}\right)^j(\gamma_6+\rho_{\rm A})^j\gamma_6^i{\rm e}^{-\frac{(\beta_1\eta_{\rm A}+\beta_2\eta_{\rm A})\gamma_6}{\rho_{\rm A}}}. \tag{62} \end{align} Then, taking expectation on $\gamma_6$, the outage probability lower bound can be written as \begin{align}P_{\rm OA}^{\rm HL} =&1- \frac{\beta_4^{m_4}{\rm e}^{-\beta_2\eta_{\rm A}}}{\Gamma(m_4)}\sum_{i=0}^{m_1-1}\frac{1}{i!}\left(\frac{\beta_1 \eta_{\rm A}}{\rho_{\rm A}}\right)^i\sum_{j=0}^{m_2-1}\frac{1}{j!}\left(\frac{\beta_2 \eta_{\rm A}}{\rho_{\rm A}}\right)^j\sum_{t=0}^{j}{j\choose t}\rho_{\rm A}^{j-t} \\ &\times\int_1^{\infty}\gamma_6^{t+i}{\rm e}^{-\frac{(\beta_1\eta_{\rm A}+\beta_2\eta_{\rm A})\gamma_6}{\rho_{\rm A}}}(\gamma_6-1)^{m_4-1}{\rm e}^{-\beta_4(\gamma_6-1)}d\gamma_6. \tag{63} \end{align} With a change of variable, and then applying the binomial expansion, the above expression can be computed as \begin{align}P_{\rm OA}^{\rm HL} =&1- \frac{\beta_4^{m_4}{\rm e}^{-\frac{\beta_1\eta_{\rm A}+\beta_2\eta_{\rm A}}{\rho_{\rm A}}-\beta_2\eta_{\rm A}}}{\Gamma(m_4)}\sum_{i=0}^{m_1-1}\frac{1}{i!}\left(\frac{\beta_1 \eta_{\rm A}}{\rho_{\rm A}}\right)^i\sum_{j=0}^{m_2-1}\frac{1}{j!}\left(\frac{\beta_2 \eta_{\rm A}}{\rho_{\rm A}}\right)^j\sum_{t=0}^{j}{j\choose t}\rho_{\rm A}^{j-t} \\ &\times\sum_{p=0}^{t+i}{t+i\choose p}\int_0^{\infty}\gamma_6^{m_4+p-1}{\rm e}^{-\frac{(\beta_1\eta_{\rm A}+\beta_2\eta_{\rm A})\gamma_6}{\rho_{\rm A}}-\beta_4\gamma_6}{\rm d}\gamma_6. \tag{64} \end{align} To this end, the desired result can be obtained by solving the integral.

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• Figure 1

(Color online) System model, where the solid lines denote the information transmission channels, while the dash lines denote the loop back interference channels.