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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

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  • ReceivedJan 31, 2020
  • AcceptedMay 27, 2020
  • PublishedJun 2, 2021

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.

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