Energy-aware two-way relaying networks under imperfect hardware: optimal throughput design and analysis

Abstract

In this paper, we propose a new formula for achieving optimal throughput in energy-aware cooperative networks with generic time and power energy harvesting protocol, namely time power switching based relaying (TPSR). Especially, this investigation analyzes the impact of imperfect hardware at the relay node and the destination node in the two-way relaying networks (TWRN). This analysis enables us to derive the closed-form expressions of outage probabilities of signal-to-noise and distortion ratio (SNDR) at the destination nodes under the effect of hardware impairments. Interestingly, the optimal policy of joint wireless information and energy transfer is designed to maximize the system throughput by finding the optimal time switching and power splitting fractions in the proposed TPSR protocol. An important achievement is that the proposed optimal design offers the maximum throughput of system when we consider the trade-off between throughput and time-power factors in energy harvesting protocol by both numerical method and simulation. Numerical results provide practical insights into the performance of energy-aware TWRN under hardware impairments. Monte-Carlo method is also deployed to corroborate the accuracy of analytical derived expressions.

Appendices

Appendix A: Proof of Lemma 1

Proof

It should be noted that $\rho _1$ and $\rho _2$ are two independent random variables. By using the law of total probability to condition on $\rho _1$ or $\rho _2$ thanks to their equivalent roles, hereafter we thus compute the probability to condition on ${\rho _1}$ as follows

$$\begin{aligned}&I \mathop {=}\limits ^{\Delta } \Pr \left( \frac{\rho _1\rho _2}{m\rho _1^2 + m\rho _1\rho _2 + n\rho _1 + k} \le \lambda \right) \nonumber \\&\quad = 1 - \int \limits _0^\infty \Pr \left( \frac{\rho _1\rho _2}{m\rho _1^2 + m\rho _1\rho _2 + n\rho _1 + k} > \lambda \left| \rho _1 \right. \right) f_{\rho _1}(x)dx\nonumber \\ \end{aligned}$$

(40)

In (43), the first integral part can be further expressed by

$$\begin{aligned}&\Pr \left( \frac{\rho _1\rho _2}{m\rho _1^2 + m\rho _1\rho _2 + n\rho _1 + k} > \lambda \left| \rho _1 \right. \right) \nonumber \\&\quad = \left\{ {\begin{array}{ll} {1 - F_{\rho _2}\left( \frac{\lambda \left( m\rho _1^2 + n\rho _1 + k \right) }{\rho _1\left( 1 - \lambda m \right) } \right) ,}&{}{}\\ {0,}&{}{} \end{array}\begin{array}{ll} {if\;\lambda < \frac{1}{m}}\\ {if\;\lambda \ge \frac{1}{m}} \end{array}} \right. \end{aligned}$$

(41)

Applying (2), (3), and (41)–(43), we have a new formula for case of $\lambda \le 1/m$

$$\begin{aligned} I= & {} 1 - \int \limits _0^\infty {{e^{ - \frac{1}{{{\varOmega _2}}}\frac{{\lambda \left( {m{x^2} + nx + k} \right) }}{{x\left( {1 - \lambda m} \right) }}}}} \frac{1}{{{\varOmega _1}}}{e^{ - \frac{x}{{{\varOmega _1}}}}}dx\nonumber \\= & {} 1 - \frac{1}{{{\varOmega _1}}}{e^{ - \frac{{n\lambda }}{{{\varOmega _2}\left( {1 - \lambda m} \right) }}}}\int \limits _0^\infty {{e^{ - \left( {\frac{{\lambda m}}{{{\varOmega _2}\left( {1 - \lambda m} \right) }} + \frac{1}{{{\varOmega _1}}}} \right) x - \frac{{\lambda k}}{{{\varOmega _2}\left( {1 - \lambda m} \right) }}\frac{1}{x}}}} dx\nonumber \\ \end{aligned}$$

(42)

Finally, to obtain (23), we use (3.324.1) given in [26].

Appendix B: Proof of Lemma 2

Proof

Invoking the outage probability expression given below

$$\begin{aligned}&I \mathop {=}\limits ^{\Delta } \Pr \left( \frac{\rho _1\rho _2}{m\rho _1^2 + m\rho _1\rho _2 } \le \lambda \right) \nonumber \\&\quad = 1 - \int \limits _0^\infty \Pr \left( \frac{\rho _1\rho _2}{m\rho _1^2 + m\rho _1\rho _2 } > \lambda \left| \rho _1 \right. \right) f_{\rho _1}(x)dx \end{aligned}$$

(43)

We first calculate the conditional probability as

$$\begin{aligned}&\Pr \left( \frac{\rho _1\rho _2}{m\rho _1^2 + m\rho _1\rho _2 } > \lambda \left| \rho _1 \right. \right) \nonumber \\&\quad = \left\{ {\begin{array}{ll} {1 - F_{\rho _2}\left( \frac{\lambda m\rho _1}{1 - \lambda m } \right) ,}&{}{}\\ {0,}&{}{} \end{array}\begin{array}{ll} {if\;\lambda < \frac{1}{m}}\\ {if\;\lambda \ge \frac{1}{m}} \end{array}} \right. \end{aligned}$$

(44)

Similarly, the outage probability in case of $\lambda < 1/m$ can be expressed as

$$\begin{aligned} I = 1 - \int \limits _0^\infty {{e^{ - \frac{1}{{{\varOmega _2}}}\frac{{\lambda mx}}{{1 - \lambda m}}}}} \frac{1}{{{\varOmega _1}}}{e^{ - \frac{x}{{{\varOmega _1}}}}}dx \end{aligned}$$

(45)

To this end, the desired result can be obtained after some simple algebraic manipulations.