Bidirectional Communication in Full Duplex Wireless-Powered Relaying Networks: Time-Switching Protocol and Performance Analysis

Abstract

In this paper, we consider two-way full-duplex relay with energy harvesting system, in which the relay node harvests transmitted power from the two source nodes to forward signals to destinations. We also analyze the relay network model with decode-and-forward protocol for information cooperation and time switching-based relaying protocol for power transfer. In particular, the outage probability, average throughput and optimal energy harvesting time of novel scheme in simultaneous wireless information and power transfer are presented. We obtain analytic closed-form expressions for both tight bounded and asymptotic of outage probability and average throughput of system. Using numerical and analytic simulations, the performances of different situations are presented and discussed more clearly. The results show that the better performance is achieved in case of worse self-interference. As important result, the better self interference cancellation leads to superior system performance.

Appendices

Appendix 1: Proof of Lemma 1

In Lemma 1, the integral formula can be obtained by using [25, vol. 4, eq. (1.1.2.3)] and re-expressed as

$$\begin{aligned} \begin{aligned} \psi \left( {\lambda ,\kappa ,\varepsilon } \right)&\buildrel \varDelta \over = \int \limits _0^\infty {\left( {\kappa x + \varepsilon } \right) \times \mathrm{K}_2 \left( {2\sqrt{\kappa x + \varepsilon } } \right) \exp \left( { - \lambda x} \right) dx} \\&= \mathrm{{exp}}\left( {\frac{{\varepsilon \lambda }}{\kappa }} \right) \left\{ {\underbrace{\kappa \int \limits _0^\infty {x\mathrm{K}_2 \left( {2\sqrt{\kappa x} } \right) \exp \left( { - \lambda x} \right) dx} }_{\psi _1 }} \right. \left. { - \underbrace{\frac{1}{\kappa }\int \limits _0^\varepsilon {y\mathrm{K}_2 \left( {2\sqrt{y} } \right) \exp \left( { - \frac{{\lambda y}}{\kappa }} \right) dy} }_{\psi _2 }} \right\} . \\ \end{aligned} \end{aligned}$$

(25)

The first term in (25) can be fulfilled by applying [25, vol. 4, eq. (3.16.2.4)]

$$\begin{aligned} \begin{aligned} \psi _1&= \kappa \int \limits _0^\infty {x\mathrm{K}_2 \left( {2\sqrt{\kappa }\sqrt{x} } \right) \exp \left( { - \lambda x} \right) dx} \\&= \frac{{\kappa ^2 }}{{\lambda ^3 }}\exp \left( { - \frac{\kappa }{\lambda }} \right) \varGamma \left( { - 2,\frac{\kappa }{\lambda }} \right) . \\ \end{aligned} \end{aligned}$$

(26)

The second term in (25) is difficult for closed-form because of the Bessel function and exponential function, hence we make an approximation value as follow

$$\begin{aligned} \psi _2&= \frac{1}{\kappa }\int \limits _0^\varepsilon {y\mathrm{K}_2 \left( {2\sqrt{y} } \right) \exp \left( { - \frac{{\lambda y}}{\kappa }} \right) dy} \\&\mathop = \limits ^{(a)} \sum \limits _{m = 0}^\infty {\frac{{\left( { - \lambda } \right) ^m }}{{\kappa ^{m + 1} }}\int \limits _0^\varepsilon {\frac{{y^{m + 1} }}{{m!}}\mathrm{K}_2 \left( {2\sqrt{y} } \right) dy} } \\&\mathop = \limits ^{(b)} \frac{1}{2}\sum \limits _{m = 0}^\infty {\frac{{\left( { - \lambda } \right) ^m \varepsilon ^{m + 1} }}{{m!\kappa ^{m + 1} }}} G_{1,3}^{2,1} \left( {\varepsilon \left| {\begin{array}{*{20}c} {2 - m} \\ {2,0,1 - m} \\ \end{array}} \right. } \right) , \\ \end{aligned}$$

(27)

where (a) is fulfilled by using the M-th order Taylor series approximation of the exponential function in [24, eq. (1.211.1)]; step (b) can be expressed by using eq. [24, eq. (6.592.2)]. Finally, substituting (26) and (27) into (25) the Lemma 1 is derived.

Appendix 2: Proof of Proposition 1

Putting $X = {\left| {{h_i}} \right| ^2}$ , $Y = {\left| {{h_j}} \right| ^2}$ and setting $U = \frac{{{{\left| {{h_i}} \right| }^2}}}{{{{\left| {{h_i}} \right| }^2} + {{\left| {{h_j}} \right| }^2}}} = \frac{X}{{X + Y}}$ and $V = {\left| {{h_i}} \right| ^2} + {\left| {{h_j}} \right| ^2} = X + Y$. It can be observed that the quantity of U is between zero and one, i.e. $0 \le U \le 1$. Also, it can be easily shown that $X = UV$ and $Y = V\left( {1 - U} \right)$. Now, we can rewrite the SINRs of $i \rightarrow r$ hop and $r \rightarrow j$ as below

$$\begin{aligned} {\gamma _{i \rightarrow r}} = \frac{{{\gamma _s}{{\left| {{h_i}} \right| }^2}}}{{\mu {\gamma _s}\left( {{{\left| {{h_i}} \right| }^2} + {{\left| {{h_j}} \right| }^2}} \right) {{\left| {{f_r}} \right| }^2} + 1}} \approx \frac{U}{{\mu {{\left| {{f_r}} \right| }^2}}} = \gamma _{i \rightarrow r} ^ {\mathrm{{asym}}}, \end{aligned}$$

(28)

and

$$\begin{aligned} {\gamma _{r \rightarrow j}} = \frac{{\mu {\gamma _s}V{{\left| {{g_j}} \right| }^2}}}{{{\gamma _s}{{\left| {{f_j}} \right| }^2} + 1}}. \end{aligned}$$

(29)

It is noted that, in this analyzing performance, we consider the asymptotic SINR of $i \rightarrow r$ hop. Hence, this CDF expression is lower bounded.

The Jacobian of the transformation from $\left( {U,V} \right)$ back to $\left( {X,Y} \right)$ can be computed as

$$\left| J \right| = \left| {\begin{array}{ll} {\frac{{\partial X\left( {u,v} \right) }}{{\partial u}}}& \quad {\frac{{\partial X( {u,v} )}}{{\partial v}}}\\ {\frac{{\partial Y( {u,v} )}}{{\partial u}}}& \quad {\frac{{\partial Y( {u,v} )}}{{\partial v}}} \end{array}} \right| = \left| {\begin{array}{ll} v& \quad u\\ { - v}& \quad {1 - u} \end{array}} \right| = v.$$

(30)

Since assumed that ${\lambda _{h_a}} = {\lambda _{h_b}} = \lambda _h$ the joint PDF of two R.Vs, X and Y, can be revealed as

$$\begin{aligned} {f_{X,Y}}\left( {x,y} \right) = {f_X}\left( x \right) \times {f_Y}\left( y \right) = \frac{1}{{{\lambda _h ^2}}}\exp \left( { - \frac{{x + y}}{\lambda _h }} \right) . \end{aligned}$$

(31)

Since the transformation is invertible, applying the change of variable formula, yield

$$\begin{aligned} {f_{U,V}}\left( {u,v} \right) = \left| J \right| {f_{X,Y}}\left( {u,v} \right) = \frac{v}{{{\lambda _h ^2}}}\exp \left( { - \frac{v}{\lambda _h }} \right) . \end{aligned}$$

(32)

And the PDF of V is determined as

$$\begin{aligned} \begin{aligned} {f_V}\left( v \right)&= \int \limits _0^1 {{f_{U,V}}\left( {u,v} \right) du} \\&= \frac{v}{{{\lambda _h ^2}}}\exp \left( { - \frac{v}{\lambda _h }} \right) . \end{aligned} \end{aligned}$$

(33)

Similarly, the PDF of U is given as

$$\begin{aligned} \begin{aligned} {f_U}\left( u \right)&= \int \limits _0^\infty {{f_{U,V}}\left( {u,v} \right) dv} \\&= \int \limits _0^\infty {\frac{v}{{{\lambda _h ^2}}}\exp \left( { - \frac{v}{\lambda _h }} \right) dv} = 1. \end{aligned} \end{aligned}$$

(34)

Then substituting (28) and (29) into (18), we get

$$\begin{aligned} {F_{{\gamma _i}}}\left( t \right)&= \Pr \left\{ {\min \left( {\frac{U}{{\mu {{\left| {{f_r}} \right| }^2}}},\frac{{\mu {\gamma _s}V {{\left| {{g_i}} \right| }^2}}}{{{\gamma _s}{{\left| {{f_i}} \right| }^2} + 1}}} \right)< t} \right\} \\&= 1 - \left[ {1 - \underbrace{\Pr \left\{ {\frac{U}{{\mu {{\left| {{f_r}} \right| }^2}}}< t} \right\} }_{{J_1}}} \right] \times \left[ {1 - \underbrace{\Pr \left\{ {\frac{{\mu {\gamma _s}V{{\left| {{g_i}} \right| }^2}}}{{{\gamma _s}{{\left| {{f_i}} \right| }^2} + 1}} < t} \right\} }_{{J_2}}} \right] . \end{aligned}$$

(35)

First, we evaluate

$$\begin{aligned} \begin{aligned} {J_1}&= \Pr \left\{ {\frac{U}{{\mu {{\left| {{f_r}} \right| }^2}}} < t} \right\} \\&= \int \limits _0^1 {\exp \left( {\frac{{ - u}}{{\mu {\lambda _{{f_r}}}t}}} \right) du} \\&= \mu {\lambda _{{f_r}}}t\left( {1 - \exp \left( {\frac{{ - 1}}{{\mu {\lambda _{{f_r}}}t}}} \right) } \right) . \end{aligned} \end{aligned}$$

(36)

In order to compute the second term in (35), we first put $Z = V{\left| {{g_i}} \right| ^2}$. The CDF of variable Z can be calculated as

$$\begin{aligned} {F_Z}\left( z \right)&= \Pr \left\{ {V{{\left| {{g_i}} \right| }^2} < z} \right\} = \int \limits _0^\infty {{F_{{{\left| {{g_i}} \right| }^2}}} \left( {\frac{z}{v}} \right) {f_V}\left( v \right) dv} \\&= \int \limits _0^\infty {\left\{ {1 - \exp \left( { - \frac{z}{{{\lambda _g}v}}} \right) } \right\} \frac{v}{{\lambda _h^2}}\exp \left( { - \frac{v}{{{\lambda _h}}}} \right) } dx\\&= 1 - \frac{{2z}}{{{\lambda _g}{\lambda _h}}}{\mathrm{K}_2}\left( {2\sqrt{\frac{z}{{{\lambda _g}{\lambda _h}}}} } \right) . \end{aligned}$$

(37)

Now, the second term in (35), $J_2$, can be computed as

$$\begin{aligned} \begin{aligned} {J_2}&= \Pr \left\{ {\frac{{\mu {\gamma _s}Z}}{{{\gamma _s} {{\left| {{f_i}} \right| }^2} + 1}} < t} \right\} \\&= \int \limits _0^\infty {{F_Z}\left( {\frac{{t{\gamma _s}x + t}}{{\mu {\gamma _s}}}} \right) {f_{{{\left| {{f_i}} \right| }^2}}}\left( x \right) dx} \\&= \frac{1}{{{\lambda _{{f_i}}}}}\int \limits _0^\infty {\left\{ {1 - \frac{2}{{{\lambda _g}{\lambda _h}}}\left( {\frac{{t{\gamma _s}x + t}}{{\mu {\gamma _s}}}} \right) {\mathrm{K}_2} \left( {2\sqrt{\frac{1}{{{\lambda _g}{\lambda _h}}}\left( {\frac{{t{\gamma _s}x + t}}{{\mu {\gamma _s}}}} \right) } } \right) } \right\} \exp \left( {\frac{{ - x}}{{{\lambda _{{f_i}}}}}} \right) dx} \\&= 1 - \frac{1}{{{\lambda _{{f_i}}}}} \psi \left( {\frac{1}{{{\lambda _{{f_i}}}}},\frac{t}{{\mu {\lambda _g}{\lambda _h}}},\frac{t}{{\mu {\gamma _s}{\lambda _g}{\lambda _h}}}} \right) , \end{aligned} \end{aligned}$$

(38)

where the last equality can be obtained by applying Lemma 1. Finally, substituting (36) and (38) into (35) and some simple manipulation, the Proposition 1 is derived exactly.