Performance Analysis of MIMO SWIPT Relay Network with Imperfect CSI

Abstract

In this paper, we consider a dual-hop multiple-input multiple-output (MIMO) simultaneous wireless information and power transfer (SWIPT) relay network, where the source node (SN) uses a transmission antenna selection (TAS) scheme to concurrently send information and energy to the single-antenna relay node (RN). This helps to utilize the harvested energy to forward the received signal. In addition, the destination node (DN) employs the maximum ratio combining (MRC) scheme to process this forwarded signal. The performance of this MIMO SWIPT relay system is investigated in imperfect channel state information (CSI) condition. Specifically, we derive the exact and approximate closed-form expressions for the outage probability, the average capacity, and the symbol error probability (SEP). This is the first time the exact and approximate formulas for the SEP of the energy harvesting networks are investigated. The Monte Carlo simulation results are provided to demonstrate the relevance of the developed analytical results, showing that the system’s performance is significantly impacted by the CSI imperfection, the number of antennas, and the energy harvesting duration.

Appendices

Appendix A

Aiming at the closed-form formulations for $I_{2}^{\text {Exact}}$ and $I_{2}^{\text {Appro}}$, we start by determining the corresponding constituent integral as follows: Substituting (23)–(29) and carrying out some manipulations, we have

$$\begin{array}{@{}rcl@{}} I_{2}^{\text{Appro}} &= &\frac{{a\sqrt b} }{{2\sqrt \pi} }{\sum\limits_{n = 1}^{{N_{S}}} {\sum\limits_{k = 0}^{N_{\mathrm{D}} - 1} \binom{N_{\text{S}}}{n}}} \frac{{n{{\left( { - 1} \right)}^{n + 1}}}}{{{\lambda_{1}}\delta} }2{\left( {\frac{{\phi {\lambda_{1}}\delta} }{{n{\lambda_{2}}}}} \right)^{\frac{{1 - k}}{2}}}\\ && \times \int\limits_{0}^{\infty} {\frac{{{e^{- bx}}}}{{\sqrt x} }{x^{\frac{{1 - k}}{2}}} {{\left( {\frac{{\phi x}}{{{\lambda_{2}}}}} \right)}^{k}}{ {\mathcal{K}}_{1 - k}}\left( {2\sqrt {\frac{{n\phi x}}{{{\lambda_{1}}{\lambda_{2}}\delta} }}} \right)dx}.\\ \end{array} $$

(39)

With the help of [32, 6.643.3], $I_{2}^{\text {Appro}}$ is then obtained as in Eq. 32.

Similarly, for the case of $I_{2}^{\text {Exact}}$, substituting the exact CDF in Eqs. 24–29, it hence is recast as

$$\begin{array}{@{}rcl@{}} {I_{2}^{\text{Exact}}} &= &\frac{{a\sqrt b} }{{2\sqrt \pi} }\!\!\!\int\limits_{0}^{\infty}{\frac{{{e^{- bx}}}}{{\sqrt x} }\sum\limits_{n = 1}^{{N_{S}}}{\sum\limits_{k = 0}^{N_{\mathrm{D}} - 1} {\sum\limits_{q = 0}^{{\mathcal{N}}_{t}}{\frac{{n{{\left( { - 1} \right)}^{n + 1}}}}{{q!k!}}{{\left( {\frac{\phi} {{{{\Omega}_{2}}}}} \right)}^{k + q}}}} } } \\ && \times \binom{N_{\text{S}}}{n} \frac{{{P_{S}}^{k + q - 1}{{\left( { - 1} \right)}^{q}}}}{{{{\Omega}_{1}}\delta} }x{E_{k + q}}\left( {\frac{{nx}}{{{{\Omega}_{1}}\delta {P_{S}}}}} \right)dx.\\ \end{array} $$

(40)

With the help of [42, 5.1.45], and using ${E_{n}}\left (z \right ) = {z^{n - 1}}{\Gamma } \left ({1 - n, z} \right )$, $I_{2}^{\text {Exact}}$ can then be recast as

$$\begin{array}{@{}rcl@{}} {I_{2}^{\text{Exact}}}& = &\frac{{a\sqrt b} }{{2\sqrt \pi} }\sum\limits_{n = 1}^{N_{\mathrm{S}}} {\sum\limits_{k = 0}^{N_{\mathrm{D}} - 1} {\sum\limits_{q = 0}^{{\mathcal{N}}_{t}}{\frac{{n{{\left( { - 1} \right)}^{n + 1}}{{\left( { - 1} \right)}^{q}}}}{{q!k!}}}} } \\ &&\times \binom{N_{\text{S}}}{n}\frac{{1}}{{{{\Omega}_{1}}\delta} }{\left( {\frac{\phi} {{{{\Omega}_{2}}}}} \right)^{k + q}}{\left( {\frac{n}{{{{\Omega}_{1}}\delta} }} \right)^{k + q - 1}} \\ &&\times \underbrace {\int\limits_{0}^{\infty} {{e^{- bx}}{x^{k + q + \frac{1}{2} - 1}}{\Gamma} \left( {1 - k - q,\frac{{nx}}{{{{\Omega}_{1}}\delta {P_{S}}}}} \right)dx} }_{{\mathcal{I}}\left( x \right)}.\\ \end{array} $$

(41)

For calculating ${\mathcal {I}}\left (x \right )$, [32, 6.455.1] is applied, and after some manipulations, we have

$$ {\mathcal{I}}\left( x \right) = \frac{{{\alpha^{v}}{\Gamma} \left( {\mu + v} \right)}}{{\mu {{\left( {\alpha + \beta} \right)}^{\mu + v}}}}{}_{2}{F_{1}}\left( {1, \mu + v;\mu + 1; \frac{\beta} {{\alpha + \beta} }} \right). $$

(42)

Using the [32, 3.361.2], we then obtain the closed-form formulation of ${I_{2}^{\text {Exact}}}$.

Appendix B

Starting from statistical expectation formulation ${\mathbb {E}}\{\gamma _{\text {e2e}}\} = \int \limits _{0}^{\infty } \left [1- F_{\gamma _{\text {e2e}}}(x)\right ]dx$ with the CDF of (23), after some manipulations, we can rewrite ${\mathbb {E}}\{\gamma _{\text {e2e}}\}$ as

$$\begin{array}{@{}rcl@{}} {\mathbb{E}}\{\gamma_{\text{e2e}}\}&= &{\sum\limits_{n = 1}^{{N_{\text{S}}}} {\sum\limits_{k = 0}^{N_{\mathrm{D}} - 1} {\binom{N_{\text{S}}}{n}\frac{1}{{k!}}}} } \frac{{n{{\left( { - 1} \right)}^{n + 1}}}}{{{\lambda_{1}}\delta} } {\left( {\frac{\phi} {{{\lambda_{2}}}}} \right)^{k}}\\ &&\times 2{\left( {\frac{{\phi {\lambda_{1}}\delta} }{{n{\lambda_{2}}}}} \right)^{\frac{{1 - k}}{2}}}\int\limits_{0}^{\infty} {{x^{\frac{{k + 1}}{2}}}{{\mathcal{K}}_{1 - k}}\left( {\sqrt {\frac{{4n\phi x}}{{{\lambda_{1}}{\lambda_{2}}\delta} }}} \right)} dx.\\ \end{array} $$

(43)

Let $\sqrt x = u$ and using [32, 6.561.16], we then have $\mathbb {E}\{\gamma _{\text {e2e}}\}$ expressed as

$$\begin{array}{@{}rcl@{}} {\mathbb{E}}\left\{ {{\gamma_{{\text{e2e}}}}} \right\} &=&{\sum\limits_{n = 1}^{N_{\mathrm{S}}} {\sum\limits_{k = 0}^{N_{\mathrm{D}} - 1} {\binom{N_{\text{S}}}{n}}} }2{\left( {\frac{{{\lambda_{2}}\delta} }{n}} \right)^{\frac{{1 - k}}{2}}}{2^{k + 2}}{\left( {\frac{{1 - \alpha} }{{2\alpha \eta {P_{S}}}}} \right)^{\frac{{k + 1}}{2}}}\\ && \times {\left( {\frac{{8n{\lambda_{2}}\left( {1 - \alpha} \right)}}{{\delta 2\alpha \eta {P_{S}}}}} \right)^{\frac{{ - 3 -k}}{2}}}{\Gamma} \left( {\frac{1}{2}} \right){\Gamma} \!\left( \!{1 + \frac{1}{2}} \right). \end{array} $$

(44)

Substituting (44)–(36), we then have the average capacity formulation of Eq. 37.

Similarly, the average end-to-end SNR is given by

$$\begin{array}{@{}rcl@{}} {\mathbb{E}}\left\{ {{\gamma_{{\text{e2e}}}}} \right\} &= &\sum\limits_{n = 1}^{{N_{\text{S}}}}{\sum\limits_{k = 0}^{N_{\mathrm{D}} - 1} { \frac{1}{{k!}}}} \frac{{n{{\left( { - 1} \right)}^{n + 1}}}}{{{\lambda_{1}}\delta} }\\ &&\times\sum\limits_{q = 0}^{{\mathcal{N}}_{t}}{\frac{{{{\left( { - 1} \right)}^{q}}{P_{S}}^{k + q - 1}}}{{q!}}} \!\!\!{\left( {\frac{\phi} {{{\lambda_{2}}}}} \right)^{k + q}}\\ &&\times\int\limits_{0}^{\infty} {x{E_{k + q}}\left( {\frac{{nx}}{{{P_{S}}{\lambda_{1}}\delta} }} \right)dx}. \end{array} $$

(45)

With the help of [42, 5.1.45], we can rewrite (45) as

$$\begin{array}{@{}rcl@{}} {\mathbb{E}}\left\{ {{\gamma_{{\text{e2e}}}}} \right\}&=&\sum\limits_{n = 1}^{{N_{\text{S}}}} {\sum\limits_{k = 0}^{N_{\mathrm{D}} - 1} {\sum\limits_{q = 0}^{{\mathcal{N}}_{t}}{\frac{{{{\left( { - 1} \right)}^{q}}n{{\left( { - 1} \right)}^{n - 1}}}}{{{\lambda_{1}}\delta k!q!}}}} } \\ &&{\times\binom{N_{\text{S}}}{n}\left( {\frac{\phi} {{{\lambda_{2}}}}} \right)^{k + q}}{\left( {\frac{n}{{{\lambda_{1}}\delta} }} \right)^{k + q - 1}}\\ &&\times\int\limits_{0}^{\infty} {{x^{k + q}}{\Gamma} \left( {1 - \left( {k + q} \right),\frac{{nx}}{{{P_{S}}{\lambda_{1}}\delta} }} \right)dx}. \end{array} $$

(46)

Thank to the help of [42, 6.5.37], and after some modifications, we have

$$\begin{array}{@{}rcl@{}} \mathbb{E}\{ \gamma_{\text{e2e}}\}& =& \sum\limits_{n = 1}^{N_{\mathrm{S}}}\sum\limits_{k = 0}^{N_{\mathrm{D}} - 1} \binom{N_{\mathrm{S}}}{n}\frac{n(-1)^{n - 1}}{\lambda_{1}\delta k!}\left( \frac{n}{\lambda_{1}\delta} \right)^{- 2} \\ &&\times \sum\limits_{q = 0}^{{\mathcal{N}}_{t}}\frac{(-1)^{q}}{q!(k + q + 1)}\left( \frac{1}{P_{\mathrm{S}}} \right)^{- k - q - 1}\left( \frac{\phi} {\lambda_{2}}\right)^{k + q}.\\ \end{array} $$

(47)

Replacing (47)–(36), we then obtain the average capacity as in Eq. 38.