Energy Harvesting for Cooperative Cognitive Radio Networks

Abstract

In this paper, we analyze the performance of cooperative cognitive radio networks where the secondary nodes harvest energy from radio frequency signals. Our analysis takes into interference aspect: the secondary source and relays transmit only when they generate low interference to primary receiver ($P_R$). Besides, we analyze the signal to interference plus noise ratio at secondary relays and destination taking into consideration primary interference. To reach higher data rates, harvesting duration is optimized in this paper.

Appendices

Appendix 1

Let $X_{1}$ and $X_{2}$ be exponential r.v. The CDF of $X=X_{1}X_{2}$ is given by

$$\begin{aligned} P_{X}(x)=P(X_{1}X_{2}\le x)=\int _{0}^{+\infty }P\left( X_{1}\le \frac{x}{y} \right) \sigma _{2}e^{-\sigma _{2}y}dy. \end{aligned}$$

(51)

We deduce

$$\begin{aligned} P_{X}(x)&= \int _{0}^{+\infty }\left[ 1-e^{-\sigma _{1}\frac{x}{y}}\right] \sigma _{2}e^{-\sigma _{2}y}dy \nonumber \\&= 1-\int _{0}^{+\infty }e^{-\sigma _{1}\frac{x}{y}}\sigma _{2}e^{-\sigma _{2}y}dy \end{aligned}$$

(52)

We have

$$\begin{aligned} \int _{0}^{+\infty }e^{-\frac{c}{y}}e^{-\frac{y}{d}}dy=2\sqrt{\frac{c}{d}} K_{1}\left( 2\sqrt{\frac{c}{d}}\right) . \end{aligned}$$

(53)

We use (52) and (53) with $c=\sigma _{1}x$ and $d=\frac{1}{ \sigma _{2}}$, we obtain

$$\begin{aligned} P_{X}(x)=1-2\sqrt{\sigma _{1}\sigma _{2}x}K_{1}\left( 2\sqrt{\sigma _{1}\sigma _{2}x}\right) . \end{aligned}$$

(54)

We deduce the PDF

$$\begin{aligned} p_{X}(x)=-\frac{\sqrt{\sigma _{1}\sigma _{2}}}{\sqrt{x}}K_{1}\left( 2\sqrt{ \sigma _{1}\sigma _{2}x}\right) -2\sqrt{\sigma _{1}\sigma _{2}x}K_{1}^{{\prime }}\left( 2\sqrt{\sigma _{1}\sigma _{2}x}\right) \frac{\sqrt{\sigma _{1}\sigma _{2}}}{ \sqrt{x}}. \end{aligned}$$

(55)

Using

$$\begin{aligned} K_{1}^{^{\prime }}(z)=-K_{0}(z)-\frac{K_{1}(z)}{z}, \end{aligned}$$

(56)

we obtain

$$\begin{aligned} p_{X}(x)=2\sigma _{1}\sigma _{2}K_{0}\left( 2\sqrt{\sigma _{1}\sigma _{2}x}\right) . \end{aligned}$$

(57)

Appendix 2

$$\begin{aligned} P\left( \Gamma _{SR_{k}}\le x\right) =P\left( \frac{eX_{4}X_{5}}{f+X_{3}}\le x\right) =P\left( X_{4}X_{5}\le x\frac{X_{6}}{e}\right) , \end{aligned}$$

(58)

where

$$\begin{aligned} X_{6}=f+X_{3}. \end{aligned}$$

(59)

Since $X_{3}$ is exponential r.v., the PDF of $X_{6}$ is given by as

$$\begin{aligned} f_{X_{6}}(u)=\alpha _{3}e^{-\alpha _{3}(u-f)},u\ge f \end{aligned}$$

(60)

Therefore, we have

$$\begin{aligned} P\left( \Gamma _{SR_{k}}\le x\right) =\int _{f}^{+\infty }P\left( X_{4}X_{5}\le x\frac{y}{ e}\right) f_{X_{6}}(y)dy, \end{aligned}$$

(61)

Using the results of “Appendix 1”, we have

$$\begin{aligned} P\left( \Gamma _{SR_{k}} \le x\right)&= \int _{f}^{+\infty }\left[ 1-2\sqrt{\alpha _{4}\alpha _{5}x\frac{y}{e}}K_{1}\left( 2\sqrt{\alpha _{4}\alpha _{5}x\frac{y }{e}}\right) \right] \alpha _{3}e^{-\alpha _{3}(y-f)}dy, \nonumber \\&= 1-\alpha _{3}e^{\alpha _{3}f}2\sqrt{\alpha _{4}\alpha _{5}\frac{x}{e}} \int _{f}^{+\infty }\sqrt{y}K_{1}\left( 2\mu \sqrt{y}\right) e^{-\alpha _{3}y}dy \nonumber \\&= 1-\alpha _{3}e^{\alpha _{3}f}2\sqrt{\alpha _{4}\alpha _{5}\frac{x}{e}} \int _{0}^{+\infty }\sqrt{y}K_{1}\left( 2\mu \sqrt{y}\right) e^{-\alpha _{3}y}dy \nonumber \\&+\alpha _{3}e^{\alpha _{3}f}2\sqrt{\alpha _{4}\alpha _{5}\frac{x}{e}} \int _{0}^{f}\sqrt{y}K_{1}\left( 2\mu \sqrt{y}\right) e^{-\alpha _{3}y}dy \end{aligned}$$

(62)

where

$$\begin{aligned} \mu =\sqrt{\frac{\alpha _{4}\alpha _{5}x}{e}} \end{aligned}$$

We have

$$\begin{aligned} \int _{0}^{+\infty }\sqrt{y}K_{1}\left( 2\mu \sqrt{y}\right) e^{-\alpha _{3}y}dy=\frac{e^{ \frac{\mu ^{2}}{2\alpha _{3}}}}{2\mu \alpha _{3}}W_{-1,0.5}\left( \frac{\mu ^{2}}{ \alpha _{3}}\right) , \end{aligned}$$

(63)

where $W_{\mu ,\nu }(x)$ is the Whittaker function.

The CDF of $\Gamma _{SR_{k}}$ is given by

$$\begin{aligned} F_{\Gamma _{SR_{k}}}(x)&= 1-\alpha _{3}e^{\alpha _{3}f}2\sqrt{ \alpha _{4}\alpha _{5}\frac{x}{e}}\frac{e^{\frac{\mu ^{2}}{2\alpha _{3}}}}{ 2\mu \alpha _{3}}W_{-1,0.5}\left( \frac{\mu ^{2}}{\alpha _{3}}\right) \nonumber \\&+\alpha _{3}e^{\alpha _{3}f}2\sqrt{\alpha _{4}\alpha _{5}\frac{x}{e}} \int _{0}^{f}\sqrt{y}K_{1}\left( 2\mu \sqrt{y}\right) e^{-\alpha _{3}y}dy. \end{aligned}$$

(64)