Routing with Energy Harvesting and Adaptive Transmit Power for Cognitive Radio Networks

Abstract

In this article, we suggest routing protocols with Energy Harvesting and adaptive transmit power for cognitive radio networks. the secondary source and relays harvest energy from wireless signal transmitted by node A. the transmitted power of secondary nodes is adapted so that interference to primary receiver ($P_R$) lower than interference threshold I. We suggest optimal routing that activates the best path between source and destination. Suboptimal routing is also considered where the network is decomposed in many subnetworks then the best path is activated in each subnetwork. One hop routing is also investigated where the best relay is selected in each subnetwork.

Appendices

Appendix 1: Proof of Proposition 1, CDF of Ratio of Two Exponential r.v.

Let X and Y be two exponential r.v. with respective mean 1/a and 1/b. Let $Z=\frac{X}{Y}$, we have

$$\begin{aligned} P(Z\le t)=P\left( Y\ge \frac{X}{t}\right) . \end{aligned}$$

Since $P(Y\ge y)=e^{-by}$, we deduce

$$\begin{aligned} P(Z\le t)=E(e^{-b\frac{X}{t}})=\int _{0}^{+\infty }ae^{-ax}e^{-\frac{bx}{t} }dx=\frac{at}{at+b}. \end{aligned}$$

Appendix 2: Proof of Proposition 2, PDF and CDF of Product of Two Exponential Random Variables

Let $Y_{1}$ and $Y_{2}$ be two exponential r.v. with respective mean 1/$\lambda _{1}$ and 1/$\lambda _{2}.$

The CDF of the product of two exponential rv $Y=Y_{1}Y_{2}$ is given by

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

(36)

We deduce

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

(37)

We have

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

(38)

We use (31) and (32) with $c=\lambda _{1}x$ and $d=\frac{1}{ \lambda _{2}}$, we obtain

$$\begin{aligned} P_{Y}(x)=1-2\sqrt{\lambda _{1}\lambda _{2}x}K_{1}(2\sqrt{\lambda _{1}\lambda _{2}x}). \end{aligned}$$

(39)

Taking the derivative of CDF, we obtain the PDF as

$$\begin{aligned} p_{Y}(x)=-\frac{\sqrt{\lambda _{1}\lambda _{2}}}{\sqrt{x}}K_{1}(2\sqrt{ \lambda _{1}\lambda _{2}x})-2\sqrt{\lambda _{1}\lambda _{2}x}K_{1}^{^{\prime }}(2\sqrt{\lambda _{1}\lambda _{2}x})\frac{\sqrt{\lambda _{1}\lambda _{2}}}{ \sqrt{x}}. \end{aligned}$$

(40)

Using

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

(41)

we obtain

$$\begin{aligned} p_{Y}(x)=2\lambda _{1}\lambda _{2}K_{0}(2\sqrt{\lambda _{1}\lambda _{2}x}), \end{aligned}$$

(42)

Appendix 3

We have

$$\begin{aligned} P\left( \gamma _{SR_{1,k}}^{(1)}<x\right) =P\left( \frac{Y_{1}}{Y_{2}}< \frac{x}{a}(b+cY_{3})\right) . \end{aligned}$$

(43)

Let $Y_{4}=\frac{x}{a}(b+cY_{3}),$ we have

$$\begin{aligned} P(Y_{4}<u)=P\left( Y_{3}<\frac{a}{cx}u-\frac{b}{c}\right) =F_{Y_{3}}\left( \frac{a}{cx}u- \frac{b}{c}\right) \end{aligned}$$

(44)

We deduce the Probability Density Function (PDF) of $Y_{4}$

$$\begin{aligned} f_{Y_{4}}(u)=\frac{a}{cx}f_{Y_{3}}\left( \frac{a}{cx}u -\frac{b}{c}\right) \end{aligned}$$

(45)

Therefore, we can write

$$\begin{aligned} P\left( \gamma _{SR_{1,k}}^{(1)}<x\right)&= \int _{\frac{bx}{a}}^{+\infty }P \left( \frac{Y_{1}}{Y_{2}}<u\right) \,f_{Y_{4}}(u)du \nonumber \\&= \int _{\frac{bx}{a}}^{+\infty }P\left( \frac{Y_{1}}{Y_{2}}<u\right) \frac{a\lambda _{3}}{cx}e^{-\frac{a\lambda _{3}}{cx}u}e^{\frac{b \lambda _{3}}{c}}du \end{aligned}$$

(46)

where

$$\begin{aligned} \lambda _{i}=\frac{1}{E(Y_{i})} \end{aligned}$$

(47)

Using the results of “Appendix 1”, we can write

$$\begin{aligned} P\left( \gamma _{SR_{1,k}}^{(1)} <x\right)&= \frac{a\lambda _{3}}{cx}e^{\frac{b\lambda _{3}}{ c}}\int _{\frac{bx}{a}}^{+\infty }\frac{\lambda _{1}u}{\lambda _{1}u+\lambda _{2} }e^{-\frac{a\lambda _{3}}{cx}u}du \nonumber \\&= 1-\frac{a\lambda _{3}\lambda _{2}}{cx}e^{\frac{b\lambda _{3}}{c}}\int _{\frac{ bx}{a}}^{+\infty }\frac{1}{\lambda _{1}u+\lambda _{2}}e^{-\frac{a\lambda _{3}}{ cx}u}du \end{aligned}$$

(48)

Let $v=\lambda _{1}u+\lambda _{2},$ we deduce

$$\begin{aligned} P\left( \gamma _{SR_{1,k}}^{(1)} <x\right)&= 1-\frac{a\lambda _{3}\lambda _{2}}{cx\lambda _{1} }e^{\frac{b\lambda _{3}}{c}}\int _{\frac{b\lambda _{1}x}{a}+\lambda _{2}}^{+\infty }\frac{1}{v}e^{-\frac{a\lambda _{3}}{cx}(\frac{v-\lambda _{2}}{ \lambda _{1}})}dv \nonumber \\&= 1-\frac{a\lambda _{3}\lambda _{2}}{cx\lambda _{1}}e^{\frac{b\lambda _{3}}{c} }E_{i}\left( \left( \frac{b\lambda _{1}x}{a}+\lambda _{2}\right) \frac{a\lambda _{3}}{cx}\right) e^{\frac{ a\lambda _{3}}{cx}\frac{\lambda _{2}}{\lambda _{1}}} \end{aligned}$$

(49)

where $E_{i}(x)$ is the exponential integral function defined as

$$\begin{aligned} E_{i}(x)=\int _{x}^{+\infty }\frac{e^{-u}}{u}du. \end{aligned}$$

(50)

Appendix 4

$$\begin{aligned} P\left( \gamma _{SR_{1,k}}^{(2)}\le x\right) =P\left( \frac{eY_{1}Y_{5}}{f+Y_{3}}\le x\right) =P\left( Y_{1}Y_{5}\le x\frac{Y_{6}}{e}\right) , \end{aligned}$$

(51)

where

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

(52)

Since $Y_{3}$ is exponentially distributed, the PDF of $Y_{6}$ is expressed as

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

(53)

Therefore, we have

$$\begin{aligned} P\left( \gamma _{SR_{1,k}}^{(2)}\le x\right) =\int _{f}^{+\infty }P\left(Y_{1}Y_{5}\le x\frac{y }{e}\right)f_{Y_{6}}(y)dy, \end{aligned}$$

(54)

Using “Appendix 2”, we have

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

(55)

We use the following result

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

(56)

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

We finally obtain

$$\begin{aligned} P\left( \gamma _{SR_{1,k}}^{(2)} \le x\right)&= 1-\lambda _{3}e^{\lambda _{3}f}2\sqrt{ \lambda _{1}\lambda _{5}\frac{x}{e}}\frac{e^{\frac{\beta ^{2}}{2\lambda _{3}}}}{ 2\beta \lambda _{3}}W_{-1,0.5}\left( \frac{\beta ^{2}}{\lambda _{3}}\right) \nonumber \\&\quad +\,\lambda _{3}e^{\lambda _{3}f}2\sqrt{\lambda _{1}\lambda _{5}\frac{x}{e}} \int _{0}^{f}\sqrt{y}K_{1}(2\beta \sqrt{y})e^{-\lambda _{3}y}dy. \end{aligned}$$

(57)

where the last term is computed using MATLAB.