Performance of energy-efficient cooperative cognitive radio system over erroneous Nakagami-m and Weibull fading channels

Abstract

Cognitive radio (CR) is developed as one of the important techniques to improve the utilization of the radio spectrum. A CR node shares the radio spectrum with a licensed primary user opportunistically. In this paper, we study the performance of an energy-efficient cooperative cognitive radio system (CCRS) in the presence of noise plus different fading conditions under channel error constraints. Two fading environments namely, Nakagami-m and Weibull fading are considered and performance characteristics are evaluated. More precisely, Every CR node uses the similar energy detectors to sense a primary user, and forwards their knowledge to fusion center (FC) as one bit binary information. Different hard-decision fusion operations are carried out at FC to take the global decision about the status of the primary user. The analytical and simulation models related to noise and fading for calculating missdetection performance, total error performance, and sensing time and throughput/energy efficiency trade-off are developed. Analytical frameworks are validated via computer based simulations. For comparison purposes, we also investigate the performance of CCRS in Rayleigh fading with and without channel errors constraints. Further, the performance CCRS is investigated for various hard-decision fusion rules under several parameters of the system. Finally, an optimal values of the number of CR nodes and sensing threshold for different fusion rules, and sub-optimal hard-decision fusion rule are also investigated. The energy efficiency is enhanced by 19% when CCRS is operated over Nakagami-m fading environment as compared to Rayleigh fading environment for a fixed parameter values of the network.

Appendix

1.1 Proof of (10)

The proof of (10) is provided here. From (7), ${\mathcal {Q}}$-function can be rewritten as ${\mathcal {Q}}(x) = \frac{1}{2} erfc \left( {\frac{x}{{\sqrt{2} }}} \right)$, where erfc(.) represents complementary error function and can be expressed as [32]:

$$erfc(x) = 1 - \frac{2}{{\sqrt \pi }}\sum\limits_{{z = 0}}^{\infty } {\frac{{( - 1)^{z} x^{{2z + 1}} }}{{2^{{z + \frac{1}{2}}} z!\left( {2z + 1} \right)}}} $$

(33)

Using (33), ${\mathcal {Q}}$-function can be obtained as [32]:

$${\mathcal{Q}}(t) = \frac{1}{2} - \frac{1}{{\sqrt{\pi }}}\sum\limits_{{z = 0}}^{\infty } {\frac{{( - 1)^{z} t^{{2z + 1}} }}{{2^{{z + \frac{1}{2}}} z!\left( {2z + 1} \right)}}}.$$

(34)

where $t={\left[ {\frac{(\lambda /\sigma _u^2)}{ {1 + \gamma }} - 1} \right] \sqrt{\frac{\tau _s {f_s}}{2}} }$. Further simplification of (8) after inserting (9) and (34) in (8), $\bar{P}_d$ can be expressed as

$$\begin{aligned} {{\bar{P}}_{d,i}^{Naka}}&= \frac{1}{2} \left( {\frac{m}{{\bar{\gamma } }}} \right) ^m\frac{1}{\Gamma (m)}\underbrace{\int _0^\infty \gamma ^{m - 1}\exp \left( { - \frac{{m\gamma }}{{\bar{\gamma } }}} \right) d\gamma }_{{I_1}} \\ &\quad-\frac{1}{\sqrt{\pi }} \left( {\frac{m}{{\bar{\gamma } }}} \right) ^m\frac{1}{\Gamma (m)}\sum \limits _{z = 0}^\infty \frac{{{{( - 1)}^z{(\tau _s {f_s} )^{z + \frac{1}{2}}}}}}{{{2^{2z + 1}}(2z + 1)z!}} \\&\quad\underbrace{ \times \int _0^\infty {\left( {\frac{{\lambda /{\sigma ^2}}}{{1 + \gamma }} - 1} \right) ^{2z+1}} \gamma ^{m - 1}\exp \left( { - \frac{{m\gamma }}{{\bar{\gamma } }}} \right) d\gamma }_{{I_2}} \end{aligned}$$

(35)

The solution to the first integral part, $I_1$ in (35) can be obtained using [33, (3.351.3)] as

$$\begin{aligned} I_1 &= \frac{1}{2} \left( {\frac{m}{{\bar{\gamma } }}} \right) ^m\frac{1}{\Gamma (m)}\int _0^\infty \gamma ^{m - 1}\exp \left( { - \frac{{m\gamma }}{{\bar{\gamma } }}} \right) d\gamma , \\ &= \frac{1}{2}. \end{aligned}$$

(36)

The solution to second integral part, $I_2$ in (35) can be obtained as follows, from $I_2$

$$\begin{aligned} {\left( {\frac{{\lambda /\sigma _u^2}}{{1 + \gamma }} - 1} \right) ^{2z + 1}} = \sum \limits _{m = 0}^{2z + 1} {{{( - 1)}^{2z + m + 1}}} \left( {\begin{array}{c}2z+1\\ m\end{array}}\right) {\left( {\frac{{\lambda /\sigma _u^2}}{{1 + \gamma }}} \right) ^m}. \end{aligned}$$

(37)

Then $I_2$ can be rewritten as

$$\begin{aligned} {I_2}&= \sum \limits _{m = 0}^{2z + 1} {{{( - 1)}^{2z + m + 1}}} \left( {\begin{array}{c}2z+1\\ m\end{array}}\right) \\ &\quad\times \int _0^\infty {{{\left( {\frac{{\lambda /\sigma _u^2}}{{1 + \gamma }}} \right) }^m}} \gamma ^{m - 1}\exp \left( { - \frac{{m\gamma }}{{\bar{\gamma } }}} \right) d\gamma \end{aligned}$$

(38)

The resultant integral in (38) is difficult to solve because of the complexity involved and an alternate approach (variable substitution method [34, 35]) is adopted here by substituting $\gamma$ with $\tan \theta$ and $d{\gamma }$ with ${{\sec }^2}\theta d\theta$ which gives an integral with definite limits from the integral with infinite limits and can go for numerical evaluation. Based on this method, (38) can be rewritten as

$$\begin{aligned} {I_2}&= \sum \limits _{m = 0}^{2z + 1} {{{( - 1)}^{2z + m + 1}}} \left( {\begin{array}{c}2z+1\\ m\end{array}}\right) \\ &\quad \times \int _0^{\pi /2} {{{\left( {\frac{{\lambda /\sigma _u^2}}{{1 + \tan \theta }}} \right) }^m}} {\tan \theta } ^{m - 1} \\ &\quad\exp \left( { - \frac{{m\tan \theta }}{{\bar{\gamma } }}} \right) {{\sec }^2}\theta d\theta . \end{aligned}$$

(39)

1.2 Proof of (12)

The proof of (12) is provided here. The expression for ${{\bar{P}}_{d,i}^{Wieb}}$ at ith CR node can be derived by substituting (7) and (11) in (8):

$$\begin{aligned} {{\bar{P}}_{d,i}^{Wieb}}&= \frac{v}{2}{\left[ {\frac{{\Gamma (p)}}{{\bar{\gamma } }}} \right] ^{\frac{v}{2}}}\underbrace{\int _0^\infty {\gamma ^{\frac{v}{2} - 1}}\exp \left[ { - {{\left\{ {\frac{{\gamma \Gamma (p)}}{{\bar{\gamma } }}} \right\} }^{\frac{v}{2}}}} \right] d\gamma }_{{I_1}} \\ &\quad-\frac{v}{2}{\left[ {\frac{{\Gamma (p)}}{{\bar{\gamma } }}} \right] ^{\frac{v}{2}}}\sum \limits _{z = 0}^\infty \frac{{{{( - 1)}^z{(\tau _s {f_s} )^{z + \frac{1}{2}}}}}}{{{2^{2z + 1}}(2z + 1)z!}} \\ &\quad\underbrace{ \times \int _0^\infty {{{\left( {\frac{{\lambda /\sigma _u^2}}{{1 + \gamma }}} \right) }^m}} {\gamma ^{\frac{v}{2} - 1}}\exp \left[ { - {{\left\{ {\frac{{\gamma \Gamma (p)}}{{\bar{\gamma } }}} \right\} }^{\frac{v}{2}}}} \right] d\gamma }_{{I_2}} \end{aligned}$$

(40)

The solution to the first integral part, $I_1$ in (40) can be obtained using [33, (3.326.2)] as

$$\begin{aligned} I_1 &= \frac{v}{2}{\left[ {\frac{{\Gamma (p)}}{{\bar{\gamma } }}} \right] ^{\frac{v}{2}}} {\int _0^\infty {\gamma ^{\frac{v}{2} - 1}}\exp \left[ { - {{\left\{ {\frac{{\gamma \Gamma (p)}}{{\bar{\gamma } }}} \right\} }^{\frac{v}{2}}}} \right] d\gamma }, \\ &= \frac{1}{2}. \end{aligned}$$

(41)

The $I_2$ part can be modified as

$$\begin{aligned} {I_2}&= \sum \limits _{m = 0}^{2z + 1} {{{( - 1)}^{2z + m + 1}}} \left( {\begin{array}{c}2z+1\\ m\end{array}}\right) \\ &\quad\times \int _0^\infty {{{\left( {\frac{{\lambda /\sigma _u^2}}{{1 + \gamma }}} \right) }^m}} {\gamma ^{\frac{v}{2} - 1}}\exp \left[ { - {{\left\{ {\frac{{\gamma \Gamma (p)}}{{\bar{\gamma } }}} \right\} }^{\frac{v}{2}}}} \right] d\gamma \end{aligned}$$

(42)

The integral part in (42) is not easy to solve, following the same variable substitution approach presented in Sect. 2.1, an alternative expression for (42) can be obtained as

$$\begin{aligned} {I_2}&= \sum \limits _{m = 0}^{2z + 1} {{{( - 1)}^{2z + m + 1}}} \left( {\begin{array}{c}2z+1\\ m\end{array}}\right) \\&\quad\times \int _0^{\pi /2} {{{\left( {\frac{{\lambda /\sigma _u^2}}{{1 + \tan \theta }}} \right) }^m}} {\tan \theta ^{\frac{v}{2} - 1}} \\ &\quad\exp \left[ { - {{\left\{ {\frac{{\tan \theta \Gamma (p)}}{{\bar{\gamma } }}} \right\} }^{\frac{v}{2}}}} \right] {{\sec }^2}\theta d\theta . \end{aligned}$$

(43)