On the performance of cooperative cognitive networks with proactive relay selection

Abstract

This paper provides a general outage analysis framework for cooperative cognitive networks with proactive relay selection over non-identical Rayleigh fading channels and under both maximum transmit power and interference power constraints. We firstly propose an exact closed-form outage probability expression, which is then exploited for determining the diversity order and coding gain for proactive relay selection scenarios as well as deriving system performance limits at either large maximum transmit power or large maximum interference power. The derived performance metrics bring several insights into system performance behavior without the need of time-consuming Monte-Carlo simulations. Various results confirm the validity of the proposed derivations and show that cooperative cognitive networks with proactive relay selection incur performance saturation and their performance depends considerably on the number of involved relays. In addition, cooperative cognitive networks are significantly better than dual-hop counterparts without any cost of system resources.

Appendices

Appendix 1: Proof of Lemma 1

We rewrite $\mathcal {G}_1$ in (21) as

$$\begin{aligned} {\mathcal {G}_1} = \frac{{{{\uplambda }_{SD}}}}{\tau }\int \limits _{x = \tau }^\infty {\int \limits _{{y_1} = 0}^\infty {\ldots \int \limits _{{y_K} = 0}^\infty {x\prod \limits _{k = 1}^K {\left( {1 + \frac{{\min \left( {y_k^{ - 1}\tau ,1} \right) }}{{{\uplambda }_{S{R_k}}^{ - 1}{{\uplambda }_{{R_k}D}}\tau }}x} \right) } \left( {\prod \limits _{k = 1}^K {\frac{{{{\uplambda }_{{R_k}D}}}}{{\min \left( {y_k^{ - 1}\tau ,1} \right) }}{f_{{Y_k}}}\left( {{y_k}} \right) d{y_k}} } \right) {f_X}\left( x \right) dx} } }. \end{aligned}$$

(26)

Using the equality $\prod \limits _{k = 1}^K {\left( {1 + {a_k}} \right) } = 1 + \sum \limits _{u = 1}^{K - 1} {\sum \limits _{{s_1} = 1}^{K - u + 1} {\sum \limits _{{s_2} = {s_1} + 1}^{K - u + 2} { \cdots \sum \limits _{{s_u} = {s_{u - 1}} + 1}^K {\prod \limits _{l = 1}^u {{a_{{s_l}}}} } } } } + \prod \limits _{k = 1}^K {{a_k}}$, one obtains

$$\begin{aligned} {\mathcal {G}_1} = \frac{{{{\uplambda }_{SD}}}}{\tau }\int \limits _0^\infty {\ldots \int \limits _0^\infty {\left( {{\mathcal {G}_{1\emptyset }} + \sum \limits _{u = 1}^{K - 1} {\sum \limits _{{s_1} = 1}^{K - u + 1} {\sum \limits _{{s_2} = {s_1} + 1}^{K - u + 2} { \cdots \sum \limits _{{s_u} = {s_{u - 1}} + 1}^K {{\mathcal {G}_{1u}}} } } } + {\mathcal {G}_{1K}}} \right) } } \prod \limits _{k = 1}^K {\left( {\frac{{{{\uplambda }_{{R_k}D}}{f_{{Y_k}}}\left( {{y_k}} \right) }}{{\min \left( {y_k^{ - 1}\tau ,1} \right) }}d{y_k}} \right) }, \end{aligned}$$

(27)

where

$$\begin{aligned} {\mathcal {G}_{1i}}= & {} \left\{ {\prod \limits _{m \in {T_i}} {\frac{{{{\uplambda }_{S{R_m}}}}}{{{{\uplambda }_{{R_m}D}}\tau }}\min \left( {\frac{\tau }{{{y_m}}},1} \right) } } \right\} \int \limits _\tau ^\infty {{x^{\left| {{T_i}} \right| + 1}}{f_X}\left( x \right) dx} \nonumber \\= & {} \left\{ {\prod \limits _{m \in {T_i}} {\frac{{{{\uplambda }_{S{R_m}}}}}{{{{\uplambda }_{{R_m}D}}\tau }}\min \left( {\frac{\tau }{{{y_m}}},1} \right) } } \right\} {e^{ - {{\uplambda }_{SL}}\tau }}\left( {\sum \limits _{l = 0}^{\left| {{T_i}} \right| + 1} {\frac{{\left( {\left| {{T_i}} \right| + 1} \right) !{\tau ^l}}}{{l!{\uplambda }_{SL}^{\left| {{T_i}} \right| + 1 - l}}}} } \right) . \end{aligned}$$

(28)

It is noted that the integral in (28) is completely solved in closed-form by integrating by parts. Plugging the above into (27), one obtains (22) where

$$\begin{aligned} {{\tilde{\mathcal {G}}}_{1i}} = \int \limits _0^\infty {\ldots \int \limits _0^\infty {{\mathcal {G}_{1i}}} } \prod \limits _{k = 1}^K {\left( {\frac{{{{\uplambda }_{{R_k}D}}{f_{{Y_k}}}\left( {{y_k}} \right) }}{{\min \left( {y_k^{ - 1}\tau ,1} \right) }}d{y_k}} \right) } \end{aligned}$$

(29)

Substituting (28) in the above, we reduce (29) to

$$\begin{aligned} {{\tilde{\mathcal {G}}}_{1i}} = \frac{{{e^{ - {{\uplambda }_{SL}}\tau }}}}{{{\tau ^{\left| {{T_i}} \right| }}}}\left( {\sum \limits _{l = 0}^{\left| {{T_i}} \right| + 1} {\frac{{\left( {\left| {{T_i}} \right| + 1} \right) !{\tau ^l}}}{{l!{\uplambda }_{SL}^{\left| {{T_i}} \right| + 1 - l}}}} } \right) \left( {\prod \limits _{m \in {T_i}} {{{\uplambda }_{S{R_m}}}} } \right) \prod \limits _{q \in \bar{T}_i} {{{\uplambda }_{{R_q}D}}\int \limits _0^\infty {\frac{{{f_{{Y_q}}}\left( {{y_q}} \right) }}{{\min \left( {y_q^{ - 1}\tau ,1} \right) }}d{y_q}} }. \end{aligned}$$

(30)

Using the fact that ${f_{{Y_q}}}\left( {{y_q}} \right) = {{\uplambda }_{{R_q}L}}{e^{ - {{\uplambda }_{{R_q}L}}{y_q}}}$, one can solve the integral in (30) in closed-form as follows

$$\begin{aligned} \int \limits _0^\infty {\frac{{{{\uplambda }_{{R_q}L}}{e^{ - {{\uplambda }_{{R_q}L}}x}}}}{{\min \left( {{x^{ - 1}}\tau ,1} \right) }}dx} = \frac{{{{\uplambda }_{{R_q}L}}}}{\tau }\int \limits _\tau ^\infty {x{e^{ - {{\uplambda }_{{R_q}L}}x}}dx + \int \limits _0^\tau {{{\uplambda }_{{R_q}L}}{e^{ - {{\uplambda }_{{R_q}L}}x}}dx} } = 1 + \frac{{{e^{ - {{\uplambda }_{{R_q}L}}\tau }}}}{{{{\uplambda }_{{R_q}L}}\tau }}, \end{aligned}$$

(31)

where the second integral is solved by integrating by parts.

Inserting (31) into (30), the closed form of $\tilde{\mathcal {G}}_{1i}$ can be expressed as (23), which completes the proof.

Appendix 2: Proof of Lemma 2

We rewrite $\mathcal {G}_2$ in (21) as

$$\begin{aligned} {\mathcal {G}_2}= & {} {{\uplambda }_{SD}}\int \limits _0^\tau {{f_X}\left( x \right) dx} \prod \limits _{k = 1}^K {\int \limits _0^\infty {\left( {{{\uplambda }_{S{R_k}}} + \frac{{{{\uplambda }_{{R_k}D}}}}{{\min \left( {y_k^{ - 1}\tau ,1} \right) }}} \right) } } {f_{{Y_k}}}\left( {{y_k}} \right) d{y_k}\nonumber \\= & {} {{\uplambda }_{SD}}\int \limits _0^\tau {{f_X}\left( x \right) dx} \prod \limits _{k = 1}^K {\left( {{{\uplambda }_{S{R_k}}}\int \limits _0^\infty {{f_{{Y_k}}}\left( {{y_k}} \right) d{y_k}} + {{\uplambda }_{{R_k}D}}\int \limits _0^\infty {\frac{{{f_{{Y_k}}}\left( {{y_k}} \right) }}{{\min \left( {y_k^{ - 1}\tau ,1} \right) }}d{y_k}} } \right) } \end{aligned}$$

(32)

It is straightforward to infer that $\int \limits _0^\tau {{f_X}\left( x \right) dx} = 1 - {e^{ - {{\uplambda }_{SL}}\tau }}$ and ${\int \limits _0^\infty {{f_{{Y_k}}}\left( {{y_k}} \right) d{y_k}} }=1$ while $\int \limits _0^\infty {\frac{{{f_{{Y_k}}}\left( {{y_k}} \right) }}{{\min \left( {y_k^{ - 1}\tau ,1} \right) }}d{y_k}} = 1 + \frac{{{e^{ - {{\uplambda }_{{R_k}L}}\tau }}}}{{{{\uplambda }_{{R_k}L}}\tau }}$ with the aid of (31). Plugging all these results in (32), one obtains (24) which completes the proof.