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Security Performance Analysis of Underlay Cognitive Networks with Helpful Jammer Under Interference from Primary Transmitter

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Abstract

Interference from primary transmitters in underlay cognitive networks is practically neither neglected nor integrated into noise terms although most literature inversely addressed. This paper firstly employs a helpful jammer to secure information transmission of a secondary transmitter. Then, its efficacy through secrecy outage probability under practical considerations consisting of exponentially distributed interference from primary transmitter, peak transmission power limitation, interference power limitation, and Rayleigh fading channels are analytically assessed. Finally, results are provided to illustrate significant security performance improvement thanks to exploiting the jammer while considerable security performance degradation due to interference from primary transmitter.

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Notes

  1. This can be alternatively interpreted that jamming signal is transmitted in the null space to SD or shared with SD (e.g., use the seed of the jamming signal generator in a secure manner). Such interpretations are widely accepted in open literature (e.g., [17,18,19,20, 22, 27,28,29] and references therein).

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Acknowledgements

This research is funded by Vietnam National Foundation for Science and Technology Development (NAFOSTED) under grant number 102.04-2017.01

Author information

Authors and Affiliations

  1. Ho Chi Minh City University of Technology, VNU-HCM, 268 Ly Thuong Kiet Str., Dist. 10, Ho Chi Minh City, Vietnam

    Khuong Ho-Van & Thiem Do-Dac

  2. Thu Dau Mot University, 6 Tran Van On Str., Phu Hoa Ward, Thu Dau Mot City, Binh Duong Province, Vietnam

    Thiem Do-Dac

Authors
  1. Khuong Ho-Van
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Correspondence to Khuong Ho-Van.

Appendices

Appendix A: Proof of Lemma 1

The conditional CDF of γd given Es is defined as

$$ {F_{{\gamma_{d}}}}\left( {\left. x \right|{E_{s}}} \right) = \Pr \left\{ {\left. {{\gamma_{d}} \le x} \right|{E_{s}}} \right\}. $$
(43)

Plugging (3) into (43) and performing some algebraic manipulations, one obtains

$$ \begin{array}{ll} &{F_{{\gamma_{d}}}}\left( {\left. x \right|{E_{s}}} \right) = \Pr \left\{ {\left. {\frac{{{E_{s}}{{\left| {{g_{sd}}} \right|}^{2}}}}{{{E_{p}}{{\left| {{g_{pd}}} \right|}^{2}} + {\sigma^{2}}}} \le x} \right|{E_{s}}} \right\} \\ &= \Pr \left\{ {\left. {{{\left| {{g_{sd}}} \right|}^{2}} \le \frac{{x\left( {{E_{p}}{{\left| {{g_{pd}}} \right|}^{2}} + {\sigma^{2}}} \right)}}{{{E_{s}}}}} \right|{E_{s}}} \right\}\\ &= {{\Xi}_{{{\left| {{g_{pd}}} \right|}^{2}}}}\!\left\{ {\Pr \!\left\{ {\left. {{{\!\left| {{g_{sd}}} \right|}^{2}} \!\le \frac{x}{{{E_{s}}}}\!\left( {{\!E_{p}}{{\left| {{g_{pd}}} \right|}^{2}} \!+ {\sigma^{2}}} \right)} \!\right|{{\left| {{g_{pd}}} \right|}^{2}},{E_{s}}} \!\right\}} \right\}\!. \end{array} $$
(44)

Due to \({g_{\texttt {tr}}} \sim \mathcal {C}\mathcal {N}\left ({0,{\alpha _{\texttt {tr}}}} \right )\), |gtr|2 follows the exponential distribution. Therefore, its CDF and PDF are respectively given by \({F_{{{\left | {{g_{\texttt {tr}}}} \right |}^{2}}}}\left (x \right ) = 1 - {e^{- x/{\alpha _{\texttt {tr}}}}}\) and \({f_{{{\left | {{g_{\texttt {tr}}}} \right |}^{2}}}}\left (x \right ) = {e^{- x/{\alpha _{\texttt {tr}}}}}/{\alpha _{\texttt {tr}}}\), x ≥ 0. Given this fact, one can simplify Eq. 44 as

$$ \begin{array}{ll} {F_{{\gamma_{d}}}}\!\left( {\left. x \right|{E_{s}}} \right) \!\!&=\! {{\Xi}_{{{\left| {{g_{pd}}} \right|}^{2}}}}\!\left\{ {\!1 - \left. {{e^{- \frac{x}{{{E_{s}}{\alpha_{sd}}}}\left( {{E_{p}}{{\left| {{g_{pd}}} \right|}^{2}} + {\sigma^{2}}} \right)}}} \!\right|\!{{\left| {{g_{pd}}} \right|}^{2}}\!,\!{E_{s}}} \right\} \\ &= 1 - \int\limits_{0}^{\infty} {{e^{- \frac{x}{{{E_{s}}{\alpha_{sd}}}}\left( {{E_{p}}y + {\sigma^{2}}} \right)}}{f_{{{\left| {{g_{pd}}} \right|}^{2}}}}\left( y \right)dy} \\ &= 1 - \int\limits_{0}^{\infty} {{e^{- \frac{x}{{{E_{s}}{\alpha_{sd}}}}\left( {{E_{p}}y + {\sigma^{2}}} \right)}}\frac{1}{{{\lambda_{pd}}}}{e^{- \frac{y}{{{\lambda_{pd}}}}}}dy} \\ &= 1 - {\left( {\frac{{{E_{p}}{\alpha_{pd}}x}}{{{E_{s}}{\alpha_{sd}}}} + 1} \right)^{- 1}}{e^{- \frac{{x{\sigma^{2}}}}{{{E_{s}}{\alpha_{sd}}}}}}. \end{array} $$
(45)

Taking the derivative of \({F_{{\gamma _d}}}\left ({\left . x \right |{E_{s}}} \right )\) with respect to x, simplifying the result and using compact notations in Eqs. 23, 24, 25, one obtains \({f_{{\gamma _d}}}\left ({\left . x \right |{E_{s}}} \right ) = \frac {d}{{dx}}{F_{\gamma _{d}}}\left ({\left . x \right |{E_{s}}} \right )\) as exactly shown in Eq. 22, completing the proof.

Appendix B: Proof of Lemma 2

The conditional CDF of γw given Es according to its definition is

$$ {F_{{\gamma_{w}}}}\left( {\left. x \right|{E_{s}}} \right) = \Pr \left\{ {\left. {{\gamma_{w}} \le x} \right|{E_{s}}} \right\}. $$
(46)

Plugging Eq. 4 into Eq. 46 and performing some algebraic manipulations, one obtains

$$ \begin{array}{ll} &{F_{{\gamma_{w}}}}\left( {\left. x \right|{E_{s}}} \right) = \Pr \left\{ {\left. {\frac{{{E_{s}}{{\left| {{g_{sw}}} \right|}^{2}}}}{{{E_{j}}{{\left| {{g_{jw}}} \right|}^{2}} + {E_{p}}{{\left| {{g_{pw}}} \right|}^{2}} + {\sigma^{2}}}} \le x} \right|{E_{s}}} \right\} \\ &= \Pr \left\{ {\left. {{{\left| {{g_{sw}}} \right|}^{2}} \le \frac{x}{{{E_{s}}}}\left( {{E_{j}}{{\left| {{g_{jw}}} \right|}^{2}} + {E_{p}}{{\left| {{g_{pw}}} \right|}^{2}} + {\sigma^{2}}} \right)} \right|{E_{s}}} \right\} \\ &= {{\Xi}_{{{\left| {{g_{jp}}} \right|}^{2}},{{\left| {{g_{jw}}} \right|}^{2}},{{\left| {{g_{pw}}} \right|}^{2}}}}\!\left\{ {\!1 - \left. {{e^{- \frac{{x\left( {{E_{j}}{{\left| {{g_{jw}}} \right|}^{2}} + {E_{p}}{{\left| {{g_{pw}}} \right|}^{2}} + {\sigma^{2}}} \right)}}{{{E_{s}}{\alpha_{sw}}}}}}} \right|{E_{s}}} \right\} \\ &= 1 - {e^{- \frac{{x{\sigma^{2}}}}{{{E_{s}}{\alpha_{sw}}}}}}\underbrace {{{\Xi}_{{{\left| {{g_{pw}}} \right|}^{2}}}}\left\{ {{e^{- \frac{{x{E_{p}}}}{{{E_{s}}{\alpha_{sw}}}}{{\left| {{g_{pw}}} \right|}^{2}}}}} \right\}}_{\mu} \times \\ &\underbrace {{{\Xi}_{{{\left| {{g_{jp}}} \right|}^{2}}}}\left\{ {\left. {\underbrace {{{\Xi}_{{{\left| {{g_{jw}}} \right|}^{2}}}}\left\{ {\left. {{e^{- \frac{{x{E_{j}}}}{{{E_{s}}{\alpha_{sw}}}}{{\left| {{g_{jw}}} \right|}^{2}}}}} \right|{E_{s}}} \right\}}_{\kappa} } \right|{E_{s}}} \right\}}_{\zeta}. \end{array} $$
(47)

Quantities (κ, μ, ζ) are computed as follows. First of all, κ can be expressed in closed-form as

$$ \begin{array}{ll} \kappa &= {{\Xi}_{{{\left| {{g_{jw}}} \right|}^{2}}}}\left\{ {\left. {{e^{- \frac{{x{E_{j}}}}{{{E_{s}}{\alpha_{sw}}}}{{\left| {{g_{jw}}} \right|}^{2}}}}} \right|{E_{s}}} \right\} \\ &= \int\limits_{0}^{\infty} {{e^{- \frac{{x{E_{j}}}}{{{E_{s}}{\alpha_{sw}}}}y}}{f_{{{\left| {{g_{jw}}} \right|}^{2}}}}\left( y \right)dy} \\ &= \int\limits_{0}^{\infty} {{e^{- \frac{{x{E_{j}}}}{{{E_{s}}{\alpha_{sw}}}}y}}\frac{1}{{{\alpha_{jw}}}}{e^{- \frac{y}{{{\alpha_{jw}}}}}}dy} \\ &= {\left( {\frac{{x{E_{j}}{\alpha_{jw}}}}{{{E_{s}}{\alpha_{sw}}}} + 1} \right)^{- 1}}. \end{array} $$
(48)

Imitating the procedure of deriving κ, one can straightforwardly infer that

$$ \mu = {\left( {\frac{{x{E_{p}}{\alpha_{pw}}}}{{{E_{s}}{\alpha_{sw}}}} + 1} \right)^{- 1}}. $$
(49)

By inserting Eq. 14 into Eq. 48 and then statistically averaging the result over a random variable |gjp|2, one obtains the exact closed form of ζ as

$$ \begin{array}{ll} \zeta &= {{\Xi}_{{{\left| {{g_{jp}}} \right|}^{2}}}}\left\{ {{{\left[ {\frac{{x{\alpha_{jw}}}}{{{E_{s}}{\alpha_{sw}}}}\min \left( {\frac{T}{{2{{\left| {{g_{jp}}} \right|}^{2}}}},{{\tilde E}_{j}}} \right) + 1} \right]}^{- 1}}} \right\} \\ &= \underbrace {\int\limits_{0}^{\frac{T}{{2{{\tilde E}_{j}}}}} {{{\left( {\frac{{x{\alpha_{jw}}{{\tilde E}_{j}}}}{{{E_{s}}{\alpha_{sw}}}} + 1} \right)}^{- 1}}{f_{{{\left| {{g_{jp}}} \right|}^{2}}}}\left( y \right)dy} }_{\tilde \zeta } \\ &+ \underbrace {\int\limits_{\frac{T}{{2{{\tilde E}_{j}}}}}^{\infty} {{{\left( {\frac{{x{\alpha_{jw}}}}{{{E_{s}}{\alpha_{sw}}}}\frac{T}{{2y}} + 1} \right)}^{- 1}}{f_{{{\left| {{g_{jp}}} \right|}^{2}}}}\left( y \right)dy} }_{\bar \zeta } \end{array} $$
(50)

The first term in Eq. 50 is easily solved as

$$ \tilde \zeta = {\left( {\frac{{x{\alpha_{jw}}{{\tilde E}_{j}}}}{{{E_{s}}{\alpha_{sw}}}} + 1} \right)^{- 1}}\left( {1 - {e^{- \frac{T}{{2{{\tilde E}_{j}}{\alpha_{jp}}}}}}} \right). $$
(51)

while the second term is further simplified as

$$ \begin{array}{ll} \bar \zeta &= \int\limits_{\frac{T}{{2{{\tilde E}_{j}}}}}^{\infty} {\left[ {1 - \frac{{x{\alpha_{jw}}T}}{{2{E_{s}}{\alpha_{sw}}}}{{\left( {y + \frac{{x{\alpha_{jw}}T}}{{2{E_{s}}{\alpha_{sw}}}}} \right)}^{- 1}}} \right]\frac{1}{{{\alpha_{jp}}}}{e^{- \frac{y}{{{\alpha_{jp}}}}}}dy} \\ &= \int\limits_{\frac{T}{{2{{\tilde E}_{j}}}}}^{\infty} {\frac{1}{{{\alpha_{jp}}}}{e^{- \frac{y}{{{\alpha_{jp}}}}}}dy - \frac{{x{\alpha_{jw}}T}}{{2{E_{s}}{\alpha_{sw}}{\alpha_{jp}}}}\int\limits_{\frac{T}{{2{{\tilde E}_{j}}}}}^{\infty} {\frac{{{e^{- \frac{y}{{{\alpha_{jp}}}}}}}}{{y + \frac{{x{\alpha_{jw}}T}}{{2{E_{s}}{\alpha_{sw}}}}}}dy} }. \end{array} $$
(52)

The first integrand in Eq. 52 is easily computed and the second one can be expressed in terms of the Ei(⋅) function, resulting in

$$ \begin{array}{ll} &\bar \zeta = {e^{- \frac{T}{{2{{\tilde E}_{j}}{\alpha_{jp}}}}}} \\ &+ \frac{{x{\alpha_{jw}}T}}{{2{E_{s}}{\alpha_{sw}}{\alpha_{jp}}}}{e^{\frac{{x{\alpha_{jw}}T}}{{2{E_{s}}{\alpha_{sw}}{\alpha_{jp}}}}}}Ei\left( { - \frac{T}{{2{{\tilde E}_{j}}{\alpha_{jp}}}} - \frac{{x{\alpha_{jw}}T}}{{2{E_{s}}{\alpha_{sw}}{\alpha_{jp}}}}} \right). \end{array} $$
(53)

Plugging Eqs. 51 and 53 into Eq. 50 and then inserting the result together with Eq. 49 into Eq. 47, one can reduce Eqs. 47 to 26 after using notations in Eqs. 27, 28, 29, 30, 31. This finishes the proof of Lemma 2.

Appendix C: Closed-form representations of some special integrands

Start with the integrand in Eq. 36. By changing variable y = x + b and then applying the definition of the Ei(⋅) function, the closed-form representation of Eq. 36 is

$$ \mathcal{J}\left( {a,b} \right) = - {e^{ab}}Ei\left( { - ab} \right). $$
(54)

Performing the integral by part to Eq. 37 and then using the result of Eq. 54, the closed-form representation of Eq. 37 is

$$ \begin{array}{ll} \mathcal{F}\left( {a,b} \right) &= \frac{1}{b} - a\int\limits_{0}^{\infty} {\frac{{{e^{- ax}}}}{{x + b}}dx} \\ &= \frac{1}{b} + a{e^{ab}}Ei\left( { - ab} \right). \end{array} $$
(55)

By using the approximation of the Ei(⋅) function in [36, Eq. (6.3.11)] as \(Ei\left (x \right ) \approx \frac {{{e^x}}}{x}\), one can obtain the closed form of Eq. 38 after some algebraic manipulations as

$$ \begin{array}{ll} \mathcal{M}\left( {b,c,g,l} \right) &\approx \int\limits_{0}^{\infty} {\frac{{{e^{- bx}}}}{{x + c}}\frac{{{e^{- gx - l}}}}{{ - gx - l}}dx} \\ &=- \frac{{{e^{- l}}}}{g}\left[ {\frac{1}{{l/g - c}}\int\limits_{0}^{\infty} {\frac{{{e^{- \left( {b + g} \right)x}}}}{{x + c}}dx} } \right. \\ &+ \left. {\frac{1}{{c - l/g}}\int\limits_{0}^{\infty} {\frac{{{e^{- \left( {b + g} \right)x}}}}{{x + l/g}}dx} } \right] \\ &=\frac{{{e^{bl/g}}}}{{cg - l}}Ei\left( { - \frac{{\left[ {b + g} \right]l}}{g}} \right) \\ &- \frac{{{e^{\left( {b + g} \right)c - l}}}}{{cg - l}}Ei\left( { - \left[ {b + g} \right]c} \right). \end{array} $$
(56)

It is noted that the integrands in the second equality of Eq. 56 are obtained with the help of Eq. 54.

Again using the approximation of the Ei(⋅) function as \(Ei\left (x \right ) \approx \frac {{{e^x}}}{x}\), one can simplify (39) as

$$ \mathcal{Y}\left( {b,c,g,l} \right) \approx \int\limits_{0}^{\infty} {\frac{{{e^{- bx}}}}{{{{\left( {x + c} \right)}^{2}}}}\frac{{{e^{- gx - l}}}}{{ - gx - l}}dx}. $$
(57)

Performing the partial fraction decomposition to \(\frac {1}{{{{\left ({x + c} \right )}^{2}}\left ({x + l/g} \right )}}\), one represents (57) in a more compact form as

$$ \begin{array}{ll} \mathcal{Y}\left( {b,c,g,l} \right) &= - \frac{{{e^{- l}}}}{g}\left( {\frac{1}{{l/g - c}}\int\limits_{0}^{\infty} {\frac{{{e^{- \left( {b + g} \right)x}}}}{{{{\left( {x + c} \right)}^{2}}}}dx} } \right. \\ &- \frac{1}{{{{\left( {c - l/g} \right)}^{2}}}}\int\limits_{0}^{\infty} {\frac{{{e^{- \left( {b + g} \right)x}}}}{{x + c}}dx} \\ &+ \left. {\frac{1}{{{{\left( {c - l/g} \right)}^{2}}}}\int\limits_{0}^{\infty} {\frac{{{e^{- \left( {b + g} \right)x}}}}{{x + l/g}}dx} } \right). \end{array} $$
(58)

Finally, exploiting the results in Eqs. 54 and 55, the closed-form representation of Eq. 39 is

$$ \begin{array}{ll} \mathcal{Y}\left( {b,c,g,l} \right) &= \frac{g} {{(l - cg)}^{2}} {e^{bl/g}}Ei\left( { - \frac{{\left[ {b + g} \right]l}}{g}} \right) - \frac{{{e^{- l}}}}{{\left( {l - cg} \right)c}} \\ &- \frac{e^{\left( {b + g} \right)c - l}} {{l - cg}}\left( {b + g + \frac{g}{{l - cg}}} \right)Ei\left( { - \left[ {b + g} \right]c} \right). \end{array} $$
(59)

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Ho-Van, K., Do-Dac, T. Security Performance Analysis of Underlay Cognitive Networks with Helpful Jammer Under Interference from Primary Transmitter. Mobile Netw Appl 25, 4–15 (2020). https://doi.org/10.1007/s11036-018-1185-x

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