Outage and energy efficiency analysis for cognitive based heterogeneous cellular networks

Abstract

Cognitive radio and small cells are the promising techniques to minimize energy consumption and satisfy the exponentially increasing data rates for the heterogeneous cellular network (HCN). In this paper, a theoretical framework is developed to calculate the outage probability of the HCN based on the opportunistic utilization of the traditional cellular bandwidth and television white space (TVWS) for the cognitive femto base stations. This work investigates overlay, underlay, mixed overlay-underlay based two tiers cognitive HCN. It also investigates the impact of the TVWS in the overlay-TVWS mixed spectrum sharing technique (SST). Tools from stochastic geometry are used to model cognitive HCN. Furthermore, the tier selection probability, average ergodic rate, area spectral efficiency (ASE), and energy efficiency (EE) of the HCN are also calculated for different SSTs. Numerical results show that mixed SST achieves a significant reduction in tier outage probability and total outage probability as compared to underlay and overlay techniques alone. It is also demonstrated that compared to the traditional single tier network, cognitive based HCN can improve the total ASE and EE of the order of $10^{2}$ and 10, respectively.

Appendices

Appendix 1: Proof of Theorem 1

When both the tiers (MBS and CFBS tier) use the same channel, it introduces cross tier and co-tier interference in the network. Let total interference is given by I = $I_{mbs-mbs}+I_{cfbs-mbs}$, where $I_{mbs-mbs}$ is co-tier interference and $I_{cfbs-mbs}$ is cross-tier interference for MBS network. If $I_{mbs-mbs}$ and $I_{cfbs-mbs}$ are independent, then the CHF of the total interference I can be written as $\psi _I(\gamma y)$ = $\psi _{I_{mbs-mbs}(\gamma y)}\cdot \psi _{I_{cfbs-mbs}(\gamma y)}.$ The CHF of I is given by [31]

$$\begin{aligned} \psi _I(\gamma y) =e^{-\pi (\gamma y)^{2/\epsilon }\left( \xi _{cfbs} \eta _{cfbs}^{2/\epsilon }+\xi _{\mathrm {mbs}} \eta _{\mathrm {mbs}}^{2/\epsilon } \right) \mathbb {E} [g^{2/\epsilon }]\varGamma \left( 1-\frac{2}{\epsilon }\right) e^{-i\pi /\epsilon }}, \end{aligned}$$

(54)

where $\mathbb {E} [g^{2/\epsilon }]$ for the Gamma distribution is derived as

$$\begin{aligned} \mathbb {E} [g^{2/\epsilon }]&=\int _0^{\infty } g^{\frac{2}{\epsilon }} f_q (g)dg \, \nonumber \\ &=\int _0^{\infty } \frac{\vartheta _q ^{-\alpha _q } e^{-\frac{g}{\vartheta _q }} g^{\alpha _q +\frac{2}{\epsilon }-1}}{\varGamma (\alpha _q )}dg \nonumber \\ &=\frac{\vartheta _q ^{-\alpha _q } \left( \frac{1}{\vartheta _q }\right) ^{-\alpha _q -\frac{2}{\epsilon }} \varGamma \left( \alpha _q +\frac{2}{\epsilon }\right) }{\varGamma (\alpha _q )} \nonumber \\ &=\frac{\vartheta _q ^{2/\epsilon } \varGamma \left( \alpha _q +\frac{2}{\epsilon }\right) }{\varGamma (\alpha _q )} \end{aligned}$$

(55)

By substituting the value of $\mathbb {E} [g^{2/\epsilon }]$ from (54), (55) can be expressed as

$$\begin{aligned} \psi _{I}(\gamma y) =e^{- y^{2/\epsilon }{\mathfrak {B}}_{\mathfrak {mc}}e^{-i\pi /\epsilon }}, \end{aligned}$$

(56)

where ${\mathfrak {B}}_{\mathfrak {mc}}=\frac{\pi \gamma ^{2/\epsilon }\left( \xi _{cfbs} \eta _{cfbs}^{2/\epsilon }+\xi _{{mbs}} \eta _{{mbs}}^{2/\epsilon } \right) \vartheta _q ^{2/\epsilon } \varGamma \left( 1-\frac{2}{\epsilon }\right) \varGamma \left( \alpha _q+\frac{2}{\epsilon }\right) }{\varGamma (\alpha _q )}.$

CHF of the Gamma distribution is given by

$$\begin{aligned} \psi _{q}(y)=(1-i\vartheta _q y)^{-\alpha _q} \end{aligned}$$

(57)

So, using (56) and (57), $\psi (y)$ can be expressed as

$$\begin{aligned} \psi (y)= \frac{e^{- y^{2/\epsilon }{\mathfrak {B}}_{\mathfrak {mc}}e^{-i\pi /\epsilon }}}{(1+i\vartheta _q y)^{\alpha _q}}. \end{aligned}$$

(58)

The imaginary part of the (58) can be written as [46]

$$\begin{aligned} \mathcal {I}\{\psi (y)e^{-i y z}\}&= \frac{e^{-y^{2/\epsilon } {\mathfrak {B}}_{\mathfrak {mc}} \cos \left( \frac{\pi }{\epsilon }\right) }}{{\left( y^2 \vartheta _q ^2+1\right) ^{\alpha _q /2}}}\nonumber \\ &\times {\sin \left( {\mathfrak {B}}_{\mathfrak {mc}} y^{2/\epsilon } \sin \left( \frac{\pi }{\epsilon } \right) - \alpha _q\tan ^{-1} (y\vartheta _q)-yz\right) } \end{aligned}$$

(59)

From (19), (59) can be expressed as

$$\begin{aligned} Pr \left[ \gamma I-q < z\right]&= \frac{1}{2}-\frac{1}{\pi }{\int _0^{\infty } \frac{e^{-y^{2/\epsilon } {\mathfrak {B}}_{\mathfrak {mc}}\cos \left( \frac{\pi }{\epsilon }\right) }}{y\left( y^2 \vartheta _q^2+1\right) ^{-\alpha _q /2}}}\nonumber \\ &\times \sin \left( {\mathfrak {B}}_{\mathfrak {mc}} y^{2/\epsilon } \sin \left( \frac{\pi }{\epsilon } \right) - \alpha _q\tan ^{-1} (y\vartheta _q)-yz\right) dy \end{aligned}$$

(60)

So, outage probability of MBS tier is given by

$$\begin{aligned} {\mathcal {P}}_{\mathrm {mc}}^{mbs}&=1-Pr \left[ \gamma I-q < 0\right] \nonumber \\ &=\frac{1}{2}+\frac{1}{\pi }{\int _0^{\infty }\frac{e^{-y^{2/\epsilon } {\mathfrak {B}}_{\mathfrak {mc}}\cos \left( \frac{\pi }{\epsilon }\right) }}{y\left( y^2 \vartheta _q^2+1\right) ^{-\alpha _q /2}}}\end{aligned}$$

(61)

$$\begin{aligned}&\times \sin \left( {\mathfrak {B}}_{\mathfrak {mc}} y^{2/\epsilon } \sin \left( \frac{\pi }{\epsilon } \right) - \alpha _q\tan ^{-1} (y\vartheta _q)\right) dy \end{aligned}$$

(62)

To obtain the exact finite range integral expression, we substitute $y = \tan (x)$ in (62). Thus, (62) can be written as

$$\begin{aligned} {\mathcal {P}}_{\mathrm {mc}}^{mbs}&= \frac{1}{2}+\frac{1}{\pi }{\int _0^{\pi /2} \frac{e^{-\tan (x)^{2/\epsilon } {\mathfrak {B}}_{\mathfrak {mc}}\cos \left( \frac{\pi }{\epsilon }\right) }}{\tan (x)\left( \tan (x)^2 \vartheta _q^2+ 1\right) ^{\alpha _q /2}}}\sec ^2 (x)\\ &\times \sin \left( {\mathfrak {B}}_{\mathfrak {mc}} \tan (x)^{2/\epsilon } \sin \left( \frac{\pi }{\epsilon } \right) - \alpha _q\tan ^{-1} (\tan (x)\vartheta _q)\right) dx \end{aligned}$$

which can be simplified as

$$\begin{aligned} {\mathcal {P}}_{\mathrm {mc}}^{mbs}&= \frac{1}{2}+\frac{2}{\pi }{\int _0^{\pi /2} \frac{e^{-\tan (x)^{2/\epsilon } {\mathfrak {B}}_{\mathfrak {mc}}\cos \left( \frac{\pi }{\epsilon }\right) }}{\sin (2x)\left( \tan (x)^2 \vartheta _q^2+1\right) ^{\alpha _q /2}}}\nonumber \\ &\times \sin \left( {\mathfrak {B}}_{\mathfrak {mc}} \tan (x)^{2/\epsilon } \sin \left( \frac{\pi }{\epsilon } \right) - \alpha _q\tan ^{-1} (\tan (x)\vartheta _q)\right) dx \end{aligned}$$

(63)

where $\gamma = T_m/\eta _{mbs} d_{mbs}^{-\epsilon } .$ $T_m$ is the SIR threshold for MBS tier.

Appendix 2: Proof of Lemma 1

Let co-tier interference is denoted by $I_{mbs-mbs}$ for MBS tier. Hence, the CHF of the total interference I is expressed by

$$\begin{aligned} \psi _{I}(\gamma y)=\psi _{I_{mbs-mbs}}(\gamma y) =e^{- ( y)^{2/\epsilon } {\mathfrak {B}}_{\mathfrak {om}}e^{-i\pi /\epsilon }}, \end{aligned}$$

(64)

where

$$\begin{aligned} {\mathfrak {B}_{\mathfrak {mo}}}=\pi (\gamma )^{2/\epsilon }\left( \xi _{{mbs}} \eta _{{mbs}}^{2/\epsilon } \right) \frac{\vartheta _q ^{2/\epsilon } \varGamma \left( 1-\frac{2}{\epsilon }\right) \varGamma \left( \alpha _q +\frac{2}{\epsilon }\right) }{\varGamma (\alpha _q )}. \end{aligned}$$

The outage probability of MBS tier with co-tier interference ${\mathcal {P}}_{\mathrm {mo}}^{mbs}$ is similar to ${\mathcal {P}}_{\mathrm {mc}}^{mbs}.$ So using (63), ${\mathcal {P}}_{\mathrm {mo}}^{mbs}$ is given by

$$\begin{aligned} {\mathcal {P}}_{\mathrm {mo}}^{mbs}&= \frac{1}{2}+\frac{2}{\pi }{\int _0^{\pi /2} \frac{e^{-\tan (x)^{2/\epsilon } {\mathfrak {B}_{\mathfrak {mo}}}\cos \left( \frac{\pi }{\epsilon }\right) }}{\sin (2x)\left( \tan (x)^2 \vartheta _q^2+1\right) ^{\alpha _q /2}}}\\ &\times \sin \left( {\mathfrak {B}_{\mathfrak {mo}}} \tan (x)^{2/\epsilon } \sin \left( \frac{\pi }{\epsilon } \right) - \alpha _q\tan ^{-1} (\tan (x) \vartheta _q)\right) dx \end{aligned}$$

where $\gamma = T_m/\eta _{mbs} d_{mbs}^{-\epsilon } .$ $T_m$ is the SIR threshold for the MBS tier.

Appendix 3: Proof of Lemma 2

When CFBS tier works in USST, it introduces co-tier and cross tier interference in the network. Thus, total interference can be expressed by I = $I_{cfbs-cfbs}$+$I_{mbs-cfbs}$, where $I_{cfbs-cfbs}$ is co-tier interference and $I_{mbs-cfbs}$ is cross-tier interference for CFBS network [30]. The CHF of the total interference I is similar to the (56). The outage probability of CFBS tier under USST ${\mathcal {P}}_{\mathrm {UL}}^{cfbs}$ is similar to ${\mathcal {P}}_{\mathrm {mc}}^{mbs}$ and it is given by

$$\begin{aligned} {\mathcal {P}}_{\mathrm {UL}}^{cfbs}&= \frac{1}{2}+\frac{2}{\pi }{\int _0^{\pi /2} \frac{e^{-\tan (x)^{2/\epsilon } {\mathfrak {B}}_{\mathfrak {ul}}\cos \left( \frac{\pi }{\epsilon }\right) }}{\sin (2x) \left( \tan (x)^2 \vartheta _q^2+1\right) ^{\alpha _q /2}}}\nonumber \\ &\times \sin \left( {\mathfrak {B}}_{\mathfrak {ul}} \tan (x)^{2/\epsilon } \sin \left( \frac{\pi }{\epsilon } \right) - \alpha _q\tan ^{-1} (\tan (x)\vartheta _q)\right) dx, \end{aligned}$$

(65)

where ${\mathfrak {B}}_{\mathfrak {ul}}=\frac{\pi \gamma ^{2/\epsilon }\left( \xi _{cfbs} \eta _{cfbs}^{2/\epsilon }+\xi _{{mbs}} \eta _{{mbs}}^{2/\epsilon } \right) \vartheta _q ^{2/\epsilon } \varGamma \left( 1-\frac{2}{\epsilon }\right) \varGamma \left( \alpha _q +\frac{2}{\epsilon }\right) }{\varGamma (\alpha _q )}$ and $\gamma = T_c/\eta _{cfbs} d_{cfbs}^{-\epsilon } .$ $T_c$ is the SIR threshold for CFBS tier.

Appendix 4: Proof of Lemma 3

In OTMSST, co-tier interference and interference from TV transmitters are considered to derive the outage probability. Let $I_{cfbs-cfbs}$ denotes co-tier interference and interference from TV transmitters is expressed by $I_{tv-cfbs}.$ So, total interference is given by $I = I_{cfbs-cfbs}+I_{tv-cfbs}.$ The derivation of ${\mathcal {P}}_{\mathrm {ot}}^{cfbs}$ is similar to ${\mathcal {P}}_{\mathrm {mc}}^{mbs}$ and it is given by

$$\begin{aligned} {\mathcal {P}}_{\mathrm {ot}}^{cfbs}&= \frac{1}{2}+\frac{2}{\pi }{\int _0^{\pi /2}\frac{e^{-\tan (x)^{2/\epsilon _{tv} } {\mathfrak {B}}_{\mathfrak {tv}}\cos \left( \frac{\pi }{\epsilon _{tv} }\right) }}{\sin (2x)\left( \tan (x)^2 \vartheta _q^2+1\right) ^{\alpha _q /2}}}\nonumber \\ &\times \sin \left( {\mathfrak {B}}_{\mathfrak {tv}} \tan (x)^{2/\epsilon _{tv} } \sin \left( \frac{\pi }{\epsilon _{tv} } \right) - \alpha _q\tan ^{-1} (\tan (x)\vartheta _q) \right) dx,\\ \text {where}\quad {\mathfrak {B}}_{\mathfrak {tv}}&= \frac{\pi \gamma ^{\frac{2}{\epsilon _{tv}}}\left( \xi _{{cfbs}} \eta _{cfbs}^{tv})^{\frac{2}{\epsilon _{tv}} } +\xi _{{tv}} \eta _{{tv}}^{\frac{2}{\epsilon _{tv}} } \right) }{\varGamma (\alpha _q )}\nonumber \\ &\times \vartheta _q ^{\frac{2}{\epsilon _{tv}} } \varGamma \left( 1-\frac{2}{\epsilon _{tv} }\right) \varGamma \left( \alpha _q +\frac{2}{\epsilon _{tv} }\right) \end{aligned}$$

and in this case $\gamma$ is given by $T_c/\eta _{cfbs}^{tv} d_{cfbs}^{-\epsilon _{tv}} (h_t^{cfbs}h_r^{cfbs})^2.$ $\xi _{{tv}}$ and $\eta _{{tv}}$ denote the density and transmitted power of TV transmitter.

Appendix 5: Proof of Theorem 2

From (19) and (36) AER for MBS tier with co-tier and cross tier interference can be written as

$$\begin{aligned} {\mathfrak {R}}_{\mathrm {mc}}^{mbs}&=-\int _0^{\infty }ln(1+w)\left( \frac{1}{2}-\frac{1}{\pi }{\int _0^{\infty } \frac{\mathcal {I}\{\psi (y)e^{-i y z}\}}{y}}dy\right) dw,\nonumber \\ &=-\int _0^{\infty }\frac{ln(1+w)}{2}dw\nonumber \\ &+\int _0^{\infty }{\int _0^{\infty }\frac{ln(1+w)\mathcal {I}\{\psi (y) e^{-i y z}\}}{\pi y}}dydw \end{aligned}$$

(66)

where $\psi (y)=\frac{e^{-y^{2/\epsilon }w^{2/\epsilon } {\mathfrak {B}_{\mathfrak {mc}}^{{\mathfrak {R}}}}e^{-i\pi /\epsilon }}}{(1+i\vartheta _q y)^{\alpha _q}}.$

$$\begin{aligned} {\mathfrak {B}_{\mathfrak {mc}}^{{\mathfrak {R}}}}=\frac{\pi \left( \xi _{cfbs} \eta _{cfbs}^{2/\epsilon }+\xi _{mbs} \eta _{mbs}^{2/\epsilon } \right) \vartheta _q ^{2/\epsilon } \varGamma \left( 1-\frac{2}{\epsilon }\right) \varGamma \left( \alpha _q+\frac{2}{\epsilon }\right) }{(\eta _{mbs} d_{mbs}^{-\epsilon })^{2/\epsilon }\varGamma (\alpha _q )} \end{aligned}$$

By substituting $-ln(1+w)=\tan (v)$ in (66), we have $w=e^{-\tan (v)}-1.$ Thus, (66) can be written as

$$\begin{aligned} {\mathfrak {R}}_{\mathrm {mc}}^{mbs}&= \int _0^{\pi /2 }\frac{\tan (v)\sec ^2(v)e^{-\tan (v)}}{2}dv\nonumber \\ &- \int _0^{\pi /2 }{\int _0^{\infty }\frac{\tan (v)\sec ^2(v)e^{-\tan (v)} \mathcal {I}\{\psi (y)e^{-i y z}\}}{\pi y}}dydv \end{aligned}$$

(67)

where $\psi (y)=\frac{e^{-y^{2/\epsilon }(e^{-\tan (v)}-1)^{2/\epsilon } {\mathfrak {B}_{\mathfrak {mc}}^{{\mathfrak {R}}}}e^{-i\pi /\epsilon }}}{(1+i\vartheta _q y)^{\alpha _q}}.$

Similar to outage analysis and using [46], (67) can be expressed as

$$\begin{aligned} {\mathfrak {R}}_{\mathrm {mc}}^{mbs}=\frac{1}{2}-\int _0^{\pi /2}{\int _0^{\infty } \frac{\tan (v)\sec ^2(v)e^{-\tan (v)}\mathcal {I}\{\psi (y)e^{-i y z}\}}{\pi y}}dydv \end{aligned}$$

To obtain the exact finite range integral expression, we substitute $y = \tan (x)$ in (68).

$$\begin{aligned} {\mathfrak {R}}_{\mathrm {mc}}^{mbs}&= \frac{1}{2}-\frac{1}{\pi }\int _0^{\pi /2 }{\int _0^{\infty }\frac{e^{-y^{2/\epsilon } {\mathfrak {B}_{\mathfrak {mc}}^{{\mathfrak {R}}}}e^{-\frac{2 \tan (v)}{\epsilon }} \left( e^{\tan (v)}-1\right) ^{2/\epsilon }\cos \left( \frac{\pi }{\epsilon }\right) }}{y\left( y^2 \vartheta ^2+1\right) ^{-\alpha /2}}} \tan (v)\sec ^2(v)e^{-\tan (v)}\nonumber \\ &\times \sin \left( {\mathfrak {B}_{\mathfrak {mc}}^{{\mathfrak {R}}}}e^{-\frac{2 \tan (v)}{\epsilon }} \left( e^{\tan (v)}-1\right) ^{2/\epsilon } y^{2/\epsilon } \sin \left( \frac{\pi }{\epsilon } \right) - \alpha \tan ^{-1} (y\vartheta )\right) dydt \end{aligned}$$

(68)

Thus, after simplification it can be written as

$$\begin{aligned} {\mathfrak {R}}_{\mathrm {mc}}^{mbs}&= \frac{1}{2}-\frac{2}{\pi }\int _0^{\pi /2 }{\int _0^{\pi /2}\frac{e^{-\tan (x)^{2/\epsilon } {\mathfrak {B}_{\mathfrak {mc}}^{{\mathfrak {R}}}}e^{-\frac{2 \tan (v)}{\epsilon }} \left( e^{\tan (v)}-1\right) ^{2/\epsilon } \cos \left( \frac{\pi }{\epsilon }\right) }}{\sin (2x)\left( \tan (x)^2 \vartheta _q^2+1\right) ^{\alpha _q /2}}} \tan (v)\sec ^2(v)e^{-\tan (v)}\nonumber \\ &\times \sin \left( {\mathfrak {B}_{\mathfrak {mc}}^{{\mathfrak {R}}}}e^{-\frac{2 \tan (v)}{\epsilon }} \left( e^{\tan (v)}-1\right) ^{2/\epsilon } \tan (x)^{2/\epsilon }\sin \left( \frac{\pi }{\epsilon } \right) - \alpha _q\tan ^{-1} (\tan (x)\vartheta _q)\right) dxdv\nonumber \\ {\mathfrak {R}}_{\mathrm {mc}}^{mbs}&= \frac{1}{2}-\frac{2}{\pi }\int _0^{\pi /2 }{\int _0^{\pi /2} \frac{e^{-\mathcal {A} {\mathfrak {B}_{\mathfrak {mc}}^{{\mathfrak {R}}}} \cos \left( \frac{\pi }{\epsilon }\right) }}{\sin (2x)\left( \tan (x)^2 \vartheta _q^2+1\right) ^{\alpha _q /2}}} \mathcal {C}\sin \left( \mathcal {A}{\mathfrak {B}_{\mathfrak {mc}}^{{\mathfrak {R}}}} \sin \left( \frac{\pi }{\epsilon } \right) - \alpha _q\tan ^{-1} (\tan (x)\vartheta _q) \right) dxdv, \end{aligned}$$

(69)

where $\mathcal {A}=\tan (x)^{2/\epsilon }e^{-\frac{2 \tan (v)}{\epsilon }} \left( e^{\tan (v)}-1\right) ^{2/\epsilon }$, $\mathcal {C}=\tan (v)\sec ^2(v)e^{-\tan (v)}.$