Coverage Probability Analysis of D2D Communication Based on Stochastic Geometry Model

Abstract

Relaying is a common application of D2D communication, which optimizes system capacity and increases the coverage of mobile cellular networks on shared downlink resources. We established a network model of cellular base-stations and adopted the theory of stochastic geometry. Based on the model, the coverage probability analysis of the network is analyzed to select a specific user as the relay node, and the relay point uses the forwarding strategy of the decoding and forwarding. Subsequently, D2D communication can help the edge user to communicate with the base-station. The coverage probability expression of the downlink cellular network is defined, then the coverage probability of the cellular link, the base-station to the relay link, and the relay to the edge user link are derived. Simulation results show that with the increasing of density of the macro base-stations, the coverage probability of the whole network will increase and the final coverage probability will become saturated.

Appendices

Appendix 1: Proof of Theorem 1

According to the definition of Eq. (3) and SINR, the coverage probability of the BS-UE1 link can be expressed as

$$p_{cU} = {\mathbb{E}}_{r} {\mathbb{P}}\left( {SINR_{cU} \ge \beta_{{{\text{c}}U}} |r} \right) = \int\limits_{r > 0} {{\mathbb{P}}\left( {h_{o} \ge \beta_{cU} r^{\alpha } (I_{cU} + \sigma^{2} } \right)} /P_{cU} |r)f_{r} (r)dr$$

(9)

where f_r(r) is BS probability density function (PDF) [8]. By h_o ~ exp(1), we can rewrite Eq. (3) as

$$\begin{aligned} {\mathbb{P}}(h_{o} \ge \beta_{cU} r^{\alpha } (I{}_{cU} + \sigma^{2} )/P_{cU} |r) & = {\mathbb{E}}\left[ {{\mathbb{P}}\left( {h_{o} \ge \beta_{cU} r^{\alpha } (I_{{{\text{c}}U}} + \sigma^{2} )/P_{cU} |r,I{}_{cU}} \right)} \right] \\ & = {\mathbb{E}}\left[ {\exp \left( {\beta_{cU} r^{\alpha } (I_{{{\text{c}}U}} + \sigma^{2} )/P_{cU} } \right)} \right] \\ & = \exp \left( {\beta_{cU} r^{\alpha } /P_{cU} } \right)L\left( {\beta_{cU} r^{\alpha } /P_{cU} } \right) \\ \end{aligned}$$

(10)

where ${\mathcal{L}}$ is Laplace transform of I_cU. Defined by the Laplace transform and h_o ~ exp(1), it can be written as

$$\begin{aligned} {\mathcal{L}}_{{I_{{cU}} }} (s) & = \rm{\mathbb{E}}(\exp (sI_{{cU}} ) = \rm{\mathbb{E}}\left[ {\exp \left( { - s\sum\limits_{{i \in \Phi _{C} /\{ x_{0} \} }} {P_{{{\text{cu}}}} } h_{i} r_{i}^{{ - \alpha }} } \right)} \right] \\ & = \rm{\mathbb{E}}_{{\Phi _{C} }} \left[ {\prod\limits_{{i \in \Phi _{C} /\{ x_{0} \} }} {\frac{1}{{1 + sP_{{{\text{c}}U}} r_{i}^{{ - \alpha }} }}} } \right] = \exp \left( { - 2\pi \lambda _{B} \int\limits_{r}^{\infty } {\left( {1 - \frac{1}{{1 + sP_{1} h_{i} r_{i}^{{ - \alpha }} }}} \right)udu} } \right) \\ \end{aligned}$$

(11)

The final step of the above derivation is obtained from the properties of the probability generation function of the PPP, which satisfies ${\mathbb{E}}\left[ {\prod\nolimits_{{x \in\Phi }} {g(x)} } \right] = \exp \left( { - \lambda \int_{{R^{2} }}^{\infty } {(1 - g(x)){\text{d}}x} } \right)$ [14]. S is changed by $\beta_{cU} r^{\alpha } /P_{cU}$, and the interference I_cU can be further derived as

$$\begin{aligned} {\mathcal{L}}_{{I_{{cU}} }} \left( {\beta _{1} r^{\alpha } /{\text{P}}_{{cU}} } \right) & = \exp \left( { - 2\pi \lambda _{B} \int\limits_{r}^{\infty } {\left( {1 - \frac{1}{{1 + \beta _{{cU}} r^{\alpha } u^{{ - \alpha }} }}} \right)u{\text{d}}u} } \right) \\ & = \exp \left( { - 2\pi \lambda _{B} \int\limits_{r}^{\infty } {\left( {\frac{{\beta _{{cU}} }}{{\beta _{{cU}} + (u/r)^{\alpha } }}} \right)u{\text{d}}u} } \right) \\ & = \exp ( - 2\pi \lambda _{B} \rho (\beta _{{cU}} ,\alpha )) \\ \end{aligned}$$

(12)

where $\rho (\beta_{cU} ,\alpha ){ = }\beta_{cU}^{2/\alpha } \int_{{\beta_{cu}^{ - 2/\alpha } }}^{\infty } {\frac{1}{{1 + \theta^{\alpha /2} }}} \text{d} \theta ,\theta = \frac{u}{{r\beta_{cU}^{{1/a^{2} }} }}$. And Eq. (4) can be obtained by combining Eqs. (9)–(12).

B: Proof of Theorem 2

The proof of Theorem 2 is similar to Theorem 1, except that the subscripts are different. Where $\lambda {\text{ = min(}}\lambda_{B} ,\lambda_{R} )$ value depended on p_cR.

A: Proof of Theorem 3

By the definition of Eq. (7), P_RU can be converted into

$$p_{RU} = {\mathbb{P}}\left( {\frac{{{\text{P}}_{RU} m_{o} r^{ - \alpha } }}{{I_{RU} + \sigma^{2} }} \ge \beta_{RU} } \right) = \int\limits_{0}^{R} {\mathbb{P}} (l_{o} \ge r^{\alpha } (I_{RU} + \sigma^{2} )/{\text{P}}_{RU} |r)f_{R} (r){\text{d}}r$$

(13)

In order to simplify the derivation, $I_{r} = I_{RU} + P_{RU} m_{o} r^{ - \alpha }$ is assumed to be signal transmitted from the RS, so P_RU can be further rewritten as

$$\begin{aligned} \int\limits_{0}^{R} {{\mathbb{P}}\left( {m_{o} \ge \frac{{\beta _{{RU}} r^{\alpha } (I_{{RU}} + \sigma ^{2} )}}{{(1 + \beta _{3} ){\text{P}}_{{RU}} }}|r} \right)f_{R} (r){\text{d}}r} & = \frac{2}{{R^{2} }}\int\limits_{0}^{R} {{\mathbb{E}}\left[ {\exp \left( { - \frac{{\beta _{{RU}} r^{\alpha } (I_{{RU}} + \sigma ^{2} )}}{{(1 + \beta _{{RU}} ){\text{P}}_{{RU}} }}} \right)} \right]} r{\text{d}}r \\ & = \frac{2}{{R^{2} }}\int\limits_{0}^{R} {\exp \left( { - \frac{{\beta _{{RU}} r^{\alpha } \sigma ^{2} }}{{(1 + \beta _{{RU}} ){\text{P}}_{{RU}} }}} \right)} L\left( {\frac{{\beta _{{RU}} r^{\alpha } }}{{(1 + \beta _{{RU}} ){\text{P}}_{{RU}} }}} \right)r{\text{d}}r \\ \end{aligned}$$

(14)

The above derivation uses $f_{R} (r) = 2r/{\text{R}}^{2}$ and $m_{o} \sim{ \exp }$(1). $L( \cdot )$ can be expressed as

$$L\left( {\beta_{RU} r^{\alpha } /P_{RU} } \right) = \exp ( - \pi r^{2} \lambda p_{cR} \rho (\beta_{RU} ,\alpha ))$$

(15)

where $\rho (\beta_{RU} ,\alpha ){ = }\beta_{RU}^{2/\alpha } \int_{{\beta_{RU}^{ - 2/\alpha } }}^{\infty } {\frac{1}{{1 + \theta^{\alpha /2} }}} \text{d} \theta$, $\lambda = \hbox{min} (\lambda_{B} ,\lambda_{R} )$. Combining Eqs. (13)–(15), we can get Eq. (8).