Interference and SINR coverage in spatial non-slotted Aloha networks

Abstract

In this paper, we propose two analytically tractable stochastic-geometric models of interference in ad hoc networks using pure (non-slotted) Aloha as the medium access. In contrast to the slotted model, the interference in pure Aloha may vary during the transmission of a tagged packet. We develop closed-form expressions for the Laplace transform of the empirical average of the interference experienced during the transmission of a typical packet. Both models assume a power-law path-loss function with arbitrarily distributed fading and feature configurations of transmitters randomly located in the Euclidean plane according to a Poisson point process. Depending on the model, these configurations vary over time or are static. We apply our analysis of the interference to study the signal-to-interference-and-noise ratio (SINR) outage probability for a typical transmission in pure Aloha. The results are used to compare the performance of non-slotted Aloha to slotted Aloha, which has almost exclusively been previously studied in the context of wired networks.

Appendix

1.1 Proof of Proposition 3.4

Each node X _j , j≠0 sends only two packets which can potentially interfere with the transmission of the packet of node X ₀=0, which starts at time T ₀(0)=0. These are the last transmission of (X _j ) which starts before time 0 (at time T _j (0)) and the first transmission which starts after time 0 (at time T _j (1) with T _j (0) ≤ 0 < T _j (1)). With this notation, we have

$$ I^{\text{mean}}= \sum\limits_{X_{j}\in{\Phi}}{F_{j}^{0}}h(T_{j}(0))/l(|X_{j}|)+ {F_{j}^{1}}h(T_{j}(1))/l(|X_{j}|) \,. $$

(7)

To simplify, let us denote $F_{j}(0) = {F_{j}^{0}}(0)$, $F_{j}(1) = {F_{j}^{0}}(1)$, H _j (0) = h (T _j (0)), H _j (1) = h (T _j (1)). Let us consider the marked point process

$$\tilde{\Phi}=\{X_{j},\,((F_{j}(0),F_{j}(1),H_{j}(0),H_{j}(1))): j\not=0\}\,.$$

It is an independently marked Poisson process with points {X _j : j ≠ 0} and marks {((F _j (0), F _j (1), H _j (0), H _j (1))} (independent across j, given points). Note that for given j, F _j (0), F _j (1) and the vector (H _j (0), H _j (1)) are mutually independent; however, H _j (0) and H _j (1) are not independent from each other (we describe their joint distribution below). Let us consider the following mapping of $\tilde {\Phi }$

$$\tilde{\Psi}\,=\,\left\{(F_{j}(0)/l(X_{j}))^{-1},\,\left((F_{j}(1)/l(X_{j}))^{-1}, H_{j}(0),H_{j}(1)\right)\right\}$$

considered again as a marked point process, with points {V _j : = F _j (0) /l (X _j ))⁻¹} and marks ((F _j (1)/l (X _j ))⁻¹, H _j (0), H _j (1)). By the displacement theorem(see ([2], Theorem 1.10) and [10]), it is again an independently marked Poisson point process, with intensity (of points) Λ(0, s) = a s ^2/β where $a=\frac {\lambda \pi {\mathbf E}[F^{\frac {2}{\beta }}]}{A^{2}}$. Regarding its marks, (H _j (0), H _j (1)) are identically distributed vectors (as in $\tilde {\Phi }$). However, (F _j (1)/l (X _j ))⁻¹, being independent of (H _j (0), H _j (1)), has a distribution which depends on the value of V _j . This distribution can be represented as follows: the law of (F _j (1)/l (X _j ))⁻¹ given V _j = v is equal to the law of l (R _v )/F where

$${\mathbf P}(R_{v} \leq r)= \frac{1}{{\mathbf E}[F^{\frac{2}{\beta}}]} {\mathbf E}\left[\mathbb{1}{\kern-2.5pt}{\mathrm{I}}\left(F \leq \frac{l(r)}{v}\right) F^{\frac{2}{\beta}}\right]\,,$$

with F, F′ having the same distribution of fading (hence, the same as F _j (0) and F _j (1)). Using these observations and the well-known formula for the Laplace transform of the Poisson point process, we obtain

$$\begin{array}{@{}rcl@{}} {\mathcal{L}}_{I^{\textrm{mean}}}(\xi)&=& {\mathbf E}\left[e^{-\xi I}\right]\\&=& \exp\left(\!\!-\!\int_{0}^{\infty}\! \left(\!1\!-\! {\mathbf E}\left[e^{-\xi\left(\frac{H(0)}{v}+\frac{H(1)F}{l(R_{v})}\right)}\right] \right)\!\Lambda(\mathrm{d} v)\!\right) \!\!\!\,. \end{array} $$

We focus now on the joint distribution of (H _j (0), H _j (1)). Let U[x, y] be the uniform law on [x, y] and 𝜖 ₀, 𝜖 ₁ two independent exponential variables of rate 𝜖. According to the renewal theory (see, e.g., [3, Eq. 1.4.3]), we have the following result. T(0) = U[− B,0] with probability $\frac {\epsilon B}{1+\epsilon B}$ and T(0) = −(B + 𝜖 ₀) with probability $\frac {1}{1+\epsilon B}$. T(1) = B + T(0) + 𝜖 ₁ if T(0) > − B and T(1) = 𝜖 ₁ otherwise. Thus, we have H(0)=h(−U) and H (1) = h(−U+B+𝜖 ₁) with probability $\frac {\epsilon B}{1+\epsilon B}$ and H(0)=h(−B − 𝜖 ₀) and H (1) = h (𝜖 ₁) with probability $\frac {1}{1+\epsilon B}$, where U is U[0,B]. Consequently,

$$\begin{array}{@{}rcl@{}} &&{\kern-1.2pc}{\mathbf E}\left[{\mathbf E}[e^{-\xi(\frac{H(0)}{v}+\frac{H(1)F}{l(R_{v})}|{H}(0),H(1)]}]\right]\\&=& {\mathbf E}\left[e^{-\xi \frac{H(-0)}{v}} {\mathcal{L}}_{F/l(R_{v})}(\xi H(1))\right] \\ &=& \frac{\epsilon B}{1+\epsilon B} {\mathbf E}\left[e^{-\xi \frac{h(-U)}{v}} {\mathcal{L}}_{F/l(R_{v})}(\xi h(-U\!+\!B\!+\!\epsilon_{1}))\right] \\&&+ \frac{1}{1+\epsilon B} {\mathbf E}\left[e^{-\xi \frac{h(-B-\epsilon_{0} )}{v}} {\mathcal{L}}_{F/l(R_{v})}(\xi h(\epsilon_{1}))\right] \, \\ \end{array} $$

where

$$\begin{array}{@{}rcl@{}} {\mathcal{L}}_{F/lR_{v}}(\xi) & = & {\mathbf E}\left[{\mathbf E}[e^{-\xi F/l(R_{v})}]\right] \\ & = &\frac{1}{\mathbf{E}[F^{\prime{2/\beta}}]} {\mathbf E}\left[F^{\prime 2/\beta} e^{-\xi F/l(vF^{\prime})^{1/\beta}}\right]\,. \end{array} $$

with $F^{\prime }$ being independent of F with the same distribution. Note that ${\mathcal {L}}_{F/l(R_{v})}(\xi )=\tilde {\mathcal {L}}(\xi /v)$, where $\tilde {\mathcal {L}}(\xi )=\frac {1}{\mathbf {E} [F^{ \prime 2/\beta }]}{\mathbf E}\big [ {\mathcal {L}}_{F}(\xi /F^{\prime })\big ]$. Thus, we have

$$\begin{array}{@{}rcl@{}} &&{\kern-1.1pc}{\mathbf E}\left[{\mathbf E}[e^{-\xi(\frac{H(0)}{v}+\frac{H(1)F}{l(R_{v})}|{H}(0),H(1))}]\right]\\ &=& \frac{\epsilon B}{(1+\epsilon B)B} {\int_{0}^{B}} e^{-\xi h(-u)/v} {\mathbf E}[\tilde{{\mathcal{L}}}(\xi h(-u\!+\!B\!+\!\epsilon_{1})/v)] \mathrm{d} u \\ && + \frac{1}{1+\epsilon B} {\mathbf E}[e^{-\xi h((-B-\epsilon_{0})/v)}] {\mathbf E}[\tilde{{\mathcal{L}}}(\xi h(\epsilon_{1})/v)] \\ & = & \frac{\epsilon B}{(1+\epsilon B)B} {\int_{0}^{B}} e^{-\xi h(-u)/v} {\mathbf E}[\tilde{{\mathcal{L}}}(\xi h(-u\!+\!B\!+\!\epsilon_{1})/v)] \mathrm{d} u \\ && + \frac{1}{1+\epsilon B} {\mathbf E}[\tilde{{\mathcal{L}}}(\xi h(\epsilon_{1})/v)]. \end{array} $$

We set

$$\begin{array}{@{}rcl@{}} E_{1}&=& {\mathbf E}[\tilde{{\mathcal{L}}}(\xi h(-u+B+\epsilon_{1})/v)] \\ E_{2}&=& {\mathbf E}[\tilde{{\mathcal{L}}}(\xi h(\epsilon_{1})/v)]\\ F_{1}&=&\frac{1}{B} {\int_{0}^{B}} e^{-\xi h(-u)/v} E_{1} \mathrm{d} u. \end{array} $$

Let us denote η=ξ/v and calculate

$$\begin{array}{@{}rcl@{}} E_{1}&= \epsilon {{\int}_{0}^{u}} e^{-\epsilon s} \tilde{{\mathcal{L}}}\Bigl(\frac{\eta (u-s)}{ B}\Bigr) \mathrm{d} s + e^{-\epsilon u}\\ E_{2}&= \epsilon {{\int}_{0}^{B}} e^{-\epsilon s} \tilde{{\mathcal{L}}}\Bigl (\frac{\eta (B-s)}{ B}\Bigr) \mathrm{d} s + e^{-\epsilon B}\\ \end{array} $$

and

$$\begin{array}{@{}rcl@{}} F_{1}&=&\frac{\epsilon}{B} {{\int}_{0}^{B}} e^{-\eta (1-u/B)} {{\int}_{0}^{u}} e^{-\epsilon s} \tilde{{\mathcal{L}}}\Bigl(\eta\frac{u-s}{B}\Bigr) \mathrm{d} s \mathrm{d} u \\&&+ \frac{1}{B} {{\int}_{0}^{B}} e^{-\eta(1-u/B)} e^{-\epsilon u}\mathrm{d} u\,. \end{array} $$

Using the change of variable $\frac {u-s}{B}=t$, we obtain

$$\begin{array}{@{}rcl@{}} F_{1}&=& {\int_{0}^{1}} \frac{\epsilon B}{\eta- \epsilon B}e^{\epsilon Bt-\eta}\bigl(e^{\eta-\epsilon B}-e^{t\eta-t\epsilon B }\bigr) {\mathcal{\tilde{L}}}(\eta t) \mathrm{d} t \\ &&+ \frac{e^{-\eta}}{\epsilon B -\eta }(1-e^{\eta-\epsilon B})\\ &=& \frac{\epsilon B e^{-\epsilon B}}{\eta- \epsilon B} {\int_{0}^{1}} e^{\epsilon B t} (1-e^{(\eta-\epsilon B)(t-1) }) {\mathcal{\tilde{L}}}(\eta t) \mathrm{d} t\\&&+ \frac{e^{-\eta}}{\epsilon B -\eta }(1-e^{\eta-\epsilon B}).\\ \end{array} $$

denoting

$$\begin{array}{@{}rcl@{}} F_{2}=E_{2}&=& \epsilon B e^{-\epsilon B} {\int_{0}^{1}} e^{\epsilon tB}{\mathcal{\tilde{L}}}(\eta t) \mathrm{d} t + e^{-\epsilon B}\\ \end{array} $$

we have

$$\begin{array}{@{}rcl@{}} {\mathcal{L}}_{I^{\textrm{mean}}}(\xi)&=& \exp \Big[-\int_{0}^{\infty} \big(1-\frac{\epsilon B}{(1+\epsilon B)} F_{1}\!-\! \frac{1}{(1+\epsilon B)} F_{2} \big) \Lambda(dv) \Big] \end{array} $$

which gives the result presented.