Interference calculation in asynchronous random access protocols using diversity

Abstract

The use of Aloha-based random access protocols is interesting when channel sensing is either not possible or not convenient and the traffic from terminals is unpredictable and sporadic. In this paper an analytic model for packet interference calculation in asynchronous random access protocols using diversity is presented. The aim is to provide a tool that avoids time-consuming simulations to evaluate packet loss and throughput in case decodability is still possible when a certain interference threshold is not exceeded. Moreover the same model represents the groundbase for further studies in which iterative interference cancellation is applied to received frames.

Appendix

In this section, the validity of the interference model is proven by demonstrating that the sum of all contributions is equal to 1. For simplicity, the demonstration is cut in four parts, each corresponding to the probability that the first event was (A), (B), (C) or (D) respectively and considering the union of the events for the second DP’s copy. Therefore

$$\begin{aligned}&Pr\{(A,CD)_{\tau '}\}+\sum _{x'=1}^{\tau '}Pr\{(B,BCD)_{x'}\} \nonumber \\&+ \sum _{x'=0}^{\tau '}Pr \left\{ (C,ABCD)_{{x'}} \right\} \nonumber \\&\quad +\sum _{x'=0}^{\tau '}Pr\{(D,ABCD)_{x'}\}=1 \end{aligned}$$

(29)

Any of the four terms are thoroughly discussed in the followings.

• (i)
  $$\begin{aligned} Pr\{(A,CD)_{\tau '}\}=\frac{1}{T_F'-\tau '+1} \end{aligned}$$

  (30)

  represents the case in which the first copy of the DP entirely interfered with the CC. In this case, the two possible events for the second DP’s copy are (C) and (D). This quantity has already been calculated and explained in Eq. (13).

• (ii)
  $$\begin{aligned}&\sum _{x'=1}^{\tau '}Pr\{(B,BCD)_{x'}\} = Pr\{(B,B)_{\tau '}\} \nonumber \\&\quad +\sum _{x'=1}^{\tau '-1}Pr\{(B,CD)_{x'}\} +\sum _{x'=1}^{\tau '-1}Pr\{(B,B)_{x'}\} \end{aligned}$$

  (31)

  represents the case in which the first copy of the DP partially interfered with the CC. In this case, the possible events for the second DP’s copy are three: (B), (C) and (D). Substituting the values from Eqs. (17), (19), (20) and considering the common factors

  $$\begin{aligned}&\sum _{x'=1}^{\tau '}Pr\{(B,BCD)_{x'}\} \nonumber \\&\quad = \Bigg [\frac{1}{T_F'-\tau '+1}\cdot \frac{2}{T_F'-\tau '+1-(2\tau '-1)}\Bigg ] \nonumber \\&\qquad \cdot \Bigg [(\tau '-1) \!+\! \sum _{x'=1}^{\tau '-1} \big [T_F'-\tau '+1-(2\tau '-1)-(\tau '-x')\big ] \nonumber \\&\qquad +\sum _{x'=1}^{\tau '-1} (x'-1)\Bigg ] \end{aligned}$$

  (32)

  that can be rewritten as

  $$\begin{aligned}&\sum _{x'=1}^{\tau '}Pr\{(B,BCD)_{x'}\} \nonumber \\&\quad = \Bigg [\frac{1}{T_F'-\tau '+1}\cdot \frac{2}{T_F'-\tau '+1-(2\tau '-1)}\Bigg ]\cdot \nonumber \\&\qquad \left[ (\tau '-1) + (\tau '-1)\cdot \left[ T_F'-\tau '+1-(2\tau '-1)\right] \right. \nonumber \\&\qquad \left. -\sum _{x'=1}^{\tau '-1}(\tau '-x') + \sum _{x'=1}^{\tau '-1} (x'-1)\right] \end{aligned}$$

  (33)

  Considering that

  $$\begin{aligned} -\sum _{x'=1}^{\tau '-1}(\tau '-x')=-(\tau '-1)-\sum _{x'=1}^{\tau '-1} (x'-1) \end{aligned}$$

  (34)

  Equation (36) becomes

  $$\begin{aligned}&\sum _{x'=1}^{\tau '}Pr\{(B,BCD)_{x'}\} \nonumber \\&\quad = \Bigg [\frac{1}{T_F'-\tau '+1}\cdot \frac{2}{T_F'-\tau '+1-(2\tau '-1)}\Bigg ]\cdot \nonumber \\&\qquad \times \Bigg [(\tau '-1)\cdot [T_F'-\tau '+1-(2\tau '-1)]\Bigg ] \end{aligned}$$

  (35)

  Therefore

  $$\begin{aligned} \sum _{x'=1}^{\tau '}Pr\{(B,BCD)_{x'}\}=\frac{2(\tau '-1)}{T_F'-\tau '+1} \end{aligned}$$

  (36)
• (iii)
  $$\begin{aligned}&\sum _{x'=0}^{\tau '}Pr\{(C,ABCD)_{x'\}} = Pr\{(C,A)_{\tau '}\} \nonumber \\&\quad + \sum _{x'=1}^{\tau '-1}Pr\{(C,B)_{x'}\} +Pr\{(C,CD)_{0}\} \end{aligned}$$

  (37)

  represents the case in which the first copy of the DP did not interfere with the CC and did not change the number of possible interference outcomes. In this case, the possible events for the second DP’s copy are still four: (A), (B), (C) and (D). Substituting the values from Eqs. (16), (21), (24) and considering the common factors

  $$\begin{aligned}&\sum _{x'=0}^{\tau '}Pr\{(C,ABCD)_{x'\}} = \Bigg [\frac{T_F'-\tau '+1-(4\tau '-1)}{T_F'-\tau '+1}\Bigg ] \nonumber \\&\cdot \Bigg [\frac{1+\sum _{x'=1}^{\tau '-1}2+[T_F'-\tau '+1-2(2\tau '-1)]}{T_F'-\tau '+1-(2\tau '-1)}\Bigg ] \end{aligned}$$

  (38)

  With some simple mathematical passages it can be seen that the term on the second big square brackets equals $1$, therefore

  $$\begin{aligned} \sum _{x'=0}^{\tau '}Pr\{(C,ABCD)_{x'\}}=\frac{T_F'-\tau '+1-(4\tau '-1)}{T_F'-\tau '+1}\nonumber \\ \end{aligned}$$

  (39)
• (iv)
  $$\begin{aligned}&\sum _{x'=0}^{\tau '}Pr\{(D,ABCD)_{x'}\} = Pr\{(D,A)_{\tau '}\} \nonumber \\&\quad + \sum _{x'=1}^{\tau '-1}Pr\{(D,B)_x'\} + Pr\{(D,CD)_0\} \end{aligned}$$

  (40)

  represents the case in which the first copy of the DP did not interfere with the CC but the number of possible interference outcomes is modified. In this case, the possible events for the second DP’s copy are (A), (B), (C) and (D). Substituting the values gathered from Eqs. (16), (22), (25) and considering the common factors

  $$\begin{aligned}&\sum _{x'=0}^{\tau '}Pr\{(D,ABCD)_{x'}\} \nonumber \\&\quad = \Bigg [\frac{1}{[T_F'-\tau '+1]\cdot [T_F'-\tau '+1-(2\tau '-1)]}\Bigg ] \nonumber \\&\qquad \cdot \left[ 2\tau ' + \sum _{x'=1}^{\tau '-1} 2x' + \sum _{x'=1}^{\tau '-1} 4(\tau '-x')\right. \nonumber \\&\qquad \left. \quad + \sum _{x'=1}^{\tau '-1} 2[T_F'-\tau '+1-(3\tau '+x'-1)]\right] \end{aligned}$$

  (41)

  Considering that

  $$\begin{aligned} \sum _{x'=1}^{\tau '-1} x' = \sum _{x'=1}^{\tau '-1} (\tau '-x') = \frac{(\tau ')\cdot (\tau '-1)}{2} \end{aligned}$$

  (42)

  Equation (41) can be rewritten as

  $$\begin{aligned}&\sum _{x'=0}^{\tau '}Pr\{(D,ABCD)_{x'}\} \nonumber \\&\quad = \Bigg [\frac{1}{[T_F'-\tau '+1]\cdot [T_F'-\tau '+1-(2\tau '-1)]}\Bigg ]\cdot \nonumber \\&\qquad \left[ 2\tau ' +2\frac{(\tau ')\cdot (\tau '-1)}{2} + 4\frac{(\tau ')\cdot (\tau '-1)}{2} \right. \nonumber \\&\qquad \left. +\,2\tau '\cdot [T_F'-\tau '+1-(3\tau '-1)] - 2\frac{(\tau ')\cdot (\tau '-1)}{2}\right] \nonumber \\ \end{aligned}$$

  (43)

  from which

  $$\begin{aligned}&\sum _{x'=0}^{\tau '}Pr\{(D,ABCD)_{x'}\} \nonumber \\&\quad = \Bigg [\frac{1}{[T_F'-\tau '+1]\cdot [T_F'-\tau '+1-(2\tau '-1)]}\Bigg ]\cdot \nonumber \\&\qquad \times \Bigg [2\tau ' \cdot [T_F'-\tau '+1-(2\tau '-1)]\Bigg ] \end{aligned}$$

  (44)

  and simplifying

  $$\begin{aligned} \sum _{x'=0}^{\tau '}Pr\{(D,ABCD)_{x'}\}=\frac{2\tau '}{T_F'-\tau '+1} \end{aligned}$$

  (45)

It can be seen that the sum of the terms found in Eqs. (30),(36),(39) and (45) is equal to 1. This validates Eq. (29).