Connectivity of ad hoc wireless networks: an alternative to graph-theoretic approaches

Abstract

Connectivity in wireless ad hoc and sensor networks is typically analyzed using a graph-theoretic approach. In this paper, we investigate an alternative communication-theoretic approach for determining the minimum transmit power required for achieving connectivity. Our results show that, if there is significant multipath fading and/or multiple access interference in the network, then graph-theoretic approaches can substantially underestimate the minimum transmit power required for connectivity. This is due to the fact that graph-theoretic approaches do not take the route quality into consideration. Therefore, while in scenarios with line-of-sight (LOS) communications a graph-theoretic approach could be adequate for determining the minimum transmit power required for connectivity, in scenarios with strong multipath fading and/or multiple access interference a communication-theoretic approach could yield much more accurate results and, therefore, be preferable.

Appendices

Appendix 1: Derivation of the BER Floor for the Strong LOS Case

Consider the link between the transmitter (node Tx) and the receiver (node Rx) shown in Fig. 9. For simplicity, we will first consider the scenario where there is only one interfering node within the effective interference region, which is the circular area of radius 2r ₀ around the receiver. We will also assume that r ₀ ≥ 1, which will be true for all the scenarios considered in this paper. The amplitude of the observable signal can generally be written as

$$ S_{\rm r} = S_{\rm sig} + S_i + W_{\rm therm} $$

where S _sig is the amplitude of the signal transmitted by node Tx, S _i is the amplitude of the signal from the interfering node, and W _therm is the amplitude of the additive white Gaussian noise (AWGN) with variance P _therm.

Fig. 9

A scenario where there is only one potential interfering node in the effective interference region

Assuming free space propagation loss, the received signal power observed at the receiver can be written as

$$ P_r = {\frac{\alpha P_{\rm t}}{r_0^2}} $$

where $\alpha = {\frac{G_{\rm t} G_{\rm r} c^2}{(4 \pi)^2 f_{\rm c}^2}}, G_{\rm t}$ and G _r are the transmitter and the receiver antenna gains, f _c is the carrier frequency, c is the speed of light, P _t is the transmit power, and r ₀ is the distance between the transmitter and the receiver. Assuming BPSK modulation, the amplitude of the signal transmitted by node Tx observed at the receiver can then be written as

$$ S_{\rm sig} =\begin{array}{ll}{\sqrt{\alpha {E}_{\rm b}}/r_0} & \hbox{if} +\hbox{1 is transmitted}\\ -{\sqrt{\alpha {E}_{\rm b}}/r_0} & \hbox{if} -\hbox{1 is transmitted}\end{array} $$

where $\sqrt{E_{\rm b}} \triangleq \sqrt{{P_{\rm t}}/{R_{\rm b}}}$ is the transmit bit energy and R _b is the data-rate. Let z be the distance between the interfering node and the receiver. Similarly, the amplitude of the interfering signal can be written as

$$ S_i = \left \{\begin{array}{ll}{\sqrt{\alpha E_{\rm b}}/z} &\hbox{if} +1 \hbox{is transmitted}\hfill \\ -{\sqrt{\alpha E_{\rm b}}/z} &\hbox{if }-1 \hbox{is transmitted}\hfill \\ 0 &\hbox{if the interfering node does not transmit.}\end{array}\right. $$

(34)

Assuming a simple random access MAC protocol with Poisson transmission, the probability that the interfering node will interfere with an ongoing transmission can be given as

$$ p = 1 - e^{-{\frac{\lambda L}{R_{\rm b}}}} $$

(35)

where λ is the packet transmission rate in pck/s and L is the number of bits in a packet. From (34) and (35), the probability mass function (PMF) of the amplitude of the interfering signal can then be given as

$$ P\{S_i\} =\left\{\begin{array}{ll}{\frac{1}{2}} p &\hbox{if}\quad S_i = \sqrt{ \alpha E_{\rm b} }/z \hfill \\ {\frac{1}{2}} p &\hbox{if}\quad S_i = -\sqrt{\alpha E_{\rm b}}/z \hfill \\ 1 - p &\hbox{if}\quad S_i = 0.\end{array}\right. $$

Since we are considering a binary modulation, due to symmetry we can assume that node Tx transmits “+1” (i.e., $S_{\rm sig} = \sqrt{\alpha E_{\rm b}} / r_0$). Given that there is only one potential interfering node, the bit error probability can be expressed as follows [28]:

$$ \begin{aligned} P\{\hbox{bit error}| \hbox{1 interfering node at distance}\,z\}\\ &= P\{S_{\rm r} < 0 | S_i = \sqrt{ \alpha E_{\rm b} }/z \} P\{ S_i = \sqrt{ \alpha E_{\rm b} }/z \}\\ & \quad+ P\{S_{\rm r} < 0 | S_i =-\sqrt{\alpha E_{\rm b} }/z \} P\{ S_i = -\sqrt{ \alpha E_{\rm b}}/z \}\\ &\quad +P\{S_{\rm r} < 0 | S_i = 0 \} P\{ S_i = 0 \}\\ &={\frac{p}{2}} Q\left({\frac{\sqrt{\alpha E_{\rm b}}}{\sigma}} \left( {\frac{1} {r_0}} + {\frac{1}{z}}\right) \right) + {\frac{p} {2}} Q\left({\frac{\sqrt{\alpha E_{\rm b}}}{\sigma}}\left( {\frac{1}{r_0}} - {\frac{1}{z}}\right)\right)\\ & \quad +(1-p) Q\left({\frac{\sqrt{\alpha E_{\rm b}}}{\sigma r_0}}\right) \end{aligned} $$

(36)

where $\sigma = \sqrt{FkT_0/2}, F$ is the noise figure, k = 1.38 × 10⁻²³J/K, T ₀ = 300 K is the room temperature, and $Q(x) = \int_{x}^{\infty} {\frac{1}{\sqrt{2 \pi}}} e^{-u^2/2} {\rm d} u$ is the standard Q-function.

As observed from the simulation results, at a particular node spatial density, if the transmit power is high enough, the link BER converges to a BER floor. Thus, to derive the BER floor, we take the limit, as E _b approaches ∞, of the conditional link BER in (36), obtaining:

$$ \begin{aligned} \lim_{E_{\rm b} \rightarrow \infty}P\{\hbox{bit error} | \hbox{1 interfering node at distance}\,z \}\\ &= \lim_{E_{\rm b}\rightarrow \infty} \left[ {\frac{p}{2}} Q \left({\frac{\sqrt{\alpha E_{\rm b}}}{\sigma}} \left({\frac{1}{r_0}} + {\frac{1} {z}}\right) \right) + {\frac{p}{2}} Q\left({\frac{\sqrt{\alpha E_{\rm b}}}{\sigma}} \left({\frac{1}{r_0}} - {\frac{1}{z}}\right) \right)\right.\\ &\quad\left. +(1-p) Q\left({\frac{\sqrt{\alpha E_{\rm b}}}{\sigma r_0}}\right)\right]\\ &= \lim_{E_{\rm b}\rightarrow \infty}{\frac{p}{2}}Q\left({\frac{\sqrt{\alpha E_{\rm b}}}{\sigma}} \left({\frac{1}{r_0}} - {\frac{1}{z}}\right) \right) \end{aligned} $$

(37)

where in the last passage we have used the fact that $Q(+\infty) = 0.$ Note that the argument of the Q-function in the last line in (37) can either be positive or negative, depending on the value of z. To find the total bit error probability, we have to consider all possible values of z. With a straightforward algebra, it can be shown that the PDF of z is

$$ f_Z(z) = {\frac{z}{2 r_0^2}}\quad 0 \leq z \leq 2 r_0. $$

Given that there is only one potential interfering node, the link BER floor can then be written as⁸

$$ \begin{aligned} \{\hbox{BER}_{\rm floor}^{\rm LOS}|\hbox{1 interfering node}\}\\ = &\int\lim_{E_{\rm b} \rightarrow \infty}P\{\hbox{bit error} | 1 \hbox{interf. node at distance}\,z \}f_Z(z)\hbox{d} z \\ = &\int\limits_{0}^{1} \lim_{E_{\rm b} \rightarrow \infty}{\frac{p}{2}} Q\left({\frac{\sqrt{\alpha E_{\rm b}}}{\sigma}} \left({\frac{1} {r_0}} - 1\right) \right){\frac{z}{2 r_0^2}}\hbox{d}z\\ & +\int\limits_{1}^{2 r0} \lim_{E_{\rm b} \rightarrow \infty}{\frac{p}{2}} Q\left({\frac{\sqrt{\alpha E_{\rm b}}}{\sigma}} \left( {\frac{1} {r_0}} - {\frac{1}{z}}\right) \right) {\frac{z}{2 r_0^2}}\hbox{d} z \\ = &\int\limits_{0}^{1} {\frac{p} {2}} Q\left(-\infty \right) {\frac{z}{2r_0^2}}\hbox{d} z + \int\limits_{1}^{r_0} {\frac{p}{2}} Q\left(-\infty \right) {\frac{z}{2 r_0^2}} \hbox{d} z \\ &+\int\limits_{r_0}^{2 r_0} {\frac{p}{2}} Q\left(+\infty \right) {\frac{z} {2 r_0^2}} \hbox{d} z ={\frac{p}{8}}. \end{aligned} $$

Now, let us consider the case where there are two potential interfering nodes within the effective interference region, as shown in Fig. 10. Let the interfering node 1 be at distance z ₁ from to the receiver and let the interfering node 2 be at distance z ₂ from the receiver. Following the same approach as in the case with a single interfering node, the conditional link BER floor, given z ₁ and z ₂, can be written as

$$ \begin{aligned} \lim_{E_{\rm b} \rightarrow \infty}P\{\hbox{bit error}|\hbox{2 interf. nodes at distances}\,z_1\,\hbox{and}\,z_2 \}\\ &=\lim_{E_{\rm b} \rightarrow \infty} \left[ \left({\frac{p} {2}}\right)^2 \sum_{i,j=\pm 1} Q\left({\frac{\sqrt{\alpha E_{\rm b}}}{\sigma}} \left({\frac{1}{r_0}} + {\frac{i}{z_1}} + {\frac{j}{z_2}}\right)\right)\right.\\ &\quad + {\frac{p}{2}}(1-p) \sum_{i=\pm 1} Q\left({\frac{\sqrt{\alpha E_{\rm b}}} {\sigma}} \left( {\frac{1}{r_0}} + {\frac{i}{z_1}} \right) \right)\\ & \quad + {\frac{p}{2}} (1-p) \sum_{j=\pm 1} Q\left({\frac{\sqrt{\alpha E_{\rm b}}}{\sigma}} \left({\frac{1}{r_0}} + {\frac{j} {z_2}} \right)\right)\\ & \quad \left . + (1-p)^2Q\left({\frac{\sqrt{\alpha E_{\rm b}}}{\sigma r_0}}\right) \right]. \end{aligned} $$

For small values of p (as in operative conditions with the used MAC protocol), the terms with coefficient p ² can be neglected, and the conditional link BER floor can be approximated as

$$ \begin{aligned} \lim_{E_{\rm b} \rightarrow \infty}P\{\hbox{bit error}| \hbox{2 interf. nodes at distances}\,z_1\,\hbox{and}\,z_2\}\\ &\approx \lim_{E_{\rm b} \rightarrow \infty} {\frac{p}{2}} Q\left({\frac{\sqrt{\alpha E_{\rm b}}}{\sigma}} \left({\frac{1} {r_0}} - {\frac{1}{z_1}} \right) \right)\\ &\quad+\lim_{E_{\rm b}\rightarrow \infty} {\frac{p}{2}} Q\left({\frac{\sqrt{\alpha E_{\rm b}}}{\sigma}} \left({\frac{1}{r_0}} - {\frac{1}{z_2}} \right) \right). \end{aligned} $$

(38)

To find the average BER floor with respect to all possible positions of the pairs of interfering nodes, one can integrate the expression in (38) by weighing it with the joint PDF of z ₁ and z ₂. With straightforward algebra, the link BER floor, given that there are two potential interfering nodes, can be approximated as

$$ \{\hbox{BER}_{\rm floor}^{\rm LOS}| \hbox{2 interfering nodes}\} \approx 2 \times {\frac{p}{8}}. $$

Similarly, for the scenarios with m interfering nodes, the BER floor can be approximated as

$$ \{\hbox{BER}_{\rm floor}^{\rm LOS}| m\,\hbox{interfering nodes}\}\approx m \times {\frac{p}{8}}. $$

Fig. 10

A scenario where there are two potential interfering nodes in the effective interference region

While the observed derivation leads to the computation of the BER floor for a given number of interferers, in reality, however, the number of potential interfering nodes in the effective interference region is random. Let Y be a random variable denoting the number of nodes in the effective interference region, and let M be the number of interfering nodes. Excluding the transmitter and the receiver, the number of potential interfering nodes is M = Y − 2 (given that Y ≥ 2), and the PMF of M is

$$ \begin{aligned} P\{M = m\} = &{\frac{1}{1- e^{-\rho_{\rm s} \nu} - \rho_{\rm s} \nu e^{-\rho_{\rm s} \nu}}}{\frac{\left(-\rho_{\rm s} \nu \right)^{(m+2)}}{(m+2)!}} e^{-\rho_{\rm s} \nu}\\ & m = 0, 1, 2, \dots \end{aligned} $$

where $\nu = \pi(2 r_0)^2$ is the area of the effective interference region. Finally, the overall average BER floor becomes

$$ \begin{aligned} \hbox{BER}_{\rm floor}^{\rm LOS}&\approx \sum_{m=0}^{\infty} {\frac{p}{8}}m P\{M = m\}\\ &= {\frac{p}{8}}{\mathbb{E}}[M]\\ &= {\frac{p}{8}}\left({\frac{\rho_{\rm s} \nu + 2 + (\rho_{\rm s} \nu - 2) e^{\rho_{\rm s} \nu}}{e^{\rho_{\rm s} \nu} - \rho_{\rm s} \nu -1}}\right). \end{aligned} $$

(39)

The link BER floor given in (39) can be further simplified. Since P{Y = 0} and P{Y = 1} are typically very small with the network sizes of interest, ${\mathbb{E}}[M]$ is very close to ${\mathbb{E}}[Y] - 2,$ where ${\mathbb{E}}[Y] = \rho_{\rm s} \nu.$ It can be shown that ${\mathbb{E}}[M]$ and ${\mathbb{E}}[Y] - 2$ are almost identical [28]. Thus, the BER floor can be approximated as

$$ \hbox{BER}_{\rm floor}^{\rm LOS} \approx {\frac{p}{8}} (\rho_{\rm s} \nu- 2). $$

(40)

The validity of the approximate expression (40) has been verified through Monte Carlo simulations [28].

Appendix 2: Derivation of the BER floor in the Rayleigh fading case

In this appendix, we focus on the case with multipath fading. Consider the scenario with only one interfering node in the effective interference region, as shown in Fig. 9. With multipath fading, the amplitude of the observed signal can generally be written as

$$ S_{\rm r} = X_{\rm s} S_{\rm sig} + X_{i} S_i + W_{\rm therm} $$

where X _s and X _i are Rayleigh distributed random variables. The bit error probability given that X _s = x _s, X _i = x _i , and the interfering node is at distance z relative to the receiver can be written as

$$ \begin{aligned} P\{\hbox{bit error} | X_{\rm s} = x_{\rm s}, X_{i} = x_i, Z = z \}\\ &=P\{S_{\rm r} < 0 | S_i = \sqrt{ \alpha E_{\rm b} }/z \} P\{S_i = \sqrt{ \alpha E_{\rm b} }/z \}\\ & \quad +P\{S_{\rm r} < 0 | S_i = -\sqrt{ \alpha E_{\rm b} }/z \} P\{ S_i = -\sqrt{ \alpha E_{\rm b} }/z \} \\ & \quad + P\{S_{\rm r} < 0 | S_i = 0 \} P\{ S_i = 0 \}\\ & = {\frac{p}{2}} Q\left({\frac{\sqrt{\alpha E_{\rm b}}}{\sigma}} \left( {\frac{x_{\rm s}}{r_0}} + {\frac{x_i}{z}}\right) \right) + {\frac{p}{2}} Q\left({\frac{\sqrt{\alpha E_{\rm b}}}{\sigma}} \left( {\frac{x_{\rm s}}{r_0}} - {\frac{x_i}{z}}\right) \right) \\ & \quad +(1-p)Q\left({\frac{x_{\rm s} \sqrt{\alpha E_{\rm b}}}{\sigma r_0}} \right). \end{aligned} $$

(41)

At any node spatial density, provided that the transmit power is large enough, the link BER converges to a BER floor [28]. Thus, to derive the BER floor, we take the limit, as E _b approaches $\infty,$ of the conditional link BER in (41), obtaining

$$ \begin{aligned} \lim_{E_{\rm b} \rightarrow \infty} P\{ \hbox{bit error} | X_{\rm s} = x_{\rm s}, X_{i} = x_i, Z = z \}\\ &= \lim_{E_{\rm b} \rightarrow \infty} \left[ {\frac{p}{2}} Q\left({\frac{\sqrt{\alpha E_{\rm b}}}{\sigma}} \left( {\frac{x_{\rm s}}{r_0}} + {\frac{x_i}{z}}\right) \right) \right. \\ & \quad \left. + {\frac{p}{2}} Q\left({\frac{\sqrt{\alpha E_{\rm b}}}{\sigma}} \left( {\frac{x_{\rm s}}{r_0}} - {\frac{x_i}{z}}\right) \right) +(1-p) Q\left({\frac{x_{\rm s}\sqrt{\alpha E_{\rm b}}}{\sigma r_0}} \right) \right]\\ & = \lim_{E_{\rm b} \rightarrow \infty}{\frac{p}{2}} Q\left({\frac{\sqrt{\alpha E_{\rm b}}} {\sigma}} \left( {\frac{x_{\rm s}}{r_0}} - {\frac{x_i}{z}}\right) \right) \end{aligned} $$

(42)

where in the last passage we have used the fact that $Q(+\infty) = 0.$ Note that the argument of the Q-function in the last line in (42) can either be positive or negative, depending on the values of x _s, x _i , and z. The argument of the Q-function will be negative if ${\frac{x_{\rm s}}{r_0}} - {\frac{x_i}{z}} < 0,$ or equivalently $x_i > x_{\rm s} z/r_0.$ Thus, the probability can now be written as

$$ \begin{aligned} \lim_{E_{\rm b} \rightarrow \infty} P\{\hbox{bit error}| Z = z \} \\ & = \int \int \lim_{E_{\rm b} \rightarrow \infty}{\frac{p}{2}} Q\left({\frac{\sqrt{\alpha E_{\rm b}}}{\sigma}} \left( {\frac{x_{\rm s}}{r_0}} - {\frac{x_i}{z}}\right) \right) {\frac{x_{\rm s}} {\sigma_{\rm f}^2}}\\ &\cdot e^{-x_{\rm s}^2/2\sigma_{\rm f}^2} {\frac{x_{\rm i}} {\sigma_{\rm f}^2}} e^{-x_{ i}^2/2\sigma_{\rm f}^2} \hbox{d} x_i \hbox{d} x_{\rm s}\\ &={\frac{p}{2}} \left(1 - \int\limits_{0}^{\infty} \int\limits_{0}^{x_{\rm s} z/r_0} {\frac{x_{\rm s}}{\sigma_{\rm f}^2}} e^{-x_{\rm s}^2/2\sigma_{\rm f}^2} {\frac{x_{\rm i}}{\sigma_{\rm f}^2}} e^{-x_{ i}^2/2\sigma_{\rm f}^2} \hbox{d} x_i \hbox{d} x_{\rm s} \right)\\ &= {\frac{p}{2}} \left(1 - \int\limits_{0}^{\infty} {\frac{x_{\rm s}}{\sigma_{\rm f}^2}} e^{-x_{\rm s}^2/2\sigma_{\rm f}^2} \int\limits_{0}^{x_{\rm s} z/r_0} {\frac{x_{\rm i}}{\sigma_{\rm f}^2}} e^{-x_{ i}^2/2\sigma_{\rm f}^2} \hbox{d} x_i \hbox{d} x_{\rm s}\right)\\ & = {\frac{p}{2}} \left(1 - \int\limits_{0}^{\infty} {\frac{x_{\rm s}}{\sigma_{\rm f}^2}} e^{-x_{\rm s}^2/2\sigma_{\rm f}^2} \left[ 1 - e^{-{\frac{x_{\rm s}^2 z^2 }{2 r_0^2 \sigma_{\rm f}^2}}} \right] \hbox{d} x_{\rm s} \right)\\ & = {\frac{p}{2}} \left(1 -{\frac{z^2}{z^2 + r_0^2}} \right). \end{aligned} $$

Integrating over all possible values of z, one gets

$$ \begin{aligned} \{\hbox{BER}_{\rm floor}^{\rm Ray}| \hbox{1 interfering node} \}\\ & = \int \lim_{E_{\rm b} \rightarrow \infty} P\{\hbox{bit error} | Z = z \} f_Z(z) \hbox{d} z \\ & = \int\limits_{0}^{1} {\frac{p}{2}} \left(1 - {\frac{1}{1 + r_0^2}} \right) \hbox{d} z + \int\limits_{1}^{2 r_0} {\frac{p}{2}} \left(1 - {\frac{z^2}{z^2 + r_0^2}} \right) {\rm d} z \\ & = {\frac{p}{8 (1 + r_0^2)}} + {\frac{p}{8}} \ln\left({\frac{5 r_0^2}{1 + r_0^2}}\right) \\ & = {\frac{p}{8}} \left[ {\frac{1}{1 + r_0^2}} + \ln\left({\frac{5 r_0^2} {1 +r_0^2}}\right) \right] . \end{aligned} $$

Following the same approach as considered in the case with single interfering node, the BER floor given that there are two interfering nodes and small values of p can be approximated as [28]

$$ \{\hbox{BER}_{\rm floor}^{\rm Ray}|\hbox{2 interfering nodes}\} \approx 2 \times {\frac{p}{8}} \left[ {\frac{1}{1 + r_0^2}} + \ln\left({\frac{5 r_0^2}{1 + r_0^2}}\right) \right] $$

and the overall BER floor (without conditioning on the number of interfering nodes) can be approximated as

$$ \hbox{BER}_{\rm floor}^{\rm Ray}\approx (\rho_{\rm s} \nu - 2) {\frac{p} {8}} \left[ {\frac{1}{1 + r_0^2}} + \ln\left({\frac{5 r_0^2}{1 + r_0^2}}\right) \right]. $$