Ad Hoc Wireless Networks (A Communication-Theoretic Perspective) || Appendix A: Analysis of the Inter-node Interference

Appendix A

Analysis of the Inter-nodeInterference

A.1 IntroductionIn this appendix, we outline an analysis for the evaluation of the impact of the interference onthe route BER performance of ad hoc wireless networks with regular topology, as introducedin Chapter 2. In the literature, it is commonplace to assume that the interference noise isGaussian [33]. This assumption holds, in the limit, when the number of interferers is veryhigh. However, this assumption loses part of its validity in a scenario where there is a finitenumber of nodes. Moreover, the propagation loss suggests that only the nodes which arerelatively close to the receiver will interfere significantly. Therefore, the ‘real’ interferencenoise comes from a limited number of nodes, rather than from a large number of nodes.This suggests that the Gaussian assumption might have limited validity.

In order to investigate the validity of the Gaussian assumption in more detail, wefirst propose a rigorous detection-theoretic approach for the evaluation of the average linkBER, from which the route BER can be derived. In particular, we analyze the route BERperformance in a scenario with strong line-of-sight (LOS) and, then, we extend this analysisto a scenario with strong multipath (Rayleigh fading). Next, we evaluate the link BER underthe assumption of Gaussian interference noise, and we show the limits of validity of thisapproach. The implications of these results are considered in Chapter 3.

A.2 Exact Computation of the Average Link BER in aScenario with Strong LOS

Consider the link between nodes nS and nD shown in Figure A.1. We now distinguishbetween the cases where only the nodes in Tier 1 are considered or nodes from Tiers 1 and2 are considered. In the following, Group ij denotes the subgroup j in Tier i, i.e. the firstnumber refers to the tier order and the second to the subgroup in that tier (the groups willbe characterized more precisely in the following). Likewise, nij is used to refer to a genericnumber of nodes in Group ij and nodesij is the total number of nodes in Group ij .

Ad Hoc Wireless Networks: A Communication-Theoretic Perspective Ozan K. Tonguz and Gianluigi Ferrari© 2006 John Wiley & Sons, Ltd. ISBN: 0-470-09110-X

266 Appendix A. Analysis of the Inter-node Interference

Tier 1

Tier 2

dlink

Tier 3

nS nD

Figure A.1 Regular tiered structure for an ad hoc wireless network.

A.2.1 Interference from Nodes in Tier 1In this case, there are two groups of nodes.

Group 11. The number of nodes in this group, denoted as nodes11, is equal to 3. Each ofthem generates an interference power equal to P

(11)r = Pr = αPt/d

2link. The received

bit energy is E(11)b = Eb = Pr/Rb.

Group 12. The number of nodes in this group, denoted as nodes12, is equal to 4.Each of them generates an interference power equal to P

(12)r = Pr/2 = αPt/2d2

link.

The received bit energy is E(12)b = Eb/2.

Suppose that there is only a single node from Group 11 interfering and that this happenswith probability p (an expression for which is provided later). In this case, the observable canbe written as follows:

r = csig + c1 + wthermal.

Considering binary phase shift keying (BPSK), for symmetry we can assume that ‘+1’ istransmitted, i.e. c = √

Eb. There are two possible cases: (i) either c1 = √Eb (i.e. a ‘+1’

is transmitted) or (ii) c1 = −√Eb (i.e. a ‘−1’ is transmitted). Therefore, assuming that the

threshold for bit detection is placed at 0, one can conclude that

P {bit error|a single node from Group 11 is transmitting}= P

{r < 0|c1 = √

Eb}P{c1 = √

Eb} + P

{r < 0|c1 = −√

Eb}P{c1 = −√

Eb}

= 1

2

(P{√

Eb + √Eb + wthermal < 0

} + P{√

Eb − √Eb + wthermal < 0

})= 1

2

[Q

(2√

Eb

σ

)+ Q(0)

]

A.2. Exact Computation of the Average Link BER in a Scenario with Strong LOS 267

where σ = √FkT0/2. Since Q(0) = 1

2 , one can immediately conclude that in this case thelink BER will always be higher than or equal to 1

4 , regardless of the transmit power Pt, and

limEb→∞ P {bit error|a single node from Group 11 is transmitting} = 1

4 .

Note that this result can be given an intuitive explanation. In fact,

limEb→∞ P {bit error|a single node from Group 11 is transmitting}

= limEb→∞ P {bit error|a single node from Group 11 is transmitting ‘destructively’}× P {the node transmits ‘destructively’|it is transmitting}

+ limEb→∞ P {bit error|a single node from Group 11 is transmitting ‘constructively’}× P {the node transmits ‘constructively’|it is transmitting} (A.1)

where the fact that the node from Group 11 is transmitting ‘destructively’ corresponds to thefact that sign(c1) = −sign(csig) and the fact that the node from Group 11 is transmitting‘constructively’ corresponds to the fact that sign(c1) = sign(csig). Since61

limEb→∞ P {bit error|a single node from Group 11 is transmitting ‘destructively’}

= P {the thermal noise, with zero mean, is in the opposite region w.r.t. c}= 1

2

P {a node from Group 11 transmits ‘destructively’|it is transmitting} = 12

limEb→∞ P {bit error|a single node from Group 11 is transmitting ‘constructively’} = 0

P {a node from Group 11 transmits ‘constructively’|it is transmitting} = 12

from (A.1) one can conclude that

limEb→∞ P {bit error|a single node from Group 11 is transmitting} = 1

2 × 12 = 1

4 . (A.2)

Obviously, this is true given that a node from Group 11 is transmitting, and this happens withprobability p. This probability depends on the traffic model considered, and we discuss thisissue in the following.

Consider now the case where a node from Group 11 and a node from Group 12 aretransmitting. The observable can be written as

r = csig + c1 + c2 + wthermal.

Without loss of generality, we still assume that c = √Eb, i.e. a ‘+1’ is transmitted. In this

case,

P {bit error|a node from Group 11 and a node from Group 12 are transmitting}= P {bit error|c1, c2}= P

{√Eb + c1 + c2 + wthermal < 0

}= P

{wthermal < −√

Eb − c1 − c2}

= P{wthermal >

√Eb + c1 + c2

} = Q

(√Eb + c1 + c2

σ

)61Equality (A.2) comes from the fact that if the interfering bit is the same as the transmitted bit, asymptotically

(for large values of the bit energy Eb), it follows that the probability of error goes to zero.


Table A.1 Possible scenarios with two interferers and the corresponding link BER

c1 c2 P {bit error|c1, c2}

+√Eb +√

Eb/2 Q

(√Eb + √

Eb + √Eb/2

σ

)= Q

(√Eb

σ

(2 + 1√

2

))+√

Eb −√Eb/2 Q

(√Eb + √

Eb − √Eb/2

σ

)= Q

(√Eb

σ

(2 − 1√

2

))−√

Eb +√Eb/2 Q

(√Eb − √

Eb + √Eb/2

σ

)= Q

(√Eb√2σ

)−√

Eb −√Eb/2 Q

(√Eb − √

Eb − √Eb/2

σ

)= Q

(−

√Eb√2σ

)

where the last passage is due to the fact that wthermal has an even distribution. There can befour possible cases for the interferers, shown in Table A.1, together with the correspondinglink BER. Therefore, since each of the four cases in Table A.1 has the same probability (equalto 1

4 ), one obtains

P {bit error|a node from Group 11 and a node from Group 12 are transmitting}= 1

22

[Q

(√Eb

σ

(2 + 1√

2

))+ Q

(√Eb

σ

(2 − 1√

2

))+Q

(√Eb√2σ

)+ Q

(−

√Eb√2σ

)].

Since limx→−∞ Q(x) = 1, it follows that

limEb→∞ P {bit error|a node from Group 11 and a node from Group 12 are transmitting} = 1

4 .

In this case as well, one can write

limEb→∞ P {bit error|a node from Group 11 and a node from Group 12 are transmitting}

= P {both nodes from Group 11 and Group 12 interfere ‘destructively’}= P {sign(c1) = −sign(csig)}P {sign(c2) = −sign(csig)}= 1

2 × 12 = 1

4

where we have used the fact that the interfering nodes are independent.


Denoting as BERlink the average link BER and generalizing the previous analysis, onecan write:

BERlink =nodes11∑n11=0

nodes12∑n12=0

P {bit error|n11 nodes are transmitting from Group 1

and n12 nodes are transmitting from Group 2}× P {n11 nodes are transmitting from Group 1

and n12 nodes are transmitting from Group 2}.(A.3)

Taking into account the independence between the nodes and assuming that each node hasprobability p of transmitting, it follows that

P {n11 nodes are transmitting from Group 11 and n12 nodes are transmitting from Group 12}=(

nodes11

n11

)(nodes12

n12

)pn11+n12(1 − p)1−n11−n12 .

At this point, the conditional probability on the right-hand side of (A.3) needs to be evaluated.In order to compute this probability, one needs to evaluate all possible combinations in whichthe two nodes can transmit. Since we are considering binary transmissions, there are 2n11+n12

possibilities, all of which are equally likely. All of these cases can be distinguished bycharacterizing the signs of the interfering signals. In order to describe all of these scenarios,we introduce, for notational conciseness, the notation termij (h, k), where

• i −→ refers to the tier (up to this point, there is only Tier 1);

• j −→ refers to the Group ij within the tier, within which there are nodesij nodes;

• h = 0, . . . , nodesij −→ indicates how many nodes from Group ij are transmitting;

• k = 1, . . . , 2h −→ is an index of the possible ways in which h nodes can transmit –since we are considering binary transmissions, the total number of ways in which thenodes can transmit is 2h.

The quantity termij (h, k) will be used to characterize the overall interfering signal, bycombining the signs of the component interfering signals. In particular, recalling that forGroup 11 there can be at most nodes11 = 3 interfering nodes and for Group 12 there can beat most nodes12 = 4 interfering nodes, we define:

term11(0, 20) = 0

term11(1, 1) = +1 = +1

term11(1, 21) = −1 = −1

term11(2, 1) = +1 + 1 = +2


term11(2, 2) = +1 − 1 = 0

term11(2, 3) = −1 + 1 = 0

term11(2, 22) = −1 − 1 = −2

term11(3, 1) = +1 + 1 + 1 = +3

term11(3, 2) = +1 + 1 − 1 = +1

...

term11(3, 23) = −1 − 1 − 1 = −3.

Considering term11(2, 1), term11(2, 2), term11(2, 3) and term11(2, 4), for instance, it is easyto see that when there are two nodes transmitting in Tier 1 coming from Group 11, there arefour different ways the interference can manifest itself. The two nodes interfere constructively(the cases with term11(2, 1) and term11(2, 4) with different polarities), and destructively(the cases with term11(2, 2) and term11(2, 3)) where the total interference is zero. Similarly,

term12(h, i) = term12(h, k) h = 0, . . . , 3 k = 1, . . . , 2h

term12(4, 1) = + 1 + 1 + 1 + 1 = +4

term12(4, 2) = + 1 + 1 + 1 − 1 = +2...

term12(4, 24) = − 1 − 1 − 1 − 1 = −4.

According to the considered definitions, it always holds that termij (h, 1) = +h (all the h

interfering nodes from Group j of Tier i are interfering constructively) and termij (h, 2h) =−h (all the h interfering nodes from Group j of Tier i are interfering constructively againwith different polarity). Therefore, the first probability on the right-hand side of (A.3) can beexpressed as follows:

P {bit error|n11 nodes are transmitting from Group 1

and n12 nodes are transmitting from Group 2}

=2n11∑i11=1

2n12∑i12=1

P {bit error|n11 nodes in the i11th configuration

and n12 nodes in the i12th configuration}× P {i11th and i12th configurations}︸︷︷︸

12n11+n12

.

Assuming, for instance, that the transmitted signal is a ‘+1’, it follows that

P {bit error|n11 nodes in the i11th configuration and n12 nodes in the i12th configuration}= P

{√Eb + term11(n11, i11)

√E

(11)b + term12(n12, i12)

√E

(12)b + wthermal < 0

}= P

{wthermal >

√Eb + term11(n11, i11)

√E

(11)b + term12(n12, i12)

√E

(12)b

}


= Q

√Eb + term11(n11, i11)

√E

(11)b + term12(n12, i12)

√E

(12)b

σ

.

Noting that E(11)b = Eb and E

(12)b = Eb/2, one finally obtains


nodes12∑n12=0

(nodes11

n11

)(nodes12

n12

)pn11+n12

· (1 − p)nodes11+nodes12−n11−n121

2n11+n12

·2n11∑i11=1

2n12∑i12=1

Q

[√Eb

σ

(1 + term11(n11, i11) + term12(n12, i12)√

2

)].

(A.4)

A.2.2 Interference from Nodes in Tiers 1 and 2

In this case, considering Tiers 1 and 2 in Figure A.1, one can distinguish the following fivegroups of nodes.


(11)r = Pr = αPt/d

2link. The received

bit energy is E(11)b = Eb = Pr/Rb.


(12)r = Pr/2 = αPt/2d2

link.



(21)r = Pr/4 = αPt/4d2

link.



(22)r = Pr/(22 + 1) = αPt/5d2

link.



(22)r = Pr/(22 + 22) = αPt/8d2

link.



Extending the combinatorial analysis considered in subsection A.2.1, the followingexpression for the average link BER can be derived:


nodes12∑n12=0

nodes21∑n21=0

nodes22∑n22=0

nodes23∑n23=0

(nodes11

n11

)(nodes12

n12

)(nodes21

n21

)·(

nodes22

n22

)(nodes23

n23

)pn11+n12+n21+n22+n23

· (1 − p)nodes11+nodes12+nodes21+nodes22+nodes23−n11−n12−n21−n22−n23

· 1

2n11+n12+n21+n22+n23

2n11∑i11=1

2n12∑i12=1

2n21∑i21=1

2n22∑i22=1

2n23∑i23=1

· Q

[√Eb

σ

(1 + term11(n11, i11) + term12(n12, i12)√

2

+ term21(n21, i21)

2+ term22(n22, i22)√

5+ term23(n23, i23)

2√

2

)](A.5)

where

term21(h, i) = term22(h, i) = term22(h, i) = term12(h, i) h = 0, . . . , 4 i = 1, . . . , 2h

term22(5, 1) = + 1 + 1 + 1 + 1 + 1 = +5

term22(5, 2) = + 1 + 1 + 1 + 1 − 1 = +3...

term22(5, 25) = − 1 − 1 − 1 − 1 − 1 = −5

term22(6, 1) = + 1 + 1 + 1 + 1 + 1 + 1 = +6

term22(6, 2) = + 1 + 1 + 1 + 1 + 1 − 1 = +4...

term22(6, 26) = − 1 − 1 − 1 − 1 − 1 − 1 = −6

term22(7, 1) = + 1 + 1 + 1 + 1 + 1 + 1 + 1 = +7...

term22(7, 27) = − 1 − 1 − 1 − 1 − 1 − 1 − 1 = −7

term22(8, 1) = + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 = +8...

term22(8, 28) = − 1 − 1 − 1 − 1 − 1 − 1 − 1 − 1 = −8.

The route BER performance considering the interference contribution from the nodes inboth Tiers 1 and 2 is shown in Figure A.2. Comparing the results shown in Figure A.2 withthose shown in Figure 3.5 and relative to the interference analysis conducted consideringonly the interfering nodes in Tier 1, one can conclude that the contribution of the nodes inTier 2 to the overall interference is negligible with respect to the interference due to nodes inTier 1. In other words, the entire interference is basically due to the nodes in Tier 1. In order


10-3

10-2

10-1

100

ρS [m

-2]

10-9

10-8

10-7

10-6

10-5

10-4

10-3

10-2

10-1

100

BERroute

L=1000, λ=0.5 pck/s

with average Pint

L=1000, λ=0.05 pck/s

L=10000, λ=0.5 pck/s

Gaussian int. noise

(L=1000 λ=0.5 pck/s)

Figure A.2 Route BER as a function of the node spatial density, considering the interferenceoriginated from nodes in both Tier 1 and Tier 2. The main parameters used in the computationare: Pt = 0.2 µW, Gt = Gr = fl = 1, Rb = 2 Mb/s, F = 6 dB and N = 1000.

to better understand the impact of nodes in Tier 2, in the following subsection we considerthe BER performance assuming that only the nodes in Tier 2 are interfering. In other words,this would correspond to a scenario where nodes in Tier 1 are ‘shut down’ during the centrallink transmission. Note that in Figure A.2 the route BER under the Gaussian assumption forthe interference noise is also shown. Section A.5 will be devoted to the evaluation of theconditions of validity for the Gaussian approximation for the interference noise.

A.2.3 Interference from Nodes in Tier 2By reasoning as in the previous subsections, it is easy to conclude that the link BER whenonly the nodes from Tier 2 are interfering has the following expression:


nodes22∑n22=0

nodes23∑n23=0

(nodes21

n21

)(nodes22

n22

)(nodes23

n23

)pn21+n22+n23

· (1 − p)nodes21+nodes22+nodes23−n21−n22−n231

2n21+n22+n23

·2n21∑i21=1

2n22∑i22=1

2n23∑i23=1

Q

[√Eb

σ

(1 + term21(n21, i21)

2+ term22(n22, i22)√

5

+ term23(n23, i23)

2√

2

)]. (A.6)


10-3

10-2

10-1

100

ρS [m

-2]

10-9

10-8

10-7

10-6

10-5

10-4

10-3

10-2

10-1

100

BERrouteL=1000, λ=0.5 pck/s

(L=1000 λ=0.5 pck/s)

L=1000, λ=0.05 pck/s

L=10000, λ=0.5 pck/s

with average Pint

Gaussian int. noise

Figure A.3 Route BER as a function of the node spatial density, considering the interferenceoriginated from nodes in Tier 2. The main parameters used in the computation are: Pt =0.2 µW, Gt = Gr = fl = 1, Rb = 2 Mb/s, F = 6 dB and N = 1000.

The route BER in this case is shown in Figure A.3. As one can see from the resultsin Figure A.3, eliminating the interference from nodes in Tier 1 leads to a significantperformance improvement. This suggests that a MAC protocol able to temporarily shut downnodes Tier 1 might be able to significantly limit the interference. More generally, this suggeststhat eliminating the interference from neighboring nodes might be sufficient to improve theperformance.

A.2.4 Simulation ScenarioIn order to verify the results predicted by the analysis, it is possible to consider a simplesimulation scenario which involves only the nodes of the first tier and is aimed at evaluatingthe average link BER.

Suppose that we enumerate the interfering nodes of the Tier 1 as shown in Figure A.4.Recalling the notation introduced in section A.2.1, nodes n2, n4 and n6 constitute Group 11,whereas the remaining interfering nodes belong to Group 12. Assuming that all nodes arestatistically independent,62 one can visualize the transmission activity of each node as anarray (of sufficient length), denoted as transmission array, such that:

• the array index is related to a discrete-time scale (i.e. each array position correspondsto a bit time, since we are considering BPSK);

62The assumption of statistical independence between the nodes is obviously not accurate in a multi-hop networkcommunication scenario, where nodes belong to multi-hop routes. However, the correlation existing between thenodes is likely to reduce the interference, since a node waiting for packets from the previous node of the routecannot transmit.


nS nD

n1 n2 n3

n4

n6 n5n7

Figure A.4 First tier nodes.

• an array position can be either filled (the node is transmitting a bit) or not filled(the node is not transmitting).

Assuming that the packet length L is fixed, the transmission array presents groups of L

consecutive filled positions. These groups of filled positions are separated by a number ofpositions which depend on the traffic characteristics at each node. For instance, assumingthat all nodes transmit with a randomized Poisson distribution63 with parameter λ and thatthe data-rate is Rb (i.e. each bit interval has duration 1/Rb), we assume that two consecutivepackets are separated, on average, by 1/λ

1/Rb= Rb

λpositions.

At this point, we can build a matrix, denoted as transmission matrix, such that each rowcorresponds to the transmission array of a node. It is, therefore, a matrix formed by eight rows(one corresponding to nS and seven corresponding to the interfering nodes from n1 to n7), andan ‘indefinite’ number of columns. From an implementation viewpoint, one needs to considera sufficient number of columns, i.e. a sufficiently long simulation time, such that the MonteCarlo simulation to evaluate the link BER converges. In order to guarantee a low memoryoccupance of the simulation program, it is possible to consider a partial matrix construction.We first consider Kpartial columns, i.e. we describe the evolution of the nodes over Kpartialconsecutive bit intervals. Then, we discard this matrix and we overwrite a new matrix relativeto the next Kpartial bit intervals of the network. Assuming packets of length L = 3 (for easeof pictorial description), the transmission matrix in a scenario with Rb/λ = 5 could look likeFigure A.5. Note that a transmitted bit could be either a +1 (denoted by a circle in Figure A.5)or a −1 (denoted by a square in Figure A.5). Each bit can be either a +1 or a −1 with thesame probability. A (partial) transmission matrix (with Kpartial columns) can be generated byextracting the first bit of the next packet with exponentially distributed distance from the lastbit of the previous packet, and then filling the L consecutive positions (by selecting randomlybetween +1 or −1).

After generating a partial transmission matrix with Kpartial columns as explained at theend of the previous paragraph, one has to evaluate the corresponding link BER. In order todo this, for each filled bit position in the row corresponding to node nS (we are implicitlyneglecting the propagation time between nodes nS and nD), one checks for the rows in whichthere is the same position filled. Therefore, the received signal will be c

sigk + c

(i1)k + c

(i2)k +

· · · , where csigk = ±√

Eb is the transmitted bit, c(i1)k is the transmitted bit from the i1th

63It can be shown that this is a pessimistic assumption given that the packet generation has a Poisson distribution.


n1

nS

n2

n3

n4

n5

n6

n7

= +1

= −1

Rb/λ

· · ·...

k

Figure A.5 Possible transmission matrix in a scenario with L = 3 and Rb/λ = 5

interfering node (i1 ∈ {1, . . . , 7}), c(i2)k is the transmitted bit from the i2th interfering node

(i2 ∈ {1, . . . , 7}), etc. The energy of each interfering bit c(ij )

k will depend on the group to

which the ij th node belongs: according to the notation in section A.2.1, it will be E(11)b = Eb

for the nodes of Group 11 and E(12)b = Eb/2 for the nodes of Group 12. Generating randomly

a Gaussian random variable nk, the observable at the receiver at epoch k can be written as

rk = csigk + c

(i1)k + c

(i2)k + · · · + nk.

Since we are considering uncoded transmission, the receiver will decide for the kth bitaccording to the following rule:

csigk =

{+1 if rk > 0

−1 if rk < 0.

At this point, comparing the decided bits {csigk } with the true bits {csig

k } it is possible to estimatethe link BER through a Monte Carlo approach. By considering a sufficiently large number oftransmitted bits, i.e. a sufficiently large number of partial transmission matrices, it is possibleto reliably estimate the link BER as

BERlink = number of bit in errors

total number of transmitted bits.

A.3 Exact Computation of the Average Link BER in aScenario with Strong Multipath (Rayleigh Fading)

We now extend the analysis for a scenario with strong LOS, to a scenario with strongmultipath, i.e. a scenario where communications are affected by Rayleigh fading.

A.3. Exact Computation of the Average Link BER with Multipath 277

A.3.1 Interference from Nodes in Tier 1As described in subsection A.2.1, there are two groups of nodes: Group 11 with nodes11 = 3nodes and Group 12 with nodes12 = 4 nodes.

Suppose that there is only a single node from Group 11 interfering with probability p.In the presence of non-selective Rayleigh fading over all communication paths, the observablecan be written as follows:

r = f csig + f1 c1 + wthermal (A.7)

where f is the fading affecting the transmitted signal csig, f1 is the fading term affectingthe interfering signal c1 and wthermal is the thermal noise with variance Eint. Since weare considering a scenario with slow Rayleigh fading, the fading coefficients f and f1are Gaussian random variables with zero mean and variance σ 2

fad. Since c1 = ±√Eb, the

term f1 c1 in (A.7) is still a Gaussian random variable with variance Eb σ 2fad. Therefore, the

observable (A.7) can be equivalently rewritten as:

r = f csig + n(1)tot (A.8)

where n(1)tot � f1 c1 + wthermal is a Gaussian random variable with variance Eint + Eb σ 2

fad(the fading is independent of the thermal noise). One concludes that the fact that theinterfering signal is a ‘+1’ or a ‘−1’ is not relevant, and the corresponding link BER is

BERRayleighlink = 1

2

(1 −

√σ 2

fad Eb

Ethermal + σ 2fad Eb + σ 2

fad Eb

)(A.9)

where the assumption of perfect estimation of the phase of the fading term f = a ejθ hasbeen used – this is realistic, since we are considering slow fading [33].

The combinatorial analysis proposed in subsection A.2.1 can be extended. In fact, itsimplifies, since now it is not important to distinguish between the possible combinationsof the interfering bits (as observed in the previous paragraph, this is due to the presenceof Gaussian fading coefficients with zero mean). More precisely, in a scenario with n11interfering nodes from Group 11 and n12 interfering nodes from Group 12, the observablecan be equivalently rewritten as

r = f csig + n(n11+n12)tot (A.10)

where n(n11+n12)tot is a Gaussian random variable with variance Eint + σ 2

fad(n11 E(11)b + n12

E(12)b ). The link BER, denoted as BERRayleigh|n11+n12

link , can then be written as

BERRayleigh|n11+n12link = 1

2

1 −√√√√ σ 2

fad Eb

Ethermal + σ 2fad(n11 E

(11)b + n12 E

(12)b ) + σ 2

fad Eb

= 1

2

1 −√√√√ σ 2

fad Eb

Ethermal + σ 2fad Eb(n11 + 1

2n12) + σ 2fad Eb

= 1

2

(1 −

√αρSPt

FkT0Rb + αρSPt(n11 + 12n12 + 1)

)(A.11)


where, in the last passage, we have used the assumption (consistent with that considered inChapter 2) that E[a2] = σ 2

fad = 1.The average link BER in a scenario with Rayleigh fading thus becomes

BERRayleighlink =

nodes11∑n11=0

nodes12∑n12=0

(nodes11

n11

)(nodes12

n12

)pn11+n12

· (1 − p)nodes11+nodes12−n11−n12 BERRayleigh|n11+n12link . (A.12)

A.3.2 Interference from Nodes in Tiers 1 and 2In a scenario with n1j interfering nodes from Group 1j , j = 1, 2, and n2� interfering nodesfrom Group 2�, � = 1, 2, 3, the observable can be equivalently rewritten as

r = f csig + n(n11+n12+n21+n22+n23)tot (A.13)

where n(n11+n12+n21+n22+n23)tot is a Gaussian random variable with variance Eint + σ 2

fad

(n11 E(11)b +n12 E

(12)b +n21 E

(21)b +n22 E

(22)b +n23 E

(23)b ), where E

(11)b , . . . , E

(23)b are defined

as in the previous section. The link BER, denoted as BERRayleigh|n11+n12+n21+n22+n23link , can then

be written as

BERRayleigh|n11+n12+n21+n22+n23link

= 1

2

1 −√√√√ σ 2

fad Eb

Ethermal + σ 2fad(

∑2j=1 n1j E

(1j)

b + ∑3�=1 n2� E

(2�)b ) + σ 2

fad Eb

= 1

2

(1 −

√σ 2

fad Eb

Ethermal + σ 2fad Eb(n11 + n12

2 + n214 + n22

5 + n238 ) + σ 2

fad Eb

)

= 1

2

(1 −

√αρSPt

FkT0Rb + αρSPt(n11 + n122 + n21

4 + n225 + n23

8 + 1)

)(A.14)

where, in the last passage, we have used the assumption (as in the previous subsection) thatE[a2] = σ 2

fad = 1.The average link BER in a scenario with Rayleigh fading becomes

BERRayleighlink =

nodes11∑n11=0

nodes12∑n12=0

nodes21∑n21=0

nodes22∑n22=0

nodes23∑n23=0

(nodes11

n11

)(nodes12

n12

)(nodes21

n21

)·(

nodes22

n22

)(nodes23

n23

)pn11+n12+n21+n22+n23

· (1 − p)nodes11+nodes12+nodes21+nodes22+nodes23−n11−n12−n21−n22−n23

· BERRayleigh|n11+n12+n21+n22+n23link . (A.15)

A.3.3 Interference from Nodes in Tiers 1, 2 and 3In order to evaluate the impact of the interference originating from the nodes in Tiers 1, 2 and3, one can directly extend the previous computations. In particular, the distributions of the

A.3. Exact Computation of the Average Link BER with Multipath 279

nodes in Tiers 1 and 2 has already been described. Considering Tier 3 in Figure A.1, one candivide the nodes in this tier in the following groups.


(31)r = Pr/9 = αPt/9d2

link.



(32)r = Pr/(32 + 32) =

αPt/18d2link. The received bit energy is E

(31)b = Eb/18.


(33)r = Pr/(32 + 22) = αPt/13d2

link.



(34)r = Pr/(32 + 12) = αPt/10d2

link.


Following the approach in the previous subsection, the average link BER in a scenariowith Rayleigh fading becomes

BERRayleighlink =

nodes11∑n11=0

nodes12∑n12=0

nodes21∑n21=0

nodes22∑n22=0

nodes23∑n23=0

nodes31∑n31=0

nodes32∑n32=0

·nodes33∑n33=0

nodes34∑n34=0

(nodes11

n11

)(nodes12

n12

)(nodes21

n21

)(nodes22

n22

)·(

nodes23

n23

)(nodes31

n31

)(nodes32

n32

)(nodes33

n33

)(nodes34

n34

)· pn11+n12+n21+n22+n23+n31+n32+n33+n34

· (1 − p)tot−nodes1−2−3−n11−n12−n21−n22−n23−n31−n32−n33−n34

· BERRayleigh|n11+n12+n21+n22+n23+n31+n32+n33+n34link (A.16)

where tot − nodes1−2−3 = ∑2i=1 nodes1i + ∑3

i=1 nodes2i + ∑4i=1 nodes3i is the total

number of nodes in the three tiers and

BERRayleigh|n11+n12+n21+n22+n23+n31+n32+n33+n34link

= 1

2

{1 − (αρSPt)

1/2[FkT0Rb + αρSPt

(n11 + n12

2+ n21

4+ n22

5

+ n23

8+ n31

9+ n32

18+ n33

13+ n34

10+ 1

)]−1/2}

. (A.17)

The route BER, evaluated considering (i) only the nodes from Tier 1; (ii) the nodes fromTiers 1 and 2; and (iii) the nodes from Tiers 1, 2 and 3 is shown in Figure A.6, in a scenariowith N = 300 nodes, Pt = 1 mW, Gt = Gr = fl = 1, F = 6 dB, and two possible valuesfor Rb (2 Mb/s and 100 kb/s).


109

108

107

106

105

104

103

102

101

100

S [m

2]

104

103

102

101

100

BERrouteRb=2 Mb/s

Rb=100 kb/s

Tier 1

Tiers 1 and 2Tiers 1, 2 and 3

Figure A.6 Route BER as a function of the node spatial density, considering the interferenceoriginated from nodes in Tiers 1, 2 and 3. The main parameters used in the computations are:Pt = 1 mW, Gt = Gr = fl = 1, F = 6 dB, λ = 0.5 pck/s, L = 1000 b/pck and N = 300.The considered values for the data-rate Rb are 2 Mb/s and 100 kb/s.

A.4 LOS and Multipath (Rice Fading)In this case, the fading coefficient affecting each signal is a Gaussian random variable withnon-zero mean. The evaluation of the average link BER can be carried out by consideringa hybrid approach between those corresponding to scenarios with strong LOS and strongmultipath:

• a part of the interfering signal (the LOS component) affects the useful transmittedsignal;

• a part of the interfering signal (the multipath component) can be incorporated togetherwith the thermal noise.

The analysis is not straightforward and the interested reader is encouraged to pursue it further.However, one can anticipate that the BER performance will lie somewhere between thattypical of a scenario with strong LOS and that typical of a scenario with strong multipath.The proximity to either one of these limiting cases will be determined by the value of theRice constant K: the higher is K , the more similar is the performance to that of a scenariowith strong LOS and vice versa.

A.5 Gaussian Assumption for the Interference NoiseWhile, as shown in the previous sections, the interference noise in the considered ad hocwireless networking scenarios is not Gaussian, from the results in Figure A.2 one can see that

A.5. Gaussian Assumption for the Interference Noise 281

the Gaussian assumption for the interference noise gives an accurate approximation of theexact route BER performance for node spatial densities below a critical value.

We now outline the interference analysis under the Gaussian assumption for the interfer-ence noise. This assumption is motivated by invoking the applicability of the central limittheorem (CLT). In Chapter 3, it is shown that the results obtained by using the followingassumptions are not always correct. In other words, in Chapter 3 it is shown that the Gaussianassumption is valid only in particular scenarios. We will first consider (i) a scenario withstrong LOS; and we will then extend the analysis to (ii) a scenario with strong multipath(Rayleigh fading); and (iii) to a scenario with LOS and multipath (Rice fading).

The discrete-time observable at the end of a link with no fading can be written as follows:

r = c + wthermal + wint (A.18)

where c is the received signal, wthermal is an additive white Gaussian noise (AWGN) randomvariable with variance Pthermal and wint is the interference noise, independent of wthermal. TheGaussian assumption consists in modeling wint as a zero-mean Gaussian random variable,with variance Pint � E[w2

int]. In other words, the observable r in (A.18) can be rewritten as

r = c + wtot (A.19)

where wtot is a Gaussian random variable with variance Pthermal + Pint. Therefore, one canevaluate the average link bit error rate (BER) through a standard detection theory approachfor uncoded communications, given the particular modulation format. In the case of binarymodulation, the link signal-to-noise ratio (SNR) can be written as

SNRlink = Pr

Pthermal + Pint(A.20)

where Pr is the received power. In the case of free-space propagation, in Chapter 2 it isshown that Pr = αPtρS, where ρS is the node spatial density. In a scenario with uncodedbinary phase shift keying (BPSK), the considered link BER can then be written as:

BERGausslink = Q

(√2Pr

Pthermal + Pint

)(A.21)

where Q(x) �∫∞x

1√2π

e−y2/2 dy. We underline that BERGausslink is not an average link BER,

but, rather, the link BER evaluated in correspondence to an average interference power (basedon the considered Gaussian assumption for the interference noise).

In a scenario with frequency non-selective fading, the observation model becomes

r = f c + wthermal + wint (A.22)

where f = aejθ is a Gaussian random variable. In this case as well, keeping the assumptionthat wint can be modeled as Gaussian, it is possible to derive expressions for the link BER,depending on the statistical characterization of fading (Rayleigh or Rice).

Assuming that interfering signals are uncorrelated, the interference noise power there-fore is

E{w2int(t)} =

∑j

E

{[s

int ja (t)

]2}

.


At this point, since the thermal noise is independent of the interference noise, thedefinition of the link SNR (averaged with respect to fading) is a direct extension of thatconsidered in the Chapter 2:

SNRintlink = E[a2]Ebit

Ethermal + Eint(A.23)

where E[a2] is the fading mean square value, Ebit is the bit energy, Ethermal is the thermalnoise energy and Eint is the interference energy. Ebit and Ethermal are given by

Ebit = Pr

Rb= αPt

d2linkRb

Ethermal = Pthermal

B= FkT0.

Assuming antipodal transmission (as will be done later), since interfering nodes transmitindependently, the total interfering signal is likely to be characterized by an autocorrelationfunction Rint(τ ) = E[wint(t)w

∗int(t + τ )] which is significantly different from zero only for

τ � 0. Therefore, assuming that the interference process is white and that the equivalentnoise bandwidth is equal to the transmission bandwidth,64 one can write:

Eint = Pint

B.

Therefore, the link SNR in (A.23) can be equivalently rewritten as follows:

SNRintlink = E[a2]Pr

(Rb/B)(Pthermal + Pint). (A.24)

In the case of BPSK binary phase shift keying, which will be considered in the remainder ofthis chapter, Rb = B, so that the expression (A.24) can be further simplified as follows:

SNRintlink = E[a2]Pr

Pthermal + Pint= E[a2]αρSPt

FkT0Rb + Pint. (A.25)

Note that the proposed approach can be straightforwardly extended to communicationscenarios with high-order modulation formats, as considered in [40].

A.5.1 Route Bit Error RateAssuming that the number of interfering signals is sufficiently large, by invoking the centrallimit theorem [83], one can assume that wint(t) is Gaussian. Therefore, the total noise process,formulated as

wtotal(t) = wthermal(t) + wint(t)

is Gaussian, with zero mean and variance Ethermal +Eint. Denoting, as done in Chapter 2, theinstantaneous link SNR ratio as , where

= a2Pr

Pthermal + Pint.

64Note that the assumption that the transmit bandwidth is equal to the equivalent noise bandwidth is true for theconsidered modulation format. For a general modulation format, however, this has to be carefully reconsidered.


If the attenuation a due to fading is constant, then the link BER could be written as [33]

BERlink() = Q(√

2)

= 1√2π

∫ ∞√

2

e−x2/2 dx. (A.26)

In order to obtain the average (with respect to fading) link BER, assuming that the statisticsof a and, consequently, of are known at the receiver, one can write:

BERlink =∫ ∞

0BERlink() p() d

where p() depends on the particular fading environment. We now outline the expressions ofthe link BER in the three scenarios considered at the end of subsection 2.3.2: (i) strong line-of-sight (LOS) (i.e. no fading); (ii) strong multipath fading (a has a Rayleigh distribution);and (iii) LOS with multipath (a has a Rice distribution).

Strong LOS

In the case of strong LOS, i.e. of transmission over an AWGN channel, the link BER can bewritten as [33]

BERAWGNlink = Q

(√2 SNRAWGN

link

)= Q

(√2

Ebit

Ethermal

)= Q

(√2aρSPt

FkT0Rb + Pint

).

(A.27)

The following expression for the BER at the end of a multi-hop route with an average numberof hops can then be obtained:

BERroute = 1 −[

1 − Q

(√2αρSPt

FkT0Rb + Pint

)]nh

. (A.28)

Strong Multipath Fading

In this case, the link BER can be written as follows [33]:

BERRayleighlink = 1

2

(1 −

√σ 2

fad Ebit

Ethermal + σ 2fad Eint + σ 2

fad Ebit

)

= 1

2

(1 −

√αρSPt

FkT0Rb + Pint + αρSPt

)(A.29)

where Eint is the interfering signal energy in a scenario with strong LOS and, in the lastpassage, we have used the assumption (consistent with that considered in Chapter 2) thatE[a2] = σ 2

fad = 1. Similarly to the previous case with strong LOS, in this case as well theevaluation of the route BER is straightforward.


Multipath and LOS

In this case, the link BER can not be given a closed-form expression. However, a computa-tionally efficient expression is the following [45]:

BERRicelink =

∫ π/2

0

(1 + K) sin2(θ)

SNRRicelink + (1 + K) sin2(θ)

· exp

[− KSNRRice

link

SNRRicelink + (1 + K) sin2(θ)

]dθ (A.30)

where

SNRRicelink = (1 + K)

σ 2fadPr

Pthermal + Pint(A.31)

and K � a20/σ 2

fad. The parameter K quantifies the ratio between the non-fading (direct LOS)and fading components, with mean powers equal to a2

0 and σ 2fad, respectively. Assuming,

consistently with Chapter 2, that a0 = 1, the link SNR can be equivalently rewritten as

SNRRicelink = 1 + K

K

Pr

Pthermal + Pint. (A.32)

A.5.2 Interference PowerWe now evaluate the average interference power Pint, depending on the considered MACprotocol. We refer to the link transmission scenario shown in Figure A.1.

RESGO MAC Protocol

Assuming that, at node nD, the interfering signal s�(t) from node n� is an ergodic randomprocess and its power corresponds to the mean square value E{s2

� (t)} , it follows that

E{s2� (t)}

= E{s2� (t)|bit interference from node n�}P {bit interference from node n�}

+ E{s2� (t)|no bit interference from node n�}P {no bit interference from node n�}

= αPt

d2�

(1 − e−λDpck) + 0 e−λDpck = αPt

d2�

(1 − e−λDpck). (A.33)

In order to evaluate the total interference power in this case, we simply have to add theinterference powers coming from all the nodes in the network. Referring to the tiered structurein Figure A.1, after simple manipulations it is possible to derive the following expression forthe interference power:

Pint = αPtρS(1 − e−λDpck) A(N) (A.34)

where

A(N) �[

imax∑i=1

(6

i2+ 8

i−1∑j=1

1

i2 + j2

)− 1

]. (A.35)

The quantity A(N) depends on the geometry of the node distribution – preliminary resultssuggest, however, that the difference between various regular geometries is negligible.


RESLIGO MAC Protocol

Proceeding, as in Chapter 2, to evaluate the total interference power by taking into accountthe square grid topology, it can be shown that the total interference power on the j th bit withthe RESLIGO MAC protocol can be written as

Pint,j = αPtρS ′C(j,N, λ, ρS, Rb) (A.36)

where

′C(j,N, λ, ρS, Rb) �

{imax∑i=1

[4

i2

(1 − e−λ min{2iτlink,τlink+j/Rb}

)+ 2

i2

(1 − e−λ min{2√

2iτlink,τlink+j/Rb})

+ 8i−1∑s=1

1

i2 + s2

(1 − e−λ min{2

√i2+s2τlink,τlink+j/Rb}

)]−

(1 − e−λ min{2τlink,τlink+j/Rb}

)}. (A.37)

It is clear that a larger maximum vulnerable interval leads to an increased probabilityof interference. In order to make a statistical (average) analysis, we focus on the centralbit of the transmitted packet (j = L/2 and j/Rb = Dpck/2). It can be shown that, if thenumber of nodes in the network is not extremely large (for instance, lower than 14 000), thenmin{2τ�, τlink + j/Rb} = 2τ� for any interfering node n�. Therefore, the general expressionfor the interference power can be simplified as follows:

Pint � αPtρS C(N, λ, ρS) (A.38)

where

C(N, λ, ρS) �{

imax∑i=1

[4

i2(1 − e−λ2iτlink) + 2

i2(1 − e−λ2

√2iτlink)

+ 8i−1∑s=1

1

i2 + s2

(1 − e−λ2

√i2+s2τlink

)]− (1 − e−λ2τlink)

}.

(A.39)

Note that, in this case, the interference power does not depend on the particular bit in thepacket.

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Ad Hoc Wireless Networks (A Communication-Theoretic Perspective) || Appendix A: Analysis of the Inter-node Interference