Authors: Andrea Zanella, Michele Zorzi zanella@dei.unipd.it Presenter: Nicola Bui

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Analysis of the Capture Probability in Wireless Systems with Multi-Packet Reception Capabilities and Successive Interference Cancellation. Authors: Andrea Zanella, Michele Zorzi zanella@dei.unipd.it Presenter: Nicola Bui. Scenario. TX 1. P 1. P n. TX n. P 2. P 3. P j. RX. - PowerPoint PPT Presentation

Text of Authors: Andrea Zanella, Michele Zorzi zanella@dei.unipd.it Presenter: Nicola Bui

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Authors: Andrea Zanella, Michele Zorzi zanella@dei.unipd.itPresenter: Nicola BuiAnalysis of the Capture Probability in Wireless Systems with Multi-Packet Reception Capabilities and Successive Interference Cancellation

1ScenarioICC 2011Kyoto (Japan) 5-9 June 2011TX1TX2TX3TXjTXnPjPnP1P2P3RX

gj > b j-th signal is correctly decoded (capture)gj < b j-th signal is collided (missed)Aggregate interferenceMulti Packet ReceptionICC 2011Kyoto (Japan) 5-9 June 2011Enabling Multi Packet Reception (MPR) can bring in several benefits [1]higher transmission efficiency due to channel diversitylarger system capacity thanks to multi-user detectionsimpler channel access schemesMPR can be enabled by means ofSignal spreading (CDMA)bbReconstruct and cancel signal j from the overall received signal[1] Wang&Garcia-Luna-Aceves,INFOCOM08Open questionsICC 2011Kyoto (Japan) 5-9 June 2011How system parameters impact on capture probability?Number of simultaneous transmissions (n)Statistical distribution of the receiver signal powers (Pi)Capture threshold (b)Max number of SIC iterations (K)Interference cancellation ratio (z)What performance gain can be expected from MPR? How many SIC iterations shall we account for? The answer & the problemICC 2011Kyoto (Japan) 5-9 June 2011The answer: compute the capture probabilityCn(r;K)=Pr[r signals out of n are capture within at most K SIC iterations]The problem: computing Cn(r;K) is difficult because the SINRs are all coupled!!!E.g.Computation of Cn(r;k) becomes more and more complex as the number (n) of signals increases

State of the artICC 2011Kyoto (Japan) 5-9 June 2011Narrowband (b>1), No SIC (K=0)Can decode at most one signal at a time[Zorzi&Rao,JSAC1994,TVT1997] derive the probability Cn(1;0) that one signal is captured

Wideband (bb are simultaneously decoded and cancelledCancellation of signal j leaves residual interference power zPjDecoding is iterated up to K+1 times, unless no signal is decoded in an iterationICC 2011Kyoto (Japan) 5-9 June 2011Assumptions*Decoding model*Gray assumptions can be relaxed Notation: reception set and vectorICC 2011Kyoto (Japan) 5-9 June 2011n : number of overlapping signalsr : overall number of decoded signalsh ={0,1,,K}: SIC iteration Uh: set of signals decoded at the hth SIC iterationUk+1: set of missed signals at the end of the reception processr=[r0,r1,,rk,rk+1]: reception vectorrh=|Uh|, rk+1=|Uk+1|= n-r

r ={ r0, , rh , . rk, rk+1 }TX1TX2TXjTXrTXr+1TXnU={ U0, , Uh, . Uk, Uk+1 }decodedmissedNotation: aggregate powerSet of signal powers for users in Uh

Aggregate power of users in Uh

Overall sign. power at the h-th decoding cycle

VisuallyICC 2011Kyoto (Japan) 5-9 June 2011TX1TX2TXjTXrTXr+1TXnP={ P0, , Ph , . Pk, Pk+1 }

G={ G0, , , . Gk, Gk+1 }

L={ L0, , , . Lk, Lk+1 }

zzzDerivation of the capture probability expressionICC 2011Kyoto (Japan) 5-9 June 2011

Step 1: a bit of combinatorial analysisICC 2011Kyoto (Japan) 5-9 June 2011

Ordered probability distributionCombinatorial coefficientStep 2: express decoding probability in terms of PjICC 2011Kyoto (Japan) 5-9 June 2011Signals in Uh are decoded at the h-th SIC iteration ifwere not decoded at previous iterationsverify capture condition after h SIC iterationsMathematically

Considering all k SIC iterations

where

Step 3: lets start conditioningICC 2011Kyoto (Japan) 5-9 June 2011The capture threshold at each SIC iteration are

Conditioning on {Gh=gh} the capture thresholds becomes deterministic

Then, we can write (we omit g in the argument of lh)

Aggregate power of signals in Ui

PDF of the random vector G evaluated in g=[g0,...,gk+1]k+2 nested integralsStep 4: swap termsICC 2011Kyoto (Japan) 5-9 June 2011Applying Bayes rule we get

The issue now is to compute this conditional probability

iid

l-1=-iidStep 5: aux variables help decoupling termsICC 2011Kyoto (Japan) 5-9 June 2011Each Gh is the aggregate power of the signals in Uh given that they are in the interval (lh-1,lh] We then define

where ah,i(u,v) are iid with PDFHence, for any given g, we have

Fourier TransformStep 6: put all pieces togetherICC 2011Kyoto (Japan) 5-9 June 2011Number of nested integrals grows linearly with number K of SIC iterations, not with nEquation can be computed for large values of n, provided that the number of SIC iterations remains reasonable (56)Central limit theorem can be invoked for sufficiently large rh

ThroughputExact and approximate expresionsICC 2011Kyoto (Japan) 5-9 June 2011

ThroughputICC 2011Kyoto (Japan) 5-9 June 2011Sn(k): average number of signals captured by a system wit collision size n and at most K SIC iterationsExact expression:

Approximate (iterative) expression

Where is the approximate mean number of signals decoded at the h-th SIC iteration

Approximate mean number of captures: first receptionICC 2011Kyoto (Japan) 5-9 June 2011Iteration h=0: number of undecoded signals n0=nCompute capture thresholdApproximate capture conditionMean number of decoded signalsResidual interference power

Approximate mean number of captures: first receptionICC 2011Kyoto (Japan) 5-9 June 2011Iteration h>0: number of undecoded signals:Compute capture thresholdApproximate capture conditionMean number of decoded signalsResidual interference power

Residual interf.Interf. from undecoded signalsCase studyICC 2011Kyoto (Japan) 5-9 June 2011

Rayleigh fadingKyoto (Japan) 5-9 June 2011Exponential distribution of the received power Pj

Fourier Transform of the auxiliary rv a(u,v)

Mean value of a(u,v)

Capture probability distributionICC 2011Kyoto (Japan) 5-9 June 2011Multiple SIC (K>1): capture probability keeps improving, but gain reducesOne SIC (K=1): likely to decode 410 signals, double capture capabilities!!!No SIC (K=0): likely to decode 25 signalsn=20, b=0.1, z=0.1K increasesSIC in highly congested scenarioICC 2011Kyoto (Japan) 5-9 June 2011n=60, b=0.1, z=0.1SIC does not yield any significant performance gainHigh probability of missing all the signals SIC is not performed at all!!!Throughput vs nICC 2011Kyoto (Japan) 5-9 June 2011b=0.1, z=0.1exactapproximateK increasesHigh congestionLow congestionMax SIC gain ~500%Approximation is quite good!Max SIC gain analysisICC 2011Kyoto (Japan) 5-9 June 2011Max SIC throughput grows almost linearly with 1/bh(k) does not change much when varying bSIC gain strongly depends on residual interference factor zThe less residual interference, the larger the SIC gainFor K>1/z, SIC gain is negligible Empirical conjecture

DiscussionICC 2011Kyoto (Japan) 5-9 June 2011We proposed a novel approach for computing the probability Cn(r;K) of capturing r out of n signals with at most K SIC iterationsExponential complexity in Kbut, nicely scalable with nWe provided a quite good approximate expression of the throughput Extremely light computationWe applied the method to study the system performance when varying the parametersSIC can be very effective, bringing large throughput gainbut cannot do much in case of high interference (n>>1/b)Max SIC throughput grows almost linearly with 1/bResidual interference has strong impact on SIC performanceMax gain is approached when K~1/z (empirical observation)We are now investigating whether the method can be used to design effective MAC and scheduling algorithms

0 2 4 6 8 10 12 14 16 18 200

0.05

0.1

0.15

0.2

0.25

0.3

0.35

0.4

0.45

Number of captured signal (r)

C n(r;

K) n

=20

K=0K=1K=2K=3K=4K=5

0 0.5 1 1.5 2 2.5 3 3.5 40

0.1

0.2

0.3

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0.5

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0.8

Number of captured signal (r)

C(s) n

(r;K)

K=0K=1K=2K=3K=4

100 101 1020

2

4

6

8

10

12

14

16

Number of overlapping signals (n)

Thro

ughp

ut

K=0K=1K=2K=3K=4K=5

0 2 4 6 8 10 12 14 16 18 200

0.5

1

1.5

2

2.5

3

Number of SIC iterations (K)

Norm

alize

d m

axim

um th

roug

hout

(!(K

))

b=0.01b=0.02b=0.04b=0.06b=0.08b= 0.1b= 0.2

z=0.05

z=0.2

z=0.1