• Nenhum resultado encontrado

Order detection and blind of 2x1 MISO channels

N/A
N/A
Protected

Academic year: 2018

Share "Order detection and blind of 2x1 MISO channels"

Copied!
4
0
0

Texto

(1)

ORDER DETECTION

AND BLIND IDENTIFICATION OF 2

x 1

MISO CHANNELS

Carlos

Estevcio

R.

Fernandes,'

Pierre

Comon

and

Ge'rard

Favier

13S

Laboratory-CNRS,

University

of Nice

Sophia

Antipolis,

France.

{cfernand,

favier, pcomon}@i3s.unice.fr

ABSTRACT

Inthispaper, we investigate theuse ofoutput4th-order

cu-mulantstodetect the number ofsourcesignalson a

multiple-input single-output (MISO) communications channel and

blindlyidentify their respective channel coefficients. More

particularly,

we areinterested in thecaseoftwo sources. The proposed cumulant-based order detectionprinciple allowsus

forrecovering the longest channel order and its coefficients.

A similarprocedure is applied fordetecting thepresence of

asecond source aswell as estimating its associated channel order and coefficients. Whenonly estimated cumulants are

available,we implementtwo hypothesistestsbasedonthree proposed test-statistics. Computer simulations illustrate the performances obtained with the methodsproposedfor order detection and channel identification.

Index Terms- Order

detection,

MISO

channels,

Blind

identification, Higher-order statistics.

1. INTRODUCTION AND PROBLEM DESCRIPTION

We are interested in detecting the number of source

sig-nals and identifying channel parameters in the context ofa

multiple-input single-output(MISO)communicationsystem.

The transmitters are supposed to use the same carrier

fre-quencyand are located far apartfrom each other, hence the physical channels are different. The following assumptions

hold:

Al The input signals

sp

[n], for everyintegerp C

[1, P],

are mutually independent and temporally i.i.d. (inde-pendentlyandidenticallydistributed).

A2 The random variables

sp,

p C

[1, P],

have unknown

non-Gaussian distributions withnon-zerokurtoses.

A3 The channellengthsareboundedbyaknown value L.

Theoutput signal is the result ofalinear combination of the discrete-time input signals, s

[n]

C Cp. After baud rate

sampling,the receivedsignal

x[n]

canbe writtenasfollows:

p Lp

x[n]

= , ,

hp[l]sp[n

-1],

p=l 1=0

(1)

PhDscholarship grantedfrom CAPES agency, Braziliangovernment.

where

hp[l],

with I C [0,

Lp]

andp C [1, P], arethe channel parameters associated with the userp and

hp[l]

= 0, Vl f

[0,

Lp],

where

Lp

< L, Vp C [1, P]. Let usdefine the chan-nel vector

hp

=

[hp(0),

....,

hp(Lp)]T

with p C [1, P]. From

the observation ofx[n] only,ourgoalsare to: PI: Detectthe number ofsources

P;

P2: Determine the channellengths

Lp

for eachp e [1,P]; P3: Identify thevectors

hp

for eachpC [1, P].

The case of overdetermined mixtures (more sensors

than sources) has been exhaustively treated in the

litera-ture, including static mixtures [1] as well as dynamic ones

[2]. On the other hand, underdetermined mixtures have

onlyrecently received attention. Our contribution is based

on cumulant-matching as in [3]. Due to assumption Al, the 4th-order cumulants of x[n], defined as C(i,j,k) A

cum{x [n], x[n+i],x*[n+ j],x[n+k] }, canbeexpressed in terms of the marginal cumulant contributions of each

source as: p

C(i,j,k) =Z Cp(i,j, k),

p=1

(2)

inwhichCp(t,j,k)dependsonthe unknown channel param-eters

hp[l]

andcanbe writtenasfollows [3,4]:

Lp

Cp(ij,k) oyp

,h*[I]hp[I

+

Zi]h*[I

+

']hp[I

+

k],

1=0

wherep e [1, P]and

-ysp

stands for the kurtosis of thesource

sp[n].

Since

hp[l]

=0, VlI

[0,

Lp],

wehave

Cp(i,

j,

k)

=0 V i,j, k> LP. The numberPofsources(users) is unknown and assumedto be boundedby two, i.e. P < 2, sothatwe

consider a MISO channel with two inputs (P = 2) and a

singleoutput,assumingtransmission channelsareof different

orders,sothat L1 > L2. Ifasinglesourceispresent(P = 1) the model still holds withL2 0 andh2

(0)

=0. Therefore,

weget:

C(i, j,

LI)

=C1 (i, ,

LI)

= h1

[0]

hi

[i] h1 [j]

hi [LI].

(3)

2. CUMULANT-BASED ORDER DETECTION

Fromequation (3),it isstraightforwardtoobtain:

C(i,

k,

LI)

h1

[j] -C(j,

k,

LI)

h1

[i]

=

0,

(4)

(2)

with 0 < i < j < LI and0 < k < L1. From equation (4)

we can get a setof L1(L1+1)2/2equations with L1+ 1

un-knowns, whichcanbe rewritten inamatrix formasfollows:

Ch1 = O, (5)

where 0 is the null-vector of dimensionL1

(Li

+ 1)2/2 and C E

C(LI(Li+1)2/2)x(Li1+)

A solution to (5) is obtained by computing the right singularvectorof C associated with the smallestsingular value [3]. That solution is optimal in the total least squares(TLS)sense. To avoid the trivial solution

we canimposeaunit-norm constraint

(E,,

|h1[n] 2 = 1).

4.5 4 -5 35

33 2

25 m 2 , 1.5

10

0.5-3 4

Tested orders (I) 5 6

Fig. 1. Profile of the largestSVor1(f) ofC(f) for f [1,L]

with L1 =4,L2 =2andL =6.

Let usconsiderachannel order testing variable fC [1, L],

L > L1 > L2, and build L cumulant matrices

C(f)

C

C(f(f+1)2/2)x(f+1)

satisfying (4) with f in the place ofL1.

Thisyields

C(f)h1

= d, where d = 0if f > L1, as in(5), and d 74 0otherwise. That is the main idea behindour chan-nel order detector. Denoting byor

(f)

> ... >

Orf+I(f)

the singular values (SV) of

C(f),

we notethat for f > L1,

C(f)

is the null matrix andclearlywehave:

or1(f)

= .. =

7f+1(f)

=0. (6)

search a rupture point in the profile ofagiven SV of

C(f),

e.g. 7i(f) for a fixed i e [1, £]. Actually, it would be

inter-estingto includenotonlyone SV

cri(f)

butas many as

pos-sible. Ideally, for each f C [1, L],the f largest singular val-ues or (f),. ..

.,

o+1

(f) of matrix C (f) should be taken into

accountfor the computation ofasingletestvariabletj(f)

con-taining information about the whole dynamic ofC(f); a de-cision threshold 'thr mustbe establishedsothati7(f) > T/t1hr

only if f<L1.

3.1. Test variables

Three testvariables

TI(f)

areproposed in Table 1. Note that TIA

(f)

andi/B

(f)

consist ofgeometricalmeans involving the

SVof C(f). Onthe otherhand,

rqC

(f)canbedirectly derived from C(f) without singular value decomposition (SVD)

com-putations, since

r/c

(f) = C(?) F;this factrepresents a great advantage ofTIC (f) overT/A (f) and /B

(i)-Table 1. Computation oftestvariables.

f

\1/f

TIA(rQ)= ?) i(tZ(- ) )

i=l J

n vi(f)

TIB(f)

v-

(

1+1

ICMt

=

A/E

Jvi(f)l2

Vi=l

InordertodetectL1,wesearcharuptureintheprofile of thetestvariables,asfollows:

1. Initialize thelength variableasf= L;

2. Estimate theoutputcumulantsC(i,k,

£)

for0 < i < f and0 < k < f and build matrix

C(f)

(as explainedin section2);

On the other hand, ifwe take f = L1 we get

ui(LI)

7

0, i = 1,...

,LI,

with

71(LI)

> ... >

ULi(LI)

and (L1±1(L1) = 0. Therefore,theflargestSV ofC(f) onlytake

non-zerovalues for f < L1 andwe candetect the order L1 of the longestchannelby lookingfor thenon-zerovalues of

oi

(f),

i C

[1,e],fortestedorders

f C

[1, L].

Toillustratethis reasoning, fig. 1 plots the largest SV, or1(f), of the cumulant matrices C(f) for values of f in [1, L], withL = 6, L1 = 4 and L2 = 2. Perfectknowledgeof the cumulant valueswas assumed. Thefigure clearlyshows that or1

(f)

= O, V f > 5 and the firstnon-zerovalue is reached for f=L1 =4.

3. LONGEST CHANNEL IDENTIFICATION

Inpractice, cumulantsareaffectedby estimationerrors sothat

(6) is no longer valid for f > L1. Our problem is thento

3. Compute the SVD of

C(f),

rI(f), ....,

O7+-i

(f) itssingular values;

and denote

4. Denoteby

vf

e

Cf

+1 the singularvectorofC(f)

asso-ciated with

oe+I

(f). This is theoptimal solution of (5) intheTLS sense[3];

5. Determine the testvariable

T/A(f), T/B(f)

or

Tc/(1)

de-fined in Table 1;

6. Decreasefby 1unit andrepeat steps 2 to 5until f= 1.

In step 4, we construct a collection ofvectors, one of whichcorrespondstothe actual longestchannelvectorh1 C

CL,+1. Decision about which is the good one should be taken basedontheprofileof thetestvariables

i/A(i), i/B(f)

or

TIC(f).

A decision threshold

T/thr

mustbe established so

that

TI(f)

>

TtLhr

ifandonlyif f < L1. Areasonable choice

III-754

r T

n)

(3)

for the decision threshold is thegeometricalmeanof thetest

variable for f C [1, L],sothat:

L 1/L

Ttthr =

fl1

Ti

(7)

where

T(f)

takes the values of eitherT'A

(C),

T1B

(Q)

orqc

()

3.2.

HI/H2

hypothesis test

Basedonthe variabletj(f),the proposed hypothesistest

con-siders thefollowing hypotheses:

* Hypothesis H1: f>

LI

* Hypothesis H2: f =

LI

We assume that H1 holds truefor f = L andtest

TI(M)

withrespect to athresholdTlthr for f =L,. ..,1. We accept H1 if

rq(f)

<

lqthr,

otherwisewereject it andacceptH2, thus determining the orderTO = L1 of the longest channel. After that, step 4of thepreviously described procedureyields:

h1i

=vfO: TO=

Li.

(8)

Inthenextsection,wepropose amethodtoestimate the order L2 of the shortest channel and its coefficientvectorh2, if it exists. We also discuss thecaseof channels ofsameorder.

4. SHORTEST CHANNEL IDENTIFICATION

Recalling from section (1) that

Cp(i,

j,k) =0,V k >

Lp,

we

notethatC2(ij,

k)

= 0 for k > L2 and hence

C(i,j,

k)

C(i,

j,k),Vk

> L2. Fork=L2,weget:

C2(i, j, L2) =

ys2

h2[0]h2[i]h2[]h2[L2] (9)

Fromdefinition(2) withP=2, we canestimate themarginal cumulant contribution of the source

s2[n]

as C2(i,i,

k)

C(i, j,

k)-

C

(i,

j,

k),

where C1

(i,

j,

k)

is obtainedas fol-lows from the channelparametersh1[1] estimated in(8):

Li

C (i,

j,k)

=

'Ys

E

h[l]hi

[I

+ i]hl + I]hl +

k].

1=0

Therefore, the marginal cumulant C2(i, j,

k)

appears to

beagoodmetrictodetect thepresence(orabsence)ofa

sec-ondsourcewith shorter channel. If sucha source ispresent, C2(i,j,

k)

will be non-zero for somevalues of k C

[1, LI].

Precisely, it equals zero for k > L2 and it is non-zero for k < L2. Hence, a new test variable

p(ii;)

can be setup to

detect thepresenceof the secondsource sothat:

{

p(Q)

=

0,

E >

L2

p(Kc)

+4

O, K <

L2-Wewill define thetestvariableas

pQ(/)

=

C2Q(K)

I2,

i.e. the

squaredFrobeniusnormof matrixC2

(/)

C

C(±+1)

x(I+1) in

100

o0 2 3 4

Tested orders (K) 5 6

Fig. 2. Testvariable

p(Qi)

for ic C [1, L] withL1 =4, L2 =2

andL =6.

which the element inposition(i, j)equalsC2 (i-1, j-1,ii), i,jC

[1,

/iC+1]

Figure2illustrates the detection of L2usingp(ii;)forlc C [1, L]. The channels considered herearethe same asthose of section2 (LI =4,L2 =2andL =6). Perfectknowledgeof the cumulant valueswas assumed. Thefigureclearly shows that the firstnon-zerovalue of

p(Qi)

is reachedat c =L2 =2.

4.1. H3/H4 hypothesis test

Let us nowconsider themorerealisticcasewhere theoutput

cumulants are notknown and needto be estimated. Inthat

case,the estimationerrorscorruptingC(i,j,k) imply inaccu-racies in the estimation ofthelongest channelh1, which leads

to abad reconstruction of C1

(i, j, k).

This means that

cu-mulative errors areinvolved in the calculation ofC2(i, j,k). Hence, (10) is no longer valid and another hypothesis test mustbe performed in order to detect the length L2 of the shortest channel. Forthispurpose, adecision threshold pthr

mustbe defined sothat

p(Q)

> Pth, only if/-; < L2. Here again,thegeometricalmean seems tobeagoodchoice:

/L II1L

Pthr

(ft

p(k)) (11)

k=I

and thehypothesistestis definedsothat:

* Hypothesis H3: EC > L2

* Hypothesis H4: C =L2

Startingwith /; = L1 andsuccessively decreasing

Ki,

we

ac-ceptH3when

p(Q)

< Pthr, otherwiseweacceptH4 andget anestimate ofL2. Ifthetestvariableneverreaches the

thresh-old,thenweconclude thatasinglesourceispresent. Once we have determined L2,we can useC2

(i, L2)

to estimate the parameter vector h2 of the shortest channelby solvingalinearsystemsimilarto(5)intheTLSsense.

4.2. Channels with identical lengths

In the case where L1 = L2, our approach is no longer

valid. Equation (3) becomes

C(i,

j,

L

1)

= Ci

(i,

j, L

1)

+

(4)

C2(i, j,

LI),

sothat theoutputcumulant contains information

on both channels. Hence, the estimation method described in section 3 willnotresult in a fair estimate ofhi and (8) doesnotholdtrue. Nevertheless, the hypothesistest HI/H2 canstill be usedto determineLI. Inaddition,we candetect thepresenceof the second sourceand determine whether the channels have thesame ordifferent orders. Proceedingtothe estimation ofthelongest channelasdescribed in section 3,we can use

h,

tocompute

p(Qc)

as described earlier in this

sec-tion. Thenwe canimplement thetest onthe shortestchannel, which will fail forhypothesis H3 atic = LI, indicating that both channelsareofsameorder. Therefore, the proposedtests HI/H2 andH3/H4canalsobe used for thecaseof channels ofsamelength. However, we cannotidentify the channel pa-rametersinthiscase. Infact, there isno means to seethesum

oftwo linearprocesses as ascalar linearprocess [5].

5. COMPUTER SIMULATIONS

The results shown in this sectionwere obtained from com-puterexperiments simulating the MISO system (1)with the

two following channels: hi = [1.0, 0.61 -1.18i,

-0.59-0.53i,-0.82 + 0.21i1T and h2 [1.0,-0.29

-0.33i,-0.62-0.73i1T sothat LI = 3 and L2 = 2. We

assumedL = 5and the 4th-order cumulantswereestimated fromanoutput sequenceof50000datasamples. Noadditive noise ispresent.

Table2. Longestchannel detectiontest:

P(tj

<

Tlthr)

T1 =TjA

T1

=TjB

T1

=TjC

£=5 100%o

100%7o

100%7o

f=4 luu% 100% luu%

£=3 2.7% 0.88% 6.26%

Success rate 97.3% 99.1% 93.7%

Figure 3 shows thehistogram plots for the testvariable

hqA,

built from thesample outputcumulants over 5000 Monte

Carloruns. Similarplotsfor

riB

and

n/c

were also obtained. Table2 shows theprobability of

nq(1)

<

Tlthr

highlighting the

success rate of the LI detection. Forf < 2all the

probabili-tieswere zero.Notethat

n/B

gives the bestprobability of good decisionsonthe channel orderLI withonly

0.88%

probabil-ity of false alarm. Onthe otherhand, although

n/c

yieldsthe

poorestperformance among the three considered criteria, it could be worthwhileconsideringto useit since it doesnot

re-quireany SVDcomputationandcanbe deriveddirectlyfrom

C(f)

F. Inthis case, the threshold level could be revised inordertoadjust the sensitivity of the detector(e.g., reduc-ing

n/thr

ofan amountof

10%

increases the success rate to 98.2%.).

6. CONCLUSION

Inthispaper, amethodusingtheoutput4th-order cumulants has beenproposedto detect the number ofsourcesand

iden-IL

50l

150 150

100 100

0( 2 4 6 8 °( 2 4 6 8

men(A.)=3273 me (A )=7.0

Fig. 3. Histogram plots for

qA

(mean value Ofrlthr is 2.19).

tify theirrespective transmission channels fora 2 x 1

com-municationsystem. Thesameprinciple applies forany P x 1 MISO channel. The proposed order detection principle is based on the search for the highest value ofatest-variable

sothat the cumulant matrix isnon-zero. Afterextracting the contributions of that channel from the output cumulants, a

similar procedure is applied for detecting thepresence ofa

secondsource aswellasits associated channel order and

co-efficients. Becausethere isno means to see thesum oftwo

linearprocesses as ascalar linearprocess, we cannotestimate

parametersof channels ofsameorder.However,it would also be possible to detect the number ofsources via a linearity

test [5]. Considering estimated cumulants, two hypothesis

testsbasedonthreeproposedtest-statistics have been

imple-mented. Someperspectives includea tensor approach based

on a testof thetensorrank. Theasymptotic distribution ofthe proposedtestvariables also needstobeinvestigated.

Acknowledgement

The authors would liketothank theanonymousreviewers for their relevant and useful remarks.

7. REFERENCES

[1] P. Comon, "Independent Component Analysis, a new con-cept?,"SignalProcessing,Elsevier,vol.36,no.3,pp.287-314,

apr. 1994, Specialissue onHigher-Order Statistics.

[2] E. Moulines, P. Duhamel, J. F. Cardoso, and S. Mayrargue,

"Subspace methods for the blind identification ofmultichannel

FIRfilters,"IEEETrans.SigProc., vol.43,no.2,pp.516-525,

feb. 1995.

[3] P. Comon,

"MA

identificationusingfourth order cumulants,"

Signal Processing, Elsevier,vol.26,no. 3,pp. 381-388, 1992. [4] D. R.Brillinger,and M.Rosenblatt, "Computationand

interpre-tation of kth-orderspectra," Spectral AnalysisofTimeSeries, Wiley,B.Harris(ed.),NewYork, USA,pp. 189-232, 1967.

[5] M.Hinich, "TestingforGaussianityandlinearityof astationary

timeseries," Journal ofTimeSeriesAnalysis,vol.3, no.3,pp. 169-176, 1982.

Imagem

Fig. 1. Profile of the largest SV or1 (f) of C(f) for f [1, L]
Fig. 2. Test variable p(Qi) for ic C [1, L] with L1 = 4, L2 = 2 andL = 6.
Table 2. Longest channel detection test: P(tj &lt; Tlthr)

Referências

Documentos relacionados

The probability of attending school four our group of interest in this region increased by 6.5 percentage points after the expansion of the Bolsa Família program in 2007 and

In an active construction such as (1a) below, for example, its external argument is realized as the syntactic subject, bearing nominative Case and triggering

In order to test the hypothesis of D. albiventris participating in the seed dispersal of disturbed environments, we performed an experiment to investigate its

In this context, the present study analyzed the fish fauna of a first-order rural stream to test the hypothesis of differences in the presence, abundance and biomass of

In order to compare the average concentrations of alpha- and gamma-tocopherol between brands and types of analyzed oils the test of Variance Analysis was performed (ANOVA),

A Conferência das Nações Unidas sobre Meio ambiente e Desenvolvimento (CNUMAD) de 1992 definiu desertificação como sendo “a degradação da terra nas regiões áridas,

No grupo de indivíduos saudáveis foram incluídos os indivíduos não relacionados que não apresentavam acidente trombótico com mais de 42 anos de idade – subgrupo saudáveis + 42