Correction Exam 2013 2014

7/23/2019 Correction Exam 2013 2014

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Corrected Exam 2013-2014

SAR-B12: Estimation and detection.

1 Radiomobile propagation: Fast fading

1.1 Observation and physical model.

1.1.1 Relationship betweenx and y.

We have:

x =Nk=1

rk. cos(k)

y =Nk=1

rk. sin(k)

withrk and k assumed independents. The E[x.y] can be written as:

E[x.y] =E

Nk=1

rk. cos(k) Nn=1

rn. sin(n)

=

Nk=1

Nn=1

E[rnrk] E[sin (n)cos(k)] = 0

because E [sin (k) cos(n)] = 0 k, n

1.1.2 Mean ofx and y.

E[x] =E

Nk=1

rk. cos(k)

=

Nk=1

E[rk] E[cos(k)] = 0

because rk andk are independents.

As kis uniformly distributed in {0, 2},E[cos (k)] = 0, consequently E[x] =Nk=1

E[rk] E[cos(k)] =

0idem for E[y] = 0.

1.1.3 Variance de x et y.

E

(x E[x])2

= Ex2

= E

N

k=1

rk. cos(k)

2= Nk=1

E

r2k

.E

cos2 (k)

= N2E

r2k

Ey2

= N2Er2k

= Ex2

1.1.4 Transformation from Cartesian to polar coordinates.

a) Give the relationship between r andx, y (we have assmued that r 0). x = r. cos()y = r. sin()

consequentlyr =

x2 + y2 withr 0.b) = arctan (y/x)

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c) From pr(r, ) = det(J)pxy(x, y)with J=

xr

x

yr

y

, we thus have

det (J) =det

cos r sin sin r cos

= r

pr,(r, ) = r 12

exp x2

22 12 exp y2

22= r

22exp

r222

d) pr(r) =

20

r

22exp

r2

22

d= r

2exp

r

2

22

1.2 Parametric Estimation

a) L(r, ) =pr(r|) = 12N

Ni=1

ri

exp

Ni=1

r2i22

b) (r, ) = ln (L(r, )) =N. ln 2

+N

i=1

ri 122

N

i=1

r2i .

c) (r,)2

= N2 + 2(22)2

Ni=1

r2i then 2ML=

12N

Ni=1

r2i .

d) T(r) =Ni=1

r2i is a sufficient statistic.

e) Starting from the probability distribution function (??), and from theNeyman-Fisher factorisationtheorem

pr,2(r, 2) =

Nk=1

rk2

exp

r2k22

=Nk=1

r2k

exp 1

22

Nn=1

r2k

2N

= h(r) g

T(r) , 2

T(r) is a sufficient statistic.

1.3 Performance analysis.

a) 2ML is unbiased because E

2ML

2

= E

12N

N

i=1r2i

2

= 12N

N

i=1E

r2i

2 = 0

b) V ar

2ML

= 14N2

Nk=1

var

r2k

=

4

N.

1.4 Cramer-Rao bound.

a)

CRB(2) = 1

E

lnpr,(R,)

2

2 = 1E2 lnpr,(R,)2

2

= 4N

b) 2

ML

is unbiased and its variance is equal to the Cramer-Rao bound, 2

ML

is then an efficientestimator.

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2 Estimator performances.

Assume that Y1, . . . , Y n are independent and identically distributed random variables with each Yihaving the following probability density function:

pY (y|) = y2

23exp

y

, y >0where > 0 is the parameter of interest. It is known that E [Yi] = 3 andV ar [Yi] = 3

2 for eachi= 1, . . . , n.

a) MOM = 13n

ni=1 yi=

13m1

(b) L(y, ) =

ni=1

y2i23

exp

ni=1

yi

.

(c) the log-likelihood function is l(y, ) = ln (L(y, )) =ni=1

ln

y2i2

3n ln () 1

ni=1

yi,

consequently the maximum of likelihood estimator of :

ML= 13n

ni=1

yi= MOM

.

(d) E

MOM

= E

ML

= 13n

ni=1

E [yi] = n3

3n =

(e) V ar

MOM

= 2

3n

(f) Show thatni=1

Yi is a sufficient statistic for the estimation of .

(g) C RB () = 2

3n .

(h) MOM and ML are efficient estimators.

3 MIMO communication system

a) IfNR > NT the linear problem is overdetermined, when NR = NT the linear problem is exactlydetermined and underdetermined ifNR < NT.

b) Condition on the range: R (y) R (H), or on the rank: rank

y H

= rank(H)

A first approach to solve this problem will be a Least Square (LS) approximation solution. Thissolution, namedsLS, minimises the following cost function:

J= Hs w2 = (y Hs)T (y Hs) (1)

c) The LS solution is sLS=

HTH1HTy.

The second approach consists to establish the Maximum of Likelihood Estimator of the transmittedsymbols sML.

d)

L(y, s) =p(y|s) = (2)NR/2

det

2INRNR1/2

exp12(y Hs)

T2INRNR1 (y Hs)= (2)NR/2 NR exp

122 y Hs

2

e) L(y, s) = ln (L(y, s)) =NR2 ln(2) NRln () 122

y Hs2.

f) sML=

HT

H1

HT

y

g) The matrix: H =

HTH1

HT is named pseudo-inverse. It may be computed by the QR

method.

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Correction Exam 2013 2014