CRBanalysis

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Derivation of CRLBs for the correlated interferer

models

Sean ORourke

March 31st, 2010

1 Correlated Interference Model

Assume that an arbitrary m element array recieves a single scaled and delayedreplica of a known signal s(t), corrupted by an interferer with an unknown (andpossibly complex) signal y(t). The sampled baseband array output is modeledas the m 1 complex vector

x[n] = as(nTs 0) + by(nTs) + e[n] , (1)

where Ts is the sampling period; a, b CIm are the spatial signatures of the

direct-path arrival and the interferer; 0 is the direct path time of arrival; ande[n] comprises additive (white Gaussian) noise. Let us further suppose thatthe interferers waveform is correlated with the delayed replica of s(t) to someextent; that is, we can decompose yu(t) as

y(t) = s(t o) + yu(t) , (2)

where E[yu(t)y

u(t)] = 2b and E[yu(t)s

(to)] = 0 We note that 2b is a measure

of how uncorrelated y(t) is with the desired signal, and in general is dependenton 0.

Using this, we can now form Equation 1 as

x[n] = (a + b) s(nTs 0) + byu(nTs) + e[n] (3)

x[n] = (a + b) s(nTs 0) + z[n] (4)

x[n] = us(nTs 0) + z[n] (5)

where the covariance matrix of the new noise and uncorrelated interferenceterm z[n] is Q = E z[n]zH[n] = 2

nIm

+ 2b

bbH. We refer to Equation 4 asthe unstructured model. If we assume that the array is calibrated such thatthe response to the direct-path signal, say a0, is known to within a complexconstant 0, we obtain the form

x[n] = (0a0 + b) s(nTs 0) + z[n] (6)

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where the covariance matrix of z[n] is identical to the previous case. We shallrefer to Equation 6 as the structured model.

If we assume that z[n] CI N(0, Q), then x[n] CI N(, Q) where [n] =(a + b) s(nTs 0) for the unstructured model, and [n] = (0a0 + b) s(nTs 0) for the structured model. Since our model is Gaussian, we can use theSlepian-Bangs formulation [?] to calculate the Fisher information matrix, andthus the Cramer-Rao bounds for each of our models. The formulation statesthat, for a given parameter vector = [1 p] CI

p,[n] CI N([n], Q), the(i, j)th element of the associated Fisher information matrix is:

FIMij = 2Re

Nn=0

[n]

i

HQ1

[n]

j

+Tr

Re

Q1

Q

i

Q1

Q

j

(7)

Then, the Cramer-Rao bound for each parameter in can be found along the

diagonal of FIM().

2 CRB Derivation for the unstructured model

For this case, our parameter vector is u = [Re{a}, Im{a}, 0, Re{b}, Im{b}, 2b ,

2n].

Note that for this parameter vector, the structure of our FIM is:

FIM =

Faa fa... Fab fa2

bfa2

n

fa F... fb F2

bF2

n

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

Fba fb... Fbb fb2

b

fb2n

f2ba F2

b

... f2bb F2

b2b

F2b2n

f2na F2

n

... f2nb F2

n2b

F2n2n

(8)

where, for example,

Faa =

FaRaR FaRaIFaIaR FaIaR

fa =

f0aR f0aI

with aR = Re{a}, aI = Im{a}. Each of these submatrices contain the appro-priate elements from the Slepian-Bangs formulation above.

With this structure in mind, we form the derivatives of the mean and co-variance matrix needed for the formulation. From inspection, we see that [n]

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has no dependence 2b or 2n, nor does Q have any dependence on a or 0. Thus:

[n]2b

= [n]

2n= 0(2m+2)1

Q

aR,i=

Q

aI,i= 0m1

Q

0= 0mm

where aR,i (aI,i) is the ith element of the real (imaginary) part of the vectora. Note that the derivative of Q with respect to a is element-wise, because wechoose not to define a 3-D structure to handle the derivative of a matrix withrespect to a vector. These mean that the blocks of the FIM that have derivativeswith respect to a, 0 contain only the first term in Equation 7; similarly, blocks

with derivatives w.r.t. 2

b ,

2

n contain only the second term.The derivatives of[n] are:

[n]

aR=

[n]

bR= s(nTs 0)Imm (9)

[n]

aI=

[n]

bI= s(nTs 0)Imm (10)

[n]

0= (a + b)d(nTs 0) (11)

where d(t) =ds(t)dt

. Likewise, the derivatives of Q are:

Q

bR,k= 2b beTk + ekbH (12)

Q

bI,k= 2b

beTk ekb

H

(13)

Q

2n= Imm (14)

Q

2b= bbH (15)

where ek is the kth m-dimensional unit vector (e.g., a 1 at position k and zeroselsewhere). Given these derivatives, we can form the constitutent blocks of theFIM. We note that any block whose subscripts are reversed (say, 0aR) is thetranspose of the original block (say, aR0).

2.1 Direct path FIM components (Faa, fa, fa, F)

For the upper diagonal blocks in Equation 8 (listed in this subsections title),the zero derivatives discussed above lead to these blocks requiring only the first

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term of Equation 7. The blocks are:

Faa =

Re {F1} Im {F1}Im {F1} Re {F1}

(16a)

fa =

Re {f3}Im {f3}

(16b)

fa =

Re {f3}T

Im {f3}T

(16c)

F = Re{F2} = F2 (16d)

where

F1 = 2Q1s(0)

2(17a)

F2 = 2(a + b)HQ1(a + b)d(0)

2(17b)

f3 = 2Q1

(a + b)sH

(0)d(0) (17c)

are used for notational ease. In Equation 16d, note that Q (and thus Q1) is apositive semidefinite matrix, thus any quadratic form xHQ1x, x CI m, x = 0is necessarily real and non-negative. Hence, Re{F2} = F2.

2.2 Cross term FIM components (Fab, fa2b

, fa2n, fb, F2b , F2n)

The blocks for this section of the FIM are:

Fab = Faa (18a)

fa2b

= fa2n

= 02m1 (18b)

fb = fa (18c)

F2b

= F2n

= 0 (18d)

We obtain zero (a zero vector) for F2b

, F2n

(fa2b

, fa2n

) because each of theparameters zeroes out one term in the Slepian-Bangs formulation. The cor-respondence of the other blocks with blocks from the previous subsection is aresult of the mean depending on both a and b. As mentioned previously, theblocks Fba, f2

ba, f2na, fb, F2b, and F2n are the transposes of the blocks listed

in Equation 18, so they are not listed explicitly.

2.3 Interference and noise FIM components (Fbb, fb2b

, fb2n,F

2

b2

b

, F2

b2n

, F2n

2n

, etc.)

So far, calculation of the FIM blocks has only required the first term of theSlepian-Bangs formulation, because of terms that only existed in the mean;hence, the derivative of Q with respect to these terms was always zero. Withthese blocks, we require either both terms (in the case of Fbb) or the secondterm only (for the rest of the blocks) of Equation 7. This requires more carefulderivations, so we shall address certain blocks in their own subsections.

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2.3.1 Calculation of Fbb

First, we consider Fbb. As described above, this block uses both terms of theSlepian-Bangs formulation. For ease of notation, let us write this block as thesum of two matrices, one formed from each term of Equation 7:

Fbb = M + C (19)

where M is the 2m 2m matrix from the term of mean derivatives, and C isthe 2m 2m matrix from the term of covariance derivatives. Each of these arefurther subdivided into 4 blocks as follows:

Fbb =

FbRbR FbRbIFbIbR FbIbR

(20a)

M =

MbRbR MbRbIMbIbR MbIbR

(20b)

C =

CbRbR CbRbICbIbR CbIbR

(20c)

where the subscripts R and I indicate the real and imaginary parts of a vector,respectively.

Since the derivatives of the mean with respect to the elements of b are thesame as those with respect to a, it is clear that

M =

Re {F1} Im {F1}Im {F1} Re {F1}

(21)

with F1 is defined as in Equation 16a.In order to derive the form for C, we require the use of several properties

from matrix algebra and calculus. First, we use the following property of thetrace operator [?]:

Tr(ABCD) = vec(D)T(A CT) vec(BT) (22)

where A, B, C, D are conformable matrices, vec(A) stacks the columns of Ainto a single vector, and is the Kronecker product. Furthermore, note thatfor a rank-one matrix, say uvH, its vectorization is:

vec(uvH) = v u (23)

Given this, we can write Equations 121315 Using the second term of the Slepian-Bangs formulation, we can write the (k, l)th element of CbRbR as:

[CbRbR ]k,l = Tr

Q1

Q

bR,kQ1

Q

bR,l

(24)

= vec Q

bR,lT

(Q1

Q1T

) vec Q

bR,k

T(25)

3 CRB Derivation for the structured model

Here, our parameter vector is s = [Re{0}, Im{0}, 0, Re{b}, Im{b}, 2b ,

2n]

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