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Detection, Estimation modulation theory part 1 - Vantress H.

Vantress H. Detection, Estimation modulation theory part 1 - Wiley & sons , 2001. - 710 p.
ISBN 0-471-09517-6
Download (direct link): sonsdetectionestimati2001.pdf
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We discuss the binary hypothesis testing version of the general Gaussian problem in detail in the text. The Af-hypothesis and the estimation problems are developed in the problems. The basic model for the binary detection problem is straightforward. We assume that the observation space is W-dimensional. Points in the space are denoted by the N-dimen-sional vector (or column matrix) r:
r =
(319)
Under the first hypothesis Hx we assume that is a Gaussian random vector, which is completely specified by its mean vector and covariance matrix. We denote these quantities as
-\ l#l) ~mxl
) mi2 A m1
-E(rN |tfa)_ ~m1N_
?[|] =
The covariance matrix is
A ?[(r - س - 41^)1 tfx] iK12 iK13 lK2i \K22 '
-lK-m
We define the inverse of Ki as Qx
Qi 1
Q.K, = KM = I,
(320)
Ki
A'v
(321)
(322)
(323)
98 2.6 The General Gaussian Problem
where I is the identity matrix (ones on the diagonal and zeroes elsewhere). Using (320), (321), (322), and (318), we may write the probability density of on #b
prlHl() = ^,!1/]-1 exp [-mT ~ m1r)Q1(R - mO]. (324)
Going through a similar set of definitions for tf0, we obtain the probability density
/>r|0(R|tf0) = [(2*)2|Ko|*]-i exp [-i(RT - m0r)Qo(R - )]. (325) Using the definition in (13), the likelihood ratio test follows easily:
* ^ ' [-KR7 - m/K^R - in,)] ( } - () |KX|* exp [-KRr - m/)Q0(R - )] < 7?'
(326)
Taking logarithms, we obtain
T ~ W) Qo(R - ) - i(Rr - in/) Qj(R - mi)
I In >7 + In |Ki| - In |K0| 4 y*.
(327)
We see that the test consists of finding the difference between two quadratic forms. The result in (327) is basic to many of our later discussions. For this reason we treat various cases of the general Gaussian problem in some detail. We begin with the simplest.
2.6.1 Equal Covariance Matrices
The first special case of interest is the one in which the covariance matrices on the two hypotheses are equal,
K, = K0 A K, (328)
but the means are different.
Denote the inverse as Q:
Q = K 1. (329)
Substituting into (327), multiplying the matrices, canceling common terms, and using the symmetry of Q, we have
(m/ - m0T)QR ^ In 7) + -Km/Qiih - m/Qmo) '*. (330)

We can simplify this expression by defining a vector corresponding to the difference in the mean value vectors on the two hypotheses:
س _ ni0. (331)
Then (327) becomes
Equal Covariance Matrices 99
or, equivalently,
/(R) A RTQ > /*.
Ho
(332)
(333)
The quantity on the left is a scalar Gaussian random variable, for it was obtained by a linear transformation of jointly Gaussian random variables. Therefore, as we discussed in Example 1 on pp. 36-38, we can completely characterize the performance of the test by the quantity d2. In that example, we defined d as the distance between the means on the two hypothesis when the variance was normalized to equal one. An identical definition is,
[E(l\) - E{I\H0)f ~ Var (/|#0) '
Substituting (320) into the definition of /, we have E(l\Hx) = ^س
and
(335)
(336)
E(l I #0) = AmTQm0.
Using (332), (333), and (336) we have
Var [/|#0] = ?{[AmTQ(R - m0)][(RT - m0T)Q Am]}. (337)
Using (321) to evaluate the expectation and then (323), we have
Var [/|#0] = AmTQ Am. (338)
Substituting (335), (336), and (338) into (334), we obtain
d2 = AmTQ Am
(339)
Thus the performance for the equal covariance Gaussian case is completely determined by the quadratic form in (339). We now interpret it for some cases of interest.
Case 1. Independent Components with Equal Variance. Each r{ has the same variance o2 and is statistically independent. Thus
and
= a21
(340)
(341)
100 2.6 The General Gaussian Problem
Substituting (341) into (339), we obtain
d2 = = = Am7 = ; 2
(342)
or
d =
||
(343)
We see that d corresponds to the distance between the two mean-value vectors divided by the standard deviation of R,. This is shown in Fig. 2.32. In (332) we see that
/ = -4 AmTR.
(344)
Thus the sufficient statistic is just the dot (or scalar) product of the observed vector R and the mean difference vector .
Case 2. Independent Components with Unequal Variances. Here the rt are
statistically independent but have unequal variances. Thus
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