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8/10/2019 Response of Bank of Correlators to Noisy Input
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Response of bank of correlators
to noisy input
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Analyzer for generating the set of signal
vectors si.
0
i=1,2,....,M( ) ( ) , (5.6)
j=1,2,....,M
T
ij i js s t t dt
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Conversion of the Continuous AWGN
Channel into a Vector Channel Suppose that the si(t) is
not any signal, but
specifically the signal at
the receiver side,
defined in accordance
with an AWGN channel:
So the output of thecorrelator can be
defined as:
( ) ( ) ( ),
0 t T (5.28)
i=1,2,....,M
ix t s t w t
i0
x ( ) ( )
= ,
j 1, 2,....., (5.29)
T
j
ij i
x t t dt
s w
N
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deterministic quantity random quantity
contributed by the
transmitted signal si(t)
sample value of the
variable Wi due to noise
0 ( ) ( ) (5.30)
T
ij i is s t t dt
0( ) ( ) (5.31)
T
i iw w t t dt
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Now,
Consider a randomprocess x1(t),with x1(t),asample function which isrelated to the received
signalx(t)as follows: Here we get:
1
( ) ( ) ( ) (5.32)N
j i
j
x t x t x t
1
1
( ) ( ) ( ) ( )
= ( ) ( )
= ( ) (5.33)
N
ij j j
j
N
j j
j
x t x t s w t
w t w t
w t
which means that the sample function x1(t) depends only on
the channel noise!
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The received signal can be expressed as:
1
1
( ) ( ) ( )
( ) ( ) (5.34)
N
j i
j
N
j i
j
x t x t x t
x t w t
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Statistical Characterization
The received signal (output of the correlator ) is arandom signal. To describe it we need to usestatistical methodsmean and variance.
The assumption is : We have assumed AWGN, so the noise is Gaussian, so
X(t) is a Gaussian process and being a Gaussian RV, X jisdescribed fully by its mean value and variance.
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Mean Value
Let Wj, denote a random variable, represented by
its sample value wj, produced by the jth correlator
in response to the Gaussian noise component w(t).
So it has zero mean (by definition of the AWGNmodel)
=
= [ ]
= (5.35)
j
j
x j
ij j
ij j
x ij
E X
E s W
s E W
s
then the mean of
Xjdepends only onsij:
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Variance
Starting from the definition,we substitute using 5.29and 5.31
2
2
2
var[ ]
= ( )
= (5.36)
ix j
j ij
j
X
E X s
E W
0
( ) ( ) (5.31)T
i iw w t t dt 2
0 0
0
= ( ) ( ) ( ) ( )
= ( ) ( ) ( ) ( ) (5.37)
i
T T
x j j
T T
j i
o
E W t t dt W u u du
E t u W t W u dtdu
2
0
0
= ( ) ( ) [ ( ) ( )]
= ( ) ( ) ( , ) (5.38)
i
T T
x i j
o
T T
j i w
o
t u E W t W u dtdu
E t u R t u dtdu
Autocorrelation function ofthe noise process
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It can be expressed as:
(because the noise is
stationary and with a
constant power
spectral density)
0R ( , ) ( ) (5.39)
2
w
Nt u t u
After substitution for
the variance we get:
2 0
0
20
0
= ( ) ( ) ( )2
= ( ) (5.40)2
i
T T
x i j
o
T
j
Nt u t u dtdu
Nt dt
And since j(t) has
unit energy for thevariance we finally
have:
2 0= for all j (5.41)
2ix
N
Correlator outputs, denoted by Xjhave varianceequal to the power spectral density N0/2 of thenoise process W(t).
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Properties
Xjare mutually uncorrelated
Xjare statistically independent (follows from above
because Xjare Gaussian) and for a memory less
channel the following equation is true:
1
( / ) ( / ), i=1,2,....,M (5.44)j
N
x i x j i
j
f x m f x m
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Xjare mutually uncorrelated
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Define (construct) a vector X of N random variables, X1, X2,XN, whose elements are independent Gaussian RVwith
mean values sij, (output of the correlator, deterministic partof the signal defined by the signal transmitted) and varianceequal to N0/2 (output of the correlator, random part,calculated noise added by the channel).
then the X1
, X2
, XN
, elements of X are statisticallyindependent
So, we can express the conditional probability of X, givensi(t)(correspondingly symbol mi) as a product of theconditional density functions (fx) of its individual elementsfxj.
NOTE: This is equal to finding an expression of the probabilityof a received symbolgiven a specific symbol was sent,assuming a memoryless channel
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that is:
1
( / ) ( / ), i=1,2,....,M (5.44)j
N
x i x j i
j
f x m f x m
where, the vectorxand thescalar xj, are sample
values of the random vector Xand the random
variable Xj.
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Vector xandscalar xj
are sample values of
the random vector X
and the randomvariable Xj
Vector x is called
observation vector
Scalar xjis called
observable element
1
( / ) ( / ), i=1,2,....,M (5.44)j
N
x i x j i
j
f x m f x m
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Since, each Xjis Gaussian with mean sjand
variance N0/2
/ 2 2
0
0
j=1,2,....,N1( / ) ( ) exp ( ) , (5.45)
i=1,2,....,MjN
x i j ijf x m N x sN
we can substitute in 5.44 to get 5.46:
/ 2 2
0
10
1
( / ) ( ) exp ( ) , i=1,2,....,M (5.46)
NN
x i j ij
jf x m N x sN
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If we go back to the formulation of the received
signal through a AWGN channel
1
1
( ) ( ) ( )
( ) ( ) (5.34)
N
j i
j
N
j i
j
x t x t x t
x t w t
The vector that wehave constructed fully
defines this part
Only projections of the noise onto
the basis functions of the signal set
{si(t)M
i=1affect the significant
statistics of the detection problem
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Finally,
The AWGN channel, is equivalent to an N-
dimensional vector channel, described by the
observation vector
, 1,2,....., (5.48)ix s w i M
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