Unformatted text preview: m process. EECS 455 (Univ. of Michigan) Fall 2012 October 3, 2012 55 / 93 Lecture Notes 7 KarhunenLoeve Expansion
By KarhunenLoeve expansion
∞ n(t ) = nm ϕi (t )
m =0 where ni are Gaussian random variables with mean 0 variance N0 and
2
E [nm nk ] = 0, m = k . Thus nm and nk are independent. Since
{ϕm (t ); m = 0, 1, ...} is a complete orthonormal set and we assume
sj (t ) has ﬁnite energy we have
∞ sj (t ) = N −1 sj ,m ϕm (t ) =
m =0 sj ,m ϕm (t ).
m =0 This last equality is because we only need a ﬁnite (N ≤ M )
orthonormal waveforms to represent a set of M signals. Equivalently
sj ,i = 0 for i ≥ N .
EECS 455 (Univ. of Michigan) Fall 2012 October 3, 2012 56 / 93 Lecture Notes 7 KarhunenLoeve Expansion
Thus
∞ Hj : r (t ) = (sj ,m + nm )ϕm (t )
m =0 rm = sj ,m + nm ,
Deﬁne
∆ Λj ,i (L) = m = 0, 1, 2, ... pj (r0 , r1 , . . . , rL )
.
pi (r0 , r1 , . . . , rL )
∆ Λj ,i (r (t )) = lim Λj ,i (L)
L→∞ where rm is Gaussian with mean sj ,m variance N0 /2.
EECS 455 (Univ. of Michigan) Fall 2012 October 3, 2012 57 / 93 Lecture Notes 7 KarhunenLoeve Expansion
pj (rm ) = 1
1
√
exp − (rm − sj ,m )2
N0
N0 π
L L pj (rm ) = pj (r ) = m=0 m=0 ( N0 π )−1 exp − 1
N0 L Λj ,l (L) = pjL (r )
plL (r ) = m=0
L m=0 = =
EECS 455 (Univ. of Michigan) L m=0 (rm − sj ,m )2 L
1
( N0 π )−1 exp − N0
1
( N0 π )−1 exp − N0 exp − 1
N0 exp − 1
N0 m=0
L m=0 (rm − sj ,m )2
(rm − sl ,m )2 L m=0 2
2
[rm − 2rm sj ,m + sj2,m − rm + 2rm sl ,m − sl2,m ] L m=0 [sj2,m − sl2,m + 2rm (sl ,m − sj ,m )] . Fall 2012 October 3, 2012 58 / 93 Lecture Notes 7 KarhunenLoeve Expansion
If we take the limit as L → ∞ we get
Λj ,l (r (t )) = exp − Λj ,l (r (t )) = exp − 1
(E − El + 2(r , sl − sj )] .
N0 j 1
[(sj , sj ) − (sl , sl ) + 2(r , sl ) − 2(r , sj )] .
N0 or equivalently
1
[sj 2 − sl 2 + 2(r , sl − sj )]
N0
1
= exp − [r − sj 2 − r − sl 2 ]
N0 Λj ,l (r (t )) = exp − EECS 455 (Univ. of Michigan) Fall 2012 October 3, 2012 59 / 93 Lecture Notes 7 Optimum Receiver Principles Claim:
The optimum decision rule to decide among M equally likely possible
transmitted signals for additive white Gaussian noise is to choose i if
si − r EECS 455 (Univ. of Michigan) 2 = min 0≤ j ≤ M − 1 Fall 2012 sj − r 2 . October 3, 2012 60 / 93 Lecture Notes 7 Demodulator
ϕ∗ (t )
0
×
∗ (t )
ϕ1
r (t ) r (t )ϕ∗ (t )dt
0 r0 × r (t )ϕ∗ (t )dt
1 r1
Find si with
smallest
r − si 2 ϕ∗ −1 (t )
N
× EECS 455 (Univ. of Michigan) r (t )ϕ∗ −1 (t )dt rN −1
N Fall 2012 October 3, 2012 61 / 93 Lecture Notes 7 Example: M equal energy signals
Now consider the optimum receiver for M ary equally likely signals and
the associated error probability. Assume the M signals are equienergy
signals and equiprobable. The decision rule derived previously for
AWGN in this case simpliﬁes to
Decide Hi if si − r 2 = min 0≤ j ≤ M − 1 sj − r 2 . Now since the M signals are equienergy we can write this as
sj − r 2 = sj 2 − 2(sj , r ) + r 2 .
The ﬁrst term a...
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 Fall '08
 Stark
 Normal Distribution, Univ. of Michigan

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