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M 0 this last equality is because we only need a nite

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Unformatted text preview: m process. EECS 455 (Univ. of Michigan) Fall 2012 October 3, 2012 55 / 93 Lecture Notes 7 Karhunen-Loeve Expansion By Karhunen-Loeve 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 Karhunen-Loeve 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 Karhunen-Loeve 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 Karhunen-Loeve 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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