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

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

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...
View Full Document

This note was uploaded on 02/12/2014 for the course EECS 455 taught by Professor Stark during the Fall '08 term at University of Michigan.

Ask a homework question - tutors are online