Communication with AWGN Interference

Transcription

1 Communcaton wth AWG Interference m {m } {p(m } Modulator s {s } r=s+n Recever ˆm AWG n m s a dscrete random varable(rv whch takes m wth probablty p(m. Modulator maps each m nto a waveform sgnal s m=m <==> s = s M- he transmtted sgnal s s dsturbed by an AWG process n wth power spectral densty = /2. he receved sgnal s equal to r = s+n. he recever produces an estmate ˆm of the message m

2 m {m } {p(m} Modulator s {s } r=s+n Recever ˆm AWG n Problem: a How to desgn an optmal recever to mnmze the error probablty P P( mˆ m e. Assume that m s ndependent of n. p(m and {s } are known to the recever. b What s the performance of modulaton schemes, and how to desgn the transmtted sgnal?

3 Bayes Decson heory Markov chan: hree random varables,y,z, are sad to form a Markov chan (->Y->Z f the cononal pmf (or pdf of Z gven Y and depends only on Y,.e., p(z y,x=p(z y for any x,y and z

4 Message space Observaton space Decson space m m m 2 m m M- x m m m 2 m m M- m=m wth prob. p(m Cononal probablty p(x m decson ˆm Decson s are based on the observaton. he orgnal message s nvsble to the recever. he decson s a functon of,.e., here s a loss L(m, m when m=m, and the decson s equal to m Our goal s to desgn an optmal decson functon g( so that the average loss s a mnmum. ˆm m->-> forms a Markov chan ˆm mˆ = g(

5 M M = = M M Bayes Crteron E[ L( m, mˆ ] = L( m, m p( m, m = L( m, m p( m p( x m p( mˆ = m x dx = = M M = L( m, m p( x p( m x p( mˆ = m x dx = ˆ Snce m=g( = = M M p( x[ L( m, m p( m x p( mˆ = m x] dx = = f m= g ( x p( mˆ = m x = otherwse M E[ L( m, mˆ ] = p( x[ L( m, g ( x p( m x] dx = Bayes decson: Gven x, choose g(xє{m,m, m M- } so that M = Lm (, g( x p( m x s mnmzed.

6 M- Bayes Rule g( x = m ff Lm (, m p( m x = M- = Lm (, m p( m x for all k k M- Let C( x, m = L( m, m p( m x, M- = C(x,m denotes the cost assocated wth decson mˆ = m Implementaton: Compute C(x,m x Compute C(x,m Select the least mˆ = m where C(x,m C(x,m k for all k Compute C(x,m 2

7 Applcatons of Bayes Rule to Communcatons In dgtal communcatons, the standard loss functon s defned as f m= m Lm (, m = otherwse he average loss E[ Lmm (, ˆ ] = P( mˆ m Pe Furthermore M- C ( x, m = L( m, m p( m x = M = p( m x = p( m x =, Mnmzaton of C(x, m over all m s now equvalent to maxmzaton of p(m x. he cononal prob. of m gven =x s called the a posteror. Accordngly, the Bayes rule reduces to the maxmum a posteror probablty (MAP rule.

8 Maxmum a Posteror Probablty Rule Gven x, choose g(xє{m, m M- } so that p(g(x x s maxmzed,.e., Implementaton: g( x = m ff p( m x = max p( m x p(m x Compute x p(m x and select mˆ = m the where largest p( m x = max p( m x p(m M- x

9 Maxmum Lkelhood (ML Decson Rule In Bayes rule, we want to maxmze p( m x = p( m p( x m p( x If all messages m are equally lkely,.e., the maxmzaton of p(m x s equvalent to the maxmzaton of p(x m. he MAP decson rule reduces to the maxmum lkelhood decson rule. ML rule: Gven x, choose g( x = m ff p( x m = max p( x m

10 Suffcent Statstcs Message space Observaton space Decson space m m m 2 m m M- p(x,y m (x,y ˆm m m m 2 m m M- m Observaton (, Y decson Suppose our observaton now s a par of rvs and Y. Accordng to Bayes rue g( x, y = m ff C(( x, y, m = mn C(( x, y, m M- = = where C(( x, y, m Lm (, m p( m x, y

11 Suppose m->->y forms a Markov chan. p( m, x, y p( m p( x m p( y xm, p( m x, y = = p( x, y p( x, y p( m p( x m p( y x p( m, x = = = p( m x p( x, y p( x M- =Lm m p m x C xm = C(( x, y, m (, ( = (, he cost functon does not depend on the observaton y. In ths case, we say s a suffcent statstc wth respect to (,Y. heorem 3.. (Suffcent Statstc heorem: If m->->y forms a Markov chan, then s a suffcent statstc wth respect to (,Y, and the observaton Y can be dsregarded as far as the desgn of otmal recever s concerned. It s nformaton theoetcally equvalent to I(m;Y = I(m;

12 Vector Communcaton wth AWG Interference m {m } {p(m} ransmtter s=(s,s 2, s {s } r=s+n n=(n,n 2, n Recever ˆm he random message m takes m wth probablty p(m. When m=m, the vector s =(s,s 2, s s transmtted. m=m <==> s = s he transmtted random vector s s dsturbed by an adve Gaussan vector n where n~(, σ2i x I x s an -dmensonal dentty matrx. he receved vector r=(r,r 2, r = s + n = (s +n, s 2 +n 2,, s +n Assume that m s ndepent of n, we want to mnmze the error probablty P Pm ( ˆ m e

13 Message space Observaton space Decson space m m m 2 m m M- p(r m r m m m 2 m m M- m=m wth prob. p(m r decson ˆm When m = m, s = s, r = s + n = s + n. Snce n~(, σ 2 I x, gven m, r s a Gaussan random vector wth mean s and covarance matrx σ 2 I x. 2σ 2 p(r m = p(r,,r m = exp{ r s } 2 ( 2 2πσ r s 2 = (r -s (r -s 2

14 MAP Rule Appled to Vector Communcaton Gven r, choose mˆ = m ff p( m r = max p( m r p( m p( r m p( m p( r m = max p( r p( r p( m p( r m = max p( m p( r m ln p( m + ln p( r m = max[ln p( m + ln p( r m ] 2 2 ln p( m r s 2 = max[ln p( m r s ] 2 2σ 2σ σ ln p( m r + r s s = max[ σ ln p( m r + r s s ] 2 2 r s+ B = max[ r s+ B ] = = 2 2 where r s rs k k, B σ ln p( m k= 2 s, M

15 Implementaton of MAP Recever r s s s 2 B B Compute and select the largest mˆ = m where B 2 Correlaton Recever r s + B = m ax[ r s + B ]

16 Mnmum Dstance Decson Rule If all messages are equally lkely, then the decson rule becomes ˆ 2 2 choose m=m ff r s = mn[ r s ] r s = mn[ r s ] Exampes, 2 on page 3 Furthermore, f all sgnal vectors s have the same energy, we get B =B, the decson rules becomes choose m=m ˆ ff r s = max r s All decson boundares pass through the orgn. Example 3 on page 4

17 Performance of the Correlaton Recevers M Pe=p(mˆ m = p(mˆ = m = p( m p(mˆ = m m D R = Compute Pe wth ntegrals n the observaton space Let (the -dmensonal Eucldean space be the decson regon for m, M-. hen M D= R and mˆ = m ff r D. = P( mˆ = m m = p( r m dr D = exp 2 2σ D 2 ( 2πσ 2 r s dr Example 4 on p. 5 Compute Pe wth ntegrals n the comparator nput space when <M Let r s+ B, M- P(m=m ˆ m = P( = max m

18 Waveform Communcaton wth AWG Interference m {m } {p(m } Modulator s {s } r=s+n Recever ˆm AWG n m s a dscrete random varable (rv whch takes m wth probablty p(m. Modulator maps each m nto a waveform sgnal s m=m <==> s = s M- he transmtted sgnal s s dsturbed by an AWG process n wth power spectral densty = /2. he receved sgnal s equal to r=s+n. he recever produces an estmate ˆm of the message m We want to mnmze the error probablty P P( mˆ m e

19 Sgnal Space Defnton: A set of functons Φ, (tє[,], s sad to be an orthornormal bass f f φ ( t φ ( t = = f he ntegral can be denoted by < Φ, Φ >, s also called the nner product of Φ and Φ. heorem 3.3.: Gven a set of sgnals s, M-, there s an orthornomal bass Φ,, wth M such that each sgnal s can be unquely represented as s ( t = sφ ( t, M- = where s =< s ( t, φ ( t >= s ( t φ ( t

20 Sgnal Space (cntd s can be vewed geometrcally a pont n the -dmensonal sgnal space spanned by the othornormal bass {Φ }. hs pont s gven by s =(s,s 2, s S s the proecton of s onto the unt functon Φ. s =< s ( t, φ ( t > s ( t s (, 2, = s s s s ( t = sφ ( t =

21 s Sgnal Space (cntd Φ Φ 2 s s 2 Φ s s Φ s 2 Φ 2 s s Φ

22 Invarance Property Inner Product Energy Dstance < s ( t, s ( t >= s ( t s ( t = = = k k = l= = l= = sφ ( t sklφl ( t = l= kl s s s s φ ( t φ ( t φ ( t φ ( t kl l = s s = s s s k ( ( t = s s = s s ( t s ( t = s s k l 2 Example p22

23 Reducton of Waveform Channels to Equvalent Vector Channels Φ s +n Φ r=s +n Φ 2 s 2 +n 2 Φ 2 r Recever m Φ s +n Φ + - r 2 Decomposton of r nto r and r 2 s =< st (, φ ( t >= st ( φ ( t n =< nt (, φ ( t >= nt ( φ ( t r ( t = sφ ( t + nφ ( t r ( t = r( t r ( t 2 = =

24 Let s= ( s, s s 2 nt ɶ( = nφ ( t = nt ˆ( = nt ( - nt ɶ( Snce st ( = sφ ( t = and m= m st ( = s ( t s= s r ( t = st ( + nt ɶ( r ( t = r( t r ( t 2 = nt ( nt ɶ( = nt ˆ( r contans the sgnal component plus part of the AWG n; r 2 contans only the remanng part of the nose n m {m } {p(m } Modulator s {s } Waveform Channel r r 2

25 Reducton of Waveform Channels to Equvalent Vector Channels (cntd We can show that r 2 s ndependent of m and r n the case of AWG channel (stronger than Markov chan. Applyng the suffcent statstc theorem, we can dsregard r 2 m {m } {p(m } Modulator nt ɶ( s {s } r =s+ nt ɶ( nt ɶ( S, s, r and can be regarded as vectors and random vectors n the -dmensonal space spanned by the othornormal bass {Φ }. s =< s ( t, φ ( t > s ( t s (, 2, = s s s s ( t = sφ ( t = s =< s ( t, φ ( t > s ( t s = ( s, s 2, s s ( t = s φ ( t =

26 Reducton of Waveform Channels to Equvalent Vector Channels (cntd r=< r ( t, φ ( t > r ( t = st ( + nt ɶ( r= ( r, r2, r = s+ n r ( t = rφ ( t n =< nɶ ( t, φ ( t > = = nɶ ( t n= ( n, n 2, n nɶ ( t = nφ ( t herefore the waveform system s equvalent to the vector system m {m } {p(m} ransmtter s=(s,s 2, s {s } r=s+n n=(n,n 2, n n=(n, n 2,,n ~(, /2I x

27 Correlaton Recever n the Wavefrom System r Φ Φ 2 Φ r r 2 r Weghtng Matrx = rs r s r s r s M B B B M- Compare and select the largest ˆm 2 B= ln p( m s = ln p( m s ( t 2 2

28 Alternatve Implementaton of the Correlaton Recever r s = r s = s r ( t φ ( t = = = r ( t [ s φ ( t ] = = r ( t s ( t r s s s M- rs = B B B M- Compare and select the largest ˆm

29 Matched Flter Recever Defnton: Gven a sgnal Φ, t, a flter wth the mpulse h s called matched to Φ f h s a delayed, tme reversed verson of Φ, h= Φ(-t = Φ(-(t-. he response y of the matched flter h= Φ(-t n response to the nput x s gven by + + yt ( = x( τ ht ( τ dτ= x( τφ ( t+ τ dτ + In patcular, y( τ = x( τφτ ( dτ= x( τφτ ( dτ y(τ s equal to the correlaton between x and Φ s (-t + r s (-t s M- Sample at t= + + B B B M- Compare and select the largest ˆm

30 Computaton of P e n the Waveform Case he computaton of error probablty can be carred out n the correspondng sgnal space. Example on p36