Detección Óp+ma: Algoritmo de Viterbi. (solo para dar una idea general)

Transcripción

1 Detección Óp+ma: Algoritmo de Viterbi (solo para dar una idea general)

2 K = N Cambio de notación A[k] =x k p[k] =h k q[k] =y k z[k] =g k o[n] =) Salida sin ruido

3 Cambio de notación Vamos a considerar que tenemos M símbolos en la constelación

4 Maximum likelihood (ML) sequence detection (ML) Sequence to be detected: L symbols (M L possible sequences) A =[A[0], A[],, A[L ]] T Channel: p =[p[0], p[],, p[k]] T Sufficient statistic: N q = K + L observations q =[q[0], q[],, q[n q ]], N q = K + L o[0] =p[0] A[0] + p[] A[ ] + p[2] A[ 2] + p[k] A[ K] o[] =p[0] A[] + p[] A[0] + p[2] A[ ] + p[k] A[ K] o[k + L ] =p[0] A[K + L ] + p[] A[K + L 2] + + p[k ] A[L] + p[k] A[L ] Additional necessary information: A[ ], A[ 2],, A[ K] y A[L], A[L + ],, A[L + K] MMC (UC3M) Digital Communications Receivers: Viterbi 4 / 39

5 Maximum likelihood (ML) sequence M L possible sequences a i =[a i [0], a i [],, a i [L ]] T, i = 0,,, M L Most likely sequence: Â = a i =[a i [0], a i [],, a i [L ]] T that maximizes the likelihood f q A (q a i ) f q A (q a j ), j = 0,,, M L, 8j 6= i MMC (UC3M) Digital Communications Receivers: Viterbi 5 / 39

6 Maximum likelihood (ML) sequence Likelihood: f q A (q a i )= N q Y n=0 f q[n] A (q[n] a i ) Conditional distribution for each observation f q[n] A (q[n] a i )=N o i [n], 2 z = 8 < p exp 2 z : 2 2 z q[n] KX p[k] a i [n k] k=0 2 9 = ; Complete Likelihood f q A (q a i )= Most likely sequence  = arg mín a i (2 2 z ) N q/2 exp N q X n=0 ( 2 2 z Nq X n=0 METRIC! q[n] o i [n] 2, o i [n] = q[n] o i [n] 2 ) KX p[k] a i [n k] MMC (UC3M) Digital Communications Receivers: Viterbi 6 / 39 k=0

7 Example: 2-PAM K =, L = 3 Symbols constellation: A[n] 2 {±} Channel: p[n] = [n]+0,5 [n ], K = Sequence to be detected/estimated: A =[A[0], A[], A[2]], L = 3 Sufficient statistic for detection: q =[q[0], q[], q[2], q[3]] q[ ] =A[ ]+0,5 A[ 2]+z[ ] q[0] = A[0] + 0,5 A[ ] + z[0] q[] = A[] + 0,5 A[0] + z[] q[2] = A[2] + 0,5 A[] + z[2] q[3] = A[3] + 0,5 A[2] + z[3] q[4] =A[4]+0,5 A[3] Premise: The values A[ ] =A[3] =+ are known Problem: detect the sequence of 3 symbols with q[0] =,4 - q[] = 0,4 - q[2] =0,6 - q[3] =,6 MMC (UC3M) Digital Communications Receivers: Viterbi 7 / 39

8 Detection: Comparison wrt to the noiseless outputs q[0] =,4 - q[] = 0,4 - q[2] =0,6 - q[3] =,6 Noiseless outputs, o[n], generated by the 8 possible sequences A[0] A[] A[2] o[0] o[] o[2] o[3] Metric 0,5,5,5 +0,5 0,44 + 0,5,5 +0,5 +,5 4,84 + 0,5 +0,5 0,5 +0,5 6, ,5 +0,5 +,5 +,5 5,24 + +,5 0,5,5 +0,5 5, ,5 0,5 +0,5 +,5 0, ,5 +,5 0,5 +0,5 6, ,5 +,5 +,5 +,5 4,44 MMC (UC3M) Digital Communications Receivers: Viterbi 8 / 39

9 Detection: Comparison wrt to the noiseless outputs q[0] =,4 - q[] = 0,4 - q[2] =0,6 - q[3] =,6 Noiseless outputs, o[n], generated by the 8 possible sequences A[0] A[] A[2] o[0] o[] o[2] o[3] Métrica 0,5,5,5 +0,5 0,44 + 0,5,5 +0,5 +,5 4,84 + 0,5 +0,5 0,5 +0,5 6, ,5 +0,5 +,5 +,5 5,24 + +,5 0,5,5 +0,5 5, ,5 0,5 +0,5 +,5 0, ,5 +,5 0,5 +0,5 6, ,5 +,5 +,5 +,5 4,44 Most similar sequence (maximum likelihood): + + MMC (UC3M) Digital Communications Receivers: Viterbi 9 / 39

10 Complexity! M L sequences! MMC (UC3M) Digital Communications Receivers: Viterbi 0 / 39

11 State [n] The noiseless output is a finite state machine o[n] =A[n] p[0]+ KX p[k] A[n k] k= Definition of state [n] =[A[n ], A[n 2],, A[n K]] T Number of possible states is M K Dependencies Con la otra notación, recordamos: o[n] =f K (A[n], = N [n]) ESTADO=MEMORIA o[n] =g( [n], [n + ]) [n + ] =f ( [n], A[n]) MMC (UC3M) Digital Communications Receivers: Viterbi / 39

12 <latexit sha_base64="hf9pcjxzfo4ydzxi250fxglkvwg=">aaaca3icbva9t8mwehx4lowrwnjfokjchaoeiqfbbqtjkqitlibicz3gqmnhtgnuuqcw/golaybw/gqb/wa3zqattzrp6b073d0lu0avtuva2fxaxlltbrwxt/y3nqu7ozekjfjtfwsmjcdecnckceuppqrtiojskjg2uhgyuy374huvpbrpuyjn6a+pxhfsbspqfsfx/2uppyiynfpywdfluhdfgqgldmn+wj4dxxclidbvpb5avbezhlcneyiau8x06nyopkwzkvo5miqqidcfeizylbdl55mnrvdakd0ycwmkazhrf0/kkffqmismm0e6vrpewpzp8zidnfo55wmmccftrvhgobzwnajsuumwzkndejbu3apxjctc2urwnie4sy/pe/eocdawr45rzfmijrkognwcbxwaprgersaczb4bm/gfbxzt9al9w59tfsxrgjmd/yb9fkdknaxug==</latexit> <latexit sha_base64="hf9pcjxzfo4ydzxi250fxglkvwg=">aaaca3icbva9t8mwehx4lowrwnjfokjchaoeiqfbbqtjkqitlibicz3gqmnhtgnuuqcw/golaybw/gqb/wa3zqattzrp6b073d0lu0avtuva2fxaxlltbrwxt/y3nqu7ozekjfjtfwsmjcdecnckceuppqrtiojskjg2uhgyuy374huvpbrpuyjn6a+pxhfsbspqfsfx/2uppyiynfpywdfluhdfgqgldmn+wj4dxxclidbvpb5avbezhlcneyiau8x06nyopkwzkvo5miqqidcfeizylbdl55mnrvdakd0ycwmkazhrf0/kkffqmismm0e6vrpewpzp8zidnfo55wmmccftrvhgobzwnajsuumwzkndejbu3apxjctc2urwnie4sy/pe/eocdawr45rzfmijrkognwcbxwaprgersaczb4bm/gfbxzt9al9w59tfsxrgjmd/yb9fkdknaxug==</latexit> <latexit sha_base64="hf9pcjxzfo4ydzxi250fxglkvwg=">aaaca3icbva9t8mwehx4lowrwnjfokjchaoeiqfbbqtjkqitlibicz3gqmnhtgnuuqcw/golaybw/gqb/wa3zqattzrp6b073d0lu0avtuva2fxaxlltbrwxt/y3nqu7ozekjfjtfwsmjcdecnckceuppqrtiojskjg2uhgyuy374huvpbrpuyjn6a+pxhfsbspqfsfx/2uppyiynfpywdfluhdfgqgldmn+wj4dxxclidbvpb5avbezhlcneyiau8x06nyopkwzkvo5miqqidcfeizylbdl55mnrvdakd0ycwmkazhrf0/kkffqmismm0e6vrpewpzp8zidnfo55wmmccftrvhgobzwnajsuumwzkndejbu3apxjctc2urwnie4sy/pe/eocdawr45rzfmijrkognwcbxwaprgersaczb4bm/gfbxzt9al9w59tfsxrgjmd/yb9fkdknaxug==</latexit> Salida sin ruido Diagrama de estado A[n] o[n] [n] [n + ] A[n]! x n o[n]! h n x n ESTADO=MEMORIA

13 State diagram - Example A A[n] 2 {±}, p[n] = [n]+ 2 [n ] Noiseless output o[n] =A[n]+ 2 A[n ] State [n] =A[n ], [n + ] =A[n] State diagram ,5,5 + 0,5 0,5 MMC (UC3M) Digital Communications Receivers: Viterbi 3 / 39

14 State diagram - Example B A[n] 2 {±}, p[n] = [n]+ 2 [n ]+ 4 [n 2] o[n] =A[n]+ 2 A[n ]+ 4 A[n 2] [n] =[A[n ], A[n 2]] T, [n + ] =[A[n], A[n ]] T, 6 +,, + +, +? -? 6 -,75 + 0,25 0,25 +,25,25 +,75 0,75 + 0,75 MMC (UC3M) Digital Communications Receivers: Viterbi 4 / 39

15 Trellis diagram - Example A Example: A[n] 2 {±}, p =[0,5] T State: [n] =A[n ] Labels: A[n], o[n] = KX p[k] A[n k] k=0 + x q + +,5 qqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqq x 0, ,5 x q qqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqq x,5 MMC (UC3M) Digital Communications Receivers: Viterbi 5 / 39

16 Trellis diagram - Example B [+, +] s + +,75 s 0, ,75 [, +] s,25 + +,25 [+, ] s 0,75 + +,25 s s [, ] s s [n],75 [n + ] MMC (UC3M) Digital Communications Receivers: Viterbi 6 / 39

17 Ejemplo - Algoritmo de Viterbi

18 Símbolos transmi+dos Estado inicial Estado final Conociendo lo transmi+do p[n] =02 [n] 03 [n ] + 02 [n 2] El o[n] arriba lo daba en problema para hallar el canal p[n] Salida sin ruido

19 Ejercicio 9 e) obs q[0] = 0, ( 05 0) ( 05 ( 03)) 2 A n A n 2 +, , + +,,

20 Ejercicio 9 e) obs q[0] = 0,5 q[] = 0,2 +, , ,,

21 Ejercicio 9 e) obs q[0] = 0,5 q[] = 0,2 +, , , 025,

22 Ejercicio 9 e) obs q[0] = 0,5 q[] = 0,2 q[2] = 0,3 +, , , , 029

23 Ejercicio 9 e) obs q[0] = 0,5 q[] = 0,2 q[2] = 0,3 +, , , ,

24 Ejercicio 9 e) obs q[0] = 0,5 q[] = 0,2 q[2] = 0,3 +, , , ,

25 Ejercicio 9 e) obs q[0] = 0,5 q[] = 0,2 q[2] = 0,3 +, , , ,

26 Ejercicio 9 e) obs q[0] = 0,5 q[] = 0,2 q[2] = 0,3 q[3] = 0,7 +, , , ,

27 Ejercicio 9 e) obs q[0] = 0,5 q[] = 0,2 q[2] = 0,3 q[3] = 0,7 +, , , ,

28 Ejercicio 9 e) obs q[0] = 0,5 q[] = 0,2 q[2] = 0,3 q[3] = 0,7 +, , , ,

29 Ejercicio 9 e) obs q[0] = 0,5 q[] = 0,2 q[2] = 0,3 q[3] = 0,7 +, , , ,

30 Ejercicio 9 e) obs q[0] = 0,5 q[] = 0,2 q[2] = 0,3 q[3] = 0,7 q[4] = 0,5 +, , , ,

31 Ejercicio 9 e) obs q[0] = 0,5 q[] = 0,2 q[2] = 0,3 q[3] = 0,7 q[4] = 0,5 +, , , ,

32 Ejercicio 9 e) obs q[0] = 0,5 q[] = 0,2 q[2] = 0,3 q[3] = 0,7 q[4] = 0,5 +, , , ,

33 Ejercicio 9 e) obs q[0] = 0,5 q[] = 0,2 q[2] = 0,3 q[3] = 0,7 q[4] = 0,5 +, , , ,

34 Ejercicio 9 e) obs q[0] = 0,5 q[] = 0,2 q[2] = 0,3 q[3] = 0,7 q[4] = 0,5 q[5] = 0,2 +, , , ,

35 Ejercicio 9 e) obs q[0] = 0,5 q[] = 0,2 q[2] = 0,3 q[3] = 0,7 q[4] = 0,5 q[5] = 0,2 +, , , ,

36 Ejercicio 9 e) obs q[0] = 0,5 q[] = 0,2 q[2] = 0,3 q[3] = 0,7 q[4] = 0,5 q[5] = 0,2 +, , , ,

37 Ejercicio 9 e) obs q[0] = 0,5 q[] = 0,2 q[2] = 0,3 q[3] = 0,7 q[4] = 0,5 q[5] = 0,2 +, , , ,

38 Ejercicio 9 e) obs q[0] = 0,5 q[] = 0,2 q[2] = 0,3 q[3] = 0,7 q[4] = 0,5 q[5] = 0,2 q[6] =0,5 +, , , , A[6] = + A[7] = + cabecera

39 Ejercicio 9 e) obs q[0] = 0,5 q[] = 0,2 q[2] = 0,3 q[3] = 0,7 q[4] = 0,5 q[5] = 0,2 q[6] =0,5 +, , , , A[6] = + A[7] = + cabecera

40 Ejercicio 9 e) obs q[0] = 0,5 q[] = 0,2 q[2] = 0,3 q[3] = 0,7 q[4] = 0,5 q[5] = 0,2 q[6] =0,5 +, , , , A[6] = + A[7] = + cabecera

41 Ejercicio 9 e) obs q[0] = 0,5 q[] = 0,2 q[2] = 0,3 q[3] = 0,7 q[4] = 0,5 q[5] = 0,2 q[6] =0,5 +, , , , A[6] = + A[7] = + cabecera

42 Ejercicio 9 e) obs q[0] = 0,5 q[] = 0,2 q[2] = 0,3 q[3] = 0,7 q[4] = 0,5 q[5] = 0,2 q[6] =0,5 q[7] =0, +, , , , A[6] = + A[7] = + cabecera

43 Ejercicio 9 e) obs q[0] = 0,5 q[] = 0,2 q[2] = 0,3 q[3] = 0,7 q[4] = 0,5 q[5] = 0,2 q[6] =0,5 q[7] =0, +, , , , A[6] = + A[7] = + cabecera

44 Ejercicio 9 e) obs q[0] = 0,5 q[] = 0,2 q[2] = 0,3 q[3] = 0,7 q[4] = 0,5 q[5] = 0,2 q[6] =0,5 q[7] =0, +, , , , A[6] = + A[7] = + cabecera

45 Ejercicio 9 e) obs q[0] = 0,5 q[] = 0,2 q[2] = 0,3 q[3] = 0,7 q[4] = 0,5 q[5] = 0,2 q[6] =0,5 q[7] =0, +, , , , secuencia detectada A[0] =- A[] =- A[2] =- A[3] =- A[4] =- A[5] =- A[6] = + A[7] = + mejor camino cabecera