1 / 20

Log-Likelihood Algebra

 = - L (d). L (d). +. L (d). 0 = 0. +. Log-Likelihood Algebra. Sum of two LLRs L(d 1 ) + L(d 2 )  L ( d 1  d 2 ) = log exp[L(d 1 )] + exp [L(d 2 )] 1 + exp[L(d 1 )].exp [L(d 2 )]

mika
Download Presentation

Log-Likelihood Algebra

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1.  = - L (d) L (d) + L (d) 0 = 0 + Log-Likelihood Algebra Sum of two LLRs L(d1) + L(d2) L (d1 d2 ) = log exp[L(d1)] + exp [L(d2)] 1 + exp[L(d1)].exp [L(d2)]  (-1) . sgn [L(d1)]. sgn [L(d2)] . Min ( |L(d1)| , |L(d2)| )

  2. Iterative decoding example • 2D single parity code • di di= pij

  3. Iterative decoding example • Estimate Lc(xk) • = 2 xk /2 • assuming 2= 1

  4. Leh( d1) = [Lc ( x 2) + L(d2) ] Lc ( x 12 ) = new L( d1 ) Leh( d2) = [Lc ( x 1) + L(d1) ] Lc ( x 12 ) = new L(d2) Leh( d3) = [Lc ( x 4) + L(d4) ] Lc ( x 34 ) = new L(d3) Leh( d4) = [Lc ( x 3) + L(d3) ] Lc ( x 34 ) = new L(d4) + + + + Iterative decoding example • Compute • Le( dj ) = [Lc ( x j) + L(dj ) ] Lc ( x ij) +

  5. Lev( d1) = [Lc ( x 3) + L(d3) ] Lc ( x 13 ) = new L( d1 ) Lev( d2) = [Lc ( x 4) + L(d4) ] Lc ( x 24 ) = new L(d2) Lev( d3) = [Lc ( x 1) + L(d1) ] Lc ( x 13 ) = new L(d3) Lev( d4) = [Lc ( x 2) + L(d2) ] Lc ( x 24 ) = new L(d4) + + + + ^ ^ ^ L( di ) = Lc ( x i) + Leh(di) + Lev( dj ) Iterative decoding example After many iterations the LLR is computed for decision making

  6. First Pass output

  7. Parallel Concatenation Codes • Component codes are Convolutional codes • Recursive Systematic Codes • Should have maximum effective free distance • Large Eb/No maximizing minimum weight codewords • Small Eb/No optimizing weight distribution of the codewords • Interleaving to avoid low weight codewords

  8. {uk} + L-1 L-1 uk =  g1i dk-i (mod 2) ; i=1 vk =  g2i dk-i (mod 2) ; i=1 {dk} dk-1 dk-2 dk + {vk} Non - Systematic Codes - NSC G1 = [ 1 1 1 ] G2 = [ 1 0 1 ]

  9. {dk} {uk} ak-1 ak-2 ak + + {vk} L-1 ak = dk +  gi’ ak-i (mod 2) ; gi’ = g1i if uk = dk i=1 g2i if vk = dk Recursive Systematic Codes - RSC

  10. NSC RSC Trellis for NSC & RSC 00 00 a = 00 a = 00 11 11 11 11 b = 01 b = 01 00 00 10 10 c = 10 c = 10 01 01 01 01 d = 11 d = 11 10 10

  11. + + ak-1 ak-1 ak-2 ak-2 ak ak + + {v2k} {v1k} Concatenation of RSC Codes {dk} {uk} Interleaver {vk} { 0 0 …. 0 1 1 1 0 0 …..0 0 } { 0 0 …. 0 0 1 0 0 1 0 … 0 0 }  produce low weight codewords in component coders

  12. APP  Joint Probability ki,m = P { dk = i, Sk = m / R1 N } State at time k Received sequence From time 1 to N APP  P { dk = i / R1 N } =  ki,m ; i = 0,1 for binary m  k0,m  k1,m  k0,m  k1,m  m m m m Likelihood Ratio  ( dk ) =  Log Likelihood Ratio  L( dk ) = Log Feedback Decoder

  13.         Feedback Decoder • MAP Rule  dk =1 ; L(dk) > 0 dk =0 ; L(dk) < 0 ^ ^ L( dk) = Lc ( x k) + L(dk) + Le( dk ) L1( dk ) = [Lc ( x k) + Le1(dk ) ] L2( dk ) = [ f{L1 ( dn) }n k + Le2(dk ) ]

  14.   Le2( dk ) L2( dk ) xk  L1( dn ) L1( dk ) De-Interleaver DECODER 1 DECODER 2 De-Interleaver Interleaver y1k y2k yk  dk Feedback Decoder 

  15. Modified MAP Vs. SOVA • SOVA  • Viterbi Algorithm acting on soft inputs over forward path of the trellis for a block of bits • Add BM to SM  compare  select ML path • Modified MAP  • Viterbi Algorithm acting on soft inputs over forward and reverse paths of the trellis for a block of bits • Multiply BM & SM  Sum in both directions  best overall statistic

  16. {uk} dk-1 dk-2 dk {dk} + {vk} MAP Decoding Example 00 a = 00 11 b = 10 00 11 01 c = 01 10 01 d = 11 10

  17. Branch Metric  ki,m = P { dk = i, Sk = m , Rk } = P { Rk / dk = i, Sk = m } . P {Sk = m / dk = i } . P { dk = i } ki,m = P { xk / dk = i, Sk = m } . P { yk / dk = i, Sk = m } . { ki / 2L } P {Sk = m / dk = i } = 1 / 2 L = ¼ ; P { dk = i } = 1 / 2 ; MAP Decoding Example • d = { 1, 0, 0 } • u = { 1, 0, 0 }  x = { 1., 0.5, -0.6 } • v = { 1, 0, 1 }  y = { 0.8, 0.2, 1.2 } • Apriori probabilities  1 = 0 = 0.5

  18. ki,m = ki,m = Assuming Ak = 1 2 =1 ; ki,m = 0.5 exp { xk . uki + yk . Vki,m } { ki / 2L } (1/2 ) exp { - (xk – uki )2 /(2 2 ) }dxk { Ak ki } exp { (xk . uki )+ (yk . Vki,m )/  2 } ki,m = P { xk / dk = i, Sk = m } . P { yk / dk = i, Sk = m } . { ki / 2L } MAP Decoding Example For AWGN channel : . (1/2 ) exp { - (xk – vki,m )2 /(2 2 ) }dyk

  19. ki,m = 0.5 exp { xk . uki + yk . Vki,m } 1 k+1m=  ki,b(j,m) kb(j,m) J=0 Subsequent steps • Calculate branch metric • Calculate forward state metric • Calculate reverse state metric 1 km=  kj,m k+1f(j,m) J=0

  20.  km k1,m k+1f(1,m)  km k0,m k+1f(0,m) m m Subsequent steps • Calculate LLR for all times  Log Likelihood Ratio  L( dk ) = Log • Hard decision based on LLR

More Related