7.5 Performance of Convolutional Codes General Comments

Transcription

1 7.5 Performance of Convolutional Coes General Comments Convolutional coes expan the banwith for a fixe information rate, or lower the information rate for a fixe banwith. We hope for some error rate improvement in return. The facts that o the SNR of coe bits γ c =γ b k n is poorer than that of uncoe information bits, o an some of the n coe bits may be the same on ifferent branches leaving a state, even for a binary ( k = 1) coe, suggest that performance may be worsene, compare to uncoe transmission. owever the trellis constraint length puts a lower limit on the length of error events, so that the coe can buil up istance over the whole error event, not just over one branch. This is the source of the error reuction properties of convolutional coes. We ll examine performance of both soft an har ecision inputs, an compare them to uncoe transmission

2 7.5. Soft ecision Input [P8..] 7.5- The shift register of a convolutional encoer etermines the length of error events. The (,1,) example coe o If two inputs iffer only at a single branch time, their corresponing state sequences will be ifferent until that input ifference is flushe out of the trellis. The shortest error event is therefore K branches, irrespective of the output connections. o Verify this with the trellis of the (,1,) encoer. Note that CCs have no parallel transitions (although TCM an CPM may have them): 7.5-

3 Following the metho of Section 7.., we base our error analysis on the pairwise error probability. Consier the two paths in some error event 7.5- o The ifference between the signals s c s is zero at coe bits where e they agree an ± at coe bits where they isagree. The square Ec Eucliean istance is therefore = 4E c where is the amming istance between the two signals. o The pairwise error probability is then (with R c for coe rate): (,) P ec Q Q = γ = γ = Q γ R n k ( c ) b ( b c ) o The minimum amming istance over all error events is enote. o For our sample (,1,) trellis, the shortest error event is branches an has = 6, an the next is 5 branches long, with = 8. Looks as though = 6 an its P (,) e c Q( 4 b ) = γ, B better than uncoe. 7.5-

4 7.5-4 Next, we account for the number of events of various lengths. A binary convolutional coe is uniform, so the istances an error probabilities are inepenent of the transmitte path. Might as well consier it to be the allzeros sequence. o Aapting the analysis of Section 7.., the probability of an error event starting at a given state is union boune by [ E] N P = a Q( γ br c ) Pr ( ) ( ) ( ) = = where Rc N ( ). = k n is the coe rate an a ( ) is Proakis notation for o We nee the istance spectrum of the coe to calculate the error event probability. etermine the istance spectrum by search, or by the transfer function (Section 7.5.). For our (,1,) example, we have a (6) = 1, a (8) =, a (10) = 4, etc. o At high SNR, the other terms are much smaller than the [ E] a Q( γ brc ) Pr ( ) term, so 7.5-4

5 7.5-5 Usually, we are more intereste in the BER than the error event probability. We have to weight each pairwise event probability with the number of information bit errors it causes, as in Section 7... Again, sort by amming istance, an assume that all error events with amming istance have the same number f ( ) of information bit errors. o Assume that all error events with amming istance have the same number f ( ) of information bit errors. Then the BER is boune by ( γ ) Pb a( ) f( ) Q brc = o Like the istance spectrum, f ( ) must be etermine by search or the transfer function. For our (,1,) trellis, a = 6 ( = ) event causes only one bit error, an the = 8 event causes two bit errors. Proakis gives tables like the one below (with no istance spectrum): 7.5-5

6 o Compare the ominant (i.e., ) pairwise probability with the error probability of uncoe binary transmission. For the rate ½, K = 6 coe, = 8. Then the pairwise probability is Q( γ b R c fre e) = Q( 8γ b ) a 6 B boost from uncoe transmission. On the other han, that s only one of the error events The Transfer Function [P8..1] For small coes, the transfer function is an effective way to enumerate the paths of various amming istances an numbers of information bit errors. Start with the uniformity of convolutional coes. We can assume that the transmitte path was all zeros. The amming istance between it an an error sequence is just the amming weight of the error sequence. For our sample (,1,) coe, that s how we interpret the branch weights: 7.5-6

7 Now use a polynomial representation of error paths from one state to another one path of weight path of weight 6, of weight 4 Rather like z-transforms or moment generating functions. If a path extens by another branch an as, say, another two 1 bits to the weight, it is equivalent to multiplication by. Reraw the trellis to represent error events starting at time 0: they epart from state a (since correct is all zeros); branches are weighte with their contribution to overall weight; a return to state a allows no further epartures (an absorbing state), so rename it as state e. Verify that this error trellis is a correct representation of the (,1,) coe: w for 7.5-7

8 Trace the evolution of weights: o At time 0, polynomial X (0) = 1 (no errors yet). a o At time 1, Xc (1) = o At time, Xb () 4, = X () = o an out to time 5 (polynomials shown in boxes): 5 o For error events of length 5 branches or fewer, we have at noe e: 1 event of weight 6 events of weight 8 1 event of weight

9 Represent the transitions among states b, c, by a matrix T: Xb( m+ 1) 0 Xb( m) Xc( m+ 1) = 0 0 Xc( m) Verify this. X ( 1) m+ 0 X( m) or x( m+ 1) = Tx( m), with X (1) 0 b x (1) = Xc (1) = X (1) o The contribution Xe( m ) at time m is Xb ( m 1 ), or X ( 1) 0 b m m Xe( m) = 0 0 Xc( m 1) 0 0 = T X ( m 1) 0 o To capture all error events, form 0 i Xe = Xe( m) = 0 0 T m= i= = 0 0 ( I T) 0 where the secon line assumes convergence o If inversion is unappealing, just calculate enough terms in the sum to capture error events of relatively small weight 7.5-9

10 o Matrix inversion (love that symbolic processor) gives Xe = = o Since this is X ( ) X ( ), call it the transfer function of the coe e 6 T( ) = 1 a o We can go backwars from this with a series expansion to enumerate the number of paths with specific weight between state a an state e: = T( ) = = a Connect it with union boun on probability of an error event: [ E] a Q( γbrc ) Pr ( ) = ( 4 b ) ( 16 b ) 4 ( 0 b ) = Q γ + Q γ + Q γ + since the coe rate is 1/

11 A shortcut to the transfer function: o Work with the error state iagram, the counterpart to the error trellis. Note that state a is split into a (a source only) an e (a sink only). For the (,1,) coe it is The outputs of the states are the complete path polynomials; for example, Xb = Xb ( m ). m= 1 o The relations among the complete path polynomials are therefore c = a + b b c X X X c X = X + X X = X + X X = X e b o Joint solution for the transfer function X e X a gives 6 T( ) = 1 Series expansion gives the multiplicities a in aq ( γ br c )

12 To get more value from the transfer function, go back to error event probability Pr E a ( ) Q γ R (use for amming istance) [ ] ( = b c ) an upper-boun the Q function with an exponential ( ) γbr c γbrc [ ] ( ) 1 Q x e x : = e = = Pr E a e = a e = T( ) γ brc Very nice! For our (,1,) example coe, [ E] 6 6γbRc 1 1 e T γ R = e b c γ 1 γbr = e c 1 Pr ( ) = = brc ( e ) This is a payoff for the work of getting the transfer function. We can tighten it up a bit with the exact expression for Q function π 1 x ( ) exp π 0 sin ( θ ) Qx = θ Since this is also exponential, π π 1 γbr c 1 γbr c Pr[ E] a exp exp θ= a = π 0 sin ( θ) π θ 0 sin ( ) = θ π 1 γbrc = T exp θ π 0 sin ( θ ) 7.5-1

13 7.5-1 Now that we know how the transfer function works, we can use it to count other properties of error events notably the probability of information bit error. o Reraw the error state iagram, tagging braches with another ineterminate N raise to a power equal to the number of info bit errors on the branch. For a binary k = 1 coe, the label will be either N (for an error) or 1 (for no error). Counts info bit errors. Also labele with J on every branch, so we can count the length in branches of error events (if we care). o Solve these equations for Xe X a: c X = JN X + JN X b Xb = JXc + JX a c X = JN X + JN X X = J X e b o Gives T(, N, J) 6 X e J N X a 1 JN (1 J ) = =

14 Now we can boun the probability of bit error o For BER, we on t care about length of an error event in branches, so set J = 1, to get (, ) T N 6 N N = = = a N 1 f ( ) where f ( ) is the number of info bit errors in a pattern of weight. o Next, ifferentiate wrt N an set N = 1 T(, N) = a f( ) = β N N = 1 = = where of weight. β =a f( ) is the number of information bit errors in patterns o Therefore the BER is obtaine by 1 ( ) P β Q γ R β e b b c = = γ b R c = 1 T(, N) N N= 1, = e γbrc o For our coe, T(, N) N = 6 ( 1 N ), so P b e γ γb ( e ) 1 b

15 o Compare this (,1,) coe with uncoe transmission BER, uncoe an emo convolutional coe 0.1 Prob of bit error SNR per bit gamma_b (B) uncoe BPSK boun 0.5 exp(-gamma_b) uncoe BPSK, Q function (,1,) CC by T() with exp boun (,1,) CC by T() with Q fn _ lower boun The improvement comes at a cost: three times the banwith! See emos on the website for automatic calculation of transfer function ar ecision Input [P8..4] Now we obtain the error probabilities if the receiver makes har ecisions on the coe bits before passing them to the MLSE algorithm:

16 As usual for a union boun, we start with the pairwise error probability. Assume the all-zero sequence is the correct one o Remember that we set up the VA to choose the path with minimum amming istance from the input ecisions c ˆ jm. o Consier an error path with amming istance from the correct path, so the correct an erroneous paths iffer in positions. Take the case where is o. c = e = If there are ( 1) or fewer coe bit errors in these positions, then ĉ is closer to the correct sequence. If there are ( + 1) or more coe bit errors in these positions, then an we have a pairwise error. Therefore k P ( ) = p (1 p) k= ( + 1) k ĉ k is closer to the erroneous sequence, o. where p= Q( γ c ) is the probability of coe bit error

17 o Now take the case where is even. c = e = If there are 1 or fewer coe bit errors in these positions, then ĉ is closer to the correct sequence. If there are + 1 or more coe bit errors in these positions, then we have a pairwise error. If there are ĉ is closer to the erroneous sequence an errors, it is equiistant from the two alternative sequences call it either way, ranomly. Therefore k k 1 P ( ) = p (1 p) p (1 ) k= + 1 k + p, even. o Our bouns are then straightforwar: [ ] Pr E a P ( ) an P β P ( ) = b = As in the soft input case, it woul be goo to have a boun on the pairwise probability that is exponential in, so we get even more mileage out of the transfer function. Proakis gives a Chernoff boun on P ( ): [ ] P ( ) 4 p (1 p ) o ence the error rates are boune by

18 [ E] T = 4 (1 ) Pr ( ) p p an P b T(, N) N N= 1, = 4 p(1 p) o For our emo coe, T( ) = an 1 6 T(, N) N = 6 ( 1 N ), so an Pr [ E] 6 64 p (1 p) 1 4 p(1 p) = 4 p(1 p) = 1 ( γ ) b c ( γb c ) Q( brc ) Q( brc ) 64Q R 1 Q R = 1 4 γ 1 γ P b 6 = = ( 1 ) = 4p( 1 p) ( p) ( p) ( 1 8p 1 ) ( ) ( ) ( γb c ) ( γb c ) ( b c ) ( b c ) ( 1 8Q γ R 1 Q γ R ) 64 p 1 64Q R 1 Q R

19 Comparison shows that har ecisions cost about.5 B, compare with soft ecisions, for this coe. That makes it only 0.5 B or so better than uncoe transmission! The performance loss will be ifferent for other coes, but to B is typical. (Caution the legen is reverse: shoul be otte lines for har ecision.) There is much more in this section of Proakis, an it is well worth reaing