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Copyright Cambridge University Press 2003. On-screen viewing permitted. Printing not permitted. http://www.cambridge.org/0521642981You can buy this book for 30 pounds or $50. See http://www.inference.phy.cam.ac.uk/mackay/itila/ for links.576 48 — Convolutional Codes and Turbo CodesSystematic recursiveThe filter shown in figure 48.1c is similar to the nonsystematic nonrecursivefilter shown in figure 48.1b, but it uses the taps that formerly made up g (a)to make a linear signal that is fed back into the shift register along with thesource bit. The output t (b) is a linear function of the state vector as before.The other output is t (a) = s, so this filter is systematic.A recursive code is conventionally identified by an octal ratio, e.g., figure48.1c’s code is denoted by (247/371) 8 .Equivalence of systematic recursive and nonsystematic nonrecursive codesThe two filters in figure 48.1b,c are code-equivalent in that the sets of codewordsthat they define are identical. For every codeword of the nonsystematicnonrecursive code we can choose a source stream for the other encoder suchthat its output is identical (and vice versa).To prove this, we denote by p the quantity ∑ kκ=1 g(a) κ z κ , as shown in figure48.3a and b, which shows a pair of smaller but otherwise equivalent filters.If the two transmissions are to be equivalent – that is, the t (a) s are equal inboth figures and so are the t (b) s – then on every cycle the source bit in thesystematic code must be s = t (a) . So now we must simply confirm that forthis choice of s, the systematic code’s shift register will follow the same statesequence as that of the nonsystematic code, assuming that the states matchinitially. In figure 48.3a we havewhereas in figure 48.3b we havet (a) = p ⊕ z nonrecursive0 (48.1)z recursive0 = t (a) ⊕ p. (48.2)z 2z 2〈dz 1✲⊕✻z 0✲ t(b)〈d ✛ s✲ ❄ ⊕ ✲ ❄ p ⊕ ✲ t(a)(a) (5, 7) 8z 1✲⊕✻z 0✲ t(b)〈d 〈d✻✲✲ ❄ ⊕ ✲ ✛ ❵p⊕ s(b) ( )1, 5 78t (a)Figure 48.3. Two rate-1/2convolutional codes withconstraint length k = 2:(a) non-recursive; (b) recursive.The two codes are equivalent.Substituting for t (a) , and using p ⊕ p = 0 we immediately findz recursive0 = z nonrecursive0 . (48.3)Thus, any codeword of a nonsystematic nonrecursive code is a codeword ofa systematic recursive code with the same taps – the same taps in the sensethat there are vertical arrows in all the same places in figures 48.3(a) and (b),though one of the arrows points up instead of down in (b).Now, while these two codes are equivalent, the two encoders behave differently.The nonrecursive encoder has a finite impulse response, that is, ifone puts in a string that is all zeroes except for a single one, the resultingoutput stream contains a finite number of ones. Once the one bit has passedthrough all the states of the memory, the delay line returns to the all-zerostate. Figure 48.4a shows the state sequence resulting from the source strings =(0, 0, 1, 0, 0, 0, 0, 0).Figure 48.4b shows the trellis of the recursive code of figure 48.3b and theresponse of this filter to the same source string s =(0, 0, 1, 0, 0, 0, 0, 0). Thefilter has an infinite impulse response. The response settles into a periodicstate with period equal to three clock cycles.⊲ Exercise 48.1. [1 ] What is the input to the recursive filter such that its statesequence and the transmission are the same as those of the nonrecursivefilter? (Hint: see figure 48.5.)
Copyright Cambridge University Press 2003. On-screen viewing permitted. Printing not permitted. http://www.cambridge.org/0521642981You can buy this book for 30 pounds or $50. See http://www.inference.phy.cam.ac.uk/mackay/itila/ for links.48.2: Linear-feedback shift-registers 577(a)11100100transmit 0 0 0 0 1 1 1 0 1 1 0 0 0 0 0 0source 0 0 1 0 0 0 0 011Figure 48.4. Trellises of therate-1/2 convolutional codes offigure 48.3. It is assumed that theinitial state of the filter is(z 2 , z 1 ) = (0, 0). Time is on thehorizontal axis and the state ofthe filter at each time step is thevertical coordinate. On the linesegments are shown the emittedsymbols t (a) and t (b) , with starsfor ‘1’ and boxes for ‘0’. Thepaths taken through the trelliseswhen the source sequence is00100000 are highlighted with asolid line. The light dotted linesshow the state trajectories thatare possible for other sourcesequences.1001(b)00transmit 0 0 0 0 1 1 0 1 0 1 0 0 0 1 0 1source 0 0 1 0 0 0 0 01110Figure 48.5. The source sequencefor the systematic recursive code00111000 produces the same paththrough the trellis as 00100000does in the nonsystematicnonrecursive case.0100transmit 0 0 0 0 1 1 1 0 1 1 0 0 0 0 0 0source 0 0 1 1 1 0 0 0
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