Table Of ContentError-Trellis Construction for Tailbiting
Convolutional Codes
Masato Tajima Koji Okino
Graduate School of Science and Engineering Information Technology Center
University of Toyama University of Toyama
3190 Gofuku, Toyama 930-8555, Japan 3190 Gofuku, Toyama 930-8555, Japan
Email: [email protected] Email: [email protected]
2
1
t=0 t=1 t=2 t=3 t=4 t=5
0 Abstract—In this paper, we present an error-trellis construc- (00) 000 000 000 000 000
tion for tailbiting convolutional codes. A tailbiting error-trellis
2 111 111 111 111 111
is characterized by the condition that the syndrome former
n startsandendsinthesamestate. Weclarifythecorrespondence
a between code subtrellises in the tailbiting code-trellis and error
J subtrellises in the tailbiting error-trellis. Also, we present a (01) 001 001 001 001 001
110 110 110 110 110
1 construction of tailbitingbackward error-trellises. Moreover, we
2 obtain the scalar parity-check matrix for a tailbiting convolu-
011 011 011 011 011
tional code. The proposed construction is based on the adjoint-
] obviousrealizationofasyndromeformeranditsbehaviorisfully (10) 100 100 100 100 100
T
used in the discussion.
I
s. I. INTRODUCTION 010 010 010 010 010
c (11)
In this paper, we always assume that the underlying field 101 101 101 101 101
[
is F = GF(2). Let G(D) be a generator matrix of an (n,k)
1 convolutional code C. Let H(D) be a corresponding r ×n Fig.1. Tailbiting code-trellis basedonG1(D).
v
parity-check matrix of C, where r = n−k. Both G(D) and
8
H(D) are assumed to be canonical [1], [5]. Denote by L the
2
4 memory length of G(D) (i.e., the maximum degree among i.e.,β0 =βN (βk istheencoderstateattimek).Supposethat
4 the polynomials of G(D)) and by M the memory length of Tt(bc) has Σ0 initial (or final) states, then it is composedof Σ0
1. H(D). Then H(D) is expressed as subtrellises, each having the same initial and final states. We
0 callthesesubtrellisestailbitingcodesubtrellises.Forexample,
H(D)=H +H D+···+H DM. (1)
2 0 1 M atailbitingcode-trellisoflengthN =5basedonthegenerator
1 Consider a terminated version of C with N trellis sections. matrix
:
v That is, each codeword is a path starting from the all-zero G (D)=(1,1+D2,1+D+D2) (3)
1
i state at time t = 0 and ending in the all-zero state at time
X
t = N. In this case, C is specified by the following scalar is shown in Fig.1. Since Σ0 = 4, this tailbiting code-trellis
ar parity-check matrix [1], [6]: is composed of 4 code subtrellises. In Fig.1, bold lines
correspond to the code subtrellis with β =β =(1,0).
H 0 5
0 On the other hand, it is reasonable to think that an error-
H1 H0 trellis T(e) for the tailbiting convolutional code C can
... H1 ... equallybtebconstructed.Inthiscase,eacherrorsubtrellistsbhould
... ... ... ... have the same initial and final states like a code subtrellis. In
Hscalar = HM ... ... ... H0 (2) this paper, taking this property into consideration, we present
HM ... ... H1 an error-trellis construction for tailbiting convolutionalcodes.
... ... ... We also clarify the correspondence between code subtrellises
... ... in T(c) and error subtrellises in T(e). In this relationship, we
HM seetthbatdualstates(i.e.,syndrome-tfbormerstatescorresponding
with size (N +M)r×Nn (blanks indicate zeros). to encoder states) play an important role. Also, a kind of
Tailbitingis a techniquebywhicha convolutionalcodecan superpositionrule associated with a syndromeformeris used.
be used to construct a block code without any loss of rate Next, we present a construction of tailbiting backward error-
[4], [7], [10]. Let C be a tailbiting convolutional code with trellises. Using the backwarderror-trellis,each tailbiting error
tb
an N-section code-trellis T(c). The fundamental idea behind pathisrepresentedintime-reversedorder.Moreover,wederive
tb
tailbiting is that the encoderstarts and ends in the same state, the general structure of the scalar parity-check matrix for a
tailbiting convolutional code. Similar to a scalar generator B. Dual States
matrix, it is shown that the obtained scalar parity-check
The encoder states can be labeled by the syndrome-former
matrix has a cyclic structure. In general, unlike code-trellises, states (i.e., dual states [2]). The dual state β∗ corresponding
k
error-trellises enable decoding with remarkably low average to the encoder state β is obtained by replacing e in σ by
k k k
complexity [1]. Hence, we think an error-trellis construction y =u G(D) (u is the information at time k). We have
k k k
presented in this paper is very important.
∗
β = (y ,···,y ,y )
k k−M+1 k−1 k
II. SYNDROME FORMERHT(D) HMT ... 0 0
A. Adjoint-Obvious Realization of a Syndrome Former ... ... ... ...
× . (10)
Consider the adjoint-obvious realization (observer canon- H2T ... HMT 0
ical form [2], [3]) of the syndrome former HT(D) (T H1T ... HMT−1 HMT
means transpose). Let e = (e(1),e(2),···,e(n)) and ζ = Example 1: Consider the parity-check matrix
k k k k k
(ζk(1),ζk(2),···,ζk(r)) be the input error at time k and the 1+D D 1+D
correspondingoutputsyndromeattimek,respectively.Denote H1(D)=(cid:18) D 1 1 (cid:19) (11)
byσ(q) thecontentsofthememoryelementsintheabovereal-
kp corresponding to G (D). H (D) is expressed as
ization.Here,thecontentsofthememoryarraycorresponding 1 1
to the syndrome bit ζ(q) are labeled with q. For any fixed q, 1 0 1 1 1 1
k H (D) = + D
σ(q) corresponds to the memory element which is closest to 1 (cid:18) 0 1 1 (cid:19) (cid:18) 1 0 0 (cid:19)
k1
theqthoutputofthesyndromeformer(i.e.,ζ(q)).Ifamemory △
k = H +H D. (12)
elementismissing,thecorrespondingσ(q) issettozero.Using 0 1
kp
σ(q), the syndrome-formerstate at time k is defined as Hence (M = 1), the dual state corresponding to the encoder
kp state βk =(uk−1,uk) is obtained as follows.
σk =△ (σk(11),···,σk(r1),···,σk(1M),···,σk(rM)). (4) β∗k = ykH1T
1 1
(Remark: The effective size of σ is equal to the overall
k = (y(1),y(2),y(3)) 1 0
constraint length of H(D).) k k k
△ 1 0
Let ξ = (ζ ,σ )T be the extended state augmented with
the syndkrome ζkk. Tkhen ξk has an expression [8], [9]: = (yk(1)+yk(2)+yk(3),yk(1))
HM HM−1 ... H1 H0 = (uk−1+uk,uk). (13)
0 HM ... H2 H1 C. Behavior of a Syndrome Former
ξk = ... ... ... ... ... Lemma 1: Let σk−1 be the syndrome-former state at time
0 0 ... HM HM−1 k−1.Here,assumethatanerrorek isinputtedtothesyndrome
0 0 ... 0 HM former and it moves to the state σ at time k. Also, assume
k
×(ek−M,ek−M+1,···,ek)T thatthe syndromeζk is outputtedaccordingto this transition.
=△ H∗×(ek−M,ek−M+1,···,ek)T . (5) (This relation is denoted as e
From this expression, we have σk−1−ζ→k σk.)
k
σ =△ (σ(1),σ(2),···,σ(M)) Similarly, assume the relation
k k k k
= (ek−M+1,···,ek−1,ek) σ′ −e→′k σ′. (14)
HMT ... 0 0 k−1 ζ′k k
× ... ... ... ... Then we have
=△ (ek−MHH+21TT1,··..·....,ekH−HMT1MT,−e1k)×H0HMT ∗∗T. (6) Proof: Froσmkt−h1e+asσsu′km−p1tζeio−kkn++→,ζet′k′kheσkre+latσio′kn.s (15)
Note that σk has an alternative expression: σ =(σ(2) ,···,σ(M),0)+e (HT,HT,···,HT ) (16)
k k−1 k−1 k 1 2 M
σk =(σ(k2−)1,···,σ(kM−1),0)+ek(H1T,H2T,···,HMT ). (7) σ′ =(σ′(2),···,σ′(M),0)+e′(HT,HT,···,HT ) (17)
k k−1 k−1 k 1 2 M
Similarly, ζ is expressed as
k hold. Hence, we have
ζk = ek−MHMT +···+ek−1H1T +ekH0T (8) σk+σ′k = (σ(k2−)1+σ′k(−2)1,···,σ(kM−1)+σk′(−M1),0)
= σ(k1−)1+ekH0T. (9) +(ek+e′k)(H1T,H2T,···,HMT ). (18)
On the other hand, using the relations Proof: From the assumption, we have
ζk = σ(k1−)1+ekH0T (19) σfinz=−y→+eσfin. (27)
ζ′ = σ′(1) +e′HT, (20) ζ
k k−1 k 0
Also, from Lemma 2,
we have
y
∗ ∗
ζk+ζ′k =(σ(k1−)1+σ′k(−1)1)+(ek+e′k)H0T. (21) β ζ−=→0β (28)
These expressions imply that is obtained. Hence, by applying Lemma 1, we have
σk−1+σ′k−1ζe−kk++→ζe′k′kσk+σ′k σfin+β∗zζ++−0y→==ζeσfin+β∗. (29)
holds.
Lemma 2: Let β and β be the initial and final states
of the code-trellis,0respectivNely. Denote by y a code path III. ERROR-TRELLISES FOR TAILBITING CONVOLUTIONAL
connecting these states. (This is denoted as CODES
y A. Error-Trellis Construction
β −→β .)
0 N Suppose that the tailbiting code-trellis based on G(D) is
Then we have definedin[0,N],whereN ≥M.Inthiscase,thecorrespond-
y
β∗−→β∗ . (22) ing tailbiting error-trellis based on HT(D) is constructed as
0ζ=0 N follows.
That is, assume that the syndrome former is in the dual state Step1: Letz ={zk}Nk=1 be a receiveddata. Denoteby σ0
β∗0 ofβ0.Inthiscase,ify isinputtedtothesyndromeformer, the initial state of the syndrome former HT(D). Let σfin(=
then it moves to the dual state β∗N of βN and the syndrome σN) be the final syndrome-former state corresponding to the
ζ =0 is outputted. input z. Note that σfin is independentof σ0 and is uniquely
Proof:Byextendingthecode-trellisinbothdirectionsby determined only by z.
L sections, if necessary, we can assume the condition Step2:Setσ toσ andinputz tothesyndromeformer.
0 fin
Here, assume that the syndrome sequence ζ = {ζ }N is
β0 =0−y→′ βL −y→βN+L −y→′′ βN+2L =0, (23) obtained. (Remark: ζk (k ≥ M +1) has been obtkaink=ed1 in
where y′ and y′′ are augmented code paths (initial and final Step 1.)
states are both 0). Hence, we can apply the standard scalar Step3: Concatenatetheerror-trellismodulescorresponding
to the syndromesζ . Then we have the tailbiting error-trellis.
parity-check matrix H (cf. (2)). Then we have k
scalar
Example2:Again,considertheparity-checkmatrixH (D).
1
y′ y y′′
β∗ =0 −→ β∗ −→β∗ −→ β∗ =0. (24) Let
0 ζ′=0 Lζ=0 N+Lζ′′=0 N+2L
z =z z z z z =111 110 110 111 000 (30)
1 2 3 4 5
Thatis,theoutputofthesyndromeformeriszeroforalltime.
In the above relation, we can note the following subsection: be the received data. According to Step 1, let us input z to
∗ y ∗ the syndrome former H1T(D). Then we have σfin = (0,0).
βLζ−=→0βN+L. (25) Next,wesetσ0 toσfin =(0,0)andinputz tothesyndrome
former. In this case, the syndrome sequence
Let z = {zk}Nk=1 be a received data. Denote by σ0 the ζ =ζ1 ζ2 ζ3 ζ4 ζ5 =00 00 10 01 11 (31)
initial state of the syndromeformer.Letσ be the syndrome-
k is obtained. The tailbiting error-trellis is constructed by con-
former state at time k correspondingto the input z. Note that
catenating the error-trellis modules corresponding to ζ . The
σ isindependentofσ ifk≥M.Also,ζ isindependentof k
k 0 k obtained tailbiting error-trellis is shown in Fig.2.
σ if k ≥M +1. In the following, we assume the condition
0
N ≥M. B. CorrespondenceBetweenCodeSubtrellisesandErrorSub-
Proposition 1: Lety beatransmittedcodepathinatailbit- trellises
ingcodesubtrelliswithβ =β =β.Also,letz =y+ebe
0 N With respect to the correspondencebetween tailbiting code
thereceiveddata,whereeisanerror.Denotebyσ (=σ )
fin N subtrellises and tailbiting error subtrellises, we have the fol-
the final syndrome-formerstate corresponding to the input z.
lowing.
Here, assume that σ is set to σ and z is inputted to the
0 fin Proposition 2: Let β (= β ) = β be the initial (final)
0 N
syndromeformer.Letζ betheoutputtedsyndrome.(Notethat
state of a tailbiting code subtrellis. Then the initial (final)
the final syndrome-formerstate is σ .) Then we have
fin state of the corresponding tailbiting error subtrellis is given
σ +β∗−e→σ +β∗. (26) by σfin+β∗.
fin fin
ζ Proof: Direct consequence of Proposition 1.
t=0 t=1 t=2 t=3 t=4 t=5 t=0 t=1 t=2 t=3 t=4 t=5
(00) 000 000 011 (00) 000 111 000
101 110 101 010
111 111 100 011 100
010 001 110 001 011
000 0 11 000 111
(01) 101 101 110 (01) 101 010 101
010 010 001 111 100 110 0 01 011 100 110
011 011 0 00 110 101 010 101 111 000 010
(10) (10)
100 100 111 001 010 001 110 100 011 001
110 110 101 011 000 111 000 010 101 111
(11) 001 001 010 100 111 (11) 100 011 001 110 100
ζ(1)ζ(2)=00 ζ(1)ζ(2)=00 ζ(1)ζ(2)=10 ζ(1)ζ(2)=01 ζ(1)ζ(2)=11 η(1)η(2)=00 η(1)η(2)=11 η(1)η(2)=01 η(1)η(2)=10 η(1)η(2)=00
1 1 2 2 3 3 4 4 5 5 1 1 2 2 3 3 4 4 5 5
Fig.2. Tailbiting error-trellis basedonH1T(D). Fig.3. Tailbiting backwarderror-trellis basedonH˜1T(D).
Example 2 (Continued): Consider the tailbiting error-trellis be the time-reversed received data. According to Step 1, let
in Fig.2. In this example, we have σ = (0,0). The us input z˜ to the syndrome former H˜T(D). Then we have
fin 1
correspondingtailbitingcode-trellisbasedonG (D)isshown σ˜ =(0,0). Next, we set σ˜ to σ˜ =(0,0) and input z˜
1 fin 0 fin
in Fig.1. In Fig.1, take notice of the code subtrellis with to the syndrome former. In this case, the syndrome sequence
initial (final) state β = (1,0) (bold lines). The dual state of
β =(1,0)iscalculatedasβ∗ =(u−1+u0,u0)=(1+0,0)= η =η1 η2 η3 η4 η5 =00 11 01 10 00 (35)
(1,0).Hence,theinitial(final)stateofthecorrespondingerror is obtained. Since M =1, we see that the correspondence
∗
subtrellisisgivenbyσ +β =(0,0)+(1,0)=(1,0)(bold
fin η = η η η η η
lines in Fig.2). 1 2 3 4 5
= ζ ζ ζ ζ ζ (36)
1 5 4 3 2
C. Backward Error-Trellis Construction
holds. The tailbiting backward error-trellis is constructed by
Let G˜(D) and H˜(D) be the reciprocal encoder and the
concatenating the error-trellis modules corresponding to η .
reciprocaldualencoder[6]associatedwithG(D),respectively. k
The obtained tailbiting backward error-trellis is shown in
Thenthe tailbitingbackwarderror-trelliscorrespondingto the
Fig.3.
original tailbiting error-trellis is constructed as follows.
Next, consider the correspondence between forward error
Step 1: Let z˜ = {z˜k}Nk=1 = {zN−k+1}Nk=1 be the time- subtrellises and backward error subtrellises. First, note the
reversed received data. Denote by σ˜ the initial state of the
0 following.
syndrome former H˜T(D). Let σ˜fin(= σ˜N) be the final Proposition 3: Let β˜ (= β˜ ) = β˜ be the initial (final)
syndrome-formerstatecorrespondingtotheinputz˜.Notethat 0 N
state of a tailbiting backward code subtrellis. Then the initial
σ˜ is independent of σ˜ and is uniquely determined only
fin 0 (final) state of the corresponding backward error subtrellis is
bySz˜te.p2:Setσ˜ toσ˜ andinputz˜tothesyndromeformer. given by σ˜fin+β˜∗.
0 fin Proof: Direct consequence of Proposition 1.
Here, assume that the syndrome sequence η = {ηk}Nk=1 is Let β˜ be the backward state corresponding to β. Then the
obtained.
forwardcodesubtrelliswithβ (=β )=βandthebackward
Remark: It is shown that ζ = {ζk}Nk=1 and η = {ηk}Nk=1 codesubtrelliswith β˜ (=β˜ 0)=β˜Ncorrespondto eachother.
have the following correspondence: 0 N
Hence, using Propositions 2 and 3, we have the following.
∗
η = η η ··· η η ··· η Proposition 4: Let σ +β be the initial (final) state of
1 2 M M+1 N fin
= ζM ζM−1 ··· ζ1 ζN ··· ζM+1. (32) a tailbiting forward error subtrellis. Then the initial (final)
state of the corresponding backward error subtrellis is given
Step3: Concatenatethe error-trellismodulescorresponding by σ˜ +β˜∗, where β˜ is the backward state of β.
fin
to the syndromes η . Then we have the tailbiting backward
k Example 3 (Continued): Consider the reciprocal encoder
error-trellis.
Example 3: Take notice of Example 2. The reciprocal dual G˜1(D)=(D2,1+D2,1+D+D2) (37)
encoder H˜1(D) associated with G1(D) is given by and the reciprocal dual encoder H˜ (D) associated with
1
H˜ (D)= 1+D 1 1+D . (33) G1(D). H˜1(D) is expressed as
1 (cid:18) 1 D D (cid:19) 1 1 1 1 0 1
H˜ (D) = + D
Let 1 (cid:18) 1 0 0 (cid:19) (cid:18) 0 1 1 (cid:19)
△
z˜=z˜ z˜ z˜ z˜ z˜ =000 111 110 110 111 (34) = H˜ +H˜ D. (38)
1 2 3 4 5 0 1
˜
Hence, the dual state corresponding to βk = (uk−1,uk) is Similarly, we have
calculated as
β˜∗k = ykH˜1T ζ2 == ((eeN−M−M+2+H2HMTMT++·····+·+e0eHN2TH)2T+)e+1He11TH+1T e+2He20TH0T
1 0 = (e1,e2,···,eN−M+2,···,eN)
= (yk(1),yk(2),yk(3)) 0 1 ×(H ,H ,0,···,0,H ,···,H )T. (45)
1 1 1 0 M 2
= (y(1)+y(3),y(2)+y(3)) The same argumentcan be applied to ζk (3≤k ≤N). Then
k k k k we see that HT is written as
= (uk−1+uk,uk−1). (39) scalar
H0T H1T ...HMT−1 HMT
Here,takenoticeoftheerrorsubtr∗elliswithinitial(final)state H0T ... ... HMT−1 HMT
(1,0)inFig.2.(Notethatσfin+β =(0,0)+(1,0)=(1,0).) ... H1T ... HMT−1...
This error subtrellis corresponds to the code subtrellis with H0T H1T ... ... HMT . (46)
initial (final) state β =(1,0) in F˜ig.1. On the other hand, the HMT H0T H1T ...HMT−1
backward˜s∗tate of β = (1,0) is β = (0,1) and its dual state HMT−1HMT H0T ... ...
becomes β =(u−1+u0,u−1)=(0+1,0)=(1,0). Hence, ... ... ... ... H1T
fbraocmkwParrodpoesrirtoiornsu4b,ttrheelliinsitiisalg(ifivneanl)bsytaσ˜teoft+heβ˜c∗or=res(p0o,n0d)in+g H1T H2T ... HMT H0T
fin By transposing this matrix, H is obtained.
(1,0)=(1,0) (bold lines in Fig.3). scalar
V. CONCLUSION
IV. Hscalar FORTAILBITING CONVOLUTIONAL CODES Inthispaper,wehavepresentedanerror-trellisconstruction
ConsiderthetailbitingconvolutionalcodeC withN trellis for tailbiting convolutional codes. A tailbiting error-trellis is
tb
sections specified by a parity-checkmatrix H(D). C can be characterizedbytheconditionthatthesyndromeformerstarts
tb
regardedasan(Nn,Nk)blockcode[4].Inthiscase,wehave and ends in the same state. We have clarified the correspon-
the following. dence between code subtrellises in the tailbiting code-trellis
Proposition 5: Assume that H(D) has the form (1). Then and error subtrellises in the tailbiting error-trellis. Also, we
the scalar parity-check matrix H for C is given by have presented a construction of tailbiting backward error-
scalar tb
trellises. Moreover, we have obtained the general structure of
H0 HM ... H2 H1 the scalar parity-check matrix for a tailbiting convolutional
H1 H0 ... ... H2 code. We see that the obtainedresults correspondto those for
... H1 ... HM ... tailbiting code-trellises in the natural manner.
HM−1 ... ... H0 HM (40)
HM HM−1... H1 H0 REFERENCES
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×(H ,0,···,0,H ,···,H )T. (44)
0 M 1
