Trellis complexity and symmetry
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Complexity measures
• State complexity: Number of states
• State space dimension profile (0, 1,…, n)
• max
• Branch complexity: Number of branches
The Wolf bound
• max  {k, n – k}
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The trellis of the dual code
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C : an [n,k] linear code, C* its dual code (an [n,n-k] linear code)
i(C) and i(C*) the state spaces at time i of C and C*, resp.
Theorem: |i(C)| = |i(C*)| (i. e. i= i* ), for all i: 0  i  n.
• Proof:
• States in i(C*) corresponds to a coset in the partition
p0,i(C*)/C*,tr0,i
• i* = k(p0,i(C*)) – k(C*,tr0,i)
• p0,i(C*) is generated by Hi, the parity check matrix for
Ctr0,i. Thus p0,i(C*) and Ctr0,i are dual codes, and p0,i(C) and
C*,tr0,i are also dual codes. Thus
• i* = k(p0,i(C*)) – k(C*,tr0,i) = (i - k(Ctr0,i) – (i – k(p0,i(C)))
= k(p0,i(C)) - k(C0,i) = k - k(Ci,n) - k(C0,i) = i.
Note: State complexities equal; branch complexities different
2
Example
[16,11] RM
[16,5] RM
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Minimal trellises
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In principle there may be different trellises corresponding to
the same code
A minimal trellis for a given code is a trellis such that is
state complexity is (’0, ’1,…, ’n) and, for any other
trellis for the code with state complexity (0, 1,…, n) it
holds that for for all i: 0  i  n, ’i  i .
A minimal trellis is unique up to isomorphism.
In general, for a given code and for a given order of the
coordinates, it is simple to find a minimal trellis (provided
the trellis itself is simple). In fact, i is given by the
dimensions of the past and future subcodes, which is fixed
for any i.
If we are free to play with the order of the coordinates the
problem of finding an optimum bit ordering is much more
challenging.
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Bit ordering
Permuting the order of the coordinates does not change
• Code rate
• Distance properties
but may change
• the dimensions of future and past subcodes at any time
instant, and may thus change the trellis complexity.
An optimum permutation
• gives the smallest possible state space at each time instant
• Hard to find;
• Known for some classes of codes, unknown for others
• may not even exist for a given code
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Branch complexity
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Total number of branches in trellis
Determines (to a large extent) the workload on a trellis based
decoding algorithm
Let Ii=2 if there is an information symbol corresponding to
output bit i, Ii =1 otherwise.
Then the branch complexity is
• i Ii 2i
Example: (8,4) RM
i 0 1 2 3 4 567
Ii 2 2 2 1 2 1 1 1
2 i 1 2 4 8 4 8 4 2
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Cyclic codes
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A linear code with the property that any rotational shift of a
codeword is also a codeword
Generated by generator polynomial
g(x) = 1 + g1x+ g2x2 +…+ gn-k-1xn-k-1 +xn-k
The corresponding generator matrix occurs in a natural TO form:
1
 1 g1 g 2 ... ... g nk 1



 1 g1 g 2 ... ... g nk 1 1

G 








1
g
g
...
...
g
1
1
2
n  k 1


7
Cyclic codes (cont.)
1
 1 g1 g 2 ... ... g nk 1



 1 g1 g 2 ... ... g nk 1 1

G 








1
g
g
...
...
g
1
1
2
n  k 1


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Time span (gi) = [i,n-k+i+1], bit span (gi) = [i,n-k+i]
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If k > n-k, then
(0, 1,…, n) = (0,1,…,n-k-1,n-k, n-k,…, n-k,n-k-1,…,1,0)
•
If k  n-k, then
(0, 1,…, n) = (0,1,…,k-1,k, k,…, k,k-1,…,1,0)
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Cyclic codes meet the Wolf bound with equality (worst possible)
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Symmetry considerations
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Assume that a TOGM has the following symmetry property:
• For each row g with bit span (g) = [a,b], there is also
a row g’ with bit span (g’) = [n-1-b,n-1-a].
Then the trellis possess a mirror symmetry,
and in particular i = n-i .
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Trellis sectionalization
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So far we have considered n-section trellises,
i. e. one code bit per trellis section.
Sectionalization combines adjacent bits and provides a trellis with
fewer time instances, in order to simplify the decoders.
Let ={t0, t1,…,t}, for   n. Delete all state spaces (and their
adjacent branches) at time instances not in , and connect every pair
of states, one in tj and one in tj+1, iff there were a path between these
states in the original trellis, label the new branches by the old path
labels.
Uniform sectionalization: all sections of the same length
Sectionalization may
• Reduce state complexity
• Increase branch complexity
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Sectionalization: Example
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Parallel decomposition
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The sectionalized trellis may be densely connected, making a
hardware implementation difficult
The motivation for parallel decomposition is to select trellis
subsections to consist of structurally identical components which
are not interconnected.
Code C with TOGM G, let g be a row of G, let G1=G\{g}.
G TOF  G1 TOF
i (C1)= i (C)-1 for ia(g), i (C1)= i (C) for i  a(g).
Imax (C) = {i: i (C) = max (C)}
Thm 9.1: If there exists a row g such that a(g)  Imax (C), then
sectionalization produces two parallel and identically structured
subtrellises, one for C1 and one for C1 +g, such that
• max (C1)= max (C) - 1, and
• Imax (C1) = Imax (C)  {i: i (C) = max (C) - 1}
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Example
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Thus, it is possible to decompose trellis in smaller independent
components
1
1
1
0
0
0
0
1
0
G'  
0

0
1
0
1
1
0
1
0
1
1
1
1
0
0
0
0
1
1
1
0

0

1
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Generalization of Thm 9.1
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R(C) = {g: a(g)  Imax (C)}
Theorem 9.2: Theorem 9.1 can be applied repeatedly:
Corollary 9.2.1: Max number of parallel isomorphic subtrellises such
that the new total space dimension does not exceed max (C) is upper
bounded by 2|R(C)|
Corollary 9.2.2: In a minimal -section trellis with section boundary
locations in ={t0, t1,…,t}, the log2 of the number of parallel
isomorphic subtrellises is equal to the number of rows in a TOGM
with active time spans containing the section boundary locations in
={t0, t1,…,t}
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Low weight subtrellises
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A code with (ordered) codeword weights 0, w1, w2, …, wm =n
The wt–weight subtrellis is obtained by pruning from the
original trellis
• every state not involved in a codeword of weight wt
• Every branch not involved in a codeword of weight wt
If necessary split the zero state at each time instant into two
states, to avoid codewords of weights larger than wt that merge
with and diverge from the zero state in the original trellis.
(Discuss)
Useful for devising a suboptimum (but simpler) decoding
algorithm
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Cartesian product
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Trellis construction for codes designed by combining several
component codes by
• Interleaving
• Product
• Direct-sum
Cartesian product of original trellises
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Interleaved codes
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C = C1*C2*…*C formed by interleaving
n-section trellises for each component code
In the trellis for the interleaved code
• The state space is the cartesian product of the
original component state spaces
• Branch in the new trellis if there is a branch in a
component trellis linking corresponding component
states
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Interleaved codes: Example
Note: This is utterly impractical – use the simpler trellises! Can decode
each code individually (except on a burst channel????)
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Product codes
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C = C1C2 formed by a product construction:
• Form a matrix where each of the k2 n1-bit rows  C1
• Then append parity bits to the columns so that each  C2
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An n1 -section trellis for the product code can be formed by
cartesian product
Again totally impractical...No; this code needs to be decoided in
full
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Direct sum construction
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Component codes C1 and C2 each of length n
Direct sum code C formed by
C = C1C2 = {u+v: u  C1, v  C2 }
Trellis: Cartesian product again
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Direct sum : example
1
0

1
1
1
0
0
0
1
0
1
1
0
1
0
0

0
1
1
1
1
0
0
0
0
1
1
1
0
0

0
1

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Suggested order of exercises
Page 390
9.1
9.3
9.4
9.7
9.11
9.13
9.14
9.15
…
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