# Linear Block Codes (3)

```II. Linear Block Codes
Last Lecture
 H Matrix and Calculation of dmin
 Error Detection Capability
 Error Correction Capability
 Error Correction & Error Detection Operation
2
Decoding and Error Correction
Let C be an (n,k) linear block code with codewords v1, v2,…,v2k
v
Channel
2k codewords
e
r
Decoder
2n n-tuples
Which vi was
sent??
Decoding: A rule to partition 2n possible received vectors into 2k
disjoint subsets D1, D2,…,D2k such that the codeword vi is
contained in the subset Di, 1≤i ≤2k
Decoding Rule:
If r ∈ Di⇒Decoding
Decision si vi
Correct Decoding: Given vi is transmitted, error pattern e does not
push r outside Di
3
Design of a Decoder
How to divide 2n vectors into 2k disjoint subsets
such that the error correcting capability is
optimal?
In other words, which n-tuples should be
designed such that they belong to a given
subset Di?
4
Standard Array
Partitioning 2n possible received vectors into 2k
disjoint subsets with each subset containing only one
codeword
D1
D2
v1 = 0
v2
e2
e2 + v 2
e3
e3 + v 2
.
.
.
.
e2n-k
e2n-k + v 2
Di
...
vi
... e2+ vi
... e3+ vi
.
.
.
.
... e2n-k + vi
D2k
...
v 2k
... e2+ v 2k
... e3+ v 2k
.
.
.
.
... e2n-k + v 2k
5
Standard Array
Partitioning 2n possible received vectors into 2k
disjoint subsets with each subset containing only one
codeword
D1
e2≠v1,v2,…,v2k
D2
v1 = 0
v2
e2
e2 + v 2
e3
e3 + v 2
.
.
.
.
e2n-k
e2n-k + v 2
Di
...
vi
... e2+ vi
... e3+ vi
.
.
.
.
... e2n-k + vi
D2k
...
v 2k
... e2+ v 2k
... e3+ v 2k
.
.
.
.
... e2n-k + v 2k
6
Standard Array
Partitioning 2n possible received vectors into 2k
disjoint subsets with each subset containing only one
codeword
D1
e3≠v1,v2,…,v2k,
e2,…,e2+v2k
D2
v1 = 0
v2
e2
e2 + v 2
e3
e3 + v 2
.
.
.
.
e2n-k
e2n-k + v 2
Di
...
vi
... e2+ vi
... e3+ vi
.
.
.
.
... e2n-k + vi
D2k
...
v 2k
... e2+ v 2k
... e3+ v 2k
.
.
.
.
... e2n-k + v 2k
7
Standard Array
Partitioning 2n possible received vectors into 2k
disjoint subsets with each subset containing only one
codeword
D1
e2n-k ≠Any
previous
element in the
array
D2
v1 = 0
v2
e2
e2 + v 2
e3
e3 + v 2
.
.
.
.
e2n-k
e2n-k + v 2
Di
...
vi
... e2+ vi
... e3+ vi
.
.
.
.
... e2n-k + vi
D2k
...
v 2k
... e2+ v 2k
... e3+ v 2k
.
.
.
.
... e2n-k + v 2k
8
Theorem
1.
2.
No two n-tuples in the same row of a standard array
are identical.
Every n-tuple appears in one and only one row.
PROOF:
1. Given a linear block code is considered. By definition of the
standard array construction, ej is not a codeword. Therefore,
ej+ vj generate distinct patterns
2. Select row l<m, assume el+vi= em+vj. Therefore,
em= el+vi+vj
em= el+vs (because vi and vj are codewords)
This means that em could be found in row l
This is a contradiction to the construction rule of the
standard array
9
Standard Array
Partitioning 2n possible received vectors into 2k
disjoint subsets with each subset containing only one
codeword
D1
e2n-k ≠Any
previous
element in the
array
D2
v1 = 0
v2
e2
e2 + v 2
e3
e3 + v 2
.
.
.
.
e2n-k
e2n-k + v 2
Di
...
vi
... e2+ vi
... e3+ vi
.
.
.
.
... e2n-k + vi
D2k
...
v 2k
... e2+ v 2k
... e3+ v 2k
.
.
.
.
... e2n-k + v 2k
10
Coset of a Code, Coset Leader
Given the standard array. The 2n-k rows are
called the cosets of the code. The first n-tuple
ej of each coset is called the coset leader
11
Error Correction
If r ∈Di⇒Decoding
Decision si v i
If the error pattern is a cosetleader⇒CORRECT DEC ODING
If the error pattern is not a cosetleader⇒FALSE DEC ODING
Assume error pattern x is in the lth coset
x = el + vi ⇒ r = v j+ x
= v j + el + v i
= el + v s
r is found in the subset Ds≠Dj ⇒ FALSE DEC ODING
A Linear block code can correct 2n-k-1 error patterns
Note that one of the coset leaders is the zero vector
12
Error Correction Methodology
The error patterns that are most likely to occur should
be used as the coset leaders
Example: A (7,4) code can correct 2(7-4)-1=7 error patterns
To correct a single error. The coset leaders should be
selected as:
e1= v1= (0 0 0 0 0 0 0)
e2 =(0 0 0 0 0 0 1)
e3 =(0 0 0 0 0 1 0)
e4 =(0 0 0 0 1 0 0)
e5 =(0 0 0 1 0 0 0)
e6 =(0 0 1 0 0 0 0)
e7 =(0 1 0 0 0 0 0)
e8 =(1 0 0 0 0 0 0)
13
Theorem
1. All 2k n-tuples of a coset have the same syndrome.
2. The syndrome for different cosets are different
PROOF:
1. (el+vi)HT=elHT+viHT=elHT (independent of vi)
2. Let ej and ei be the coset leaders. If they have the same
syndrome
ejHT+eiHT=0. Therefore,
(ej+ei) is a codeword
This contradicts the definition of a linear block code because ej
and ei are not codewords
Advantages: We can use a syndrome decoding table which
is much simpler to use than a standard array
14
Syndrome Decoding
1. Compute syndrome of r (i.e., rHT)
2. Locate the coset leader el whose syndrome is
equal to rHT
3. v*= r+el where v* is the codeword output by the
decoder
15
Weight Distribution of a Block Code
Code Word
Information
Parity Bits
0
0000
000
1
0001
101
2
0010
111
3
0011
010
4
0100
011
5
0101
110
Example:
6
0110
100
For the (7,4) code shown,
A0 = A7 = 1,
A1 = A2 = A5 = A6= 0
A3 = A4 = 7
Note that:
∑Ai=16=24 (Number of valid
code words)
7
0111
001
8
1000
110
9
1001
011
10
1010
001
11
1011
100
12
1100
101
13
1101
000
14
1110
010
15
1111
111
For an (n.k) block code:
Let Ai be the number of
codewords of weight i in
C. The numbers A0, A1, …,
An are called the
Weight Distribution
16
Weight Distribution and Probability
of Detecting an Error Pattern
For an (n,k) linear code:
 Given that the bit error probability of the physical channel is p.
The probability that an error pattern of weight j occurs is
pj(1-p)n-j
 In Total, there are nCj error patterns that have j erroneous bits.
ONLY Aj of those are NOT DETECTABLE because they
represent valid codewords
 Probability of not detecting an error pattern (Pu(E)) is:
n
Pu E    A jp j 1  p 
n j
j 1
Example:
For the (7,4) code in the previous slide,
Pu(E)=7p3(1-p)4 + 7p4(1-p)3 + p7
17
Let αi denote the number of coset leaders of weight i.
Then the numbers α0,α1,…, αn are the weight distribution
Probability of Correct Decoding of an error pattern Ps(E)
PS(E)=∑ αi pi(1-p)(n-i) where is the bit error probability of the
channel
Probability of False Decoding of an error pattern P(E)
P (E)=1-PS(E)
18
```
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