Fundamental of Channel Coding

1. Channel Coding-2
Dr. Varun Kumar
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2. Outlines
1 Numerical on Channel Coding
2 Linear Block Code
3 Conclusion
4 References
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3. Numerical on Channel Coding [1],[2]
Q. For a (6,3) code, the generator matrix
G =


1 0 0 1 0 1
0 1 0 0 1 1
0 0 1 1 1 0


For all eight data words, find the corresponding codewords, and verify
that this code is a single-error correcting code.
Ans. As per question, G = [Ik P] and n = 6, k = 3 and m = n − k = 3.
Hence
P =


1 0 1
0 1 1
1 1 0


For k = 3 total 23 data words can be generated. Also, c = dG
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4. Continued–
Data word d Codeword c
111 111000
110 110110
101 101011
100 100101
011 011101
010 010011
001 001110
000 000000
Hint: During matrix multiplication, use the modulo-2 addition operation.
Q. Find the Hamming distance between two codeword ca = 111000 and
cb = 001110.
Ans. Hamming distance
d(ca cb) = weight of (ca ⊕ cb)=weight of (110110) = 4
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5. Linear Codes
Linear block codes
A block code is a linear block code if every pair of codewords ca and cb
from the block code,
ca ⊕ cb
is also a codeword.
Decoding of Codeword:
From previous discussion cp = dP, Hence
dP ⊕ cp = d cp
P
Im
= 0
where Im is an identity matrix of order m × m, also (m = n − k). Thus
cHT
= 0 (1)
and H = [PT Im] is called as the parity check matrix.
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6. Continued–
Let r is the received codeword and c is the transmitted codeword. Due to
channel error the observed error is e. Hence
r = c ⊕ e (2)
Ex- Let r = 101101, c = 100101. Hence from above relation, e = 001000.
An element 1 in e indicate an error in the corresponding position.
Hamming distance between r and c is simply the number of 1s in e.
Making r = ci + ei , If there is no error than ei = 000000.
cHT = rHT = 0, when there is no error during reception.
Due to noise, rHT = 0 or s = rHT , where s is called the syndrome.
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7. Continued–
Mathematically syndrome can also be expressed as
s =rHT
=(ci + ei )HT
= ci HT
⊕ ei HT
=ei HT
(3)
Q. A linear (6,3) code is generated according to the generating matrix as
per above question. The receiver receives r = 100011. Determine the
corresponding data word, if the channel is a BSC and the maximum
likelihood decision is used.
Ans. We have,
s = rHT
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8. Continued–
= 1 0 0 0 1 1








1 0 1
0 1 1
1 1 0
1 0 0
0 1 0
0 0 1








= [1 1 0] (4)
c = r ⊕ e
s = [1 1 0] = eHT
We see that e = 001000 and e = 000110, 010101, 011011, 110000,
111110, 101101, 100011 also satisfy the above relation.
Suitable choice, the minimum weight emin is 001000. Hence
c = r ⊕ e = 100011 ⊕ 001000 = 101011
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9. Continued–
Decoding Table Generation
Error, e Syndrome, s
000000 000
100000 101
010000 011
001000 110
000100 100
000010 010
000001 001
100010 111
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10. Conclusion
Channel encoding ensure that how can we detect and correct the
error in information bit stream.
How the redundant bit helps for proper decoding without
re-transmission of the codeword.
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11. References
B. P. Lathi, Z. Ding et al., “Modern Digital and Analog Communication Systems /
BP Lathi, Zhi Ding.” 2010.
M. Borda, Fundamentals in information theory and coding. Springer Science &
Business Media, 2011.
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12. Thank You
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