This document discusses channel coding and linear block codes. It provides an example of a (6,3) code with a generator matrix and calculates the corresponding codewords for each data word. The document also defines linear block codes and discusses decoding received codewords using the parity check matrix and syndrome. An example decoding problem is worked through step-by-step. The document concludes that channel encoding helps detect and correct errors in information and redundant bits aid in proper decoding without retransmission.

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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