t com presentation (error correcting code)
TRANSCRIPT
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Error Correcting CodesT-Com
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How Is Data Transmitted ?
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Occurrence of Error
Causes of Error (Noise):
o Multi-path propagation of the signal.
o Interference from other communication devices
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Contd..
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Solution
Channel Coding
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Block Codes
● Divide the message into blocks each of k bits called data words.
● Add r redundant bits to each blocks to make the length n = k+r.
● Examples : Hamming Codes, Reed-Muller Codes
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Redundancy Coding● In n-redundancy coding, each data bit is encoded in n
bits.● In a 3-redundancy coding scheme, a ‘0’ data bit is
encoded as ‘000’ and a ‘1’ data bit is encoded as ‘111’.
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How does it reduce the error?
❏ The decoder is taking blocks of n bits.
❏ The decoder expects all n bits to have the same value
❏ When the n bits in a block do not have the same value, the decoder detects an error
❏ n-redundancy coding can correct up to (n-1)/2 bits in a code
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Correction Abilities of Codes
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Signal Transmission Example
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Hamming Code
❏Hamming codes can correct one-bit errors ❏ 2r >= n + r + 1 , r=redundancy bits
n=data bits
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Redundancy bits calculationr1 will take care of these bits
r2 will take care of these bits
r4 will take care of these bits
r8 will take care of these bits
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Example
d
d
d
r
d
d
d
r
d
r
r
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Error Correction using hamming code
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Reed Muller Code:R(1,3)
One of the interesting things about these codes is that
there are several ways to describe them and we shall look
at one of these-
1st order Reed Muller R(1,3) which can correct one bit
error.
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Contd..
Need a Generator matrix
Dimension is
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Encoding❏ rows in the generator matrix = Dimension
❏ So 3 vectors are there and hence total length of the code to be sent =23
❏ Let m = (m1, m2, . . . mk) be a block, the encoded message Mc is
Mc=∑i=1,kmiriwhere ri is a row of the encoding matrix of R(r, m).
❏ For example, using R(1, 3) to encode m = (0110) gives:
0 ∗(11111111) + 1 ∗(11110000) + 1 ∗(11001100) + 0 ∗(10101010) =
(00111100)
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Decoding❏ The rows of this matrix are basis vectors for the code v0, v1,v2 and v3.
❏ Any vector v of the code is a linear combination of these v =
a0v0+a1v1+a2v2 + a3v3
❏ If no errors occur, a received vector r = (y0, y1, y2, y3, y4, y5, y6,y7)
a1 =y0 +y1 =y2 +y3 =y4 +y5 =y6 +y7
a2 =y0 +y2 =y1 +y3 =y4 +y6 =y5 +y7
a3 =y0 +y4 =y1 +y5 =y2 +y6 =y3 +y7
❏ If one error has occurred in r, then when all the calculations above are
made, 3 of the 4 values will agree for each ai, so the correct value will be
obtained by majority decoding.
a0 =r + a1v1 + a2v2 + a3v3
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ExampleLet m=1101 & Suppose that v = 10100101 is received as r= 10101101. Using,
a1 =y0 +y1 =y2 +y3 =y4 +y5 =y6 +y7
a2 =y0 +y2 =y1 +y3 =y4 +y6 =y5 +y7
a3 =y0 +y4 =y1 +y5 =y2 +y6 =y3 +y7 we calculate:
a1 =1=1=0=1
a2 =0=0=1=0
a3 =0=1=1=1
a1 =1 a2 =0 a3 =1
a0 =r + a1v1 + a2v2 + a3v3
So,10101101 + 01010101 + 00001111 = 11110111.
➔ a0 = 1
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Contd..❏ Decoded data bits are:a0 a1 a2 a3 =1101
v = a0v0 + a1v1 + a2v2 + a3v3
v = 11111111 + 01010101 + 00001111 = 10100101
❏ On comparing with r=10101101 we can find the error bit which is the fifth
bit.
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References❏ Reed-Muller Error Correcting Codes by Ben Cooke:❏ http://citeseerx.ist.psu.edu/viewdoc/download?doi=10.
1.1.208.440&rep=rep1&type=pdf❏ Error Correction Code by Todd K Moon❏ Hamming Block Codes:
http://web.udl.es/usuaris/carlesm/docencia/xc1/Treballs/Hamming.Treball.pdf
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End of the presentation
ThankYou
Akshit Jain 2013124Nitin Varun 2013070