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UNIT 5Problems
1. The generator matrix (G) of a particular (7, 4) linear block code is given by G
= 1 1 1
1 0 1 0 1 1 1 1 0
1 0 0
0 1 0
0 0 1
0
0
00 0 0 1
i) Find the parity check matrix (H).ii) List all the code vectors.iii) compute the syndrome for the received vector 1101101. Is this a valid
code.(16)
Solution :
i) List all the code vectors.Code words.X = DG = [0000]
= [0000]1 1 1 1 0 1 0 1 1 1 1 0
1 0 0
0 1 0
0 0 1
0
0
00 0 0 1
= [0000] = 0000000
= [0001]
= 1100001
= [0010] = 0110010 = [0011] = 1010011 = [0100] = 1010100 = [0101] = 1100110 = [0110] = 1100001 = [0111] = 0000111
= [1000]
= 1111000
= [1001] = 0011001 = [1010] = 1001010 = [1011] = 0101011 = [1100] = 0101100 = [1101] = 1001101
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= [1110] = 0011110 = [1111] = 1111111
= 0000000,
= 1100001,
= 0110010,
= 1010011,
= 1010100,
= 0110101, = 1100110, = 0000111, = 1111000,= 0011001, = 1001010, = 0101011, = 0101100,= 1001101, = 0011110, = 1111111
ii) The parity check matrix (P).
=
1 1 1 1 0 1 0 1 1
1 1 0
1 0 0
0 1 0
0 0 1
0
0
0
0 0 0 1
=
1 1 1 1 0 1 0 1 1 1 1 0
1 0 0
0 1 0
0 0 1
0
0
00 0 0 1
Given n = 7, K = 4 where
= 1 1 1
1 0 1
0 1 11 1 0and
= 1 0 00 1 00 0 1
0
0
00 0 0 1
thus = [ ] = [ ]
= 1 0 00 1 00 0 1
= 1 1 01 0 1
1 1 1
1
1
0
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= 1 0 00 1 00 0 1
1 1 0
1 0 1
1 1 1
1
1
0
iii) Error detection:Code vector received r = 1101101
The combination for error detection r.HT
= 0. = 0[1101101] =
1 0 0
0 1 0
0 0 1
1 1 1
1 0 1
0 1 1
1 1 0
= [0 1 0]
. 0The code vector is not valid2. The generator matrix (G) of a particular (7, 4) linear block code is given by G
= 1 1 1 1 0 1 1 1 1 1 1 1
1 0 0
0 1 0
0 0 1
0
0
00 0 0 1
i) Find the parity check matrix (P).ii) List all the code vectors.iii) compute the syndrome for the received vector 1101101. Is this a valid
code.(16)
Solution :
i) List all the code vectors.
Code words.X = DG
= [0000]
= [0000]1 1 1 1 0 1 1 1 1 1 1 1
1 0 0
0 1 0
0 0 1
0
0
00 0 0 1
= [0000] = 0000000 = [0001] = 1110001 = [0010] = 1110010
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= [0011] = 0000011 = [0100] = 1010100 = [0101] = 0100101 = [0110] = 0100110
= [0111]
= 1010111
= [1000] = 1111000 = [1001] = 0001001 = [1010] = 0001010 = [1011] = 1111011 = [1100] = 0101100 = [1101] = 1011101 = [1110] = 1011110 = [1111] = 1111111 = 0000000, = 1110001, = 1110010, = 0000011, = 1010100,
= 0100101,
= 0100110,
= 1010111,
= 1111000,
= 0001001, = 0001010, = 1111011, = 0101100,= 1011101, = 1011110, = 1111111ii) The parity check matrix (P).
= 1 1 1 1 0 1 1 1 1 1 1 1
1 0 0
0 1 0
0 0 1
0
0
00 0 0 1
= 1 1 1
1 0 1 1 1 1 1 1 1
1 0 0
0 1 00 0 1
0
00
0 0 0 1
Given n = 7, K = 4 where
= 1 1 11 0 11 1 11 1 1
and
= 1 0 0
0 1 0
0 0 1
0
0
00 0 0 1
thus = [ ] = [ ]
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= 1 0 00 1 00 0 1
= 1 1 1
1 0 1
1 1 11 1 1
= 1 1 11 0 1
1 1 1
1
1
1
= 1 0 00 1 0
0 0 1
1 1 1
1 0 1
1 1 1
1
1
1
iii) Error detection:Code vector received r = 1101101The combination for error detection r.HT = 0. = 0
[1101101] =
1 0 0
0 1 0
0 0 1
1 1 1
1 0 1
1 1 11 1 1
= [0 1 1]. 0The code vector is not valid
3. The generator matrix (G) of a particular (7, 4) linear block code is given by G
=
1 1 1 1 0 1
1 1 1 1 0 1 1 0 0
0 1 0
0 0 1
0
0
00 0 0 1
i) Find the parity check matrix (P).ii) List all the code vectors.iii) compute the syndrome for the received vector 1101101. Is this a valid
code.
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Solution :
i) List all the code vectors. - 8marksCode words.X = DG
= [0000] = [0000]1 1 1 1 0 1 1 1 1 1 0 1
1 0 0
0 1 0
0 0 1
0
0
00 0 0 1
= [0000] = 0000000 = [0001] = 1010001 = [0010] = 1110010 = [0011] = 0100011
= [0100]
= 1010100
= [0101]
= 0000101
= [0110] = 0100110 = [0111] = 1110111 = [1000] = 1111000 = [1001] = 0101001 = [1010] = 0001010 = [1011] = 1011011 = [1100] = 0101100 = [1101] = 1111101 = [1110] = 1011110
= [1111]
= 0001111
= 0000000, = 1010001, = 1110010, = 0100011, = 1010100,= 0000101, = 0100110, = 1110111, = 1111000,= 0101001, = 0001010, = 1011011, = 0101100,= 1111101, = 1011110, = 0001111
ii) The parity check matrix (P). - 4marks
= 1 1 1 1 0 1 1 1 1 1 0 1
1 0 0
0 1 0
0 0 1
0
0
00 0 0 1
= 1 1 1 1 0 1 1 1 1 1 0 1
1 0 0
0 1 0
0 0 1
0
0
00 0 0 1
Given n = 7, K = 4 where
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= 1 1 11 0 11 1 11 0 1
and
= 1 0 00 1 00 0 1
00
00 0 0 1
thus = [ ] = [ ]
= 1 0 00 1 00 0 1
= 1 1 1
1 0 1
1 1 11 0 1
= 1 1 11 0 11 1 1
1
0
1
=
1 0 0
0 1 0
0 0 1 1 1 1
1 0 1
1 1 1
1
0
1
iii) Error detection: `- 4marksCode vector received r = 1101101The combination for error detection r.HT = 0. = 0
[1101101] =
1 0 0
0 1 0
0 0 1
1 1 1
1 0 1
1 1 1
1 0 1
= [0 0 1]. 0The code vector is not valid
