tut7_sol - EE 4212 Information and Coding Solution to...

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1 Input bit 0 01 11 00 11 00 01 10 10 a =00 b =10 c =01 d =11 Input bit 1 EE 4212 Information and Coding Solution to Tutorial 7 A convolution code with K =3, code rate=1/2, the connection polynomials are g 1 ( X ) = 1 + X + X 2 , g 2 ( X ) = 1 + X (a) Pictorial Diagram (b) (i) State Diagram u 2 u 1
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2 (b) (ii) Tree Diagram (b) (iii) Trellis Diagram (c) Let’s use the state diagram for encoding, start from state a . 00 11 00 10 01 01 10 00 00 10 01 01 10 00 11 00 10 01 01 10 00 00 00 00 a a a a b b b a b c c d d c d 0 1 a =00 b =10 c =01 d =11 00 00 11 t 1 t 2 t 5 00 11 t 3 t 4 00 11 11 11 11 11 10 10 01 01 00 00 00 10 10 01 01 m = 0 1 1 0 0 0 0 2 flush zeros U = 00 11 00 01 10 00 00 a a b d ca a a
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3 (d) r = 11 10 10 10, feedback decoding using the tree diagram ( L =3) First block: 11 10 10 Metric of upper 4 paths = 4, 4, 4, 4 Metric of lower 4 paths = 1 , 3, 3, 1 Take lower parth, decoded bit = 1. Second block: 10 10 10 Metric of upper 4 paths = 2, 2, 4, 4 Metric of lower 4 paths = 3, 5, 3, 1 Take lower parth, decoded bit = 1. = U ˆ 11 0 0 10 10 single-bit error = m ˆ 1 1 1 1 10 00 11 00 10 01 01 10 00 00 10 01 01 00 11 00 10 01 01 10 00 00 00 00 a a a a b b b a b c c d d c d 0 1
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4 D 2 D D D D 2
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tut7_sol - EE 4212 Information and Coding Solution to...

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