Tut4_sol - EE 4212 Information and Coding Solution to Tutorial 4 1 A(n,k Linear Block Code 0 1 G= 1 1 1 1 1 0 0 0 0 1 0 1 0 0 1 0 0 0 1 0 1 1 0 0 0

Info iconThis preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon
1 EE 4212 Information and Coding Solution to Tutorial 4 1. A ( n , k ) Linear Block Code = 1 0 0 0 1 1 1 0 1 0 0 0 1 1 0 0 1 0 1 0 1 0 0 0 1 1 1 0 G (a) n = 7 and k = 4. As the dimension of G is 4 x 7, this is a (7,4) Hamming code. (b) There are a total of 2 4 = 16 valid code vectors. m U = mG m U = mG 0000 0 0 0 0 0 0 0 1000 0 1 1 1 0 0 0 0001 1 1 1 0 0 0 1 1001 1 0 0 1 0 0 1 0010 1 1 0 0 0 1 0 1010 1 0 1 1 0 1 0 0011 0 0 1 0 0 1 1 1011 0 1 0 1 0 1 1 0100 1 01 0 1 0 0 1100 1 1 0 1 1 0 0 0101 0 1 0 0 1 0 1 1101 0 0 1 1 1 0 1 0110 0 1 1 0 1 1 0 1110 0 0 0 1 1 1 0 0111 1 0 0 0 1 1 1 1111 1 1 1 1 1 1 1 (c) = = 1 1 1 0 1 1 1 0 1 1 1 0 1 0 0 0 1 0 0 0 1 1 0 1 1 1 0 0 1 1 0 1 0 1 0 1 1 1 0 0 0 1 T H H
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
2 (d) The code can identify 2 n-k = 8 error patterns, including the all-zero pattern. Error Pattern
Background image of page 2
Image of page 3
This is the end of the preview. Sign up to access the rest of the document.

This note was uploaded on 02/26/2011 for the course EE 4011 taught by Professor Wadewe during the Spring '11 term at City University of Hong Kong.

Page1 / 3

Tut4_sol - EE 4212 Information and Coding Solution to Tutorial 4 1 A(n,k Linear Block Code 0 1 G= 1 1 1 1 1 0 0 0 0 1 0 1 0 0 1 0 0 0 1 0 1 1 0 0 0

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online