There is a tutorial video in the Google Drive under Lecture Videos called \"Hamm

Question

There is a tutorial video in the Google Drive under Lecture Videos called "Hamming Code Tutorial," about 15 minutes in length, that will take you through using the tables in this exercise. It's located here: https:// drive 0B9Ors6ez Cg TE9xL, WidhUDNHWUU 1. The first problem is inspired from the reading section assigned at the end of HW #1 from Hennessey and Patterson: Hamming Code 5.5 from Comp Org and Design (2014).pdf found in the Google Drive folder Textbook Chapters, or here: https:// drive Cg WG diRIJxdvi mc Your task is to repeat the work done on page 421, but with 63 You'll need 4 more parity bits, so the problem will flow exactly like the one on page 421. Your final solution should be entered in the last table. You should use the intermediate table to demonstrate that you know how to work through the problem. Start with the following table Data To Encode 3 4 5 6 7 8 Data Bit Position Data Bits Bit Position 1 2 3 4 5 6 7 8 9 10 11 12 p1 p2 1 4 d5 d6 d7 d8 d2 d3 d4 pl p2 p4 p8 p16 p32 p64 Perform the following steps: i. Input your data values in the "Data To Encode row in yellow. You will enter the value in binary. The goal of this table is to help you calculate the values in the vertical yellow column "Values to Embed."

Explanation / Answer

Lets consider a data which is to be encoded by hamming code

Data to be Encoded -

Now we put this data in the green " data fiels " skipping the grey cell

Now calculating the value of p1 by hamming rule :

p1 - p1 checks 1 bit and skip 1 bit and so on . so value of p1 depends upon 1,3,5,7,9,11

? 1 0 1 1 1

the number of ones are even so place p1 =0

Now calculating the value of p2 by hamming rule :

p2 ---> p2 checks 2 bit and skip 2 bit and so on . so value of p1 depends upon p2,d3,d6,d7,d10,d11

? 1 0 1 0 1

the number of ones are odd so place p2=1

Now calculating the value of p4 by hamming rule :

p4 ---> p4 checks 4 bit and skip 4 bit and so on . so value of p4 depends upon p4,d5,d6,d7,d12

? 0 0 1 0

the numbers of one are odd so place p4=1

Now calculating the value of p8 by hamming rule :

p8 ---> p8 checks 8 bit and skip 8 bit and so on . so value of p8 depends upon p8,d9,d10,d11,d12

? 1 0 1 0

the numbers of ones are even so place p8=0

Hence the final codeword wil be ---> 011100101010

1 0 0 1 1 0 1 0

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