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2001 년 6 월 2 일

변 우 성연세대학교 전기전자공학과

부호 및 정보이론 연구실

Introduction and Explanationof Exercise #5 & #6

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Table of Contents Exercise #5

– Introduction and related Theorem– Explanation of exercise #5– Discussion

Exercise #6– Introduction and related Theorem– Explanation of exercise #6– Discussion

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Exercise #5 Introduction

Suppose a binary sequence of length 31 is selected at random. Calculate the expected number of 0-runs and 1-runs of lengths 1, 2, 3, 4, and 5. Compare your result to Theorem 10.2.

Theorem 10.2 Run-length property

Approach Make a binary random sequence of length 31 by C programming Find and Calculate the number of 0-runs and 1-runs of each length Compare

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Exercise #5 Explanation(1/2)

Binary random sequence (made by C)– “0000011100001100011111010101101”

– Result

length 0- runs 1- runs

1 4 3

2 0 2

3 1 1

4 1 0

5 1 1

Totals : 7 7

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Exercise #5 Explanation(2/2)

Result to Theorem 10.2

length 0- runs 1- runs

1 4 4

2 2 2

3 1 1

4 1 0

5 0 1

Totals : 8 8

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Exercise #5 Discussion

The number of 0-runs and 1-runs different between m-sequence and arbitrary random sequence

In conclusion, Run-length property is exactly applied to m-sequence

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Exercise #6 Introduction

According to Theorem 10.1, each of the 31 different nonzero 5-grams appears once in the m-sequence of length 31 given in Example 10.1. In this problem, we ask you to consider the 6-grams of that sequence.

A. How many of the 64 possible 6-grams appear in the length 31 m-sequence of Example 10.1?

B. Show that there exists a fixed length 6 vector a with the property that the 6-gram x appears in the m-sequence if and only if x is not zero and xㆍ a=0. Find a explicitly.

Theorem 10.1 Window property

Approach This problem is related to generate the m-sequence by the linear

feedback shift registers

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Exercise #6 Explanation of A

If a repeated 6-gram exists, the period of sequence is sooner than 31, because of recurrence relation

In conclusion, the number of 6-gram is 31

Explanation of B(1/2) Vector a is a fixed length 6 vector and satisfies x ㆍ a=0 Here vector x is a 6-gram vector x = (x1, x2, x3, x4, x5, x6)

By the way, x ㆍ a = a1x1+a2x2+a3x3+a4x4+a5x5+a6x6 and this is

the recursion

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Exercise #6 Explanation of B(2/2)

Under circuit to implement the recursion x6 = x3 + x1, whose characteristic polynomial is the primitive polynomial x5+x2+1

x6 = x3 + x1 x6 + x3 + x1=0

In conclusion, a6 = 1 , a3 = 1 , a1 = 1 a = (1, 0, 1, 0, 0, 1)

m-sequenceoutput(m=5)

x6

x5 x1x2x3x4

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Exercise #6 Discussion

Check the result

except all zero

Total number of 6-grams

If a fixed length m+1 vector a of m-sequence satisfies x ㆍ a=0,

vector a is the set of coefficient of recursion

a 1 0 1 0 0 1 # of event

1 X 1 X X 0 23

1 X 0 X X 1 23

0 X 1 X X 1 23

x

0 X 0 X X 0 23- 1

234- 1=31