data and telecommunication problem

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University of DhakaDEPARTMENT OF COMPUTER SCIENCE AND ENGINEERING,

Data And Telecommunications

Presentation on:CHAPTER 10 (ERROR DETECTION AND CORRECTION)

By: Md. Al – ZihadRoll: 35CSEDU 20th Batch

To: Professor Dr. Md. Abdur RazzaqueCSEDUUniversity of Dhaka

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2Contents

• General Discussion of CRC

• Figure out the problems assigned

• Developing a perfect solution for each of them

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3Cyclic Redundancy Check (CRC)

Error detection mechanism

Simple to implement in binary hardware

Calculated by performing a modulo 2 division of the data by a generator

polynomial and recording the remainder after division.

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4Problem 27

27. Referring to the CRC-8 polynomial in Table 10.7, answer the following questions:

a. Does it detect a single error? Defend your answer. b. Does it detect a burst error of size 6? Defend your answer. c. What is the probability of detecting a burst error of size 9? d. What is the probability of detecting a burst error of size 15?

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5Resources

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6Problem 28

28. Referring to the CRC-32 polynomial in Table 10.7, answer the following questions:

a. Does it detect a single error? Defend your answer. b. Does it detect a burst error of size 16? Defend your answer. c. What is the probability of detecting a burst error of size 33? d. What is the probability of detecting a burst error of size 55?

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7Detecting Single Bit Error

Lets Assume,

Dataword = d(x)Codeword = c(x)Error = e(x)So,

Received Codeword = c(x)+e(x)Generator = g(x)

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8Detecting Single Bit Error Cont…

• If a single-bit error is caught, then xi is not divisible by g(x).

• If the generator has more than one term and the coefficient of x0

is 1, all single errors can be caught

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9Detecting Single Bit Error Cont…

Both In,

The generator has more than one term and the coefficient of x0 is 1, all single errors can be caught

and

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10Detecting Burst Error

e(x) = (x^j + . . . + x^i)Or,e(x) = x^i (x^ (j-i) + . . . + 1)

If our generator can detect a single error (minimum condition for a generator), then it cannot divide xi.

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11Detecting Burst Error Cont…

So our concern is: (x^ (j-i) + . . . + 1) / ( x^r + . . . +1)

If the division has non zero remainder, error can be detected

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• If j - i < r, the remainder can never be zero.• all burst errors with length smaller than or equal to the number

of check bits r will be detected.

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Here in problem 27(b),

• This is a 8 degree polynomial.• Number of check bit is 8

So, it will obviously detect burst error size of 6 as 6 < 8

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Here in problem 27(b),

• This is a 32 degree polynomial.• Number of check bit is 32

So, it will obviously detect burst error size of 16 as 16 < 32

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15Probability of missing detection of Burst error

• If, j - i = r

• Then, L – 1 = r

• So, L = r + 1

Here,r is polynomial size

L is for burst error size

In this case, Probability = (1/2)^(r-1)

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Here in problem 27(c),

So, L = r + 1So, it will miss to detect (1/2)^(8-1) burst errors.

• r = 8• L = 9

Probability of detecting burst error of size 9 = 1 – (1/2)^7.

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So, L = r + 1So, it will miss to detect (1/2)^(32-1) burst errors.

• r = 32• L = 33


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18Probability of missing detection of Burst error

• If, j - i > r

• Then, L – 1 > r

• So, L > r + 1

Here,r is polynomial size

L is for burst error size

In this case, Probability = (1/2)^(r)

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Here in problem 27(d),

So, L > r + 1So, it will miss to detect (1/2)^(8) burst errors.

• r = 8• L = 15


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So, L > r + 1So, it will miss to detect (1/2)^(32) burst errors.

• r = 32• L = 55


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21Lesson Learned

• All g(x) having multiple terms and X0 = 1, all single bit error will be detected

• All burst errors with L < = r will be detected.

• All burst errors with L = r + 1 will be detected with probability 1 - (1/2)^(r-1)

• All burst errors with L > r + 1 will be detected with probability 1- (1/2)^r

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ThanksAny Questions?

data and telecommunication problem

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