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05/02/2023
1
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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12Detecting Burst Error Cont…
• 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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13Detecting Burst Error Cont…
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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14Detecting Burst Error Cont…
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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16Detecting Burst Error Cont…
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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17Detecting Burst Error Cont…
Here in problem 28(c),
So, L = r + 1So, it will miss to detect (1/2)^(32-1) burst errors.
• r = 32• L = 33
Probability of detecting burst error of size 33 = 1 – (1/2)^31.
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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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19Detecting Burst Error Cont…
Here in problem 27(d),
So, L > r + 1So, it will miss to detect (1/2)^(8) burst errors.
• r = 8• L = 15
Probability of detecting burst error of size 15 = 1 – (1/2)^8.
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20Detecting Burst Error Cont…
Here in problem 28(c),
So, L > r + 1So, it will miss to detect (1/2)^(32) burst errors.
• r = 32• L = 55
Probability of detecting burst error of size 33 = 1 – (1/2)^32.
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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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22
ThanksAny Questions?