Class presentation on some problem sets from the exercise

1. University of Dhaka
DEPARTMENT OF COMPUTER SCIENCE AND ENGINEERING,
1Data And Telecommunications
Presentation on:
CHAPTER 10 (ERROR DETECTION AND CORRECTION)
By: Md. Al – Zihad
Roll: 35
CSEDU 20th Batch
To: Professor Dr. Md. Abdur Razzaque
CSEDU
University of Dhaka

2. Contents
6/7/2016
2
• General Discussion of CRC
• Figure out the problems assigned
• Developing a perfect solution for each of them

3. Cyclic 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.
6/7/2016
3

4. Problem 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?
6/7/2016
4

5. Resources
6/7/2016
5

6. Problem 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?
6/7/2016
6

7. Detecting Single Bit Error
6/7/2016
7
Lets Assume,
Dataword = d(x)
Codeword = c(x)
Error = e(x)
So,
Received Codeword = c(x)+e(x)
Generator = g(x)

8. Detecting Single Bit Error Cont…
6/7/2016
8
• 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

9. Detecting Single Bit Error Cont…
6/7/2016
9
Both In,
The generator has more than one term and the coefficient of x0 is
1, all single errors can be caught
and

10. Detecting Burst Error
6/7/2016
10
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.

11. Detecting Burst Error Cont…
6/7/2016
11
So our concern is:
(x^ (j-i) + . . . + 1) / ( x^r + . . . +1)
If the division has non zero remainder, error can be detected

12. Detecting Burst Error Cont…
6/7/2016
12
• 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.

13. Detecting Burst Error Cont…
6/7/2016
13
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

14. Detecting Burst Error Cont…
6/7/2016
14
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

15. Probability of missing detection of Burst error
6/7/2016
15
• 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)

16. Detecting Burst Error Cont…
6/7/2016
16
Here in problem 27(c),
So, L = r + 1
So, 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.

17. Detecting Burst Error Cont…
6/7/2016
17
Here in problem 28(c),
So, L = r + 1
So, 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.

18. Probability of missing detection of Burst error
6/7/2016
18
• 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)

19. Detecting Burst Error Cont…
6/7/2016
19
Here in problem 27(d),
So, L > r + 1
So, 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.

20. Detecting Burst Error Cont…
6/7/2016
20
Here in problem 28(c),
So, L > r + 1
So, 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.

21. Lesson Learned
6/7/2016
21
• 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

22. 6/7/2016
22
Thanks
Any Questions?