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Error Detection and Correction-Data Communication Systems-Assignment Solution, Exercises of Data Communication Systems and Computer Networks

Birla Institute of Technology and ScienceData Communication Systems and Computer Networks

This file contains solution to problems related Data Communication Systems. Mr. Prajin Ahuja assigned task at Birla Institute of Technology and Science. Its main points are: Error, Detection, Correction, Single, Bit, Forward, Retransmission, Hamming, Distance, Vulnerable, Codework

Typology: Exercises

2011/2012

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CHAPTER 10
Error Detection and Correction
Solutions to Odd-numbered Review Questions and Exercises
Review Questions
1. In a single bit error only one bit of a data unit is corrupted; in a burst error more
than one bit is corrupted (not necessarily contiguous).
3. In forward error correction, the receiver tries to correct the corrupted codeword;
in error detection by retransmission, the corrupted message is discarded (the
sender needs to retransmit the message).
5. The Hamming distance between two words (of the same size) is the number of
differences between the corresponding bits. The Hamming distance can easily be
found if we apply the XOR operation on the two words and count the number of 1s
in the result. The minimum Hamming distance is the smallest Hamming distance
between all possible pairs in a set of words.
7.
a. The only relationship between the size of the codeword and dataword is the one
based on the definition: n = k + r., where n is the size of the codeword, k is the
size of the dataword, and r is the size of the remainder.
b. The remainder is always one bit smaller than the divisor.
c. The degree of the generator polynomial is one less than the size of the divisor.
For example, the CRC-32 generator (with the polynomial of degree 32) uses a
33-bit divisor.
d. The degree of the generator polynomial is the same as the size of the remainder
(length of checkbits). For example, CRC-32 (with the polynomial of degree 32)
creates a remainder of 32 bits.
9. At least three types of error cannot be detected by the current checksum calcula-
tion. First, if two data items are swapped during transmission, the sum and the
checksum values will not change. Second, if the value of one data item is increased
(intentionally or maliciously) and the value of another one is decreased (intention-
ally or maliciously) the same amount, the sum and the checksum cannot detect
these changes. Third, if one or more data items is changed in such a way that the
change is a multiple of 216 โˆ’ 1, the sum or the checksum cannot detect the changes.
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1

CHAPTER 10

Error Detection and Correction

Solutions to Odd-numbered Review Questions and Exercises

Review Questions

  1. In a single bit error only one bit of a data unit is corrupted; in a burst error more than one bit is corrupted (not necessarily contiguous).
  2. In forward error correction , the receiver tries to correct the corrupted codeword; in error detection by retransmission , the corrupted message is discarded (the sender needs to retransmit the message).
  3. The Hamming distance between two words (of the same size) is the number of differences between the corresponding bits. The Hamming distance can easily be found if we apply the XOR operation on the two words and count the number of 1s in the result. The minimum Hamming distance is the smallest Hamming distance between all possible pairs in a set of words.
  4. a. The only relationship between the size of the codeword and dataword is the one based on the definition: n = k + r ., where n is the size of the codeword, k is the size of the dataword, and r is the size of the remainder. b. The remainder is always one bit smaller than the divisor. c. The degree of the generator polynomial is one less than the size of the divisor. For example, the CRC-32 generator (with the polynomial of degree 32) uses a 33-bit divisor. d. The degree of the generator polynomial is the same as the size of the remainder (length of checkbits). For example, CRC-32 (with the polynomial of degree 32) creates a remainder of 32 bits.
  5. At least three types of error cannot be detected by the current checksum calcula- tion. First, if two data items are swapped during transmission, the sum and the checksum values will not change. Second, if the value of one data item is increased (intentionally or maliciously) and the value of another one is decreased (intention- ally or maliciously) the same amount, the sum and the checksum cannot detect these changes. Third, if one or more data items is changed in such a way that the change is a multiple of 2^16 โˆ’ 1, the sum or the checksum cannot detect the changes.

Exercises

  1. We can say that (vulnerable bits) = (data rate) ร— (burst duration)

Comment: The last example shows how a noise of small duration can affect so many bits if the data rate is high.

  1. The codeword for dataword 10 is 101. This codeword will be changed to 010 if a 3-bit burst error occurs. This pattern is not one of the valid codewords, so the receiver detects the error and discards the received pattern.
  2. a. d (10000, 00000) = 1 b. d (10101, 10000) = 2 c. d (1111, 1111) = 0 d. d (000, 000) = 0 Comment: Part c and d show that the distance between a codeword and itself is 0.
  3. a. 01 b. error c. 00 d. error
  4. We check five random cases. All are in the code.
  5. We show the dataword, codeword, the corrupted codeword, the syndrome, and the interpretation of each case: a. Dataword: 0100 โ†’ Codeword: 0100011 โ†’ Corrupted: 1100011 โ†’ s 2 s 1 s 0 = 110 Change b 3 (Table 10.5) โ†’ Corrected codeword: 0100011 โ†’ dataword: 0100 The dataword is correctly found. b. Dataword: 0111 โ†’ Codeword: 0111001 โ†’ Corrupted: 0011001 โ†’ s 2 s 1 s 0 = 011 Change b 2 (Table 10.5) โ†’ Corrected codeword: 0111001 โ†’ dataword: 0111 The dataword is correctly found. c. Dataword: 1111 โ†’ Codeword: 1111111 โ†’ Corrupted: 0111110 โ†’ s 2 s 1 s 0 = 111 Change b 1 (Table 10.5) โ†’ Corrected codeword: 0101110 โ†’ dataword: 0101 The dataword is found, but it is incorrect. C(7,4) cannot correct two errors.

a. vulnerable bits = (1,500) ร— (2 ร— 10 โˆ’^3 ) = 3 bits b. vulnerable bits = (12 ร— 103 ) ร— (2 ร— 10 โˆ’^3 ) = 24 bits c. vulnerable bits = (100 ร— 103 ) ร— (2 ร— 10 โˆ’^3 ) = 200 bits d. vulnerable bits = (100 ร— 106 ) ร— (2 ร— 10 โˆ’^3 ) = 200,000 bits

I. (1st) โŠ• (2nd) = (2nd) II. (2nd) โŠ• (3th) = (4th) III. (3rd) โŠ• (4th) = (2nd) IV. (4th) โŠ• (5th) = (8th) V. (5th) โŠ• (6th) = (2nd)

  1. Figure 10.2 shows the checksum to send (0x0000). This example shows that the checksum can be all 0s. It can be all 1s only if all data items are all 0, which means no data at all.

Figure 10.1 Solution to Exercise 31

Figure 10.2 Solution to Exercise 33

Codeword

x^7 +^ x^5 +^ x^2 +^ x^ +^1

x^7 +^ x^4 +^ x^3 +^ x^ +^1 x 4 +^ x^2 +^ x^ +^1 x^11 +^ x^9 +^ x^6 +^ x^5 +x^4

x 11 +^ x 9 +x 6 + x 5 + x 4 +

x 11 +^ x 9 +^ x 8 +x 7 x 8 +x 7 + x 6 + x^5 +x 4 x 8 +^ x 6 +^ x^5 +x 4 x 7 x 7 +^ x^5 +^ x 4 +x^3 x 5 +^ x 4 +x^3 x 5 +x^3 +x^2 +x xx 4 4 ++^ xx 2 2 + (^) + xx + 1 1 1

Dataword

Sender

Quotient Divisor

Remainder

Codeword

x^7 +^ x 5 +^ x^2 +^ x^ +^1

x^4 x 3 x 1 1

x 7 + + + + x^4 +^ x^2 +^ x^ +^1 x^11 +x^9 +^ x^6 +^ x^5 +x^4 +

x 11 +^ x^9 +^ x 6 +^ x^5 +^ x 4 +

x 11 +x^9 +x^8 +x 7 x 8 + x 7 + x^6 + x^5 +x 4 x 8 +^ x^6 +^ x^5 +x 4 x^7 x 7 +^ x^5 +x^4 +x^3 xx^5 5 +x^4 ++xx 33 + x (^2) +x x 4 +^ x^2 +x x 4 x^2 x 1 0

1

1

Dataword

Quotient Divisor

Remainder

Receiver

Checksum (initial) Sum

4 5 6 7 B A 9 8

F F F F 0 0 0 0 Checksum (to send)

0 0 0 0