Fundamentals of Error Correction MCQ || Error Correction Questions and Answers

11. A linear Hamming code is used to map 4-bit messages to 7-bit code words. The encoder mapping is linear. If the message 0001 is mapped to the codeword 0000111, and the message 0011 is mapped to the codeword 1100110, then the message 0010 is mapped to

  1. 0010011
  2. 1100001
  3. 1111000
  4. 1111111

Answer.2. 1100001

Explanation

  • A linear code is an error-correcting code.
  • Any linear combination of codewords is also a codeword.
Message  Codeword
0001

0011

0010

0000111

1100110

?

0010 = (0001) ⊕ → EXOR (0011)

To get the code word, Exor the two code words.

(0000111) ⊕ (1100110)

= 1100001

 

12. Which has a continuous transmission?

  1. Asynchronous
  2. Synchronous
  3. Asynchronous & Synchronous
  4. None of the mentioned

Answer.2. Synchronous

Explanation

Synchronous has continuous transmission where as asynchronous have sporadic transmission.

Synchronous transmission is a data transfer method that is characterized by a continuous stream of data in the form of signals which are accompanied by regular timing signals which are generated by some external clocking mechanism meant to ensure that both the sender and receiver are synchronized with each other.

 

13. A (7, 4) Hamming Code contains ______ parity bits.

  1. 3
  2. 4
  3. 2
  4. 1

Answer.1. 3

Explanation

Hamming codes are linear block codes. The family of (n, k) hamming codes for q ≥ 3 is defined by the following expressions:

1. Number of Block diagrams (n) = 2q – 1

2. Number of message bits (k) = 2q – q – 1

3. Number of Parity bits (q) = (n – k)

Calculation:

Given: Hamming codes (7, 4)

So, Number of Parity bits = n – k

q = 3

Note: Minimum number of Parity bits for (n, k) family is 3.    [∵ (q ≥ 3)]

 

14. Select the INCORRECT statement about TDMA from the following options.

  1. High transmission rates
  2. Discontinuous data transmission
  3. Single carrier frequency for single user
  4. No requirement of duplexers

Answer.3. Single carrier frequency for single user

Explanation

Time-division multiple access (TDMA):

  • This is a channel access method for shared-medium networks.
  • It allows several users to share the same frequency channel by dividing the signal into different time slots.
  • The users transmit in rapid succession, one after the other, each using its own time slot. This allows multiple stations to share the same transmission medium (e.g. radio frequency channel) while using only a part of its channel capacity.
  • TDMA uses different time slots for transmission and reception thus duplexers are not required.

 

15. In synchronous transmission, the receiver must stay synchronous for

  1. 4 bits
  2. 8 bits
  3. 16 bits
  4. 9 bits

Answer.4. 9 bits

Explanation

Synchronous transmission is a data transfer method that is characterized by a continuous stream of data in the form of signals which are accompanied by regular timing signals which are generated by some external clocking mechanism meant to ensure that both the sender and receiver are synchronized with each other.

 

16. The Hamming distance between two code words C1 = [1011 0101] and C2 = [0111 1010] is

  1. 4
  2. 5
  3. 3
  4. 6

Answer.4. 6

Explanation

The Hamming distance between two codewords is the number of bit positions in which they differ.

Given:

C1 = [1011 0101]

C2 = [0111 1010]

Since the two codewords differ in 6-bit positions as indicated by bold letter bits, the hamming distance is 6.

 

17. If one of the code words of a Hamming (7, 4) code is 0001011, which of the following cannot be the valid codeword in the same group?

  1. 0011101
  2. 0101100
  3. 0011010
  4. 1110100

Answer.3. 0011010

Explanation

The minimum bits change between any valid Hamming codes is 3.

Analysis:

The change of bits between the given code: 0001011 and option C code: 0011010 is two bits change only.

So 0011010 can’t be a valid Hamming code of the same group.

 

18. How error detection and correction is done?

  1. By passing it through equalizer
  2. By passing it through filter
  3. By amplifying it
  4. By adding redundancy bits

Answer.4. By adding redundancy bits

Explanation

The error can be detected and corrected by adding additional information that is by adding redundancy bits. Forward error correction works by adding redundant bits to a bitstream to help the decoder detect and correct some transmission errors without the need for retransmission. The name forward stems from the fact that the flow of data is always in the forward direction (i.e., from the encoder to the decoder).

 

19. CDMA is a

  1. Direct Sequence Spread Spectrum system
  2. Frequency Sharing system
  3. Tie hopping system
  4. Time-sharing system

Answer.1. Direct Sequence Spread Spectrum system

Explanation

  • CDMA stands for Code Division Multiple Access.
  • CDMA (Code Division Multiple Access) uses a digital modulation technique called Direct Sequence spread spectrum which spreads voice data over a very wide spectrum using user or cell-specific pseudo-random codes.
  • It is a wireless technology used in the transmission of signals from places with high security and Noise reduction.
  • The principle of the spread spectrum is used to work with CDMA.
  • CDMA is not frequency specific to each user, instead, every channel uses the full available spectrum.
  • Each user in a CDMA system uses a different code to modulate their signal.

 

20. A digital transmission system uses a (7, 4) systematic linear Hamming code for transmitting data over a noisy channel. If three of the message-codeword pairs in this code (mi ; ci), where ci is the codeword corresponding to the ith message mi, are known to be (1 1 0 0 ; 0 1 0 1 1 0 0), (1 1 1 0 ; 0 0 1 1 1 1 0) and (0 1 1 0 ; 1 0 0 0 1 1 0), then which of the following is a valid codeword in this code?

  1. 1 0 1 1 0 1 0
  2. 0 1 1 0 1 0 0
  3. 0 0 0 1 0 1 1
  4. 1 1 0 1 0 0 1

Answer.3. 0 0 0 1 0 1 1 

Explanation

Given (7, 4) linear Hamming code.

(mi, ci) given are:

(1100 ; 0101100)

(1110 ; 0011110)

(0110 ; 1000110)

Clearly, we can observe that the first 3 bits are parity bits and the last 4 bits are message bits.

P0 P1 P2 d0 d1 d2 d3
C1 → 0 1 0 1 1 0 0
C2 → 0 0 1 1 1 1 0
C3 → 1 0 0 0 1 1 0

Observe when P0, P1, and P2 are one, and write their combination, i.e.

P0 = d0 ⊕ d1 ⊕ d3

P1 = d1 ⊕ d2 ⊕ d3

P2 = d0 ⊕ d1 ⊕ d2

Message bits are:

$\begin{array}{*{20}{c}}1&0&1&1\\{{d_0}}&{{d_1}}&{{d_2}}&{{d_3}}\end{array}$

P0 = d0 ⊕ d1 ⊕ d3 = 0

P1 = d1 ⊕ d2 ⊕ d3 = 0

P2 = d0 ⊕ d1 ⊕ d2 = 0

So the valid codeword is 0001011

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