Virtual Labs

Error Detection and Correction, Standard Array, Syndrome Decoding

1. Decode

00111

using the following standard array.

\\

a: Decoded codeword is

11000

. Explanation

Explanation

b: Decoded codeword is

01010

. Explanation

Explanation

c: Decoded codeword is

11111

. Explanation

Explanation

d: Decoded codeword is

00111

. Explanation

Explanation

2. Consider the following standard array. What is the error correction capability of this code?

\\

a: 1 Explanation

Explanation

b: 2 Explanation

Explanation

c: 3 Explanation

Explanation

d: None of the above Explanation

Explanation

3. Decode

1001011

using the following syndrome table.

\\

a: Decoded codeword is

1111111

Explanation

b: Decoded codeword is

1101001

Explanation

c: Decoded codeword is

1001011

Explanation

d: Data insufficient Explanation

Explanation

4. Consider the linear block code

\{0000, 0011, 1100, 1111 \}

and set of standard array tables given below. Select the correct option for standard array of this code for performing optimal decoding over the BSC

(p)

with

p<1/2

\\

a: Standard array 1 is correct. Explanation

Explanation

b: Standard array 2 is correct. Explanation

Explanation

c: Standard array 3 is correct. Explanation

Explanation

d: Standard arrays 2 and 3 are correct. Explanation

Explanation

5. Choose the incorrect statement.

a: Coset leaders determine the error correction capability of the code. Explanation

Explanation

b: Coset leaders determine the error detection capability of the code. Explanation

Explanation

c: Let

\mathbf{e}_4

and

\mathbf{e}_5

be the coset leaders of the 4th and 5th rows of a standard array of some linear block code respectively. Then the Hamming weight of

\mathbf{e}_4

should necessarily be less than or equal to the Hamming weight of

\mathbf{e}_5

. Explanation

Explanation

d: Let

\mathbf{e}_4

and

\mathbf{e}_5

be the coset leaders of the 4th and 5th rows of a standard array of some linear block code respectively. Then the Hamming weight of

\mathbf{e}_5

can be equal to the Hamming weight of

\mathbf{e}_4

. Explanation

Explanation