**1 Answer**

written 2.5 years ago by | • modified 2.5 years ago |

7-bits even parity Hamming code is received as **1100010**

In 7-bits Hamming code **4-bits (D3, D5, D6 & D7)** represents ** data bits** and

**3-bits (P1, P2, & P4)**are

*parity bits.*This 7-bit Hamming code is represented as follows:

Bit Position | 1 | 2 | 3 | 4 | 5 | 6 | 7 |
---|---|---|---|---|---|---|---|

Hamming Code Representation |
P1 |
P2 |
D3 |
P4 |
D5 |
D6 |
D7 |

Received Hamming Code |
1 | 1 | 0 | 0 | 0 | 1 | 0 |

To check whether the received Hamming code is correct or not let's perform the following operations:

Take the bits present at the below-mentioned places

** Parity bit-1** covers all the bits positioned at bit-position

**1, 3, 5, 7, 9, 11,.. etc.**

Therefore,

*C1 = Even_Parity(1, 3, 5, 7) = 1 0 0 0 = 1 (Odd number of 1,s, which means error is present)*

** Parity bit-2** covers all the bits positioned at

**2, 3, 6, 7, 10, 11,... etc.**

Therefore,

*C2 = Even_Parity(2, 3, 6, 7) = 1 0 1 0 = 0 (Even number of 1's, which means No error is present)*

** Parity bit-4** covers all the bits positioned at

**4–7, 12–15, 20–23,... etc.**

Therefore,

*C4 = Even_Parity(4, 5, 6, 7) = 0 0 1 0 = 1 (Odd number of 1's, which means error is present)*

Therefore,

**C = C4 C2 C1 = 1 0 1** = It represent **decimal number 5**

That means a **Single bit error** is present at the $5^{th}$ position of *received Hamming code 1100010.*

Now, remove the $5^{th}$ position single bit error.

The $5^{th}$ bit holds 0 which creates an error in a Hamming encoded 7-bit number. Hence, to remove this error 0 is replaced with 1.

Therefore,