Problem 25
Question
Determine whether the given message is a code word in the Hamming \((7,4)\) code. If it is, decode it. If it is not, correct the single error and decode the corrected message. $$ \left.\begin{array}{lllllll} (0 & 1 & 1 & 1 & 0 & 0 & 1 \end{array}\right) $$
Step-by-Step Solution
Verified Answer
Corrected message is (0, 1, 0, 0).
1Step 1: Understand the Hamming (7,4) Code
Hamming (7,4) code is a type of error-correcting code that encodes 4 data bits into a 7-bit code word by adding 3 parity bits. The idea is to use the parity bits to detect and correct single-bit errors in the 7 bits.
2Step 2: Verify Parity Check Matrix
To check if the received message is a code word, we use the parity check matrix (H). The parity check matrix for Hamming (7,4) code is:\[ H = \begin{bmatrix} 0 & 1 & 1 & 1 & 1 & 0 & 0 \ 1 & 0 & 1 & 1 & 0 & 1 & 0 \ 1 & 1 & 0 & 1 & 0 & 0 & 1 \end{bmatrix} \]
3Step 3: Multiply Message by Parity Check Matrix
Multiply the message vector \[ v = \begin{bmatrix} 0 & 1 & 1 & 1 & 0 & 0 & 1 \end{bmatrix} \] by the parity check matrix H:\[ s = v \cdot H^T \]
4Step 4: Calculate Syndrome
You multiply the message vector with the transpose of the parity check matrix, \[ s = \begin{bmatrix} 0 & 1 & 1 & 1 & 0 & 0 & 1 \end{bmatrix} \cdot \begin{bmatrix} 0 & 1 & 1 \ 1 & 0 & 1 \ 1 & 1 & 0 \ 1 & 1 & 1 \ 1 & 0 & 0 \ 0 & 1 & 0 \ 0 & 0 & 1 \end{bmatrix} = \begin{bmatrix} 0 & 1 & 1 \end{bmatrix} \] The result is called the syndrome, which in this case is non-zero \([0, 1, 1]\), indicating an error.
5Step 5: Locate and Correct the Error
The syndrome \([0, 1, 1]\) corresponds to the error in the third position of the vector using the standard position table. Correct the third bit from 1 to 0. Thus, the corrected vector is \[ (0, 1, 0, 1, 0, 0, 1) \].
6Step 6: Decode the Corrected Code Word
After correcting the error, extract the original 4 data bits from the corrected vector. The 4 data bits are the first, second, fourth, and fifth bits of the 7-bit vector: \[ (0, 1, 0, 0) \].
Key Concepts
Parity Check MatrixSyndrome CalculationError Correction
Parity Check Matrix
An essential component of Hamming code is the parity check matrix, abbreviated as "H". This matrix is used to verify whether a given message is a code word in the Hamming code or if it contains errors. For the Hamming (7,4) code specifically, the parity check matrix is structured to handle 7 bits (which includes both data and parity bits) and helps in detecting single-bit errors.
- The matrix for Hamming (7,4) is a 3x7 matrix that identifies errors based on parity relationships.
- Each row in the matrix relates to a parity equation, ensuring that any received message stays within error detection criteria when multiplied with its transpose.
Syndrome Calculation
Syndrome calculation is the process used to detect errors in a message encoded with Hamming code. After receiving a 7-bit message, you multiply it by the transpose of the parity check matrix, H. The resulting vector, known as the syndrome, indicates whether an error exists and aids in locating it.
To calculate the syndrome:
This process quickly pinpoints the exact location of errors, allowing for efficient error correction without needing to re-transmit the entire message.
To calculate the syndrome:
- Take the received message vector, for example, \( \begin{bmatrix} 0 & 1 & 1 & 1 & 0 & 0 & 1 \end{bmatrix} \).
- Multiply it with the transpose of the parity check matrix H.
This process quickly pinpoints the exact location of errors, allowing for efficient error correction without needing to re-transmit the entire message.
Error Correction
Once the syndrome calculation reveals that an error is present, the next step involves correcting this error. Hamming codes are specifically designed to allow for single-bit error correction, making them highly efficient in communication systems.
The procedure for correcting an error is straightforward:
The procedure for correcting an error is straightforward:
- Use the syndrome to determine the erroneous bit position. The syndrome corresponds to a specific column in the parity matrix, directly mapping the location of the error.
- In our example: Syndrone \( \begin{bmatrix} 0 & 1 & 1 \end{bmatrix} \) points to the third position, which is incorrect.
- Correct the error by flipping the bit in this position. Hence, the vector \( \begin{bmatrix} 0 & 1 & 1 & 1 & 0 & 0 & 1 \end{bmatrix} \) corrects to \( \begin{bmatrix} 0 & 1 & 0 & 1 & 0 & 0 & 1 \end{bmatrix} \).
Other exercises in this chapter
Problem 24
In Problems 23-26, find the eigenvalues and eigenvectors of the given nonsingular matrix \(\mathbf{A}\). Then without finding \(\mathbf{A}^{-1}\), find its eige
View solution Problem 24
In Problems 15-28, evaluate the determinant of the given matrix by cofactor expansion. $$ \left(\begin{array}{ccc} 1 & 1 & 1 \\ x & y & z \\ 2+x & 3+y & 4+z \en
View solution Problem 25
In Problems, the given matrix \(\mathbf{A}\) is symmetric. Find an orthogonal matrix \(\mathbf{P}\) that diagonalizes \(\mathbf{A}\) and the diagonal matrix \(\
View solution Problem 25
Show that if \(\mathbf{A}\) and \(\mathbf{B}\) are \(n \times n\) orthogonal matrices, then \(\mathbf{A B}\) is orthogonal.
View solution