The code is linear too since k j i x x x and k j i c c c for all i x and i c nk

# The code is linear too since k j i x x x and k j i c

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The code is linear too, since k j i x x x and k j i c c c for all i x and i c , n k j i , 1 , , 27. 30.04.2007 2 nd Midterm Determine the correct information word when corresponding Hamming codeword is received as 0110101 given with the (7,4) Hamming codewords table and syndrome equations below. Solution The codeword having the smallest hamming distance to the received word is expected to be the actual transmitted codeword (assuming that a maximum of 1 bit is in error.) The closest codeword, in our case, is 0100101. First 4 bits are the information word in (7,4) Hamming code. Then, X=0100. 5 4 2 1 1 r r r r s 6 4 3 1 2 r r r r s 7 4 3 2 3 r r r r s 0 0 0 0 0 0 0 0 0 0 1 1 1 1 0 0 1 0 0 1 1 0 0 1 1 1 0 0 0 1 0 0 1 0 1 0 1 0 1 0 1 0 0 1 1 0 1 1 0 0 1 1 1 0 0 1 1 0 0 0 1 1 0 1 0 0 1 0 0 1 1 0 1 0 1 0 1 1 0 1 1 0 1 0 1 1 0 0 0 1 1 1 1 0 1 1 0 0 1 1 1 0 0 0 0 1 1 1 1 1 1 1
151227621 DIGITAL COMMUNICATIONS 19 28. 30.04.2007 2 nd Midterm For given information word length n in bits, how many bits would a linear block code codeword minimally have? Explain using syndrome vector concept. Solution The syndrome vector must have adequate number of bits that can address all possible error combinations. That is, in a ( n,k ) code of k information bits, p parity bits and n codeword bits, e p b 1 2 where e b is the number of all allowed error combinations, must be satisfied. For single bit errors n b e . With p parity bits, one can write p linearly independent syndrome vector equations. For example, let a ( n,k ) code is designed to detect all 1 and 2 bit errors in a word, and n =10. The number of possible 2 bit errors are given by )! 2 ( 2 ! n n and 1 bit errors by n. Calculating for n=10, we find 55 10 ! 8 2 ! 10 e b , which requires 6 parity bits. This leaves 4 bits to the information word. Therefore, it is actually a (10, 4) code with 2 bits error detection. This is meaningful since (7,4) hamming code has 3 parity bits for single bit errors. Another 3 bits would help detect other erroneous bit. 29. 30.04.2007 2 nd Midterm An ensemble (A, z ) is given where A={'000', '001', '010', '011', '100'} and z =[0.5 0.22 0.2 0.05 0.03] T is given. Find the code-length per symbol before and after Huffman coding. Solution Without using Huffman codes, 3 avg L (fixed) bits/symbol. Using Huffman codes, 86 . 1 4 03 . 0 4 05 . 0 3 2 . 0 2 22 . 0 1 5 . 0 5 1 i i i avg L p L bits/symbol. 0.50 0.22 0.20 0.05 0.03 0.08 0.28 0.50 1. 0 0 1 11 10 110 111 1110 1111
151227621 DIGITAL COMMUNICATIONS 20 30. 24.05.2007 Final Exam Find the energy and power of the given periodic waveform. Solution Since the signal is periodic E , but we can talk about power 3 3 8 2 2 2 2 / 0 3 2 2 2 / 0 2 A t T A dt t T A T P T T 31. 24.05.2007 Final Exam Show that 1 and 2 are orthogonal waveforms. Solution Two waveforms are said to be orthogonal if 0 ) ( ) ( 2 1 dt t t is satisfied. Since 0 ) ( ) ( 2 1 t t everywhere, the condition is satisfied. Therefore ) ( 1 t and ) ( 2 t are orthogonal.

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• Fall '18
• Mr. Bhullar
• Hamming Code, Error detection and correction, Parity bit

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