# sol3 - c 2 We denote the distance between these two...

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EE3015 Tutorial 3 (Solution) Question 1 Number of samples = 44 kHz * 20 sec = 880000 Each sample is represented by 16 bits = 2 bytes Hence, the file size is 880000 * 2 = 1760000 bytes, or roughly 1.68 Mbytes Question 2 The rightmost column and bottom row are for parity bits. 1 0 1 0 0 1 0 1 0 0 1 0 1 0 0 1 0 1 1 1 0 0 0 1 1 Question 3 The sum is 1001110010011101 The checksum is 0110001101100010 All 1-bit errors can be detected, but 2-bit errors may go undetected. Question 4 Probability that the receiver makes a decoding error = Probability that any two bits are in error + Probability that all three bits are in error ( )( ) ( ) 6 9 6 3 3 2 3 3 10 998 . 2 10 10 997 . 2 10 10 10 1 3 × = + × = + = Question 5 a) Let the first codeword be c 1 and consider another codeword

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Unformatted text preview: c 2 . We denote the distance between these two codewords by ( ) 2 1 , c c d . By definition, . ( ) min 2 1 , d c c d ≥ Let the given bit string be x . Since we can always flip ( ) x c d , 1 bits to change c 1 into x , and then flip bits to change x into c ( 2 , c x d ) 2 , we must have ( ) ( ) ( ) 2 1 2 1 , , , c c d c x d x c d ≥ + . Combining the above two results, we have ( ) ( ) min 2 1 , , d c x d x c d ≥ + . If , then we must have ( ) 2 / , min 1 d x c d < ( ) min 2 , d c x d > . b) If the number of errors is fewer than d min / 2, then according to the result of part (a), the received bit string must be closer to the original codeword than to any other codewords....
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