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Unformatted text preview: UNIVERSITY OF SOUTHERN CALIFORNIA, FALL 2008 1 EE 550: Problem Set # 2 Solution I. MARKOV AND CHEBYSHEV INEQUALITY Let Y = ( X E { X } ) 2 , and we have E { Y } = V ar ( X ) = 2 . Thus, Pr[  X E { X } k ] = Pr[ Y k 2 2 ] E { Y } k 2 2 = 1 k 2 , where the inequality is due to Markov Inequality. II. BIT STUFFING BOOK PROBLEM 2.31 The modified destuffing rule starts at the beginning of the string and destuffs bit by bit. A zero is removed from the string if the previous six bits in the already destuffed portion of the string have the value 01 5 . For the given example, the destuffed string, with flags show underlined and removed bits shown as x s., is as follows: 011011111 x 111111011111 x 101111110 III. FLAGS BOOK PROBLEM 2.35 Stuffed bits are always s and always follow the pattern 01 5 . The initial in this pattern could be a bit in the unstuffed data string, or could itself be a stuffed bit. As in the analysis of subsection 2.5.2, we ignore the case where this initial is a stuffed bit since it is almost negligible compared with the other case (also a well designed flag detector would not allow a stuffed bit as the first bit of a flag). If a stuffed bit (preceded by 01 5 in the data) is converted by noise into a 1 , then it is taken as a flag if the next bit is and is taken as an abort if the next bit is 1 . Thus an error in a stuffed bit causes a flag to appear with probability 1 / 2 and the expected number of falsely detected flags due to errors in stuffed bits is K 2 7 . If one is less crude in the approximations, one sees that there are only K 6 places in the data stream where a stuffed bit could be inserted following 1 5 in the data; thus a more refined answer is that the expected number of falsely detected flags due to errors in stuffed bits is ( K 6)2 7 ....
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 Fall '08
 Neely

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