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soln6

# soln6 - Boston University Department of Electrical and...

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Boston University Department of Electrical and Computer Engineering ENG EC515 Digital Communication ( Fall 2009 ) Problem set 5 Assigned: Mon 9 Nov Due: Wed 18 Nov Suggested reading: Class-notes, handouts, and Ch. 5 textbook O ffi ce hours: M, W 4:00–6:00pm PHO438 Problem 5.1 Chapter 5: Problem 5.2 Solution: From Equation 5.15a we have P e exp " - K E b 2 N 0 - log e (2) !# = 2 - K E b E min - 1 = M - E b E min - 1 (a) We desire P e < 10 - 6 . From this bound, i) E b E min = 1 dB = 1 . 26. so M - 0 . 26 = 10 - 6 or M = 10 6 /. 26 = 10 24 , which is 80 bits. ii) E b E min = 3 dB = 2. So M = 10 6 which is 20 bits. ii) E b E min = 6 dB = 4. So M = 100 which is 7 bits. (b) K = RT where R = 100 bits / sec. We must, therefore, wait T seconds to build up M messages, where i) T = K / R = 800 msec ; ii) T = 200 msec ; iii) T = 70 msec ; The use of M orthogonal signals in time T says we must provide M dimensions in time T . Thus, D must exceed M / T D = (3 / 2) W > M / T = W > (2 / 3) M / T i) W > (2 / 3)10 24 / 0 . 8 10 24 Hz ; ii) W > (2 / 3)10 6 / 0 . 2 3 . 3 × 10 6 Hz ; iii) W > (2 / 3)128 / 0 . 07 1 . 2 × 10 3 Hz . Problem 5.2 Chapter 5: Problem 5.3 Solution: (a) A set of M simplex signals with energy, E s , has the same error probability as a set of M orthogonal signals with energy E s M M - 1 . Thus P e ( M - 1) Q ( r E s M ( M - 1) N 0 ) M exp( - E s M 2( M - 1) N 0 ) Now, E s = P s T and M = 2 RT = exp( RT log e (2)). Therefore,

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soln6 - Boston University Department of Electrical and...

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