Unformatted text preview: 1 I. P ROBLEM 1 II. P ROBLEM 2 2 III. P ROBLEM 3 3 IV. P ROBLEM 4 4 V. P ROBLEM 5 5 VI. P ROBLEM 6 A. Part a The capacity of this channel is zero since, (1) B. Part b (2) (3) (4) C. Part c Let only if , . Then, (5) (6) Hence, (7) (8) (9) where the last inequality is met with equality if is a deterministic function of . Now, and . Also, as (10) This shows that is a concave function in . Hence, has a maximum when the ﬁrst derivate is zero. That is, (11) It follows that, (12) . Note that is a Markov chain, since if and 6 VII. P ROBLEM 7 We follow the same footsteps of the proof of Theorem the typical set. in text. We ﬁrst derive upper and lower bounds on the size of (13) (14) (15) Hence, (16) Also, for any , such that , (17) (18) (19) Hence, (20) Now, our desired bounds are derived as follows. (21) (22) (23) Also, for sufﬁciently large, (24) (25) (26) VIII. P ROBLEM 8 We note that two transmitted symbols can not be confused at the receiver if and only if . Based on this observation, one can verify that the sequence is decodable with zero probability of error. For example, consider the codeword . The ﬁrst symbol is , can not be confused with and . Hence, we only consider the codewords for a chance of decoding error. But, those codewords have a second symbol that can not be confused with . ...
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 Spring '10
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 Derivative, ROBLEM

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