solfinal00 - University of Toronto F. R. Kschischang...

This preview shows pages 1–3. Sign up to view the full content.

University of Toronto Department of Electrical F. R. Kschischang ECE1502F — Information Theory Final Examination Solutions December 13, 2000 1 . Short Snappers (a) I ( X ; X ) = H ( X ), the entropy of X . (b) False. Although I ( X ; Y ) = H ( Y ) - H ( Y | X ), maximizing H ( Y ) does not necessarily maximize I ( X ; Y ). In fact, it may not be possible to achieve a uniform output distri- bution. An example is the binary erasure channel, where the capacity achieving input distribution yields an output distribution ((1 - ± ) / 2 ,±, (1 - ± ) / 2)), which is not uniform unless ± = 1 / 3. (c) This symmetric channel has capacity C = log 2 (3) - H (1 / 2 , 1 / 4 , 1 / 4) = log 2 (3) - (1 / 2)log 2 (2) - (1 / 4)log 2 (4) - (1 / 4)log 2 (4) = log 2 (3) - 3 / 2 . 085 bits / channel - use . (d) The channel transition matrix is M = " 1 - ± ± 0 0 ± 1 - ± # . For this to deﬁne a weakly symmetric channel, the rows must be permutations of each other (they are), and the column sums must be equal, which occurs when 1 - ± = 2 ± , i.e., when ± = 1 / 3. (e) We have C = W log 2 ± 1 + P N 0 W ² bits/s . If the SNR is 20 dB, we have P/ ( N 0 W ) = 100, so C 20 Kbps. If the SNR is 30 dB, we have P/ ( N 0 W ) = 1000, so C 30 Kbps. Thus, if the channel is as noisy as indicated, the telphone line capacity is on the order of 20 to 30 Kbps. (f) True. Let B be a code of length n and rate R < C that achieves error probability ± . (By the coding theorem, we know that such a code exists for some length n (or longer) for any ± > 0.) Form the code B 0 of length n + 1 by appending an extra symbol to each codeword of B , where the extra symbol is a one if the given codeword has an odd number of ones, and zero otherwise. Then every codeword of B 0 has an even number of ones. Now, the error probability for B 0 is no greater than ± , since one decoding strategy for B 0 would be to use a decoder for B that simply ignores the additional symbol. The rate of B 0 is nR/ ( n + 1). By choosing n large enough, we can make the rate of B 0 approach arbitrarily close to R , and hence arbitrarily close to C . Hence the given constraint does not a±ect the capacity-achieving ability of this family of codes. Page 1 of 7

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
(g) False. Consider the binary symmetric channel with zero crossover probability, which has a capacity of one bit per channel use. The given family of codes cannot achieve this rate, since the given constraint limits the number of valid sequences. Indeed, the set of valid sequences corresponds to the graph of Fig. 1. The adjacency matrix corresponding to this graph is A = 1 1 0 1 0 1 1 0 0 which has largest eigenvalue λ = 1 3 ± 1 + (19 - 3 33) 1 / 3 + (19 + 3 33) 1 / 3 ² . Thus the capacity of the constrained system is
This is the end of the preview. Sign up to access the rest of the document.

This note was uploaded on 10/24/2011 for the course ELECTRICAL ECE 571 taught by Professor Kelly during the Spring '11 term at University of Illinois, Urbana Champaign.

Page1 / 7

solfinal00 - University of Toronto F. R. Kschischang...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online