X 1 : X Table 1: Y = X X 2

Transcription

1 ECE 534: Elements of Information Theory, Fall 200 Homework 3 Solutions (ALL DUE to Kenneth S. Palacio Baus) December, 200. Problem Multiple access (a) Find the capacity region for the multiple-access channel Y = X X 2, where: X {2, 4}, X 2 {, 2} Solution: The output of the channel is given in the following table: X : X Table : Y = X X 2 Here we can see that by fixing X 2 = we can obtain a rate R = per transmission since we are able to decode two symbols at the output of the channel. Similarly, by setting X = 2 it is possible to achieve a rate R 2 =. These rates can be obtained by the following expressions, considering that X and X 2 are independent and that both have binary alphabets: R I(X ; Y X 2 ) () I(X ; X X 2 X 2) (2) H(X X 2 ) H(X X X 2, X 2) (3) H(X ) (4) (5) The third boundary is obtained by: R 2 I(X 2 ; Y X ) (6) I(X 2 ; X X 2 X ) (7) H(X 2 X ) H(X 2 X X 2, X ) (8) H(X 2 ) (9) (0) R + R 3 I(X, X 2 ; Y ) () H(Y ) H(Y X, X 2 ) (2) H(Y ) (3)

2 From table, we have Y {2, 4, 6}. uniformly distributed as follows: Then, we know that α = 2 So, In order to have H(Y ) maximized we can set Y 2 with probability α 2 p(y) = 4 with probability α 6 with probability α 2 to reach the maximum entropy so: ( H(Y ) = H 4, 2, ) =.5 4 R + R 3 H(Y ) (4).5 (5) From the behavior of the channel, shown in table, we can observe that this channel behaves like the Binary erasure multiple-access channel. Then for a fixed rate, i.e. R = we have that the transmission of X will look like noise for the transmission of X 2. Then we have a binary erasure channel with capacity 0.5 bits for the transmission of X 2. So, we obtain the capacity region shown in the following figure: (b) Suppose that the range of X is {, 2}. Is the capacity region decreased? Why or why not? The following table shows the behavior of the channel under the new conditions: X : X Table 2: Y = X X 2 Here we notice that if we set X =, then the output of the channel remains the same regardless the value of X 2. So we won t be able to recover X 2 by fixing X. The rate R 2 will 2

3 depend on the value given for X so, we must consider the probability of X to be or 2. Let X and X 2 be Bernoulli random variables with the following distributions: Then, p(x ) = p(x 2 ) = { with probability r 2 with probability r { with probability s 2 with probability s R I(X ; Y X 2 ) (6) I(X ; X X 2 X 2) (7) H(X X 2 ) H(X X X 2, X 2) (8) H(X ) = H(r) (9) (20) R 2 I(X 2 ; Y X ) (2) H(Y X ) H(Y X, X 2 ) (22) H(Y X ) (23) p(x = )H(Y X = ) + p(x = 2)H(Y X = 2) (24) p(x = 2)(H(Y X = 2) (25) rh(s) (26) Here we can set s = /2 to maximize the region such that H(s) =. The third boundary is obtained by: We need the distribution of Y, that can obtain as follows: R + R 2 I(X, X 2 ; Y ) (27) H(Y ) H(Y X, X 2 ) (28) H(Y ) (29) X X 2 Y Probability ( r)( s) 2 ( r)s 2 2 r( s) rs Table 3: Y = X X 2 with probability ( s)( r) + ( r)s = r p(y) = 2 with probability r( s) 4 with probability rs 3

4 So H(Y ) = H ( r, r( s), rs) (30) = ( r) log 2 ( r) r( s) log 2 (r( s)) rs log 2 (rs) (3) = ( r) log 2 ( r) r( s) log 2 (r) r( s) log 2 ( s) rs log 2 (r) rs log 2 (s) (32) = ( r) log 2 ( r) r log 2 (r) r [( s) log 2 ( s) + s log 2 (s)] (33) = H(r) + rh(s) (34) Here we can set s = /2 as before to maximize the region such that H(s) =. Hence: R + R 2 H(Y ) (35) H(r) + rh(s) (36) H(r) + r (37) To answer the question: Is the capacity region decreased? Why or why not?, we need to plot both capacity regions and compare them. However we can pick up some rate pairs (R, R 2 ) and see whether they are achievable for both schemes or not, so that we can have at least one argument to compare. The rate pair (H(0.8) = 0.729, 0.8) is achievable in the region for part (b), we obtain R + R 2 =.529. However, this rate is clearly outside the capacity region in part(a) since R + R 2.5. Also, we can take the rate pair (0.5, ) which is achievable in capacity region in part (a), we can see however that to achieve a rate R 2 =, we need to choose r =, then we have that R = H() = 0, therefore we have that this rate is not achievable in the capacity region of part (b). We can conclude that are some rates achievables in one region that are not in the other so the a plot will be the best way to see the differences. 4

5 2. Problem Trilingual-speaker broadcast channel. A speaker of Dutch, Spanish, and French wishes to communicate simultaneously to three people: D, S, and F. D knows only Dutch but can distinguish when a Spanish word is being spoken as distinguished from a French word; similarly for the other two, who know only Spanish and French, respectively, but can distinguish when a foreign word is spoken and which language is being spoken. Suppose that each language, Dutch, Spanish, and French, has M words: M words of Dutch, M words of French, and M words of Spanish. (a) What is the maximum rate at which the trilingual speaker can speak to D? Following example of textbook, we see that the maximum rate at which the trilingual speaker can speak to D is log 2 (M + 2) since D is able to distinguish M words from his mother tongue plus two other words spoken by the speaker, one for each foreign language. (b) If he speaks to D at the maximum rate, what is the maximum rate at which he can speak simultaneously to S? The trilingual speaker sends a word in Spanish M+2 of the time, so he can encode log 2 (M) data of Spanish at a rate of log 2 (M) M+2. (c) If he is speaking to D and S at the joint rate in part (b), can he also speak to F at some positive rate? If so, what is it? If not, why not? As we remember form part (a), the maximum rate at which the trilingual speaker is communicating to D is log 2 (M + 2) since D can recognize when a Spanish word or French word, so here, the same logic used in part (b) can be applied since the speaker will say a French word in M+2 of the time so, he can encode log 2 (M) data of French at a rate of log 2 (M) M+2. 5

6 3. Problem Multiple-access channel capacity with costs. The cost of using symbol x is r(x). The cost of a codeword x n is r(x n ) = n n r(x i). A (2 nr, n) codebook satisfies cost constraint r if n n r(x i(w)) r for all w 2 nr. (a) Find an expression for the capacity C(r) of a discrete memoryless channel with cost constraint r. We can introduce the cost to the capacity expression in a similar way that the power constraint is considered. Going back to Chapter 9 of the textbook, we can see how the power constraint is inserted in the achievability proof of the information capacity of the Gaussian channel with power constraint P, given as: C = max f(x):e[x 2 ] P I(X; Y ). Specifically it is introduced in the probability of error part computation as a new event that yields an error when the power constraint is violated. (Eq. 9.23) We can do something similar with the cost such that an error is produced when n n r(x i(w)) > r for any w. Then, the capacity expression given the constraint can be written as: C(r) = max p(x): P I(X; Y ) x p(x)r(x) r (b) Find an expression for the multiple-access channel capacity region for (X X 2, p(y x, x 2 ), Y) if sender X has cost constraint r and sender X 2 has cost constraint r 2. Here we can apply Theorem of the textbook: we have that the set of achievable rates is given by the closure of the set of all (R, R 2 ) pairs satisfying: R < I(X ; Y X 2, Q) (38) R < I(X 2 ; Y X, Q) (39) R + R 2 < I(X, X 2 ; Y Q) (40) for some choice of the joint distribution p(q)p(x q)p(x 2 q)p(y x, x 2 ) with Q 4. However we need to consider the cost constraints in the same format they were introduced in part (a) but now for the specific case of each sender: x p(x )r (x ) r (c) Prove the converse for part (b). x 2 p(x 2 )r 2 (x 2 ) r 2 We assume we have a sequence of ((2 nr, 2 nr 2 ), n) codes with probability of error going asymptotically to zero that must satisfy the cost constraints. From the text of the problem 6

7 we have that: A (2 nr, n) codebook satisfies cost constraint r if n n r(x i(w)) r for all w 2 nr. We need to extend this to our codebook for the messages: w i =, 2,..., 2 nr and w 2i =, 2,..., 2 nr 2 : n n r (x i (w i )) r (4) r 2 (x 2i (w 2i )) r 2 (42) From the proof of the converse for the Multiple-Access Channel in textbook we have that: R n R 2 n R + R 2 n I(X i ; Y i X 2i ) + ɛ n (43) I(X 2i ; Y i X i ) + ɛ n (44) I(X i, X 2i ; Y i ) + ɛ n (45) Then, we can introduce the time-sharing variable Q = i {, 2,..., n} with probability /n in order to avoid dealing with the cost constraint for each i, as it is done in the textbook pp.542 the capacity region can be written as: R n R 2 n R + R 2 n I(X q ; Y q X 2q, Q = i) + ɛ n (46) I(X 2q ; Y q X q, Q = i) + ɛ n (47) I(X q, X 2q ; Y q Q = i) + ɛ n (48) R I(X Q ; Y Q X 2Q, Q) + ɛ n (49) R 2 I(X 2Q ; Y Q X Q, Q) + ɛ n (50) R + R 2 I(X Q, X 2Q ; Y Q Q) + ɛ n (5) Now we can define the following random variables: X = X Q, X 2 = X 2Q and Y = Y Q whose distributions depend on Q in the same way as the distributions of X i, X 2i and Y i depend on i. The last three equations become: R I(X ; Y X 2 ) + ɛ n (52) R 2 I(X 2 ; Y X ) + ɛ n (53) R + R 2 I(X, X 2 ; Y ) + ɛ n (54) 7

8 for a joint distribution p(q)p(x q)p(x 2 q)p(y x, x 2 ) and Q 4. Recalling the power constraints: x p(x )r (x ) r x 2 p(x 2 )r 2 (x 2 ) r 2 The following analysis is valid for both cost constraints: r and r 2. We can check the power constraint for each i: x P (X = x )r (x ) = x P (X Q = x )r (x ) (55) = P (Q = i)p (X Q = x Q = i)r (x ) (56) x = n = n P (X i = x )r (x ) (57) x E[r (X i )] (58) This result is actually the expectation of the definition of the cost constraint extended to the new codebook (defined at the beginning of this proof: Equation 4) with respect to random message W. It has an equivalent form for the constraint in the cost r 2. (59) 8