{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

HW13s[1] - ECE 534 Elements of Information Theory Fall 2010...

Info icon This preview shows pages 1–4. Sign up to view the full content.

View Full Document Right Arrow Icon
ECE 534: Elements of Information Theory, Fall 2010 Homework 13 Solutions (ALL DUE to Kenneth S. Palacio Baus) December 1, 2010 1. Problem 15.20. Multiple access (a) Find the capacity region for the multiple-access channel Y = X X 2 1 , where: X 1 ∈ { 2 , 4 } , X - 2 ∈ { 1 , 2 } Solution: The output of the channel is given in the following table: X 1 : X 2 1 2 2 2 4 4 4 16 Table 1: Y = X X 2 1 Here we can see that by fixing X 2 = 1 we can obtain a rate R 1 = 1 per transmission since we are able to decode two symbols at the output of the channel. Similarly, by setting X 1 = 2 it is possible to achieve a rate R 2 = 1. These rates can be obtained by the following expressions, considering that X 1 and X 2 are independent and that both have binary alphabets: R 1 I ( X 1 ; Y | X 2 ) (1) I ( X 1 ; X X 2 1 | X 2 ) (2) H ( X 1 | X 2 ) - H ( X 1 | X X 2 1 , X 2 ) (3) H ( X 1 ) (4) 1 (5) R 2 I ( X 2 ; Y | X 1 ) (6) I ( X 2 ; X X 2 1 | X 1 ) (7) H ( X 2 | X 1 ) - H ( X 2 | X X 2 1 , X 1 ) (8) H ( X 2 ) (9) 1 (10) The third boundary is obtained by: R 1 + R 3 I ( X 1 , X 2 ; Y ) (11) H ( Y ) - H ( Y | X 1 , X 2 ) (12) H ( Y ) (13) 1
Image of page 1

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

View Full Document Right Arrow Icon
From table 1, we have Y ∈ { 2 , 4 , 16 } . In order to have H ( Y ) maximized we can set Y uniformly distributed as follows: p ( y ) = 2 with probability 1 - α 2 4 with probability α 16 with probability 1 - α 2 Then, we know that α = 1 2 to reach the maximum entropy so: H ( Y ) = H 1 4 , 1 2 , 1 4 = 1 . 5 So, R 1 + R 3 H ( Y ) (14) 1 . 5 (15) From the behavior of the channel, shown in table 1, we can observe that this channel behaves like the Binary erasure multiple-access channel . Then for a fixed rate, i.e. R 1 = 1 we have that the transmission of X 1 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 1 is { 1 , 2 } . Is the capacity region decreased? Why or why not? The following table shows the behavior of the channel under the new conditions: X 1 : X 2 1 2 1 1 1 2 2 4 Table 2: Y = X X 2 1 Here we notice that if we set X 1 = 1, 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 1 . The rate R 2 will 2
Image of page 2
depend on the value given for X 1 so, we must consider the probability of X 1 to be 1 or 2. Let X 1 and X 2 be Bernoulli random variables with the following distributions: p ( x 1 ) = 1 with probability 1 - r 2 with probability r p ( x 2 ) = 1 with probability 1 - s 2 with probability s Then, R 1 I ( X 1 ; Y | X 2 ) (16) I ( X 1 ; X X 2 1 | X 2 ) (17)
Image of page 3

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

View Full Document Right Arrow Icon
Image of page 4
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

What students are saying

  • Left Quote Icon

    As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

    Student Picture

    Kiran Temple University Fox School of Business ‘17, Course Hero Intern

  • Left Quote Icon

    I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

    Student Picture

    Dana University of Pennsylvania ‘17, Course Hero Intern

  • Left Quote Icon

    The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

    Student Picture

    Jill Tulane University ‘16, Course Hero Intern