{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

sol5[1] - ECE 255AN Fall 2011 Homework set 5 solutions...

Info iconThis preview shows pages 1–2. Sign up to view the full content.

View Full Document Right Arrow Icon
ECE 255AN Fall 2011 Homework set 5 - solutions Solutions to Chapter 7 problems 28. Choice of Channels (a) This is solved by using the very same trick that was used to solve problem 2.10. Consider the following communication scheme: X = ( X 1 Probability α X 2 Probability (1 - α ) Let θ ( X ) = ( 1 X = X 1 2 X = X 2 Since the output alphabets Y 1 and Y 2 are disjoint, θ is a function of Y as well, i.e. X Y θ . I ( X ; Y, θ ) = I ( X ; θ ) + I ( X ; Y | θ ) = I ( X ; Y ) + I ( X ; θ | Y ) Since X Y θ , I ( X ; θ | Y ) = 0. Therefore, I ( X ; Y ) = I ( X ; θ ) + I ( X ; Y | θ ) = H ( θ ) - H ( θ | X ) + αI ( X 1 ; Y 1 ) + (1 - α ) I ( X 2 ; Y 2 ) = H ( α ) + αI ( X 1 ; Y 1 ) + (1 - α ) I ( X 2 ; Y 2 ) Thus, it follows that C = sup α { H ( α ) + αC 1 + (1 - α ) C 2 } . Maximizing over α one gets the desired result. The maximum occurs for H 0 ( α )+ C 1 - C 2 = 0, or α = 2 C 1 / (2 C 1 + 2 C 2 ). (b) If one interprets M = 2 C as the effective number of noise free symbols, then the above result follows in a rather intuitive manner: we have M 1 = 2 C 1 noise free symbols from channel 1, and M 2 = 2 C 2 noise free symbols from channel 2. Since at each step we get to chose which
Background image of page 1

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

View Full Document Right Arrow Icon
Image of page 2
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}