Tham khảo tài liệu 'proakis j. (2002) communication systems engineering - solutions manual (299s) episode 14', kỹ thuật - công nghệ, cơ khí - chế tạo máy phục vụ nhu cầu học tập, nghiên cứu và làm việc hiệu quả | Therefore C max II Y H Y X max h p 2p To find the optimum value of p that maximizes I X Y we set the derivative of C with respect to p equal to zero. Thus VC -y- 0 vp log2 p pp ln1 2 log2 1 p 1 p 1 2 2 log2 1 p log2 p 2 and therefore 1 p 1 p 1 log2 f 2 p 4 p 5 The capacity of the channel is C h Ẹ bits transmission 5 5 Problem The capacity of the product channel is given by C max I X1X2 Y1Y2 p xi X2 However I X1X2 Y1Y2 H Y1Y2 H Y1Y2IX1X2 H Y1Y2 H Y1 X1 H Y2IX2 H Y1 H Y2 H Y1 X1 H Y2IX2 I X1 Y1 I X2 Y2 and therefore C max I X1X2 Y1 Y2 p xi X2 max I Xi Y1 I X2 Y2 p xi X2 max I X1 Y1 max I X2 Y2 p xi p x2 C1 C2 The upper bound is achievable by choosing the input joint probability density p x1 x2 in such a way that p x1 X2 p x1 p x2 where jõ x1 p x2 are the input distributions that achieve the capacity of the first and second channel respectively. p y x Problem 1 Let X X1 X2 y Y1 Y2 and p y1lx1 if x EX1 p y2 x2 if x E X2 the conditional probability density function of y and X. We define a new random variable M taking the values 1 2 depending on the index i of X. Note that M is a function of X or Y. This 258 is because X1 C X2 0 and therefore knowing X we know the channel used for transmission. The capacity of the sum channel is C max I X Y max H Y - H Y X max H Y - H Y X M p x p x p x max H Y - p M 1 H Y X M 1 - p M 2 H Y X M 2 p x max H Y - XH Y1 X1 - 1 - X H Y2IX2 p x where X p M 1 . Also H Y H Y M H M H Y M H X XH Y1 1 - X H Y2 Substituting H Y in the previous expression for the channel capacity we obtain C max I X Y p x max H X XH Y1 1 - X H Y2 - XH Y1 X1 - 1 - X H Y2X2 p x max H X XI X1 Y1 1 - X I X2 Y2 p x Since p x is function of X p x1 and p x2 the maximization over p x can be substituted by a joint maximization over X p x1 and p x2 . Furthermore since X and 1 - X are nonnegative we let p x1 to maximize I X1 Y1 and p x2 to maximize I X2 Y2 . Thus C max H X XC1 1 - X C2 A To find the value of X that maximizes C we set the derivative of
