《信息论与编码》第四章习题解答.pdf

1 4.1 4.1 a - - - - = d e d e e d d e 1 1 P 5 . 0 ) 1 ( ) 0 ( = = = = X P X P d 5 . 0 5 . 0 ) 0 ( - = = Y P d = = ) 1 (Y P d 5 . 0 5 . 0 ) 2 ( - = = Y P ) ; 1 ( ) ; 0 ( Y X I Y X I C = = = = = = 2 0 ) ( ) 0 | ( log ) 0 | ( j j p j p j p 0 0 1 1 1-e 1-d e d 0 0 1 2 1-e-d e d d 1-e-d 1 e 0 0 2 2 1 0.5 d 1 1 0.25 0.25 0.5 0.5 0 0 2 2 1-e e e 1-e 1 e 1 1-e 0 0 2 2 3/4 1/4 1 1 1/3 1/3 1/3 1/4 3/4 0 0 2 3 1 1/3 2 1 1/3 1/3 1/3 1/3 1/3 1/3 1/3 1/3 (c) (a) (b) (e) (f) (d) 2 d e e d d d d d e d e 5 . 0 5 . 0 log log 5 . 0 5 . 0 1 log ) 1 ( - + + - - - - - = ) 5 . 0 5 . 0 log( ) 1 ( log ) 1 log( ) 1 ( d d e e d e d e - - - + - - - - = b = 5 . 0 5 . 0 0 25 . 0 25 . 0 5 . 0 0 0 1 P 5 . 0 ) 2 ( ) 0 ( = = = = X P X P 0 ) ( = X P 5 . 0 ) 0 ( = = Y P 25 . 0 ) 1 ( = = Y P 25 . 0 ) 2 ( = = Y P 1 ) ( ) 0 | ( log ) 0 | ( ) ; 0 ( 2 0 = = = = j j p j p j p Y X I bit = = = 2 0 ) ( ) 2 | ( log ) 2 | ( ) ; 2 ( j j p j p j p Y X I 1 25 . 0 5 . 0 log 5 . 0 25 . 0 5 . 0 log 5 . 0 = + = bit 1 0 ) ; 1 ( = = Y X I 4.2.2 C=1 bit/ c - - - = e e e e e e 1 0 1 0 0 1 P 3 1 ) 2 ( ) 1 ( ) 0 ( = = = = = = X P X P X P ) 1 log( ) 1 ( log 3 log e e e e - - + + = C ) 1 , ( 3 log e e - - = H d = 4 3 4 1 0 3 1 3 1 3 1 0 4 1 4 3 P 5 . 0 ) 2 ( ) 0 ( = = = = X p X p 0 ) 1 ( = = X p 8 3 ) 2 ( ) 0 ( = = = = Y P Y P 8 2 ) 1 ( = = Y P = = = 2 0 ) ( ) 0 | ( log ) 0 | ( ) ; 0 ( j j p j p j p Y X I 3 1 log 4 1 2 log 4 3 + = 4 3 = bit 4 3 ) ; 2 ( = = Y X I bit = = = 2 0 ) ( ) 1 | ( log ) 1 | ( ) ; 1 ( j j p j p j p Y X I 8 / 3 3 / 1 log 3 1 8 / 2 3 / 1 log 3 1 8 / 3 3 / 1 log 3 1 + + = 6 8 log 3 1 9 8 log 3 2 + = 4 3 bit 4.2.2 4 3 = C bit/ e = 3 1 3 1 0 3 1 3 1 3 1 3 1 0 3 1 0 3 1 3 1 P 3 1 ) 2 ( ) 1 ( ) 0 ( = = = = = = X p X p X p = = = 2 0 3 0 ) ( ) | ( log ) | ( ) ( i j j p i j p i j p i p C 9 2 ) 2 ( ) 1 ( ) 0 ( = = = = = = Y p Y p Y p 3 1 ) 3 ( = = Y p 3 / 1 3 / 1 log 9 1 3 9 / 2 3 / 1 log 9 1 6 + = C 2 3 log 3 2 = bit/ f - - = d d e e 1 1 P q X p = = ) 0 ( q X p - = = 1 ) 1 ( 1 0 = = - - = 0 | 2 1 X p p Z P N = - - 0 | 2 1 | | X p p Z P N 2 2 0 | 2 1 - = p X Z N s 2 ) 2 1 ( ) 1 ( p N p p - - = 2 1 1 a a + 1 1 0 = 5 . 0 0 5 . 0 0 0 5 . 0 0 5 . 0 P C=1 bit/ a=1 = 5 . 0 5 . 0 0 0 5 . 0 5 . 0 P 5 . 0 5 . 0 1 = - = C bit/ 4.14 W 0 N 0.693 0 N 4000 0 WN S S 482 4.6.1 h 1bit 0 1 2 ) ( N E b - h h h b b WT W R 1 = = h W b T 0 h 0 0 min 693 . 0 ) ( lim N E E b = = h h b T 1bit Shannon 0 log11 b Tb S CTW NW =+= bit + = W N S W T b 0 1 log / 1 4000 0 W N S b b T E S / = 0.5 0.5 0.5 0.5 0 1 2 0 1 Y X 0.5 0.5 0.5 0.5 0 1 1+a a 1 0 Y X 12 + = W N S N WT N E b b 0 0 0 1 log 4000 4000 482 ) 4001 log( 693 . 0 4000 0 0 min = N N E E b