?X ? ? x1 ?=? ?P X( )? ?0.6
x2 ?通过一干扰信道，接收符号为Y = { y1, y2 }，信道转移矩 ? 0.4?
3.1 设信源?
?5 1?
6?
?6 ?，求：阵为?1 3
?? ?4 4?
(1) 信源X中事件x1和事件x2分别包含的自信息量；
(2) 收到消息yj (j=1,2)后，获得的关于xi (i=1,2)的信息量； (3) 信源X和信宿Y的信息熵；
(4) 信道疑义度H(X/Y)和噪声熵H(Y/X)； (5) 接收到信息Y后获得的平均互信息量。解：
1)
2)
I x( 1) =?log2 p x( 1) =?log 0.62 = 0.737 bit
I x( 2 ) =?log2 p x( 2 ) =?log 0.42 =1.322 bit
p y( 1) = p x p y( 1) ( 1 / x1)+ p x( 2 ) (p y1 / x2 ) = 0.6× +
0.4× =
0.6 p y( 2 ) = p x p y( 1) ( 2 / x1)+ p
·17·
x( 2 ) (p y2 / x2 ) = 0.6× + 0.4× =
3)
0.4
p y( 1 / x1) 5/6
I x y( 1; 1) = log2= log2 = 0.474 bit
p y( 1) 0.6
p y( 2 / x1) 1/6
I x y( 1; 2 ) = log2= log2
p y( 2 ) 0.4 p y( 1 / x2 ) 1/4 I x( 2; y1) = log2 p y( 1)
= log2
=?1.263 bit =?1.263 bit 0.6
p y( 2 / x2 ) 3/4
I x( 2; y2 ) = log2 = log2 = 0.907 bit
p y( 2 ) 0.4
i
H X() =?∑ p x( i )log p x( i ) =?(0.6log0.6 + 0.4log0.4)log2 10 = 0.971 bit symbol/
H Y( ) =?∑ p y( j )log p y( j ) =?(0.6log0.6 + 0.4log0.4)log2 10 = 0.971 bit symbol/
j
4)
H Y X(
·18·
/ )
=?
∑∑ p x p y( ) ( / x)log p y( / x)
i
j
i
j
i
j
i
5)
(0.6
= 0.715 bit symbol/ ?H X( ) + H Y X( /) = H Y( ) + H X Y( / ∴H X Y(
/
) = H X( ) + H Y X( /
)
) ?H Y( )
= 0.971+ 0.715?0.971 = 0.715 bit symbol/
I X Y(
; ) = H X( ) ?H X Y(
?2
/ 1?
) = 0.971?0.715 = 0.256 bit symbol/
?3 3??
2
3.2 设二元对称信道的传递矩阵为?1
??
?3 3?
(1) 若P(0) = 3/4, P(1) = 1/4，求H(X), H(X/Y), H(Y/X)和
I(X;Y)；
(2) 求该信道的信道容量及其达到信道容量时的输入概率分布；解：
1)
H X p x ) 0.811 bit
symbol/
H Y X(
i
j
=?
/ ) ∑∑ p x p y( i ) ( j / xi )log p y( j / xi )
=?(
= 0.918 bit symbol/
× + × + × + × )×log 102
·19·
p y
p x y p x p y x p y x
p x yp x
p y
p x y p x p y
j
p x y
x p x p y x
= 0.980 bit symbol/
H Y( ) =?∑ p y( j ) =?(0.5833×log 0.58332 + 0.4167×log 0.4167)2 I X Y(
; ) = H X(
) ?H X Y(
/
) = H Y( ) ?H Y X( /
)
H X Y( / ) = H X( ) ?H Y( ) + H Y X( / ) = 0.811?0.980 + 0.918 = 0.749 bit symbol/ I X Y( ; ) = H X( ) ?H X Y( / ) == 0.811?0.749 = 0.062 bit symbol/
2)
+
C = max (I X Y; ) = log2 m ?H mi = log 22 + (
)×log 102 = 0.082 bit symbol/
p x( i ) =
3.3 设有一批电阻，按阻值分70%是2KΩ，30%是5 KΩ；按瓦分64%是0.125W，其余是
0.25W。现已知 2 KΩ阻值的电阻中 80%是 0.125W，问通过测量阻值可以得到的关于瓦数的平均信息量是多少？解：对本题建立数学模型如下：
?X阻值? ?x1 = ΚΩ2 x2 = ΚΩ5? ?
? =?
??
?Y瓦数? ?y1 =1/8
? =?
y2 =1/ 4?
?
·20·