T1
在干扰离散信道上传输符号“1”和“0”,在传输过程中每 100 个符号发生一个错传的符号。已知 p ( 0 ) = p ( 1 ) = 1 / 2 p(0)=p(1)=1/2 p(0)=p(1)=1/2 ,信道每秒钟内允许传输 1000 个符号。求此信道的容量。解：该信道为对称离散信道，故信道容量为：
C = log s − H ( p 1 ′ , p 2 ′ , ⋯ p s ′ ) = log 2 − H ( 0.01 , 0.99 ) = 1 + 0.01 log ( 0.01 ) + 0.99 log ( 0.99 ) = 0.919 ( bit/symbol ) \begin{aligned}&C=\log s-H(p_1^{\prime},p_2^{\prime},\cdots p_s^{\prime})\\&=\log2-H(0.01,0.99)\\&=1+0.01\log(0.01)+0.99\log(0.99)\\&=0.919(\text{bit/symbol})\end{aligned} C=logs−H(p1′,p2′,⋯ps′)=log2−H(0.01,0.99)=1+0.01log(0.01)+0.99log(0.99)=0.919(bit/symbol)
即信道容量 C t C_t Ct为：C t = 0919 × 1000 = 919 _\mathrm{t}=0919\times1000=919 t=0919×1000=919 (bit/秒)
T2
求下列两个信道的容量，并加以比较。
P = [ 1 − p − ε p − ε 2 ε 0 p − ε 1 − p − ε 0 2 ε ] P=\begin{bmatrix}1-p-\varepsilon&p-\varepsilon&2\varepsilon&0\\p-\varepsilon&1-p-\varepsilon&0&2\varepsilon\end{bmatrix} P=[1−p−εp−εp−ε1−p−ε2ε002ε]
解：
(1)
此信道是准对称信道。信道矩阵中 Y 可划分成二个互不相交的子集，由于集列所组成
的矩阵
[ p ‾ − ε p − ε p − ε p ‾ − ε ] [ 2 ε 2 ε ] \begin{bmatrix}\overline{p}-\varepsilon&p-\varepsilon\\p-\varepsilon&\overline{p}-\varepsilon\end{bmatrix}\quad\begin{bmatrix}2\varepsilon\\2\varepsilon\end{bmatrix} [p−εp−εp−εp−ε][2ε2ε]
而这两个子矩阵满足对称性，因此，可直接利用准对称信道的信道容量公式进行计算。
C 1 = log r − H ( p 1 ′ p 2 ′ p 3 ′ ) − ∑ k = 1 2 N k log M k C_1=\log r-H(p_1^{\prime}p_2^{\prime}p_3^{\prime})-\sum_{k=1}^2N_k\log M_k C1=logr−H(p1′p2′p3′)−k=1∑2NklogMk
其中 r = 2 , N 1 = M 1 = 1 − 2 ε , N 2 = 2 ε , M 2 = 4 ε r=2,N_{1}=M_{1}=1-2\varepsilon,N_{2}=2\varepsilon,M_{2}=4\varepsilon r=2,N1=M1=1−2ε,N2=2ε,M2=4ε
C 1 = log 2 − H ( p ‾ − ε , p − ε , 2 ε ) − ( 1 − 2 ε ) log ( 1 − 2 ε ) − 2 ε log 4 ε = log 2 + ( p ‾ − ε ) log ( p ‾ − ε ) + ( p − ε ) log ( p − ε ) + 2 ε log 2 ε − ( 1 − 2 ε ) log ( 1 − 2 ε ) − 2 ε log 4 ε = log 2 − 2 ε log 2 − ( 1 − 2 ε ) log ( 1 − 2 ε ) + ( p − ε ) log ( p − ε ) + ( p − ε ) log ( p − ε ) = ( 1 − 2 ε ) log 2 1 − 2 ε + ( p ‾ − ε ) log ( p ‾ − ε ) + ( p − ε ) log ( p − ε ) \begin{aligned}&C_1=\log2-H(\overline{p}-\varepsilon,p-\varepsilon,2\varepsilon)-(1-2\varepsilon)\log(1-2\varepsilon)-2\varepsilon\log4\varepsilon\\&=\log2+(\overline{p}-\varepsilon)\log(\overline{p}-\varepsilon)+(p-\varepsilon)\log(p-\varepsilon)+2\varepsilon\log2\varepsilon-(1-2\varepsilon)\log(1-2\varepsilon)-2\varepsilon\log4\varepsilon\\&=\log2-2\varepsilon\log2-(1-2\varepsilon)\log(1-2\varepsilon)+(p-\varepsilon)\log(p-\varepsilon)+(p-\varepsilon)\log(p-\varepsilon)\\&=(1-2\varepsilon)\log\frac2{1-2\varepsilon}+(\overline{p}-\varepsilon)\log(\overline{p}-\varepsilon)+(p-\varepsilon)\log(p-\varepsilon)\end{aligned} C1=log2−H(p−ε,p−ε,2ε)−(1−2ε)log(1−2ε)−2εlog4ε=log2+(p−ε)log(p−ε)+(p−ε)log(p−ε)+2εlog2ε−(1−2ε)log(1−2ε)−2εlog4ε=log2−2εlog2−(1−2ε)log(1−2ε)+(p−ε)log(p−ε)+(p−ε)log(p−ε)=(1−2ε)log1−2ε2+(p−ε)log(p−ε)+(p−ε)log(p−ε)
输入等概率分布时达到信道容量。
(2)此信道也是准对称信道，采用准对称信道的信道容量公式进行计算。此信道矩阵
中 Y 可划分成两个互不相交的子集，由子集列所组成的矩阵为
[ p − ε p − ε p − ε p − ε ] [ 2 ε 0 0 2 ε ] \begin{bmatrix}p-\varepsilon&p-\varepsilon\\[1ex]p-\varepsilon&p-\varepsilon\end{bmatrix}\quad\begin{bmatrix}2\varepsilon&0\\[1ex]0&2\varepsilon\end{bmatrix} [p−εp−εp−εp−ε][2ε002ε]
这两矩阵为对称矩阵。
r = 2 , N 1 = M 1 = 1 − 2 ε , N 2 = M 2 = 2 ε r=2,N_1=M_1=1-2\varepsilon,N_2=M_2=2\varepsilon r=2,N1=M1=1−2ε,N2=M2=2ε
其中
所以
C = log r − H ( p ‾ − ε , p − ε , 2 ε , 0 ) − ∑ k = 1 2 N k log M k = log 2 + ( p ‾ − ε ) log ( p ‾ − ε ) + ( p − ε ) log ( p − ε ) + 2 ε log 2 ε − ( 1 − 2 ε ) log ( 1 − 2 ε ) − 2 ε log 2 ε = log 2 − ( 1 − 2 ε ) log ( 1 − 2 ε ) + ( p ‾ − ε ) log ( p ‾ − ε ) + ( p − ε ) log ( p − ε ) = ( 1 − 2 ε ) log 2 1 − 2 ε + 2 ε log 2 + ( p ‾ − ε ) log ( p ‾ − ε ) + ( p − ε ) log ( p − ε ) = C 1 + 2 ε log 2 \begin{aligned}&C=\log r-H(\overline{p}-\varepsilon,p-\varepsilon,2\varepsilon,0)-\sum_{k=1}^2N_k\log M_k\\&=\log2+(\overline{p}-\varepsilon)\log(\overline{p}-\varepsilon)+(p-\varepsilon)\log(p-\varepsilon)+2\varepsilon\log2\varepsilon-(1-2\varepsilon)\log(1-2\varepsilon)-2\varepsilon\log2\varepsilon\\&=\log2-(1-2\varepsilon)\log(1-2\varepsilon)+(\overline{p}-\varepsilon)\log(\overline{p}-\varepsilon)+(p-\varepsilon)\log(p-\varepsilon)\\&=(1-2\varepsilon)\log\frac2{1-2\varepsilon}+2\varepsilon\log2+(\overline{p}-\varepsilon)\log(\overline{p}-\varepsilon)+(p-\varepsilon)\log(p-\varepsilon)\\&=C_1+2\varepsilon\log2\end{aligned} C=logr−H(p−ε,p−ε,2ε,0)−k=1∑2NklogMk=log2+(p−ε)log(p−ε)+(p−ε)log(p−ε)+2εlog2ε−(1−2ε)log(1−2ε)−2εlog2ε=log2−(1−2ε)log(1−2ε)+(p−ε)log(p−ε)+(p−ε)log(p−ε)=(1−2ε)log1−2ε2+2εlog2+(p−ε)log(p−ε)+(p−ε)log(p−ε)=C1+2εlog2
输入等概率分布( P ( a 1 ) = P ( a 2 ) = 1 2 P(a_1)=P(a_2)=\frac12 P(a1)=P(a2)=21)时达到此信道容量。比较此两信道容量，可得
C 2 = C 1 + 2 ε log 2 C_2=C_1+2\varepsilon\log2 C2=C1+2εlog2
T3
设有找离散信道的传输情况分别如下图所示,求该信息的信道各量
解：
P ( Y ∣ X ) = [ 1 2 1 2 0 0 0 1 2 1 2 0 0 0 1 2 1 2 1 2 0 0 1 2 ] P(Y\mid X)=\begin{bmatrix}\dfrac{1}{2}&\dfrac{1}{2}&0&0\\0&\dfrac{1}{2}&\dfrac{1}{2}&0\\0&0&\dfrac{1}{2}&\dfrac{1}{2}\\\dfrac{1}{2}&0&0&\dfrac{1}{2}\end{bmatrix} P(Y∣X)= 210021212100021210002121
C = log s − H ( p 1 ′ , p 2 ′ , ⋯ p s ′ ) = log ( 4 ) − H ( 1 2 , 1 2 ) = log ( 4 ) + 1 2 log ( 1 2 ) + 1 2 log ( 1 2 ) = 1 ( b i t / s y m b o l ) C=\log s-H(p_1^{\prime},p_2^{\prime},\cdots p_s^{\prime})=\log(4)-H(\frac12,\frac12)=\log(4)+\frac12\log(\frac12)+\frac12\log(\frac12)=1(bit/symbol) C=logs−H(p1′,p2′,⋯ps′)=log(4)−H(21,21)=log(4)+21log(21)+21log(21)=1(bit/symbol)