?X??0?P(X)???1???4ri?111214Dmin??p(ai)?min d?ai,bj?
j?43? 1?4???r?Dmax?min??p(ai)? d?ai,bj??
j?i?1??0111??1011?? ?D????1101???1110??Dmin?0
?3333?3Dmax?min?,,,??
?4444?4R?Dmax??0
1??X???10??2. 设无记忆信源???p(X)??1/31/31/3??，接收符号AY ={1/2, 1/2}，失真矩阵?????12???D??11?。试求：Dmax和Dmin及达到Dmax和Dmin时的转移概率矩阵。
?21???1??X???10?P(X)???111? ???33??3??12??
?D???11????21???44?4Dmin?1, Dmax?min?,??
?33?3在Dmin时，
??p?bjai??1?由于?j?Ji，所以
??p?bjai?, j?Ji?10?11? ?P???22????01??在Dmax时，
由于pbjai?Pbj，所以
?????10??
?P???10????10??3. 三元信源的概率分别为p(0) = 0.4, p(1) = 0.4, p(2) = 0.2，失真函数dij为：当i = j时，dij =
0；当i ? j时，dij = 1 (i, j = 0, 1, 2)，求信息率失真函数R(D)。
12??X??0??P(X)??0.40.40.2?
?????011??
?D???101????110??5221lb5Dmin?0，R?0??H?X??52?5lb2?5lb5?lb5?0.8?1.52
R?Dmax??0
由定义知：R(D)?rsDmax?min?0.6, 0.6, 0.8??0.6
min?I(X;Y)?，I?X;Y??H?X??H?X|Y? ??平均失真度一定与试验信道的平均错误概率Pe有关，即
Pbj/ai?BDD???P(ai)P(bj/ai)d(ai,bj)i?1j?1s
??P(ai)P(bj/ai)?Pei?j根据保真度准则，应有Pe ? D 根据Fano不等式
H(X/Y) ? H (Pe) + Pe log (r–1)
?H?X|Y??H?Pe??Pelb2?H?X|Y??H?Pe??Pe?H?D??D?R?D??H?X??H?D??D
?lb5?0.8?H?D??D?lb5?0.8?D?H?D?0?D?0.6R?D???
0D?0.6?4. 设有一连续信源，其均值为0，方差为?X，熵为H(X)，定义失真函数为―平方误差‖
失真，即d(x,y)?(x?y)。证明其率失真函数满足下列关系式：
211?XH(X)?log2πeD?R(D)?log
22D22当输入信源为高斯分布时等号成立。
证明：
（1） 证明上界：
1R(D)?H(X)?log2πeD
2连续信源R(D)函数是在
??p?x|y??0?? ????p?x|y?dx?1???D?????p?x?p?x|y?d?x,y?dxdy?D?约束条件下，求平均互信息：
I?X;Y??????????p?x?p?x|y?lbp?x|y?dxdy
p?x?引入参量S和待定函数??x?。在失真不超过D时，I?X;Y?为下确界的试验信道满足
??p0?x|y??p0?y???x?eSd?x,y??????Sd?x,y?????x?p?x?edx?1 ???1???x????p?y?eSd?x,y???????0dy???D?s??????????p?x?p0?y???x?eSd?x,y?d?x,y?dxdy
R?S??SD?S??????p?x?lb??x?dx
由泛函分析中的变分法求I?X;Y?的条件极值 令x?y??, d?x,y??d?????2 由于以上规定了下确界，则
??????x?p?x?eSd?x,y?dx?1
设集合
?s????x?: ????x?p?x?eSd?x,y???dx?1,任意y?
则有R?D??sup?SD?S??s?0,??x???s??????p?x?lb??x?dx???
令??x??K?S?p?x? ?1其中K?S?????????eSd???d????
由（1）得K?S????x,y???eSd?dx?1
即K?S??????eSd???d??1
当d?????2时，且S?0， 得
??S?2?2???ed??2???eS?d???S??
由（2）（3）两式，有
R?D??SD???p?x?lbK?S???p?x?dx?SD?????p?x?lb1p?x?dx?????p?x?lb1K?S?dx?SD?H?X????p?x??lb??eSd???????d??????dx?SD?H?X??lb??eSd?????d?由对数得换底公式，有
ln2?R?D??SD?H?X??ln??eSd?????d??R 若要（1）式等号成立，则等效于（5）式等号成立。
1）2）4）5） （
（
（
（
R??D??令gS?????eSd??????????Sd???eSd???d?
d???d?
e?ed?则R??D??g???d???d?
???Sd?????S然后再求二阶导数，得
?R?????gS???d???d????gS???d???d??????????
??Ed2????E2?d????由于gS???是d???得概率密度函数
2?2????且
????gS???d??????????eSd???d?eSd???d??1
所以R???0，即（5）式右边为上凸函数，在R??0的S上确极大值，有
代入d?????得
2D??gS???d???d?
???gS?????S?eS?2
?D??由式（5）得
?S????e?2S?2d???1 2S （6）
S???ln2?R?D???1d?2?HX?ln?e2???H?X??12lne?ln?H?X??12ln2?eD??S1?H?X??12lne?2ln2D?
即R?D???H?X??12ln2?eD
21?X（2） 证明上界R(D)?log
2D设信道的传递函数的概率为：
?1?y??x?2?exp??? 2?D2??D??它是已知X?x时y的概率分布，即均值为?x，方差为?D的高斯分布，其中??1??D2。 p??y|x??1D?????????p?x?p??y|x?d?x,y?dxdy???p?x?dx????????p?x?dx??2?y?x?p??y|x?dy???2?????y??x??2xy?1????x?1????p??y|x?dy?22??y??x?2?p??y|x?dy?2xy?1????p??y|x?dy????????????p?x?dx?????22??x1???p??y|x?dy?????????????????D?2x?1????x?x?1????p?x?dx22
????D??1???x?p?x?dx2222??D??1???????x2p?x?dx2??D??1????X?D设BD??p?y|x?:D?D?，p??y|x??BD 由于R?D??Inf?I?X;Y??
所以R?D??I??X;Y?
p?x|y??BDI?X;Y??H?Y??H?Y|X?输出信号Y???X?Z?，
?H?Y??12lb2?e?D
??????于是Y时均值为零，方差为??2H?Y??12lb2?e?X?D
E?Y????E?X??E?Z???0
22EY2??2EX2?EZ2??2?X??D??X?D
2X???D的随即变量。
?当方差受限时，高斯随即变量的差熵最大，有
?2I?X;Y??1lb2?e??D?1X22lb2?e?D221?X?D1?X?lb?lb2?D2D?????
21?XR?D??I??X;Y??lb
2D当且仅当p?x?是高斯分布时，上式等号成立。
211?X综上所述，H(X)?log2πeD?R(D)?log
22Da?a|x|5. 随机变量X服从对称指数分布p(x)?e，失真函数为d (x, y) = | x – y |，求信源的
2R(D)。
pX?x??a?axe，d?x,y??x?y 2