第二章 信源熵改-习题答案(3)

?1?H0?H?H0H0?H?H0??log2?0.881log2log2?0.553log2?11.9%(3)

?1?

?44.7%H(X) > H2(X)

表示的物理含义是：无记忆信源的不确定度大与有记忆信源的不确定度，有记忆信源的结构化信息较多，能够进行较大程度的压缩。

2.17 给定声音样值X的概率密度为拉普拉斯分布p(x)?1?e??x,???x???，求Hc(X)，并

2证明它小于同样方差的正态变量的连续熵。

解： H1??|x|c(X)?????(x)logp(x)dx???????p??p(x)log2?edx ??log???p(x)dx??????|x|2?????p(x)logedx ?log2????1|x|loge??|x|???2?e??dx ?log2?????0?e??xloge??xdx其中：????x0?e?loge??xdx??????xd?e??x0loge??e??xlogx??2e??0????0e??xd?loge??x?????e??x????0??log2e?log2e?Hlog22ec(X)???log2e?log? bit/symbolm?E(X)??????1x|?x?x??p(x)?xdx????2?e??|xdx??01??2?exdx????102?e?xdx?01?x0???y??2?exdx??01(?y)??2?e?(?y)d(?y)??1?y??2?eydy?????102?e?ydy?m?????102?e??xxdx????102?e??xxdx?0?2?E??x?m?2??E(x2)????2??12??p(x)?xdx????2?e??|x|xdx??????x0?ex2dx ?????2??x?????x2????x0xde?x2?ex0???0e??dx???????0e??xdx2?2???0e?xdx ??2???x??????x??0xde???2???x???ex0??0edx????2?2?H?2?log22ec(X正态)?12log2?e??e?Hc(X)?log?

· 11 ·

2.18 每帧电视图像可以认为是由3?105个像素组成的，所有像素均是独立变化，且每像素又取128个不同的亮度电平，并设亮度电平是等概出现，问每帧图像含有多少信息量？若有一个广播员，在约10000个汉字中选出1000个汉字来口述此电视图像，试问广播员描述此图像所广播的信息量是多少（假设汉字字汇是等概率分布，并彼此无依赖）？若要恰当的描述此图像，广播员在口述中至少需要多少汉字？

解： 1)

H(X)?logn?log128?7 bit/symbolH(XN)?NH(X)?3?105?7?2.1?106 bit/symbol

2)

H(X)?logn?log10000?13.288 bit/symbolH(XN)?NH(X)?1000?13.288?13288 bit/symbol

N3)N?H(X).1?106H(X)?213.288?158037

2.19 连续随机变量X和Y的联合概率密度为：?p(x,y)??1??r2??0H(XYZ)和I(X;Y)。、 （提示：

??2log 02sinxdx???2log22）解：

· 12 ·

x2?y2?r2，求H(X), H(Y), 其他222222p(x)??r?x?r2?x2p(xy)dy??r?x1?r2?x2?r2dy?2r?x?r2 (?r?x?r)Hrc(X)????rp(x)logp(x)dx ???rp(x)log2r2?x2?r?r2dx ???r?rp(x)log2?r2dx??r?rp(x)logr2?x2dx?r2 ?log22??r?rp(x)logr?x2dx?r2 ?log12?logr?1?2log2e ?log2?r?12log2e bit/symbol其中：?r?rp(x)logr2?x2dxr?x2??2r22?r?r2logr2?xdx?4r222?r2?0r2?xlogr?xdx令x?rcos?40?r2??rsin?logrsin?d(rcos?)2??422?r2?0?rsin?logrsin?d?24???22?0sin?logrsin?d???4?2sin24?22?0?logrd????0sin?logsin?d?4??2cos2?4??cos2??logr?1?02d????2102logsin?d?

13 ·

·

?2??logr?2d??02???0logr?2cos2?d??02??2?20logsin?d??2?2???20cos2?logsin?d??logr?1?logr?2dsin2??2?2?(??2log2)???20cos2?logsin?d??logr?1??logr?1?其中：2??12?20cos2?logsin?d?elog2????20cos2?logsin?d??1?20logsin?dsin2??201??sin2?logsin?????????????????1????20?sin2?dlogsin????2?2?202sin?cos??2cos?logsin?ed??2logloglogloglog2e?2cos?d?0?2?1e?201?cos2?21d??

0?2?1212e?d??20?2log2e?2cos2?d??202e?e2212?logesin2?2p(y)??r?y2?r?y2p(xy)dx??r?y22212?r?y?r2dx?2r?y22?r2 (?r?y?r)p(y)?p(x)HC(Y)?HC(X)?log2?r?12log2e bit/symbolHc(XY)????p(xy)logp(xy)dxdyR ????p(xy)logR1?r2dxdy

?log?r2??R2p(xy)dxdy ?log2?r bit/symbolIc(X;Y)?Hc(X)?Hc(Y)?Hc(XY) ?2log2?r?log2e?log?r ?log2??log2e bit/symbol2 · 14 ·

2.21 设X?X1X2...XN是N维高斯分布的连续信源，且X1, X2, ? , XN的方差分别是

?1,?2,...,?222N，它们之间的相关系数?(XiXj)?0(i,j?1,2...,N,i?j)。试证明：N维高斯分布

的连续信源熵

Hc(X)?Hc(X1X2...XN)?1N?log2i2?e?i2

证明：相关系数??xixj??0 ?i,j?1,2,...,N, i?j?，说明X1X2...XN是相互独立的。 ?Hc(X)?Hc(X1X2...XN)?Hc(X1)?Hc(X2)?...?Hc(XN)?Hc(Xi)?12log2?e?2i?Hc(X)?Hc(X1)?Hc(X2)?...?Hc(XN) ?122log2?e?21?...?11?2log2?e?22log2?e?2N1N ?22?log2?e?ii?12.22 设有一连续随机变量，其概率密度函数?bx2p(x)???0(1) 试求信源X的熵Hc(X)；

(2) 试求Y = X + A (A > 0)的熵Hc(Y)； (3) 试求Y = 2X的熵Hc(Y)。

Hc(X)???f(x)logf(x)dx???f2RR(x)logbxdx ??logb??Rf(x)dx??Rf(x)logx2dx ??logb?2b?x2Rlogxdx解：1)33 ??logb?2baloga

9e3?FX(x)?bx3,Fba3X(a)?3?1H23?c(X)??logb?3?logae bit/symbol2)

?0?x?a?0?y?A?a?A?y?a?AFY(y)?P(Y?y)?P(X?A?y)?P(X?y?A) ??y?Abx2dx?bA3(y?A)3

f(y)?F?(y)?b(y?A)2Hc(Y)???f(y)logf(y)dy???f(y)logb(y?A)2dyRR ??logb??f(y)dy?f(y)log(y?A)2dyR?R ??logb?2b?(y?A)2log(y?A)d(y?A)R

0?x?a其他

15 ·

· ??logb??FY(y)?b32ba933loga3e bit/symbolba33(y?A),FY(a?A)?23?loga3?1

?Hc(Y)??logb?e bit/symbol3)

?0?x?a?0??0?y?2aFY(y)?P(Y?y)?P(2X?y)?P(X?yy2?ay2) ??20bxdx?b82b24y2y3f(y)?F?(y)?Hc(Y)???f(y)logf(y)dy???f(y)logRRb8ydy2 ??log ??log ??logb8b8b8??f(y)dy?R?Rf(y)logydy2??b4?Rylogydy322ba92ba9log38aea33 ??logb??FY(y)?b243logeba3a3?39?2ba33

?1y,FY(2a)?23?log?Hc(Y)??logb?e?1 bit/symbol· 16 ·

百度搜索“77cn”或“免费范文网”即可找到本站免费阅读全部范文。收藏本站方便下次阅读，免费范文网，提供经典小说综合文库第二章 信源熵改-习题答案(3)在线全文阅读。