某信息源的符号集由ABCD和E组成,设每一符号独立出现,其出现概率分别为1／41/8、1/16、3/16 5/16

第二题：
（1）最大值为2048个最小量化台阶,就是2048mV
（2）1,100,1001
极性码1,为+；
段落码100,第5段,起始电平128Δ；
段内码1001,比较电平128+（64+8）=200Δ；
最后一位是“1”,量化电平=200+8/2=204Δ；
第一题：
信息速率是每秒传输的二进制符号数（bit/s）；
（1）不同的符号等概出现时,每个二进制脉冲宽度为0.5ms,则每秒可传送1/0.5m=2000个二进制符号,信息速率为2kbit/s；

1,The Internet is the world's largest source of information, with a computer can access.、
2、Currently, about 80% of online communication are in English, but this proportion is declining

In this paper,by using the TM remote sensing images in 1991 and 2005 as the basic information source,according to the theory of landscape ecology,landscape pattern index,dynamic variation of landscape pattern of 3.3 analysis and in the study area using FRAGSTATS and its driving force.The results show that,in 1991 the changes of landscape pattern in the study area south of Hangzhou Bay area -2005 is more intense,during the period of construction land increased greatly,while the mudflat and forestland,showing a decreasing trend,the landscape fragmentation increased,evenness decreased.The disturbance of human activities on landscape type distribution of intensity increases,the distribution of landscape types in the study area is more widely.The increase of population,economic development,climate and hydrological factors are the main driving force for the evolution of landscape pattern of Hangzhou Bay area.By the end of the atmospheric environment,water environment and biodiversity research,you can see the ecological effects of this change in landscape pattern of the study area is negative.

你玩我呢!

A

小题:B
小题:D
小题:C
小题:A


小题:此项符合下定义的方法，揭示了事物的本质特征
小题:画线句子的主体是“电子图书馆”，所以陈述对象应为“电子图书馆”，但B项不甚恰当，因为作为时代背景的内容要广泛复杂得多
小题:“人机界面”的功用介绍不全面，它还可以为用户储存信息
小题:电子图书馆本身不具备这方面功能，而是人们利用这些信息为人们提供各种服务时才能提高生产率、改善生活水平。

m=1000
H（x）=16x1/32log1/32+112x1/224log1/224=6.404比特/符号
Rb=m·H（X）=1000x6.404

每个符号的平均信息量
＝1/4*log2(1/4)+1/8*log2(1/8)+1/8*log2(1/8)+3/16*log2(3/16)+5/16*log2(5/16)
=2.23bits
1小时的信息量=3600*1000*2.23=8.028Mbits
1小时可能的最大信息量＝3600×1000×log2（5）=8.352Mbits 等概率时

m=1000
H（x）=16x1/32log1/32+112x1/224log1/224=6.404比特/符号
Rb=m·H（X）=1000x6.404

单个字符的概率P1=1/5*1/2=1/10 ;连续两个字符的概率P2 = 1/20*1/2=1/40
同概率平均信息量：H(p)=∑P(x)log1/p(x)
所以,H(p)=5*1/10*log10 + 20*1/40*log40
=1.66+2.16
=3.82 bit/字符
符号周期为T
符号速率R1=1/T
平均信息速率为R2=R1*H(p)=3.82/T
符号周期自带

平均信息量：H(x)=1/2log2+2*1/8log8+1/16log16+3/16log(16/3)+5/16log(16/5)=a(自己算)
1小时的信息量：I=3600*Rb*H（x）=b(自己算)
最大信息量==>等概率分布==>H(x)=log5==>Imax=3600*Rb*log5=c
另外,log如果没写下标,都是以2为底.