There are 4 messages A,B,C and D, with probability 1/4, 1/8, 1/8 and 1/2 respectively. The occurrence of each message is independent of each other. Calculate the average information of each message
A: 1.5 bit/symbol
B: 2.5 bit/symbol
C: 2 bit/symbol
D: 1.75 bit/symbol
A: 1.5 bit/symbol
B: 2.5 bit/symbol
C: 2 bit/symbol
D: 1.75 bit/symbol
举一反三
- 信源产生a、b、c、d四个符号,各符号独立出现,这个信源可能的最大信源熵为()。若四个符号的出现概率分别为1/2、1/4、1/8、1/8,则该信源熵为() A: 4 bit/符号; 0.75 bit/符号 B: 2 bit/符号; 1.75 bit/符号 C: 4 bit/符号;1.75 bit/符号 D: 2 bit/符号;0.75 bit/符号
- 一个十六进制是____个bit A: 8 B: 4 C: 1 D: 2
- 两个点数的各种组合(无序)对的熵和平均信息量为() A: 4.337bit/symbol B: 4.517bit/symbol C: 4.566bit/s D: 5,231bit/symbol
- The transmission rate of the binary digital transmission system is 2400B,then the transmission bit rate is ____;if the system is hexadecimal, with the same symbol rate, the transmission bit rate is ____. A: 2400 bit/s, 2400 bit/s B: 2400 bit/s, 4800 bit/s C: 2400 bit/s, 9600 bit/s D: 4800 bit/s, 9600 bit/s
- 如果在已知发送独立的符号中,符号“E”出现的概率为1/8,则符号“E”所包含的信息量为<br/>()。 A: 1<br/>bit B: 2<br/>bit C: 3<br/>bit D: 4<br/>bit