下图中画出曼彻斯特编码和差分曼彻斯特编码的波形图,实际传送的比特串为()。
A: 0 1 1 1 1 0 0 1 0
B: 01 1 0 1 0 0 1 1
C: 1 0 0 1 0 1 1 0 0
D: 1 0 0 0 0 1 1 0 1
A: 0 1 1 1 1 0 0 1 0
B: 01 1 0 1 0 0 1 1
C: 1 0 0 1 0 1 1 0 0
D: 1 0 0 0 0 1 1 0 1
举一反三
- 对信码1000100100001000011000011进行HDB3编码，结果可能是（ ）。 A: -1 0 0 0 +1 0 0 -1 0 0 0 -V +1 0 0 0 +V -1 +1 -B 0 0 -V +1 -1 B: +1 0 0 0 -1 0 0 +1 0 0 0 +V -1 0 0 0 -V +1 -1 +B 0 0 +V -1 +1 C: +1 0 0 0 -1 0 0 +1 0 0 0 +1 -1 0 0 0 -1 +1 -1 +1 0 0 +1 -1 +1 D: -1 0 0 0 +1 0 0 -1 0 0 0 +V +1 0 0 0 +V -1 +1 +B 0 0 -V +1 -1
- 以下HDB3码中，哪些可以确定其中有误码 A: +1 0 0 0 -1 0 +1 -1 +1 0 0 +1 0 -1 +1 -1 0 0 -1 +1…… B: +1 0 0 -1 +1 0 0 0 +1 -1 0 0 -1 0 0 0 +1 0 0 +1 -1…… C: -1 0 0 0 -1 0 +1 0 0 0 +1 -1 +1 0 0 +1 0 0 -1 +1…… D: -1 0 +1 0 0 0 -1 +1 0 0 0 +1 -1 +1 -1 0 0 -1 +1 0 -1……
- 对HDB3码-1000+100-1000-1+1000+1-1+1-100-1+1-1进行译码，结果是（ ）。 A: 1 0 0 0 1 0 0 1 0 0 0 0 1 0 0 0 0 1 1 0 0 0 0 1 1 B: 1 0 0 0 1 0 0 1 0 0 0 1 1 0 0 0 0 1 1 0 0 0 0 0 1 C: 1 0 1 0 1 0 0 1 0 0 0 0 1 1 0 0 0 1 1 0 0 0 0 1 1 D: 1 0 0 0 1 0 0 1 0 0 0 0 1 0 0 1 1 0 0 0 0 0 0 1 1
- 编写程序，创建下列10*10的数组，数组边界全为1，里面全为0。 [[1 1 1 1 1 1 1 1 1 1] [1 0 0 0 0 0 0 0 0 1] [1 0 0 0 0 0 0 0 0 1] [1 0 0 0 0 0 0 0 0 1] [1 0 0 0 0 0 0 0 0 1] [1 0 0 0 0 0 0 0 0 1] [1 0 0 0 0 0 0 0 0 1] [1 0 0 0 0 0 0 0 0 1] [1 0 0 0 0 0 0 0 0 1] [1 1 1 1 1 1 1 1 1 1]]
- 将二进制序列编为HDB3码,其中输出正电平用“+1”表示,负电平用“-1”表示,信息代码100000000011编码后表示为( ) A: +1 -1 0 0 +1 -1 0 0 -1 0 +1<br/>-1 B: +1 0 0 0 +1 -1 0 0 -1 0 +1 -1 C: -1 0 0 0 +1 -1 0 0 -1 0 +1 -1 D: +1 0 0 0 +1 -1 0 0 +1 0 -1 +1