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
-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
A: +1 -1 0 0 +1 -1 0 0 -1 0 +1
-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
B
举一反三
- 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
- 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
- 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]]
- 已知逻辑函数式为 F = A B + B C,可列出真值表如表2中的是( ) A B C ① ② ③ ④ 0 0 0 0 0 1 0 1 0 0 1 1 1 0 0 1 0 1 1 1 0 1 1 1 0 0 0 0 1 0 1 0 0 1 1 0 1 1 0 1 0 0 0 0 0 1 0 0 0 0 0 1 1 0 1 1
内容
- 0
1 1 1 1 1 0 1 1 1 1 0 0 1 1 1 0 0 0 1 1 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]] 提示：可以先创建全为0或者全为1的数组，再通过索引和切片机制进行修改
- 2
1⊙1⊙1⊙1⊙0⊙0⊙0⊙0等于？（） A: 0⊙0⊙0⊙1 B: 1⊕1⊕1⊕1 C: 1⊙0⊙0⊙1 D: 1⊕1⊙0⊙0
- 3
全加器的真值表如下图，输出端CO代表的是。 输 入 输 出 A B CI S CO 0 0 0 0 0 0 0 1 1 0 0 1 0 1 0 0 1 1 0 1 1 0 0 1 0 1 0 1 0 1 1 1 0 0 1 1 1 1 1 1
- 4
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