题目详情

设信道带宽为 4000Hz,调制为 4 种不同的码元,根据 Nyquist 定理,理想信道的数据速率为()。

  • A.4Kb/s
  • B.8Kb/s
  • C.16Kb/s
  • D.24Kb/s

正确答案及解析

正确答案
C
解析

根据奈奎斯特定理及码元速率与数据速率间的关系,数据速率R=2W*Log2(N),可列出如下算式:

R=2*4000*Log2(4)=16000b/s=16Kb/s

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