共有4个用户进行CDMA通信。这4个用户的码片序列为：

A: (–1 –1 –1 +1 +1 –1 +1 +1)；B: (–1 –1 +1 –1 +1 +1 +1 –1)

C: (–1 +1 –1 +1 +1 +1 –1 –1)；D: (–1 +1 –1 –1 –1 –1 +1 –1)

现收到码片序列：(–1 +1 –3 +1 –1 –3 +1 +1)。问是哪些用户发送了数据？发送的是1还是0？

解答：A的内积为1，B的内积为–1，C的内积为0，D的内积为1。因此，A和D发送的是1，B发送的是0，而C未发送数据。

详细计算过程如下(公式：S*A/m)：

A： (–1 –1 –1 +1 +1 –1 +1 +1)*(–1 +1 –3 +1 –1 –3 +1 +1)/8=(+1 -1 +3 +1 -1 +3 +1 +1)/8=1

B：(–1 –1 +1 –1 +1 +1 +1 –1)*(–1 +1 –3 +1 –1 –3 +1 +1)/8=(+1 -1 -3 -1 -1 -3 +1 -1)/8=-1

C： (–1 +1 –1 +1 +1 +1 –1 –1)*(–1 +1 –3 +1 –1 –3 +1 +1)/8=(+1 +1 +3 +1 -1 -3 -1 -1)/8=0

D： (–1 +1 –1 –1 –1 –1 +1 –1)*(–1 +1 –3 +1 –1 –3 +1 +1)/8=(+1 +1 +3 -1 +1 +3 +1 -1)/8=1