Question

Q6) Suppose you have the following users whose chips are given below:

User A (1 –1 –1 1 –1 1)

User B (1 1 –1 –1 1 1)

User C (1 1 –1 1 1 –1)

Solve the following cases:

1. Station A is sending 1 and station D is trying to listen to station A transmission.
2. Station B is sending 1 and station D is trying to listen to station A transmission.
3. Stations B and C are sending 1 and station D is trying to listen to station A transmission.
4. Station B and C are sending 1 and station D is trying to listen to station B transmission.

Answer 1

3) Stations B and C are sending 1 and station D is trying to listen to station A transmission. B=( +1 +1 -1 -1 +1 +1) Station B sending 1: bit i = +1 (+1).(+1 +1 -1 -1 +1 +1)=( +1 +1 -1 -1 +1 +1) Station C sending 1: bit 1 = +1 (+1).(+1 +1 -1 +1 +1 -1)=( +1 +1 -1 +1 +1 -1) - +2 +2 -2 0 +2 After multiplexing (+2 +2 -2 0 +2 0) is send Station D listening Station A: (+1 -1 -1 +1 -1 +1).(+2+2-20+20) =(+2 -2 +2 0-20) Add all the value = 0 Finally divide by 6 = 0/6 = 0 = Silent 4) Station B and C are sending 1 and station D is trying to listen to station B transmission. B=( +1 +1 -1 -1 +1 +1) Station B sending 1: bit 1 = +1 (+1). (+1 +1 -1 -1 +1 +1)=( +1 +1 -1 -1 +1 +1) Station C sending 1: bit 1 = +1 (+1). (+1 +1 -1 +1 +1 -1)=( +1 +1 -1 +1 +1 -1) + +2 +2 -2 0 +2 After multiplexing (+2 +2-20+20) is send Station D listening Station B: (+1 +1 -1 -1 +1 +1).(+2 +2-20+20) =(+2 +2 +20+20) Add all the value = 8 Note: Station B and C have not pairwise orthogonal sequences. When we multiply two different sequences, element by element and add the result then we will get 0. This is known as inner product of two different sequences. But, here the inner product not equals to zero. For example, B.C=( +1 +1 -1 -1 +1 +1).(+1 +1 -1 +1 +1 -1)= 2