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CDMA with Noise Solution
In the class, we have considered CDMA in an ideal system. However, in reality, we have much more complicated scenario: Users will experience noise.
Consider the scenario with one sender and one receiver. The chipping sequence is (-1 -1 -1 +1 -1 +1 +1 +1). Suppose the sender sends bit ¡°1¡±, then the signal sent will be (-1 -1 -1 +1 -1 +1 +1 +1). In the channel, noise is added to the signal so that the received signal will be (-1+n1 -1+n2 -1+n3 +1+n4 -1+n5 +1+n6 +1+n7 +1+n8), where n1, …, n8 are noise terms. They are independently normally distributed with zero mean and ¦Ò2 variance, in this question, ¦Ò2 = 1. Formally, ni ~ N(0, 1). You should know the normal distribution in a prerequisite course.
After the computing the inner product at the receiver, what ¡°value¡± does the receiver derive? If the value is smaller than 0, it is decoded as -1, otherwise, it is decoded as 1. Use the provided table to find the probability that it is wrongly decoded as -1.
The tail probability (Q function) of a standard normal distribution is given in the attached q_function.pdf.
(1) Inner product/M: R=1+ (-n1- n2 -n3 -n4+n5 +n5 +n7 +n8)/8 ni ~ N(0, 1).
-ni~ N(0, 1).
ni/8 ~ N(0, 1/64).

-ni/8 ~ N(0, 1/64).
This is because if n is a Gaussian random variable with variance ¦Ò2 , a*n is a Gaussian random variable with variance a2¦Ò2
R~ N(1,1/8) = (¦Ì0, ¦Ò02)
This is because the sum of two independent Gaussian random variables is still a Gaussian random variable. See https://en.wikipedia.org/wiki/Sum_of_normally_distributed_random_variables
P(R<0)=Q(|"# ¦Ì0|)= Q(2.82)¡Ö 2.4012*10-03 ¦Ò0 It follows the definition of Q function, see https://en.wikipedia.org/wiki/Q-function