Question

Prove that, if every user code is orthogonal to every other user code in CDMA, unwanted signals can easily be filtered out. Note: you have to show me a generalize proof. Showing me examples of three or more instances or examples is NOT a proof!

Answer 1

A code for a certain user should have a good autocorrelation and should be orthogonal to other codes in CDMA.

Orthogonal in code space has the same meaning as in standard space (i.e., the three dimensional space),
Think of a system of coordinates and vectors starting at the origin, i.e., in (0,0,0). Two vectors are called orthogonal if their inner product is 0, as is the case for the
two vectors (2, 5,0) and (0, 0, 17): (2,5, 0)*(0, 0, 17) = 0 + 0 + 0 = 0. But also vectors like (3,-2, 4) and (-2, 3, 3) are orthogonal: (3,-2, 4)*(-2, 3, 3) = -6 -6 + 12 = 0. By
contrast, the vectors (1,2,3) and (4,2,-6) are not orthogonal (the inner product is -10).

Example

Two senders, A and B, want to send data. CDMA assigns the following unique and orthogonal key sequences: key Ak = 010011 for sender A, key Bk = 110101 for sender B. Sender A wants to send the bit Ad = 1, sender B sends Ba= 0. To illustrate this example, let us assume that we code a binary 0 as -1, a binary 1 as +1. We can then apply the standard addition and multiplication rules. Both senders spread their signal using their key as chipping sequence (the term 'spreading here refers to the simple multiplication of the data bit with the whole chipping sequence). In reality, parts of a much longer chipping
sequence are applied to single bits for spreading. Sender A then sends the signal A, = A *Ak = +1*(-1, +1, -1, -1, +1, +1) = (-1, +1,-1,-1, +1, +1).
Sender B does the same with its data to spread the signal with the code:
B, = B.*B = -1*(+1, +1,-1, +1,-1, +1) = (-1,-1, +1,-1, +1, -1).

Both signals are then transmitted at the same time using the same fre-quency, so, the signals superimpose in space (analog modulation is neglected in this example). Discounting interference from other senders and environmental noise from this simple example, and assuming that the signals have the same strength at the receiver, the following signal C is received at a receiver: C = A + B = (-2,0,0,-2, +2,0). The receiver now wants to receive data from sender A and, therefore,
tunes in to the code of A, i.e., applies A's code for despreading: C*A = (-2, 0, 0, -2, +2, 0)*(-1, +1,-1,-1, +1, +1) = 2 + 0 + 0 + 2 + 2 + 0 = 6.
As the result is much larger than 0, the receiver detects binary 1. Tuning in to sender B, i.e., applying B's code gives C*Bx = (-2, 0, 0, -2, +2, 0)*(+1, +1,-1, +1,-1, +1) = -2 +0+0 - 2 - 2 + 0 = -6. The result is negative, soa 0 has been detected.