Tutorial Handouts for 12/04

EIE325: Telecommunications Technologies Tutorial Exercises (12/04/2006) 1.

We will consider a simple four user CDM system, using orthogonal codes. The chipping codes (or chips) for each of the four users are four samples in length. U1 = [1 1 1 1] U1 = [1 -1 1 -1] U1 = [1 1 -1 -1] U1 = [1 -1 -1 1] Multiplication four these four chipping codes is defined with the ordinary vector dot product (that is, is [u v w x].[a b c d] = ua + vb + wc + xd). a. Show that these four vectors are orthogonal. b. Moreover, if we multiply any of these vectors by itself, what is the result? c. Now, suppose each of the four users wishes to communicate a single number (any number will do – it doesn’t have to be binary). Let a, b, c, and d denote those four numbers. What is the signal transmitted by each of the four users? d. What is the multiplexed signal? e. Demonstrate that the receiver can demultiplex the signal by multiplying your answer in part d by the chipping code of the corresponding user.

2.

Repeat Q1, replacing the four chipping codes U1, U2, U3, and U4 with the following four codes: V1 = [1 0 0 0] V1 = [0 1 0 0] V1 = [0 0 1 0] V1 = [0 0 0 1] Explain why this is exactly equivalent to ordinary synchronous TDM. I.e. synchronous TDM is a special case of CDM.

3.

[DIFFICULT] CDM can also be achieved with random rather than orthogonal sequences. But, the example I gave you in lectures (actually ) was a bit of a con. This is how it really should work. Each user is assigned a long random sequence, and the receiver maintains a copy of that sequence. a. Show that a long random sequence multiplied by itself (using the vector dot product as in question 1) is non-zero, but that two different long random sequences, multiplied together is almost zero. b. Using this fact, you can repeat the procedure of question one to encode multiple signals from several users, using the random sequences as the chipping codes. c. Implement and test this using MATLAB (or some other programming language).