Classification of CDMA Systems

1
Classification of CDMA Systems

• Synchronous transmissions of users are
  synchronized down to chip levels (difficult, but
  permits orthogonal transmission).
• Asynchronous systems
• Classification of CDMA Sequences
• Orthogonal Sequences
• Pseudonoise (PN) sequences

2
Desirable Properties of Sequences

• Good autocorrelation (for synchronization and
  synchronous CDMA)
• Low crosscorrelation (for low multiple access
  interference)
• Availability of large number of codes

3
Autocorrelation

• Degree of correspondence between a sequence and a
  phase-shifted replica of itself.

For example let N5, and
Then C(0) 5 C(1)-1-1-11-1-3
C(2)111-1-11 C(3)1-1-1111
C(4)-1-11-1-1-3 C(5)C(0)5
4
Cross correlation

• Degree of agreement between two different
  sequences.

E.g, let
Then R(0) 11-1-1-1-1
R(1)-11-11-1-1 R(2)-111113
R(3)1-1-1111 R(4)-1-11-11-1
R(5)R(0)-1
5
Properties of Random Sequences

• 1. Relative frequencies of 0 and 1 are each 1/2
• 2. Run length like coin flipping
• 1/2 with length 1
• 1/4 with length 2
• 1/8 with length 3
• 1/2n with length n
• 3. If the sequence is shifted by any number of
  elements, the resulting sequence will have equal
  number of agreements and disagreements with the
  original sequence.

6
Max Length PN Sequence Generator
1 0 0 0 1 0 1 0 1 1 1 0 1 1 1 0 1 1 0
0 1 --------- 1 0 0 0 1 0
a
-
3
i
7
Number of M Sequences
8
Gold Codes

• M-seq. 1

5
1
2
3
4
5
1
2
3
4
M-seq. 2
Sequence 1 111110001101110101000010
0101100 Sequence 2
1111100100110000101101010001110 0 shift
combination 0000000111101101111101110100010 1
shift combination 000010101011110000101000011000
1 30 shift combination10000100010001010001100011
01011
9
Properties of Orthogonal Sequences

• When two orthogonal sequences are multiplied and
  then integrated (summed), the result is zero.
  Therefore if they are used as for channelization,
  theoretically there is no MAI (except due to
  multipath and sync).
• Autocorrelation properties of orthogonal
  sequences usually are not good.

10
Walsh-Hadamard Codes

• Replace zeros by -1 and observe that
• the sum of products of two rows is zero.
• Shifted versions may not be orthogonal.

11
Walsh Sequences of Order 16

• W0 00000000 00000000
• W1 00000000 11111111
• W2 00001111 11110000
• W3 00001111 00001111
• W4 00111100 00111100
• W5 00111100 11000011
• W6 00110011 11001100
• W7 00110011 00110011

• W8 01100110 01100110
• W9 01100110 10011001
• W10 01101001 10010110
• W11 01101001 01101001
• W12 01011010 01011010
• W13 01011010 10100101
• W14 01010101 10101010
• W15 01010101 01010101

12
Variable Length Orthogonal Codes
SF1
SF2
SF4
SF8
SF spreading factor
13
Spreading For Forward Traffic Channel
Base PN code
Walsh 1
User 1 data
S
Modulation
Walsh N
User N data
14
Spreading For Reverse Traffic Channel
data dependent Walsh
PN code 1
User 1 data
Modulation
Spread
data dependent Walsh
PN code N
User N data
Modulation
Spread
15
Advantage of Power as the common
resource

• For handling mixed services and variable bit rate
  demands, allocate power to ensure that maximum
  interference is not exceeded.
• Physical channel allocation remains unchanged
  although bit rate might change. (reallocation of
  codes, time slots, frequencies etc. not needed).

16
Example of power as the common resource
multiplexing variable bit-rate users
17
DS-CDMA BER Performance
18
Simplified Capacity (CDMA)
19
Processing Gain
20
Voice activity

• If the carrier is turned off when the speaker is
  silent, the self-noise reduces from
  U Eb Rb/Ws
• to
• V U Eb Rb/Ws
• where voice activity factor V is typically 0.35
  to 0.4.

21
Example

• (a)A CDMA system uses direct-sequence BPSK
  modulation with a data rate R 5 kbps. Assume 30
  equal power data users. Ignore the thermal noise.
  Determine the minimum chip rate to obtain a BER
  of 10-5.
• BER 10-5 requires an Eb/No of approximately 10.

22
Example (ctd)

• (b) Repeat the calculations if 15 users are data
  the other 15 users are voice users with voice
  activity factor of 0.4.
• 15 voice activated users are equivalent to
  0.4x156 users, then the total is 21 users