1. UCLA Progress Report OCDMA Channel Coding
UCLA Electrical Engineering Department-Communication Systems Laboratory
UCLA Progress Report OCDMA Channel Coding
Jun Shi
Andres I. Vila Casado
Miguel Griot
Richard D. Wesel

2. OCDMA Goals: Security Improve security over assigned-wavelength WDM?
UCLA answer: Cryptographically secure wavelength hopping.
A special case of “wheat and chaff” where the other users are the chaff. (Ron Rivest 1998)
Criticism: Requires synchronization of all users.
Hey, that’s a good point!
We’ll spend the rest of our program on that…

3. OCDMA Goals: Uncoordinated Access
Can a transmitter and receiver in a large optical network communicate efficiently without coordinating with the other users in the network?
There are two types of coordination:
Synchronization at the bit or frame level.
Assignment (as in wavelength or code assignment)
Systems that are asynchronous but require customized code assignment could be complex ways of delivering what WDM already does.

4. The OR Channel
User 1
Output
User 2 +
User n
No multi-level detection (receiver detects only presence of light)
This means that the channel can be seen as an OR channel (1+ X = 1; 0 + X = X)
With proper thresholding, bit synchronization is the worst case.

5. OR Channel
Channel capacity for a two user OR channel: a union of pentagons.

6. What’s possible on the OR channel
Two-User Case
Information theory says 100%
Efficiency is always possible
But different points require
different ones densities for
Different users. Coordination!

7. What’s possible on the OR channel
Two-User Case
The blue pentagon shows the region achievable when both users are assigned the same ones density of
The equal-rate point can be achieved for any number of users without coordination.

8. Summary so far…
Information theory says that 100% efficiency is possible. (Rates sum to 1.)
However, for a particular set of rates each user needs to transmit a particular ones density.
If all n users transmit with a particular ones density: then uncoordinated transmission with perfect efficiency is possible
for a restricted set of rate-tuples including the equal-rate point.

9. The Catch
In order to have 100% efficiency, a receiver must decode all streams (either jointly or successively) …unless the streams are orthogonal.
This is in contrast to the common practice of treating the other users as noise.

10. Cost of treating other user as noise.
Two-User Case

11. Cost converges to 30.6%.

12. How to treat the other users
Some systems might try to make the other users orthogonal to the desired user.
This is hard to do properly without coordination (assignment).
A receiver may treat the other users as noise and incur approximately a 30% rate penalty (not that bad, really.)
A receiver may decode all the users either jointly or successively.

13. The Z-Channel
Whether we treat the other users as noise or decode succesively, we encounter the Z-channel.

14. Simple Two-User Code User Information Bit Codeword User 1, Rate-1/4
1100 1
0011
User 2, Rate-1/6
101010
010100

15. Simple codes for Demo (2)
Source 1 1
Rate 1/4
Source 2
Rate 1/6
Receiver 1 looks for position of 0 (which always exists)
If 1 or 2, decide 1
If 3 or 4, decide 0
Sum Rate 5/12

16. Simple codes for Demo (2)
Source 1 1
Rate 1/4
Source 2
Rate 1/6
Receiver 1 looks for position of 0 (which always exists)
If 1 or 2, decide 1
If 3 or 4, decide 0
Sum Rate 5/12

17. Simple codes for Demo (2)
Source 1 1
Rate 1/4
Source 2
Rate 1/6
Receiver 1 looks for position of 0 (which always exists)
If 1 or 2, decide 1
If 3 or 4, decide 0
Sum Rate 5/12
Receiver 2 looks for
FIRST position of 0.
If 1, 3 or 5, decide 1
If 2, 4 or 6, decide 0
Worst Case :
block i
block i+1

18. Simple codes for Demo (2)
Source 1 1
Rate 1/4
Source 2
Rate 1/6
Receiver 1 looks for position of 0 (which always exists)
If 1 or 2, decide 1
If 3 or 4, decide 0
Sum Rate 5/12
Receiver 2 looks for
FIRST position of 0.
If 1, 3 or 5, decide 1
If 2, 4 or 6, decide 0
Worst Case :
block i
block i+1

19. LDPC codes to avoid code assignment
There are essentially two elements that must be designed when building and LDPC code :
Degree distributions
Edge positioning
This group has worked a lot on the edge positioning problem, thus we have all the necessary designing tools
Also, the degree distribution design problem for symmetric channels (AWGN, BSC, BEC) has been thoroughly studied by many authors.

20. LDPC codes for asymmetric channels
Density evolution is a concept developed by Richardson et al. that helps predict the LDPC code behavior in symmetric channels
This concept can be used in the design of good degree distribution in many different ways
Wang et al. recently generalized Richardson’s result for asymmetric channels.
We took all these concepts and tried several different linear programming based algorithms, and built a program that efficiently designs degree distributions for any binary memoryless channel.
Given a channel, the program tries to maximize the rate while maintaining an acceptable performance.

21. Degree Distribution Design for Z-channels
Target alpha =
Capacity
Rates
l(x) = x x x x11

22. Max variable node degree
Code Characteristics
Max variable node degree
Total number of edges
MD 12 12
7289
MD 13 13
7305
MD 12 v2
7413
Wang
7316
RCEV 10
7182

23. Simulation of Codes on the Z Channel

24. Simulation of Codes on the Z Channel

25. Degree Distribution Design for Z-channels
Capacity
Rates
Target alpha = 0.5

26. Simulation of Rate-0.291964 LDPC code

27. LDPC code vs. Information Theory

28. Density Transformer First approach
After the density transformer each user transmits a one with probability p

29. Density Transformer
The output bits of a binary linear code (such as LDPC codes) are equally likely 1 or 0 if the information bits are also equally likely
Thus we need to change this distribution in order to avoid such a large interference between users
Transmitter: Non-uniform mapper
1) Mapper
2) Huffman Decoder
Receiver: Use soft Huffman encoder

30. Joint decoding
Joint decoding will be done on graphs that look like this one
The formulas necessary in this joint decoding have already been found and simplified

31. Future work Combine density transformer and LDPC code design.
Test in joint-decoding, successive decoding, and users-as-noise cases.
Make efficiencies close to 100% and 70% a reality for joint-decoding and single-user decoding respectively.

32. Successive Decoding: The Z-Channel
Successive decoding for n users:
User with lowest rate is decoded first
Other users are treated as noise
The decoded data of the first user is used in the decoding of the remaining users
First user sees a “Z-channel”
Where ai = 1-(1-p)n-i is the probability that at least one of the n-i remaining users transmits a 1

33. Successive Decoding: Z with erasures
Intermediate users see the following channel assuming perfect previous decoding
Where bi = 1-(1-p)i-1 is the probability that at least one of the I-1previously decoded users transmitted a 1 Xi
1 - a1 Yi bi
ai (1-bi) e bi 1 1
1-bi

34. Successive Decoding The last user sees a BEC channel a1 a1 X n Yna1 e a1 1 1
1 - a1

35. Successive Decoding For two users
User 1 sees a Z-channel
User 2 sees the following channel
When BER1 is small, the channel can be approximated to a BEC X2 Y2
1 - p
p(1-BER1)
p(BER1) e
p(1-BER1)+(1-p)BER1 1 1
(1-p)(1-BER1) + p(BER1)

36. A 3-user example ? 1 R1 1 User 1 User 2 User 3 1 1 R3 R2 User 1 1 11
User 1 1 1
User 2
User 3
Receiver

37. A 3-user example 1 ? e R1 1 User 1 User 2 User 3 1 1 R3 R2 User 1 1 11 ? e
User 1 1 1
User 2 e
User 3
Receiver

38. A 3-user example 1 ? e R1 1 User 1 User 2 User 3 1 1 R3 R2 User 1 1 11 ? e
User 1 1 1
User 2 e
User 3
Receiver

39. A 3-user example 1 R1 1 User 1 User 2 User 3 1 1 R3 R2 User 1 User 21
User 2
User 3
Receiver