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

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…

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.

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.

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

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!

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.

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.

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.

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

Cost converges to 30.6%.

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.

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

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

Simple codes for Demo (2) Source 1 1 2 3 4 1 Rate 1/4 Source 2 1 2 3 4 5 6 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

Simple codes for Demo (2) Source 1 1 2 3 4 1 Rate 1/4 Source 2 1 2 3 4 5 6 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

Simple codes for Demo (2) Source 1 1 2 3 4 1 Rate 1/4 Source 2 1 2 3 4 5 6 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

Simple codes for Demo (2) Source 1 1 2 3 4 1 Rate 1/4 Source 2 1 2 3 4 5 6 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

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.

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.

Degree Distribution Design for Z-channels Target alpha = 0.2731 0.371989 0.380872 0.389073 0.396646 0.403669 0.410180 0.416236 0.421862 0.427106 0.431998 0.436565 0.440834 0.444828 0.448566 0.452070 0.455357 0.458442 0.461342 0.464062 0.466625 0.469032 0.471299 0.473436 0.475448 0.477345 0.479132 0.480820 0.482412 0.483915 0.485333 0.486675 0.487941 0.489139 0.490271 0.491342 0.492354 0.493311 0.494217 0.495076 0.495891 0.496662 0.497393 0.498086 0.498743 0.499366 0.499957 0.500516 0.501047 0.501550 0.502027 Capacity 0.532439 Rates l(x) = 0.27571047 x + 0.15042832 x2 + 0.18575028 x3 + 0.38811080 x11

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

Simulation of Codes on the Z Channel

Simulation of Codes on the Z Channel

Degree Distribution Design for Z-channels -0.051744 -0.026240 -0.003107 0.017941 0.037158 0.054751 0.070872 0.085686 0.099326 0.111926 0.123588 0.134406 0.144449 0.153785 0.162473 0.170558 0.178103 0.185163 0.191761 0.197937 0.203717 0.209134 0.214218 0.218996 0.223478 0.227695 0.231663 0.235397 0.238912 0.242225 0.245348 0.248085 0.250882 0.253517 0.256000 0.258348 0.260559 0.262657 0.264637 0.266504 0.268278 0.269951 0.271536 0.273036 0.274453 0.275797 0.277072 0.278278 0.279422 0.280506 0.281533 0.282507 0.283431 0.284306 0.285137 0.285925 0.286673 0.287384 0.288059 0.288703 0.289321 0.289910 0.290461 0.290986 0.291486 0.291964 Capacity 0.321928 Rates Target alpha = 0.5

Simulation of Rate-0.291964 LDPC code

LDPC code vs. Information Theory

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

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

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

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.

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

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

Successive Decoding The last user sees a BEC channel a1 a1 X n Yn a1 e a1 1 1 1 - a1

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)

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

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

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

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