Turbo Code Example

1
Turbo Code Example

• EE 548
• Adv. Topics in Comm. Systems
• Shen Wan

2
Encoder

• Send 3 sub-blocks
• First sub-block is the m-bit data.
• Second sub-block is n/2 parity bits for data.
• Third sub-block is n/2 parity bits for a known
  permutation of the data.
• mn bits total
• Code rate r m/(mn)

3
Information to be sent

• m9 bits
• Data 100010101

4
Add Parity Bits

• Each parity sub-block has 3 bits.
• n6 bits
• Total 15 bits
• Code rate 60

5
D/A Modulation

• Bit 0 is modulated to amplitude -1.0, or phase
  180º.
• Bit 1 is modulated to amplitude 1.0, or phase
  0º.
• Can apply to other modulation similarly

6
Channel Noise

• Assume the change of signal due to the noise is
  normal distribution, with mean 0.0 and standard
  deviation 0.7.

7
Received Analog Signal

• The received signal is the summation of the
  transmit signal and the noise.

8
A/D without Parity Check

• If we just treat all s0.0 to be bit 1 and all
  slt0.0 to be bit 1,
• There will be three error bits.

9
Decoder

• The decoder front-end produces an integer for
  each bit in the data stream.
• This integer measures how likely the bit is a 0
  or 1.
• Therefore, it conveys more information about each
  bit than just 0 or 1.
• There will be two decoders working in parallel to
  check the two parity sub-blocks.

10
Row Decoder

• Solve Row 1
• 0000 0.7
• 0011 2.4
• 0101 0.3
• 1001 93.7
• 0110 0.0
• 1010 0.3
• 1100 2.4
• 1111 0.2

11
Row Decoder

• Solve Row 2
• 0000 0.1
• 0011 0.1
• 0101 82.3
• 1001 10.7
• 0110 0.0
• 1010 0.0
• 1100 2.1
• 1111 4.7

12
Row Decoder

• Solve Row 3
• 0000 1.3
• 0011 0.3
• 0101 0.6
• 1001 10.3
• 0110 1.3
• 1010 23.3
• 1100 52.6
• 1111 10.3

13
Column Decoder

• Solve Column 1
• 0000 0.5
• 0011 0.1
• 0101 0.0
• 1001 0.0
• 0110 28.3
• 1010 63.9
• 1100 5.5
• 1111 1.6

14
Column Decoder

• Solve Column 2
• 0000 0.4
• 0011 6.9
• 0101 80.1
• 1001 0.1
• 0110 10.4
• 1010 0.0
• 1100 0.1
• 1111 2.0

15
Column Decoder

• Solve Column 3
• 0000 0.9
• 0011 80.0
• 0101 3.1
• 1001 15.6
• 0110 0.0
• 1010 0.1
• 1100 0.0
• 1111 0.3

16
Iterative Solving

• The two decoders exchange their guess and
  likelihoods information.
• And then re-compute the likelihoods of every
  possible source bits pattern.
• Until they come to a same result.
• Just like solving crossword or sudoku.

17
Performance in Simulation

• Input frame size 10384 bits.

18
Performance in UMTS

• Frame size1530 bits. Modulation BPSK. Channel
  AWGN

19
Performance in UMTS

• 14 decoder iterations Modulation BPSK Channel
  AWGN
• Shannon limit -1.59dB