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Channel coding in Java Assign PDF 417 in Java Channel coding




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Channel coding using barcode maker for jar control to generate, create pdf417 image in jar applications. Microsoft Official Website Figure 7.10 Encoder and decoder for a parallel concatenated turbo code. Information bits u Encoder 1 Interleaver u Modulator and channel y (1) Parity bits y (2). Encoder 2 Parity bits y (3) Decoder inputs y (1). Decoder 1 (BCJR) y (2) L applet barcode pdf417 12 (u) Lin ( u). L out (u) = L in(u) + L channel (u) + L code (u) Used for final hard decisions L12(u) = Lcode(u) Extrinsic information sent to Decoder 2. Lin (u) 1 L21 ( u) L21( u) = Lcode( u) Extrinsic information sent to Decoder 1 y (1) y (3). Decoder 2 (BCJR). through the channel (we use BPSK over an AWGN channel as a running example). Thus, if the constituent encoders are rate 1/2, then the overall turbo code thus obtained is of rate 1/3. A higher rate can be achieved using the same construction simply by not transmitting some of the bits generated by the encoders.

This procedure is referred to as puncturing. Note that a punctured convolutional code can be decoded in the same manner as one without puncturing, by interpreting the bits not sent as erasures (set Lchannel b = 0 in the logarithmic BCJR algorithm). For simplicity of notation, however, we do not consider puncturing in our discussion here.

In the iterative decoding depicted in Figure 7.10. both decoders see the channel output y 1 for the information sequence.

Decoder 1 sees the channel output y 2 for the parity sequence for Encoder 1, while Decoder 2 sees the channel output y 3 for the parity sequence from Encoder 2. The decoders exchange information about the information sequence u. For a typical information bit u, the two decoders function as follows.

Channel LLRs These are computed as in the standard BCJR algorithm for both information and parity bits. For example, for a bit b sent using BPSK over AWGN, we have Lchannel b = 2Ay/ 2 , where y = Ab + N , N N 0 2 . Decoder 1 Operates using channel outputs y 1 and y 2 , and extrinsic information from Decoder 2.

Step 1 Receive extrinsic information from Decoder 2 regarding the LLRs of information bits. Use these as Lin u for BCJR algorithm (set Lin u = 0 for first iteration). Set Lin v = 0 for parity bits.

. 7.2 Turbo codes and iterative decoding Step 2 Run forward and b ackward recursions. Step 3 Compute Lcode u for each information bit u. Feed Lcode u back as extrinsic information to Decoder 2 (note that Lcode u does not depend on Lin u or Lchannel u , and hence is not directly available to Decoder 2).

Decoder 2 Operation is identical to that of Decoder 1, except that it works with the permuted information sequence, and with the received signals y 1 (permuted) and y 3 . Iteration and termination Decoders 1 and 2 interchange information in this fashion until some termination condition is satisfied (e.g.

, a maximum number of iterations is reached, or the LLRs have large enough magnitude, or a CRC check is satisfied). Then they make hard decisions based on Lout u = Lin u + Lchannel u + Lcode u for each information bit based on the output of Decoder 1. Serial concatenated codes Serial concatenation of convolutional codes is shown in Figure 7.

11. The output from the first convolutional encoder is interleaved and then fed as input to a second convolutional encoder. The output from the second encoder is then modulated and transmitted over the channel.

In the iterative decoding method depicted in Figure 7.11, only Decoder 2 sees the channel outputs. Thus, extrinsic information sent by Decoder 2 is given by Lcode + Lchannel , since Decoder 1 does not have access to the channel.

Decoder 1 employs this extrinsic information to compute Lcode , and sends it back to Decoder 2 as extrinsic information. The final decisions are based on the output LLRs Lout from Decoder 1, since the information sequence is the.
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