--- title: "Convolutional encoder c(3,1,3), input 00110" subject: "Data Communication" chapter: 8 key: "semester_5/data_communication:conv-encoder-313-00110" source: https://www.bctengineer.com/q/semester_5/data_communication/conv-encoder-313-00110 site: BCT Engineering --- # Convolutional encoder c(3,1,3), input 00110 Data Communication · Chapter 8: Information Theory and Coding - Asked in: 2076 Asoj, 10 marks - Video: [Convolution Encoder | Example of Convolution coding | AKTU 2020-21 | ITC](https://www.youtube.com/watch?v=Tn0qOkUZUD0) — Techno Tutorials ( e-Learning) - Answer checked by Claude Sonnet 5.5 on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_5/data_communication/conv-encoder-313-00110 - Whole chapter: https://www.bctengineer.com/md/semester_5/data_communication/chapter-8 ## Question Design a suitable generation matrix for a convolution code using c(3,1,3) architecture and encode input data stream of (00110). ## Short answer Encoder with g1=111, g2=101, g3=011. Input 00110 yields output 000 000 110 011 010. With two tail zeros, final sequence: 000 000 110 011 010 111 000. ## Answer A convolutional encoder **c(n, k, K) = c(3,1,3)** takes k = 1 input bit at a time and produces n = 3 output bits (rate 1/3), with constraint length K = 3 (the present bit and two previous bits are stored: 2 shift-register stages). #### Encoder design Generator sequences (taps on b, s1, s2 = present, previous, second previous bit): - g1 = 111 → out1 = b ⊕ s1 ⊕ s2 - g2 = 101 → out2 = b ⊕ s2 - g3 = 011 → out3 = s1 ⊕ s2 ```text input b ---+--->[s1]--->[s2] | | | +--> (+) <-----+ -> out1 (b+s1+s2) +--> (+) <-----+ -> out2 (b+s2) s1,s2 -> (+) -> out3 (s1+s2) ``` #### Generator matrix The block for each delay is (g1 g2 g3 taps): delay 0 → 110, delay 1 → 101, delay 2 → 111. The matrix for a 5-bit input has each row shifted by 3 columns: ```text | 110 101 111 000 000 000 000 | | 000 110 101 111 000 000 000 | G = | 000 000 110 101 111 000 000 | | 000 000 000 110 101 111 000 | | 000 000 000 000 110 101 111 | ``` Output = u·G (mod 2). In polynomial form: g1(D) = 1 + D + D², g2(D) = 1 + D², g3(D) = D + D². #### Encoding u = 0 0 1 1 0 (state = s1 s2, starting 00) | Time | Input b | State s1 s2 | out1 | out2 | out3 | Output | |---|---|---|---|---|---|---| | 1 | 0 | 00 | 0 | 0 | 0 | 000 | | 2 | 0 | 00 | 0 | 0 | 0 | 000 | | 3 | 1 | 00 | 1 | 1 | 0 | 110 | | 4 | 1 | 10 | 0 | 1 | 1 | 011 | | 5 | 0 | 11 | 0 | 1 | 0 | 010 | | 6 (flush) | 0 | 01 | 1 | 1 | 1 | 111 | | 7 (flush) | 0 | 00 | 0 | 0 | 0 | 000 | (Row 5: b=0, s1=1, s2=1 gives out1 = 0⊕1⊕1 = 0, out2 = 0⊕1 = 1, out3 = 1⊕1 = 0.) Coded sequence: **000 000 110 011 010** and, including the two tail zeros that return the encoder to state 00, **000 000 110 011 010 111 000**. **Answer: 000 000 110 011 010 (111 000 with tail)** ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. What does c(3,1,3) mean here? A. 3 inputs, 1 output, 3 stages B. 1 input bit in, 3 output bits out (rate 1/3), constraint length 3 C. Rate 3, 1 memory D. 3-bit blocks with 1 parity bit Answer: B. n = 3 outputs per k = 1 input bit with K = 3, i.e. the present bit and two stored bits. (Key point: Code parameters) 2. Which generators are used? A. g1 = 111, g2 = 101, g3 = 011 B. g1 = 100, g2 = 010, g3 = 001 C. g1 = 110, g2 = 011, g3 = 111 D. g1 = 111, g2 = 111, g3 = 111 Answer: A. out1 = b+s1+s2, out2 = b+s2, out3 = s1+s2. (Key point: Generator sequences) 3. For input 00110 starting at state 00, what is the output when the first 1 enters? A. 000 B. 111 C. 011 D. 110 Answer: D. b=1, s1=s2=0 gives out1 = 1, out2 = 1, out3 = 0 = 110. (Key point: First nonzero output) 4. What is the coded sequence for the 5 message bits? A. 000 000 110 011 010 B. 110 011 010 000 000 C. 000 110 011 010 111 D. 000 000 111 011 010 Answer: A. Outputs per bit are 000, 000, 110, 011, 010; the tail adds 111 000 to flush. 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