---
title: "Rate 1/3 convolutional encoder: encode 10011"
subject: "Communication Systems"
chapter: 8
key: "semester_6/communication_systems:conv-encoder-third-rate-10011"
source: https://www.bctengineer.com/q/semester_6/communication_systems/conv-encoder-third-rate-10011
site: BCT Engineering
---

# Rate 1/3 convolutional encoder: encode 10011

Communication Systems · Chapter 8: Error Detection and Correction Coding

- Asked in: 2076 Chaitra (CS II), 8 marks
- Video: [Convolutional Encoding Using Time Domain and transfer domain Approache | Generator matrix | encoder](https://www.youtube.com/watch?v=-CmR6Dc0cxQ) — Techno Tutorials ( e-Learning)
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_6/communication_systems/conv-encoder-third-rate-10011
- Whole chapter: https://www.bctengineer.com/md/semester_6/communication_systems/chapter-8

## Question

The 1/3 rate convolutional encoder with constraint length equal to 3 has following three generator sequences each of length 3.
(g₀⁽¹⁾, g₁⁽¹⁾, g₂⁽¹⁾) = (0, 1, 1)
(g₀⁽²⁾, g₁⁽²⁾, g₂⁽²⁾) = (1, 0, 1)
(g₀⁽³⁾, g₁⁽³⁾, g₂⁽³⁾) = (1, 1, 0)
Determine the encoded sequence for the following input message (m₀, m₁, m₂, m₃, m₄) = (1 0 0 1 1)

## Short answer

Using the generators (0,1,1), (1,0,1), (1,1,0) to compute c1, c2, c3 per input bit and flushing with two zeros, the encoded sequence is 011 101 110 011 110 011 110.

## Answer

Rate $1/3$: each input bit gives 3 output bits. Constraint length $K=3$, so the register holds $m_i,m_{i-1},m_{i-2}$. Each output is the modulo-2 convolution of the input with its generator:

$$
c_i^{(j)}=\sum_{l=0}^{2}g_l^{(j)}m_{i-l}\pmod 2
$$

With the given generators:

$$
\begin{aligned}
c_i^{(1)}&=0\cdot m_i+m_{i-1}+m_{i-2}=m_{i-1}\oplus m_{i-2}\\
c_i^{(2)}&=m_i\oplus m_{i-2}\\
c_i^{(3)}&=m_i\oplus m_{i-1}
\end{aligned}
$$

```text
 m -->[ m_i ]-->[ m_i-1 ]-->[ m_i-2 ]
        |           |           |
        |           +----(+)----+------> c1 = mi-1+mi-2
        +----------(+)----------+------> c2 = mi+mi-2
        +----(+)----+------------------> c3 = mi+mi-1
     commutator: c1 c2 c3 per input bit
```

#### Encoding

Message $(m_0\dots m_4)=10011$, then two 0s ($K-1$) to flush the register. Register starts at 00.

| $i$ | $m_i$ | $m_{i-1}$ | $m_{i-2}$ | $c^{(1)}$ | $c^{(2)}$ | $c^{(3)}$ |
|---|---|---|---|---|---|---|
| 0 | 1 | 0 | 0 | 0 | 1 | 1 |
| 1 | 0 | 1 | 0 | 1 | 0 | 1 |
| 2 | 0 | 0 | 1 | 1 | 1 | 0 |
| 3 | 1 | 0 | 0 | 0 | 1 | 1 |
| 4 | 1 | 1 | 0 | 1 | 1 | 0 |
| 5 | 0 | 1 | 1 | 0 | 1 | 1 |
| 6 | 0 | 0 | 1 | 1 | 1 | 0 |

#### Check by polynomial (transform-domain) method

$m(D)=1+D^3+D^4$; $g^{(1)}(D)=D+D^2$, $g^{(2)}(D)=1+D^2$, $g^{(3)}(D)=1+D$.

$$
\begin{aligned}
c^{(1)}(D)&=(1+D^3+D^4)(D+D^2)=D+D^2+D^4+D^6\ \Rightarrow 0110101\\
c^{(2)}(D)&=(1+D^3+D^4)(1+D^2)=1+D^2+D^3+D^4+D^5+D^6\ \Rightarrow 1011111\\
c^{(3)}(D)&=(1+D^3+D^4)(1+D)=1+D+D^3+D^5\ \Rightarrow 1101010
\end{aligned}
$$

Interleaving the three sequences bit by bit gives the same result.

**Encoded sequence: 011 101 110 011 110 011 110** (21 bits; the first 15 bits, 011 101 110 011 110, correspond to the message itself and the last 6 come from flushing.)

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. For the rate-1/3 encoder, how many output bits are produced for message 10011 plus the 2 flushing zeros?
   A. 14
   B. 21
   C. 15
   D. 35
   Answer: B. 7 input bits (5 message + K-1 = 2 zeros) times 3 outputs each gives 21 bits. (Key point: Rate and length)

2. With g^(1) = (0,1,1), what is c^(1)_i?
   A. m_i + m_(i-1)
   B. m_(i-1) + m_(i-2)
   C. m_i + m_(i-2)
   D. m_i + m_(i-1) + m_(i-2)
   Answer: B. The leading 0 means the present bit is not tapped, so c^(1) = m_(i-1) XOR m_(i-2). (Key point: Output equation from generator)

3. For m_0 = 1 with register 00, what is the first output c1 c2 c3?
   A. 111
   B. 011
   C. 101
   D. 110
   Answer: B. c1 = 0+0 = 0, c2 = 1+0 = 1, c3 = 1+0 = 1, giving 011. (Key point: First output group)

4. At i = 2 (m_2 = 0, m_1 = 0, m_0 = 1), what is the output c1 c2 c3?
   A. 101
   B. 011
   C. 110
   D. 111
   Answer: C. c1 = 0+1 = 1, c2 = 0+1 = 1, c3 = 0+0 = 0, giving 110. (Key point: Third output group)

## Related in Communication Systems

- [Convolutional encoding of 10011 with state and trellis](https://www.bctengineer.com/question.md?key=semester_6%2Fcommunication_systems%3Aconvolutional-encoder-10011) — Communication Systems, chapter 8
- [Error detection vs correction; rate ½ convolutional encoder](https://www.bctengineer.com/question.md?key=semester_6%2Fcommunication_systems%3Aerror-detect-correct-conv-half-rate) — Communication Systems, chapter 8
- [Hamming distance and weight; rate 1/3 convolutional encoder](https://www.bctengineer.com/question.md?key=semester_6%2Fcommunication_systems%3Ahamming-third-rate-convolutional) — Communication Systems, chapter 8
- [Short note: Trellis diagram of convolutional encoder](https://www.bctengineer.com/question.md?key=semester_6%2Fcommunication_systems%3Ashort-note-trellis-1010-half-rate) — Communication Systems, chapter 8
- [T1 and E1 bit rate and bandwidth; DPCM encoder](https://www.bctengineer.com/question.md?key=semester_6%2Fcommunication_systems%3At1-e1-bitrate-dpcm) — Communication Systems, chapter 5
- [PCM 7-bit at 50 Mbps: maximum message bandwidth](https://www.bctengineer.com/question.md?key=semester_6%2Fcommunication_systems%3Apcm-7bit-50mbps-bandwidth) — Communication Systems, chapter 4

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- [Convolutional encoder c(3,1,3), input 111011](https://www.bctengineer.com/question.md?key=semester_5%2Fdata_communication%3Aconv-encoder-313-111011) — Data Communication, chapter 8
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