---
title: "Convolutional vs block codes; (7,4) cyclic code for 1011"
subject: "Communication Systems"
chapter: 8
key: "semester_6/communication_systems:conv-vs-block-cyclic-7-4-1011"
source: https://www.bctengineer.com/q/semester_6/communication_systems/conv-vs-block-cyclic-7-4-1011
site: BCT Engineering
---

# Convolutional vs block codes; (7,4) cyclic code for 1011

Communication Systems · Chapter 8: Error Detection and Correction Coding

- Asked in: 2078 Chaitra, 2+5 marks
- Video: [Problem on systematic and non-systematic code vectors](https://www.youtube.com/watch?v=q3OUAXpGPCo) — PRAKASAM TUTORIALS
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_6/communication_systems/conv-vs-block-cyclic-7-4-1011
- Whole chapter: https://www.bctengineer.com/md/semester_6/communication_systems/chapter-8

## Question

Why convolution coder is better than block coder? Determine systematic and non-systematic code vector for a (7,4) cyclic hamming code for message vector {1011} with generator polynomial g(x) = 1 + X + X³.

## Short answer

Convolutional coders work on continuous streams with memory, providing better performance. For (7,4) cyclic code with message 1011 and generator g(x)=1+x+x^3, non-systematic code vector is 1111111, systematic is 1001011.

## Answer

#### Why a convolutional coder is better than a block coder

- It works on a **continuous stream**, so no need to wait for a full block; encoding delay and buffering are small.
- It has **memory**: each output depends on several past bits, so redundancy is spread over many bits and burst-like noise is handled better with interleaving.
- **Soft-decision Viterbi decoding** is easy, giving 2 to 3 dB more coding gain than hard-decision block decoding at the same rate.
- Simple shift-register encoder and good performance at low rates (used in GSM, satellite, Wi-Fi).

#### (7,4) cyclic Hamming code, $g(x)=1+x+x^3$, message 1011

Convention (Haykin): message $(m_0m_1m_2m_3)=1011$ in ascending powers, so

$$
m(x)=1+x^2+x^3
$$

Modulo-2 arithmetic.

**Non-systematic:** $c(x)=m(x)g(x)$

$$
\begin{aligned}
c(x)&=(1+x^2+x^3)(1+x+x^3)\\
&=1+x+x^3\;+\;x^2+x^3+x^5\;+\;x^3+x^4+x^6\\
&=1+x+x^2+x^3+x^4+x^5+x^6
\end{aligned}
$$

Non-systematic code vector $(c_0\dots c_6)$ = **1111111**.

**Systematic:** $c(x)=b(x)+x^{n-k}m(x)$, with $b(x)$ = remainder of $x^3m(x)/g(x)$.

$$
x^3m(x)=x^3+x^5+x^6
$$

```text
              x^3 + x^2 + x + 1
          ---------------------------
x^3+x+1 ) x^6 + x^5 +       x^3
          x^6 +       x^4 + x^3
          -------------------------
                x^5 + x^4
                x^5 +       x^3 + x^2
                -----------------------
                      x^4 + x^3 + x^2
                      x^4 +       x^2 + x
                      -------------------
                            x^3 +       x
                            x^3 +       x + 1
                            -----------------
                                          1
```

$b(x)=1$, so parity bits $(b_0b_1b_2)=100$.

$$
c(x)=1+x^3+x^5+x^6
$$

Systematic code vector $(b_0b_1b_2\,m_0m_1m_2m_3)$ = **100 1011**.

Check: $c(x)=(1+x+x^2+x^3)g(x)$, which expands to $1+x^3+x^5+x^6$, so it is a valid codeword.

| Form | Code vector |
|---|---|
| Non-systematic | 1111111 |
| Systematic | 1001011 |

## Check yourself

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

1. Which is an advantage of a convolutional coder over a block coder?
   A. It must wait for a full block before encoding
   B. It works on a continuous stream and allows easy soft-decision Viterbi decoding with extra coding gain
   C. It needs no shift registers
   D. It cannot be used with interleaving
   Answer: B. Continuous encoding with small delay and soft-decision Viterbi decoding gives 2 to 3 dB more coding gain. (Key point: Why convolutional is better)

2. With the Haykin ascending-power convention, what is m(x) for message (m0 m1 m2 m3) = 1011?
   A. x^3 + x + 1
   B. 1 + x^2 + x^3
   C. 1 + x + x^3
   D. x^3 + x^2 + 1
   Answer: B. m0 = 1, m1 = 0, m2 = 1, m3 = 1 gives 1 + x^2 + x^3. (Key point: Message polynomial)

3. For g(x) = 1 + x + x^3, what is the non-systematic code vector c(x) = m(x)g(x)?
   A. 1001011
   B. 1111111
   C. 1011100
   D. 1010001
   Answer: B. (1+x^2+x^3)(1+x+x^3) = 1+x+x^2+x^3+x^4+x^5+x^6 after mod-2 cancellation. (Key point: Non-systematic code)

4. What is the remainder of x^3 m(x) divided by g(x), and the systematic code vector (b0b1b2 m0m1m2m3)?
   A. Remainder 1, code 100 1011
   B. Remainder x, code 010 1011
   C. Remainder 0, code 000 1011
   D. Remainder 1, code 1011 100
   Answer: A. The division leaves b(x) = 1, so parity bits are 100 placed before the message in this convention. (Key point: Systematic code)

## Close variants set in other exams

- [Convolutional vs block coder; (7,4) cyclic code for 1010](https://www.bctengineer.com/question.md?key=semester_6%2Fcommunication_systems%3Aconv-vs-block-cyclic-7-4-1010) — Communication Systems, chapter 8, asked 2×

## Related in Communication Systems

- [Hamming weight and distance; (7,4) cyclic code for 1011](https://www.bctengineer.com/question.md?key=semester_6%2Fcommunication_systems%3Ahamming-weight-cyclic-1011) — Communication Systems, chapter 8, asked 2×
- [(7,4) cyclic code g(x) = 1 + x + x³ for 1011](https://www.bctengineer.com/question.md?key=semester_6%2Fcommunication_systems%3Acyclic-7-4-1x-x3-1011) — Communication Systems, chapter 8, asked 2×
- [(7,4) cyclic code with g(x) = x³ + x² + 1 for 1011](https://www.bctengineer.com/question.md?key=semester_6%2Fcommunication_systems%3Acyclic-code-7-4-1011-x3x2) — Communication Systems, chapter 8, asked 2×
- [Significance of dmin; (7,4) cyclic code for 1101 with g(x) = 1 + x + x²](https://www.bctengineer.com/question.md?key=semester_6%2Fcommunication_systems%3Admin-cyclic-7-4-1101) — Communication Systems, chapter 8
- [(7,4) cyclic code G(p) = p³ + p + 1 for 0101](https://www.bctengineer.com/question.md?key=semester_6%2Fcommunication_systems%3Acyclic-7-4-p3p1-0101) — Communication Systems, chapter 8, asked 2×