---
title: "(7,4) cyclic code g(x) = 1 + x + x³ for 1011"
subject: "Communication Systems"
chapter: 8
key: "semester_6/communication_systems:cyclic-7-4-1x-x3-1011"
source: https://www.bctengineer.com/q/semester_6/communication_systems/cyclic-7-4-1x-x3-1011
site: BCT Engineering
---

# (7,4) cyclic code g(x) = 1 + x + x³ for 1011

Communication Systems · Chapter 8: Error Detection and Correction Coding

- Asked in: 2070 Chaitra (CS II), 6 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/cyclic-7-4-1x-x3-1011
- Whole chapter: https://www.bctengineer.com/md/semester_6/communication_systems/chapter-8

## Question

The generator polynomial of a (7,4) cyclic code is g(x) = 1 + x + x³. Find the code for the message vector 1011 in a non-systematic and systematic form.

## Short answer

Non-systematic: multiply m(x)=1+x²+x³ by g(x)=1+x+x³ modulo 2, giving codeword 1111111. Systematic: divide x³m(x) by g(x); remainder 1 yields parity bits 100, codeword 1001011.

## Answer

A **cyclic code** is a linear block code in which every cyclic shift of a codeword is also a codeword. An $(n,k)$ cyclic code is generated by a polynomial $g(x)$ of degree $n-k$ that divides $x^n+1$:

- **Non-systematic encoding:** $c(x)=m(x)\,g(x)$; message bits are not visible directly in the codeword.
- **Systematic encoding:** $c(x)=x^{n-k}m(x)+\text{rem}\left[\dfrac{x^{n-k}m(x)}{g(x)}\right]$; the message appears unchanged and the remainder gives the $n-k$ parity bits.

Given $(n,k)=(7,4)$, $g(x)=1+x+x^3$, $n-k=3$. Convention (Haykin): message $(m_0m_1m_2m_3)=1011$ in ascending powers of $x$:

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

All arithmetic is modulo 2.

#### Non-systematic form

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

($x^3$ appears three times, leaving $x^3$.)

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

#### Systematic form

1. $x^{n-k}m(x)=x^3+x^5+x^6$
2. Divide by $g(x)=x^3+x+1$:

```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
```

3. Remainder $b(x)=1$, parity bits $(b_0b_1b_2)=100$.
4. $c(x)=b(x)+x^3m(x)=1+x^3+x^5+x^6$.

Check: quotient $\times\,g(x)=(1+x+x^2+x^3)(1+x+x^3)=1+x^3+x^5+x^6$. Valid.

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

| 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. In the Haykin convention, what is m(x) for message 1011?
   A. x^3 + x + 1
   B. 1 + x + x^3
   C. 1 + x^2 + x^3
   D. x^2 + x + 1
   Answer: C. Bits m0..m3 = 1,0,1,1 in ascending powers give 1 + x^2 + x^3. (Key point: Polynomial convention)

2. What is c(x) = m(x)g(x) and the vector?
   A. 1 + x^3 + x^5 + x^6, giving 1001011
   B. 1 + x + x^2 + x^3 + x^4 + x^5 + x^6, giving 1111111
   C. x^2 + x^3 + x^5, giving 0011010
   D. 1 + x^2, giving 1010000
   Answer: B. (1+x^2+x^3)(1+x+x^3) sums to all seven powers (x^3 appears three times leaving one) mod 2. (Key point: Non-systematic result)

3. What is the remainder of x^3 m(x) divided by g(x), and the parity bits?
   A. Remainder 1, parity b0b1b2 = 100
   B. Remainder x^2, parity 001
   C. Remainder x, parity 010
   D. Remainder 0, parity 000
   Answer: A. x^3+x^5+x^6 divided by x^3+x+1 gives quotient 1+x+x^2+x^3 and remainder 1. (Key point: Systematic parity)

4. What is the systematic code vector?
   A. 1011100
   B. 1111111
   C. 0011101
   D. 1001011 (parity 100 then message 1011)
   Answer: D. c(x) = b(x) + x^3 m(x) = 1 + x^3 + x^5 + x^6, arranged as parity bits then message. (Key point: Systematic vector layout)

## Close variants set in other exams

- [(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×

## Related in Communication Systems

- [Convolutional vs block codes; (7,4) cyclic code for 1011](https://www.bctengineer.com/question.md?key=semester_6%2Fcommunication_systems%3Aconv-vs-block-cyclic-7-4-1011) — 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×
- [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×
- [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×
- [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
- [Short note: Syndrome calculation in linear systematic block code](https://www.bctengineer.com/question.md?key=semester_6%2Fcommunication_systems%3Ashort-note-syndrome-calculation) — Communication Systems, chapter 8

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