--- title: "(7,4) cyclic code G(p) = p³ + p + 1 for 0101" subject: "Communication Systems" chapter: 8 key: "semester_6/communication_systems:cyclic-7-4-p3p1-0101" source: https://www.bctengineer.com/q/semester_6/communication_systems/cyclic-7-4-p3p1-0101 site: BCT Engineering --- # (7,4) cyclic code G(p) = p³ + p + 1 for 0101 Communication Systems · Chapter 8: Error Detection and Correction Coding - Asked in: 2075 Asoj (CS II), 8 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-p3p1-0101 - 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(p) = p³ + p + 1. Obtain the code vector for the code in non-systematic and systematic form with message vector 0101. ## Short answer Multiply M(p)=p^2+1 by G(p)=p^3+p+1 yields 0100111 non-systematic. Systematic encoding appends remainder 100 to message, giving 0101100. ## 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 $G(p)$ of degree $n-k$ that divides $p^n+1$: - **Non-systematic encoding:** $X(p)=M(p)\,G(p)$; message bits are not visible directly in the codeword. - **Systematic encoding:** $X(p)=p^{n-k}M(p)+\text{rem}\left[\dfrac{p^{n-k}M(p)}{G(p)}\right]$; the message appears unchanged and the remainder gives the $n-k$ parity bits. Given $(n,k)=(7,4)$, $G(p)=p^3+p+1$, $n-k=3$. Convention: leftmost message bit is the highest power. Message $0101$: $$ M(p)=0\cdot p^3+1\cdot p^2+0\cdot p+1=p^2+1 $$ All additions are modulo 2. #### Non-systematic form $$ \begin{aligned} X(p)&=M(p)\,G(p)=(p^2+1)(p^3+p+1)\\ &=p^5+p^3+p^2+p^3+p+1\\ &=p^5+p^2+p+1 \end{aligned} $$ Coefficients from $p^6$ down to $p^0$: | $p^6$ | $p^5$ | $p^4$ | $p^3$ | $p^2$ | $p^1$ | $p^0$ | |---|---|---|---|---|---|---| | 0 | 1 | 0 | 0 | 1 | 1 | 1 | **Non-systematic code vector: 0100111** #### Systematic form 1. Multiply by $p^{n-k}=p^3$: $p^3M(p)=p^5+p^3$. 2. Divide by $G(p)$: ```text p^2 ----------------- p^3+p+1 ) p^5 + p^3 p^5 + p^3 + p^2 --------------------- p^2 <- remainder ``` 3. Remainder $C(p)=p^2$, so parity bits $(p^2,p^1,p^0)=100$. 4. $X(p)=p^3M(p)+C(p)=p^5+p^3+p^2$. Check: $X(p)=p^2\,G(p)=p^5+p^3+p^2$, a multiple of $G(p)$. **Systematic code vector: 0101 100** (message bits, then check bits) | Form | Code vector | |---|---| | Non-systematic | 0100111 | | Systematic | 0101100 | ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. With the leftmost bit as the highest power, what is M(p) for message 0101? A. p^3 + p B. p^2 + 1 C. p^3 + p^2 D. p^2 + p Answer: B. 0101 means 0*p^3 + 1*p^2 + 0*p + 1 = p^2 + 1. (Key point: Message polynomial) 2. How is the non-systematic code vector formed, and what is the result? A. X(p) = M(p)G(p) = p^5 + p^2 + p + 1, i.e. 0100111 B. X(p) = M(p)G(p) = p^5 + p^3 + p^2, i.e. 0101100 C. X(p) = M(p)/G(p), i.e. 0000101 D. X(p) = M(p) + G(p), i.e. 0001101 Answer: A. (p^2+1)(p^3+p+1) = p^5+p^2+p+1 after the two p^3 terms cancel mod 2. (Key point: Non-systematic encoding) 3. What is the first step of systematic encoding for a (7,4) code? A. Divide M(p) directly by G(p) B. Multiply M(p) by G(p) C. Add G(p) to M(p) D. Multiply M(p) by p^(n-k) = p^3 and divide by G(p) to get the remainder Answer: D. Shifting the message by p^3 makes room for the 3 check bits, which are the remainder of p^3 M(p) / G(p). (Key point: Systematic encoding steps) 4. What is the systematic code vector for 0101? A. 0100111 B. 0101001 C. 0101100 D. 1000101 Answer: C. p^3 M(p) = p^5 + p^3 leaves remainder p^2, giving parity 100 appended to message 0101: 0101100. (Key point: Systematic result) ## Close variants set in other exams - [(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× ## 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× - [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 - [(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× - [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 ## Related in other subjects - [Cyclic Redundancy Code with example](https://www.bctengineer.com/question.md?key=semester_5%2Finstrumentation_ii%3Acrc-example) — Instrumentation II, chapter 3 - [Mass and energy equations for non-flow; cyclic process](https://www.bctengineer.com/question.md?key=semester_2%2Ffundamental_of_thermodynamics_and_heat_transfer%3Amass-energy-nonflow-cyclic-closed) — Fundamental of Thermodynamics & Heat Transfer, chapter 4 - [(6,3) block code words, weights, dmin](https://www.bctengineer.com/question.md?key=semester_5%2Fdata_communication%3Ablock-code-63-matrix-weights) — Data Communication, chapter 8