---
title: "Encode 10110001110 with unipolar, polar, bipolar codes"
subject: "Data Communication"
chapter: 5
key: "semester_5/data_communication:line-coding-10110001110-68c"
source: https://www.bctengineer.com/q/semester_5/data_communication/line-coding-10110001110-68c
site: BCT Engineering
---

# Encode 10110001110 with unipolar, polar, bipolar codes

Data Communication · Chapter 5: Data Encoding and Modulation

- Asked in: 2068 Chaitra, 6 marks
- Video: [Line Coding Techniques | RZ, NRZ, Unipolar, Polar, Bipolar & Manchester | Digital Communication](https://www.youtube.com/watch?v=daf9F9NC7zQ) — Engineering Funda
- Answer checked by Claude Sonnet 5.5 on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_5/data_communication/line-coding-10110001110-68c
- Whole chapter: https://www.bctengineer.com/md/semester_5/data_communication/chapter-5

## Question

Encode the following bit stream using unipolar (NRZ), polar (NRZ-L, NRZ-I, RZ, Manchester), bipolar (AMI) 10110001110.

## Short answer

Unipolar NRZ: + 0 + + 0 0 0 + + + 0. NRZ-L: + – + + – – – + + + –. NRZ-I (from –V): + + – + + + + – + – –. RZ: +0 –0 +0 +0 –0 –0 –0 +0 +0 +0 -0. Manchester: ↑ ↓ ↑ ↑ ↓ ↓ ↓ ↑ ↑ ↑ ↓. AMI: + 0 – + 0 0 0 – + – 0.

## Answer

**Conventions used:**
- Unipolar NRZ: 1 = +V, 0 = 0 V.
- NRZ-L: 1 = +V, 0 = −V (level depends on the bit value).
- NRZ-I: a 1 causes a transition at the start of the bit, a 0 keeps the previous level (starting level assumed −V).
- RZ: 1 = +V for the first half then 0; 0 = −V for the first half then 0 (returns to zero in every bit).
- Manchester: mid-bit transition; 1 = low-to-high, 0 = high-to-low.
- Bipolar AMI: 0 = zero; 1s alternate +V and −V, first 1 is +V.

```text
unipolar
   1   0   1   1   0   0   0   1   1   1   0   
+V ────┐   ┌───────┐           ┌───────────┐   
0      └───┘       └───────────┘           └───
-V                                             

NRZ-L
   1   0   1   1   0   0   0   1   1   1   0   
+V ────┐   ┌───────┐           ┌───────────┐   
0      │   │       │           │           │   
-V     └───┘       └───────────┘           └───

NRZ-I
   1   0   1   1   0   0   0   1   1   1   0   
+V ────────┐   ┌───────────────┐   ┌───┐       
0          │   │               │   │   │       
-V         └───┘               └───┘   └───────

RZ
   1   0   1   1   0   0   0   1   1   1   0   
+V ──┐     ┌─┐ ┌─┐             ┌─┐ ┌─┐ ┌─┐     
0    └─┐ ┌─┘ └─┘ └─┐ ┌─┐ ┌─┐ ┌─┘ └─┘ └─┘ └─┐ ┌─
-V     └─┘         └─┘ └─┘ └─┘             └─┘ 

Manchester
   1   0   1   1   0   0   0   1   1   1   0   
+V   ┌───┐   ┌─┐ ┌───┐ ┌─┐ ┌─┐   ┌─┐ ┌─┐ ┌───┐ 
0    │   │   │ │ │   │ │ │ │ │   │ │ │ │ │   │ 
-V ──┘   └───┘ └─┘   └─┘ └─┘ └───┘ └─┘ └─┘   └─

AMI
   1   0   1   1   0   0   0   1   1   1   0   
+V ────┐       ┌───┐               ┌───┐       
0      └───┐   │   └───────────┐   │   │   ┌───
-V         └───┘               └───┘   └───┘   
```

**Level sequences (per bit):**
- Unipolar NRZ: +, 0, +, +, 0, 0, 0, +, +, +, 0
- NRZ-L: +, −, +, +, −, −, −, +, +, +, −
- NRZ-I: + (toggled from −), +, −, +, +, +, +, −, +, −, −
- RZ: +0, −0, +0, +0, −0, −0, −0, +0, +0, +0, −0
- Manchester: ↑, ↓, ↑, ↑, ↓, ↓, ↓, ↑, ↑, ↑, ↓
- AMI: +, 0, −, +, 0, 0, 0, −, +, −, 0

The long run of three 0s (bits 5 to 7) gives a flat line in unipolar, NRZ-L (at −V), NRZ-I and AMI, which is a synchronization problem. RZ and Manchester still have transitions in every bit.

## Check yourself

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

1. What are the levels for unipolar NRZ in this answer?
   A. 1 = +V, 0 = 0 V
   B. 1 = +V, 0 = -V
   C. 1 = 0 V, 0 = +V
   D. Alternating
   Answer: A. Unipolar uses only one polarity. (Key point: Unipolar)

2. Starting at -V, what is the NRZ-I sequence for 1 0 1 1 0 0 0 1 1 1 0?
   A. +, +, -, +, +, +, +, -, +, -, -
   B. +, -, +, +, -, -, -, +, +, +, -
   C. +, +, +, -, -, +, +, -, -, +, +
   D. -, -, +, -, -, -, -, +, -, +, +
   Answer: A. Each 1 toggles; each 0 keeps the level. (Key point: NRZ-I sequence)

3. Why is the run of three 0s a problem for NRZ-L, NRZ-I, unipolar and AMI?
   A. Flat line, so no transitions for clock recovery
   B. It creates a DC offset only
   C. It doubles bandwidth
   D. It causes alias
   Answer: A. No transitions gives a synchronisation problem; RZ and Manchester still transition each bit. (Key point: Sync problem)

4. What is the AMI sequence for the bit stream?
   A. +, 0, -, +, 0, 0, 0, -, +, -, 0
   B. +, 0, +, +, 0, 0, 0, +, +, +, 0
   C. +, 0, -, -, 0, 0, 0, +, +, -, 0
   D. +, 0, -, +, 0, 0, 0, +, -, +, 0
   Answer: A. 1s alternate + - + - + -. (Key point: AMI)

## Close variants set in other exams

- [Encode 1101001 with polar NRZ, AMI, Manchester](https://www.bctengineer.com/question.md?key=semester_5%2Fdata_communication%3Aline-coding-1101001) — Data Communication, chapter 5, asked 3×
- [Encode 1001101110 with unipolar, polar, bipolar codes](https://www.bctengineer.com/question.md?key=semester_5%2Fdata_communication%3Aline-coding-1001101110) — Data Communication, chapter 5, asked 3×

## Related in Data Communication

- [Polar, unipolar, bipolar, Manchester; encode 101000111011](https://www.bctengineer.com/question.md?key=semester_5%2Fdata_communication%3Aline-coding-101000111011) — Data Communication, chapter 5
- [Encode 10101111000011 with NRZ, RZ, AMI, Manchester](https://www.bctengineer.com/question.md?key=semester_5%2Fdata_communication%3Aline-coding-10101111000011) — Data Communication, chapter 5
- [Encode 10110001110 with five line codes](https://www.bctengineer.com/question.md?key=semester_5%2Fdata_communication%3Aline-coding-10110001110) — Data Communication, chapter 5
- [Line coding: polar RZ and bipolar AMI](https://www.bctengineer.com/question.md?key=semester_5%2Fdata_communication%3Aline-coding-polar-rz-ami) — Data Communication, chapter 5
- [Encode 10010110001 with NRZ-L, NRZ-I, diff. Manchester](https://www.bctengineer.com/question.md?key=semester_5%2Fdata_communication%3Aline-coding-10010110001) — Data Communication, chapter 5
- [Encode 1010011001 with five line codes](https://www.bctengineer.com/question.md?key=semester_5%2Fdata_communication%3Aline-coding-1010011001) — Data Communication, chapter 5

## Related in other subjects

- [Line coding 1001101101 (unipolar NRZ, polar RZ, AMI, Manchester)](https://www.bctengineer.com/question.md?key=semester_6%2Fcommunication_systems%3Aline-codes-1001101101) — Communication Systems, chapter 6, asked 8×
- [Line coding 10110101 (polar NRZ, unipolar RZ, AMI, Manchester)](https://www.bctengineer.com/question.md?key=semester_6%2Fcommunication_systems%3Aline-codes-10110101) — Communication Systems, chapter 6, asked 8×
- [Line coding 100111010 (Polar RZ, bipolar NRZ, AMI, Manchester)](https://www.bctengineer.com/question.md?key=semester_6%2Fcommunication_systems%3Aline-codes-100111010) — Communication Systems, chapter 6
- [Definitions of unipolar, polar, bipolar, RZ, Manchester codes](https://www.bctengineer.com/question.md?key=semester_6%2Fcommunication_systems%3Adefine-line-codes) — Communication Systems, chapter 6, asked 2×
