--- title: "Line coding 110011000001 (unipolar RZ, Manchester, HDB3)" subject: "Communication Systems" chapter: 6 key: "semester_6/communication_systems:line-codes-110011000001" source: https://www.bctengineer.com/q/semester_6/communication_systems/line-codes-110011000001 site: BCT Engineering --- # Line coding 110011000001 (unipolar RZ, Manchester, HDB3) Communication Systems · Chapter 6: Baseband Digital Data Transmission - Asked in: 2080 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 Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_6/communication_systems/line-codes-110011000001 - Whole chapter: https://www.bctengineer.com/md/semester_6/communication_systems/chapter-6 ## Question Represent the given binary sequence 110011000001 in unipolar RZ, Manchester and HDB3 encoders. ## Short answer Unipolar RZ sends +V for first half of bit 1, 0 otherwise. Manchester uses mid-bit transition: 1=+V to -V, 0=-V to +V. HDB3 uses AMI except four consecutive zeros replaced; for 110011000001 the output is + - 0 0 + - + 0 0 + 0 -. ## Answer Bit sequence: **1 1 0 0 1 1 0 0 0 0 0 1** (12 bits), bit period $T_b$, levels $\pm V$. #### Unipolar RZ 1 is sent as +V for the first half of the bit, then 0. A 0 is sent as no pulse. **Unipolar RZ** ```text 1 1 0 0 1 1 0 0 0 0 0 1 +V ──| |─| |─| |─| |─| 0 |─| |─────────| |─| |─────────────────────| |─ ``` #### Manchester Every bit has a transition at its middle. 1 = +V then −V, and 0 = −V then +V (Lathi/Haykin convention; IEEE 802.3 uses the opposite). **Manchester** ```text 1 1 0 0 1 1 0 0 0 0 0 1 +V ──| |─| |─| |───| |─| |─| |─| |─| |─| |───| -V |─| |───| |─| |─| |───| |─| |─| |─| |─| |─ ``` #### HDB3 HDB3 is AMI, except that every run of four 0s is replaced by **000V** (if an odd number of pulses have been sent since the last substitution) or **B00V** (if even). V has the same polarity as the previous pulse. B follows the AMI rule (opposite to the previous pulse). - Bits 1 to 6 (110011): AMI gives + − 0 0 + −. - Bits 7 to 10 (0000): 4 pulses have been sent since the start, which is even, so this becomes B00V. The previous pulse is −, so B = + and V = +: + 0 0 +. - Bits 11 and 12 (01): 0 stays 0. The 1 must be opposite to the last pulse (+), so it is −. HDB3 pulses: **+ − 0 0 + − + 0 0 + 0 −** **HDB3 (half-width pulses)** ```text 1 1 0 0 1 1 0 0 0 0 0 1 +V ──| |─| |─| |─| 0 |─| |─────────| |─| |─| |─────────| |─────| |─ -V |─| |─| |─| ``` ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. In unipolar RZ, how is a 1 sent and how is a 0 sent? A. 1 is +V then -V; 0 is -V then +V B. 1 is +V for the first half bit then 0; 0 is no pulse C. 1 is +V for the full bit; 0 is -V D. 1 is alternating polarity; 0 is no pulse Answer: B. Unipolar RZ sends a half-width +V pulse for 1 and no pulse for 0. (Key point: Unipolar RZ) 2. In the convention used for this answer, how is a 1 encoded in Manchester code? A. +V for the full bit B. 0 for the full bit C. -V then +V D. +V then -V Answer: D. The answer uses 1 = +V then -V and 0 = -V then +V (Lathi/Haykin); IEEE 802.3 uses the opposite. There is always a mid-bit transition. (Key point: Manchester) 3. In HDB3, what replaces a run of four 0s? A. Always B00V B. Four alternating pulses C. 000V if an odd number of pulses since last substitution, B00V if even D. Always 000V Answer: C. HDB3 chooses 000V when the pulse count since the last substitution is odd and B00V when even, which keeps successive V pulses alternating in polarity. (Key point: HDB3 substitution) 4. For 110011000001, the first four 0s come after pulses + - + - (even count) with last pulse -. What is the substitution? A. - 0 0 + B. + 0 0 + C. 0 0 0 + D. - 0 0 - Answer: B. Even count gives B00V. B follows AMI (opposite to the last pulse -, so +) and V has the same polarity as the previous pulse (B, so +), giving + 0 0 +. (Key point: HDB3 result) ## Related in Communication Systems - [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 1011001010 (polar NRZ, polar RZ, Manchester, AMI)](https://www.bctengineer.com/question.md?key=semester_6%2Fcommunication_systems%3Aline-codes-1011001010) — Communication Systems, chapter 6, asked 8× - [Line coding 1001001101 (polar NRZ, polar RZ, Manchester, AMI)](https://www.bctengineer.com/question.md?key=semester_6%2Fcommunication_systems%3Aline-codes-1001001101) — Communication Systems, chapter 6, asked 8× - [Line coding 1101010111 (polar NRZ, polar RZ, Manchester, AMI)](https://www.bctengineer.com/question.md?key=semester_6%2Fcommunication_systems%3Aline-codes-1101010111-polar-ami) — Communication Systems, chapter 6, asked 8× - [Line coding 1101010011 (polar NRZ, polar RZ, Manchester, AMI)](https://www.bctengineer.com/question.md?key=semester_6%2Fcommunication_systems%3Aline-codes-1101010011) — Communication Systems, chapter 6, asked 8× ## Related in other subjects - [Line coding waveforms for 1101110111](https://www.bctengineer.com/question.md?key=semester_5%2Fdata_communication%3Aline-coding-1101110111) — Data Communication, chapter 5 - [Line coding and NRZ, RZ, AMI](https://www.bctengineer.com/question.md?key=semester_8%2Ftelecommunications%3Aline-coding-nrz-rz-ami) — Telecommunications, chapter 3 - [HDB3 (HD3S) line coding](https://www.bctengineer.com/question.md?key=semester_5%2Fdata_communication%3Ahdb3-coding) — Data Communication, chapter 5, asked 2× - [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