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- 2074 Chaitra (CS II) · 6 marks
Line coding 1011000010000000001 (polar NRZ, AMI, Manchester)
Short answer
Polar NRZ: 1 is +V, 0 is -V. AMI: 1 alternates +V/-V, 0 is 0 V. Manchester: 1 is +V then -V, 0 is -V then +V. Long zero runs cause DC in NRZ and timing loss in AMI, while Manchester is self-clocking.
Draw the timing diagram of Polar NRZ, AMI and Manchester for the following binary sequence 1011000010000000001.
Conventions (bit period , amplitude ):
| Code | Bit 1 | Bit 0 |
|---|---|---|
| Polar NRZ | for full bit | for full bit |
| AMI (bipolar) | and alternately | 0 |
| Manchester | then (high-to-low at mid-bit) | then (low-to-high) |
The first 1 in AMI is taken as (full-width pulses).
Bit sequence (19 bits): 1 0 1 1 0 0 0 0 1 0 0 0 0 0 0 0 0 0 1
Each character below is half a bit period.
bits 1 0 1 1 0 0 0 0 1 0 0 0 0 0 0 0 0 0 1
Polar NRZ:
+V |‾‾ ‾‾‾‾ ‾‾ ‾‾
0 |
-V | __ ________ __________________
AMI (bipolar NRZ):
+V |‾‾ ‾‾ ‾‾
0 | -- -------- ------------------
-V | __ __
Manchester:
+V |‾ ‾‾ ‾ ‾ ‾ ‾ ‾‾ ‾ ‾ ‾ ‾ ‾ ‾ ‾ ‾ ‾‾
0 |
-V | __ _ __ _ _ _ __ _ _ _ _ _ _ _ _ _
AMI levels of the five 1s (bits 1, 3, 4, 9, 19): .
Observations:
- Polar NRZ: stays at for the run of nine 0s (bits 10–18), so there are no transitions for timing recovery, and the waveform has a DC component.
- AMI: zero DC because 1s alternate; a violation of alternation shows an error. But the long run of zeros gives 0 V for 9 bits, so timing is lost (solved by B8ZS or HDB3).
- Manchester: has a mid-bit transition in every bit, including the zero run, so it is self-clocking with no DC; it needs twice the bandwidth of NRZ.
✓ Answer checked by Claude Opus 5.5 (high effort)
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