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- 2076 Asoj (CS II) · 6 marks
Line coding 1011000010000000000111
Short answer
Polar NRZ: +V for 1, -V for 0, no transitions in long zero run. Polar RZ: +V/-V for first half of bit then 0; self-clocking but double bandwidth. Manchester: +V then -V for 1, -V then +V for 0; always a mid-bit transition. Unipolar RZ: +V half-bit for 1, 0 for 0; loses timing during zeros.
Draw the timing diagram of Polar NRZ, Polar RZ, Manchester and unipolar RZ for the following binary sequence 1011000010000000000111.
Conventions (bit period , amplitude ):
| Code | Bit 1 | Bit 0 |
|---|---|---|
| Polar NRZ | for full bit | for full bit |
| Polar RZ | first half, then 0 | first half, then 0 |
| Manchester | then (high-to-low at mid-bit) | then (low-to-high) |
| Unipolar RZ | first half, then 0 | 0 for full bit |
Bit sequence (22 bits): 1 0 1 1 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 1 1 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 0 1 1 1
Polar NRZ:
+V |‾‾ ‾‾‾‾ ‾‾ ‾‾‾‾‾‾
0 |
-V | __ ________ ____________________
Polar RZ:
+V |‾ ‾ ‾ ‾ ‾ ‾ ‾
0 | - - - - - - - - - - - - - - - - - - - - - -
-V | _ _ _ _ _ _ _ _ _ _ _ _ _ _ _
Manchester:
+V |‾ ‾‾ ‾ ‾ ‾ ‾ ‾‾ ‾ ‾ ‾ ‾ ‾ ‾ ‾ ‾ ‾ ‾‾ ‾ ‾
0 |
-V | __ _ __ _ _ _ __ _ _ _ _ _ _ _ _ _ _ _ _
Unipolar RZ:
+V |‾ ‾ ‾ ‾ ‾ ‾ ‾
0 | --- - --------- --------------------- - - -
Observations:
- Polar NRZ stays at for the long run of ten 0s (bits 10–19): no transitions, so the receiver can lose bit timing.
- Polar RZ returns to zero in every bit, so every bit has an edge, but it needs twice the bandwidth of NRZ.
- Manchester has a transition at the middle of every bit even in the long zero run, so it is self-clocking and has no DC component; its bandwidth is twice that of NRZ.
- Unipolar RZ gives no pulse at all during the run of zeros, so timing is lost there, and it has a DC component.
✓ Answer checked by Claude Opus 5.5 (high effort)
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