- Asked 3/20
- 2068 Chaitra · 6 marks
Encode 10110001110 with unipolar, polar, bipolar codes
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.
Encode the following bit stream using unipolar (NRZ), polar (NRZ-L, NRZ-I, RZ, Manchester), bipolar (AMI) 10110001110.
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.
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.
✓ Answer checked by Claude Sonnet 5.5
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