- 2080 Chaitra · 6 marks
Line coding 110011000001 (unipolar RZ, Manchester, HDB3)
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 -.
Represent the given binary sequence 110011000001 in unipolar RZ, Manchester and HDB3 encoders.
Bit sequence: 1 1 0 0 1 1 0 0 0 0 0 1 (12 bits), bit period , levels .
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
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
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)
1 1 0 0 1 1 0 0 0 0 0 1
+V ──| |─| |─| |─|
0 |─| |─────────| |─| |─| |─────────| |─────| |─
-V |─| |─| |─|
✓ Answer checked by Claude Opus 5.5 (high effort)
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