- 2071 Chaitra · 3+3 marks
AMI and Manchester for 011010001
Short answer
AMI: 1s alternate +V and -V starting +V, 0=0V; pattern 0+-0+000-. Manchester: 1=low-to-high mid-bit transition, 0=high-to-low.
Draw AMI and Manchester encoding for the sequence [0 1 1 0 1 0 0 0 1].
Bit stream: 0 1 1 0 1 0 0 0 1
Conventions used (state these in the exam, as books differ): level before the first bit is low (-V); NRZ-L/RZ: 1 = +V, 0 = -V (unipolar: 0 = 0 V); NRZ-I: a 1 toggles the level at the start of the bit, a 0 keeps it; RZ: the pulse returns to zero in the middle of the bit; Manchester: 1 = low-to-high and 0 = high-to-low transition at mid-bit; differential Manchester: always a mid-bit transition, an extra transition at the start of the bit for a 0 and none for a 1; AMI: 0 = 0 V, 1s alternate +V and -V starting with +V.
AMI
0 1 1 0 1 0 0 0 1
+V |---| |---|
0 ----| | |---| |-----------|
-V |---| |---
Manchester
0 1 1 0 1 0 0 0 1
+V --| |-| |---| |---| |-| |-| |-
0 | | | | | | | | | | | |
-V |---| |-| |---| |-| |-| |---|
AMI pulses: 0 + - 0 + 0 0 0 - (the four 1s alternate in polarity; 0 = 0 V). Manchester has a transition in the middle of every bit (1 = low-to-high, 0 = high-to-low).
✓ Answer checked by Claude Sonnet 5.5
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