- 2072 Chaitra · 10 marks
Encode 10110001110 with five line codes
Short answer
RZ: 1=+V half-bit pulse returning to zero, 0=-V half-bit. NRZ-I: toggle at start of each 1, initial low gives high-low-high... NRZ-L: 1=+V, 0=-V. AMI: 1s alternate ±V starting +V, 0=0V; stream yields +0-+000-+-0. Manchester: 1=low-to-high mid-bit transition, 0=high-to-low.
Encode the bit stream 10110001110 using the following scheme: a) RZ b) NRZ-I c) NRZ-L d) AMI e) Manchester
Bit stream: 1 0 1 1 0 0 0 1 1 1 0
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.
RZ
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 |---| |---| |-------
NRZ-L
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 |---| |---| |---|
Manchester
1 0 1 1 0 0 0 1 1 1 0
+V |---| |-| |---| |-| |-| |-| |-| |---|
0 | | | | | | | | | | | | | | | |
-V --| |---| |-| |-| |-| |---| |-| |-| |-
AMI pulses: + 0 - + 0 0 0 - + - 0.
✓ Answer checked by Claude Sonnet 5.5
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