Line Coding Visualizer

How Line Coding Visualizer works

Line coding turns a bit stream into a voltage waveform, and the schemes differ in what they optimise for. Unipolar NRZ uses zero and a positive level, which is simple but leaves a large DC component. Polar NRZ-L maps 0 and 1 to opposite levels; NRZ-I instead signals a 1 by inverting the current level and a 0 by holding it, so it survives swapped wires. Both suffer at long runs of identical bits because the receiver has no edge to lock its clock to. Manchester fixes that by putting a transition in the middle of every bit — low-to-high for one value, high-to-low for the other — which makes the signal self-clocking at the price of double the bandwidth. Differential Manchester keeps the mid-bit transition for timing and encodes data as the presence or absence of a transition at the bit boundary. Bipolar AMI leaves 0 at zero volts and alternates each 1 between +V and −V, cancelling DC, but long strings of zeros still starve the clock. B8ZS and HDB3 are AMI with a scrambling rule: they replace a run of zeros with a pattern containing deliberate bipolar violations that the receiver recognises and reverses.

Pseudocode

// Polar NRZ-L: level carries the bit
nrzL(bits):  for b in bits: emit(b == 1 ? +V : -V)

// Polar NRZ-I: a 1 inverts, a 0 holds
nrzI(bits):
    level = -V
    for b in bits:
        if b == 1: level = -level
        emit(level)                       // one level per bit period

// Manchester: mid-bit transition always present
manchester(bits):
    for b in bits:
        if b == 1: emit(-V, +V)           // low then high, half a bit each
        else:      emit(+V, -V)

// Bipolar AMI: zeros idle, ones alternate polarity
ami(bits):
    lastPolarity = -V
    for b in bits:
        if b == 0: emit(0)
        else:
            lastPolarity = -lastPolarity
            emit(lastPolarity)

// B8ZS: replace eight consecutive zeros (V = violation, B = normal pulse)
//   if last pulse was positive: 0 0 0 V B 0 V B  ->  0 0 0 + - 0 - +
//   if last pulse was negative: 0 0 0 V B 0 V B  ->  0 0 0 - + 0 + -

// HDB3: replace four consecutive zeros, choosing by pulse count since
// the last substitution, so that the DC balance is preserved
//   odd  number of ones since last substitution:  0 0 0 V
//   even number of ones since last substitution:  B 0 0 V

Time and space complexity

Time

O(n) in the number of bits

Space

O(n) for the emitted waveform samples

Every scheme is a single left-to-right pass carrying at most a small amount of state — the last polarity for AMI, a zero-run counter for B8ZS and HDB3.

Worked example

Encode the bit stream 0 1 0 0 1 1 with Bipolar AMI, assuming the previous pulse was negative.

1. Bit 0Zero → hold 0 V. Last polarity still −V.
2. Bit 1One → invert from −V, so emit +V. Last polarity = +V.
3. Bit 0Zero → 0 V, polarity unchanged.
4. Bit 0Zero → 0 V, polarity unchanged.
5. Bit 1One → invert from +V, so emit −V.
6. Bit 1One → invert from −V, so emit +V.

Waveform: 0, +V, 0, 0, −V, +V. The ones alternate strictly, so the average voltage is zero — but the two zeros in the middle produced no transitions, which is exactly the clock-recovery gap B8ZS and HDB3 exist to close.

Common exam mistakes

• Mixing up NRZ-L and NRZ-I. NRZ-L encodes the bit in the level itself; NRZ-I encodes a 1 as a change of level and a 0 as no change.
• Drawing Manchester with the transition at the bit boundary. The defining transition is at the middle of the bit period; boundary transitions only appear when consecutive bits require them.
• Treating AMI zeros as a voltage. A zero is no signal at all, not the negative level.
• Forgetting that B8ZS and HDB3 substitutions deliberately violate the alternation rule. The bipolar violation is the signal to the receiver — a "correct" alternating pattern would be undetectable.
• Applying HDB3's parity rule without tracking ones since the last substitution. Choosing 000V versus B00V by the wrong parity destroys the DC balance the scheme exists to maintain.

Frequently asked questions

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