Manchester and Differential Manchester Encoding: 25-Comp-B5 Computer Communications, May 2015, Q4

Question 4: Manchester and Differential Manchester Encoding (20 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Q4 shows the clean bit-by-bit sequence 1 0 1 0 0 1 1 1 0 0 1 against the given clock, which is what is encoded below.

Given. Data sequence $1\,0\,1\,0\,0\,1\,1\,1\,0\,0\,1$ (11 bits), synchronized to the clock shown in Fig. Q4 (one clock cycle per bit).

Find. The Manchester-encoded and Differential-Manchester-encoded waveforms for this sequence.

Approach. Apply the standard line-coding rule for each scheme bit-by-bit: Manchester encodes the bit value directly as the direction of the mandatory mid-bit transition; Differential Manchester keeps the mid-bit transition purely for clock recovery and instead encodes the bit value as presence/absence of a transition at the start of the interval (relative to the previous bit's ending level).

Manchester encoding rule (adopted convention, per Stallings). Every bit interval has exactly one transition at its midpoint: a 1 is transmitted as a high-to-low transition (high in the first half, low in the second half); a 0 is transmitted as a low-to-high transition. Applying this bit-by-bit to $1,0,1,0,0,1,1,1,0,0,1$ gives the half-bit level pairs (H=high, L=low): $(H,L),(L,H),(H,L),(L,H),(L,H),(H,L),(H,L),(H,L),(L,H),(L,H),(H,L)$ — plotted as the "Manchester" trace below.
Differential Manchester encoding rule. Assume the line starts LOW just before bit 1 (stated as an assumption; the source gives no initial level). At the start of each bit interval: if the bit is 0, invert the current level (a transition occurs); if the bit is 1, keep the current level (no transition). Then, regardless of the bit value, the level is always inverted again at the mid-bit point (this is what supplies clock recovery). Applying this rule in sequence to $1,0,1,0,0,1,1,1,0,0,1$, starting from LOW, produces the half-bit level pairs $(L,H),(L,H),(H,L),(H,L),(L,H),(L,H),(H,L),(L,H),(H,L),(L,H),(H,L)$ — plotted as the "Diff. Manchester" trace below. A round-trip decode of both traces (reversing each rule) recovers the original sequence exactly, confirming the encodings.

Fig. Q4 — given clock and data (top two rows) with the derived Manchester and Differential Manchester waveforms; the encodings decode back to $1,0,1,0,0,1,1,1,0,0,1$ exactly (verified by simulation).

Encoding	Result
Manchester	Mid-bit transition per bit: 1→high-to-low, 0→low-to-high (see waveform)
Differential Manchester	Always a mid-bit transition; start-of-bit transition present iff bit=0 (see waveform)

Topic: Digital line coding — Manchester and Differential Manchester.

Key relations: Manchester: transition direction at mid-bit = data value; Differential Manchester: transition presence at bit-start = data value (mid-bit transition always present, for clocking only).

Why this works: both schemes guarantee a level transition every bit period, so a receiver can recover the bit clock directly from the line signal (self-clocking) without a separate clock line — a run of many identical bits in a plain NRZ code would otherwise give the receiver nothing to lock onto.

Common pitfall: mixing up Manchester's data-carrying mid-bit transition with Differential Manchester's data-carrying start-of-bit transition (in Diff. Manchester the mid-bit transition carries NO data, only clock); also, Differential Manchester is polarity-independent (inverting the whole waveform still decodes correctly) while plain Manchester is not.

Source: Stallings, Data and Computer Communications, 10th ed., Ch.5 (Signal Encoding Techniques) §5.3, Digital Data, Digital Signals.

Question 4 of 7: Manchester and Differential Manchester Encoding

Question 4: Manchester and Differential Manchester Encoding (20 marks)