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Differential Manchester encoding has a transition at every bit midpoint and encodes the data by the presence or absence of a boundary transition. A consistent signal polarity reversal preserves those transition decisions.

The bit rate counts data bits carried per second; the baud rate counts transmitted signal elements per second. A signal element need not carry one whole data bit. In Manchester encoding, each data bit occupies two half-bit signal elements, one high and one low. For example, choose low–high for zero and high–low for one. The compulsory transition in the middle provides a timing reference for clock recovery.

If one data bit lasts $T$ seconds, the bit rate is $1/T$, while each half-bit signal element lasts $T/2$. With this binary signal-element convention,Thus a 10 Mbit/s Manchester encoding signal uses 20 million half-bit signal elements per second. This is not a count of observed voltage transitions: some bit boundaries also contain a transition and others do not. Calling a whole two-element pattern a symbol gives a different symbol-counting convention; the physical fact is two half-bit elements per data bit.

In differential Manchester encoding, every data bit still has a compulsory transition at its midpoint for clock recovery. The data value is specified by whether a transition occurs at the beginning of the bit, rather than by the direction of the midpoint transition. One convention is beginning transition for zero and no beginning transition for one; interchanging the meanings consistently also works.

Reversing the signal polarity exchanges high and low everywhere, but preserves both the presence and the absence of transitions. A receiver using differential Manchester encoding therefore decodes the same data after a polarity reversal, provided it has acquired the timing and boundary reference. Ordinary Manchester encoding with a fixed low–high/high–low interpretation would exchange the two data values. The advantage is