Posted by John Chappel on January 18, 1998 at 13:30:59:
In Reply to: Lab1 Question posted by Jim on January 18, 1998 at 12:17:21:
Ooops! I was looking at a different version of the lab handout.
There is a no phase shift in the digram. In reality, the differential
will change some time after the data changes.
You are probably better to implement the encoder as a clocked circuit,
in which case the differential value will be shifted to the right.
Consider the Data signal to be the input to the encoder signal.
The differential is the output of the encoder, and is the
same as D+.
The differential can be driven directly by a flip flop. If the Data is
1, the flip flop retains its current value. If the Data is 0, the flip
flop is toggled. This circuit would produce a waveform in which the differential
is shifted 1 clock to the right of that shown in the lab.