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a) Taking the Qbar outputs instead of the Q outputs for a two-bit counter yields the sequence 11, 10, 01, 00, repeat. This is close. We just need to skip the 00 step. This is when the two Q outputs are 1 (since we're taking the Qbar outputs), so we need a two-bit counter which skips 11. We'll skip 11 by saying that when the output is 10, the high bit unexpectedly flips and the low bit unexpectedly doesn't. Thus, the low bit flips when the value is not 10 (as opposed to flipping always), and the high bit flips, instead of when x0, rather when x0+(x1 notx0), which equals x0+x1.

There are many other ways to do this, of course.

b) The low bit flips always, and the high bit is a constant one:


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