We did not need to construct the state table to obtain either the timing
diagram or the trace.We could, at each clock trailing edge, determine
the behavior of each flip flop. The output can then be constructed last,
since it is just the OR of the two state variables (A and B). Thus, when the
first clock edge arrives, A =B= x= 0 and, thus, JA = KA = 0, leaving
A at 0. At the same time, JB = KB = 1 and thus B toggles, that is, goes to
1. We can now shift our attention to the next clock time and repeat the
computations.
At this point, a word is in order about the initial value. For this
example, we assumed that we knew what was stored in A and B when thefirst clock arrived. That may have been achieved using a static clear
input, which was not shown to simplify this problem. In some cases, we
can determine the behavior of the system after one or two clock periods
even if we did not know the initial value. (That will be the case in the
next example.) But, in this problem, we must initialize the system. (Try
the other initial states and note that each follows a completely different
sequence over the time period shown.) Finally, for this problem (a Moore
model), the state diagram is given in Figure 6.26.