Without the clear, we do not know what Q is until after the
first trailing edge. At that point, since S 1, Q goes to 1. At
the next clock, both S and R are 0; thus, Q does not change.
Since R 1 for the next two clock times, Q goes to 0. Then
S 1, making Q 1; both are 0 leaving Q at 1; and finally
R 1, returning Q to 0. With the clear, Q goes to 0 earlier
and the first clock edge is ignored. Thus, Q remains at 0 for
the next three clock times. Then, this part behaves like the
first part. (Once the Q from the second part is the same as
that from the first, the behavior is identical.)
Clo