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Aaron Goldsmith
ELEN 21
HW 7
May 24th
1)
Part A:
Part B:
When S = R, Q’s next state will be the same
When S = R,
Q’s next state won’t change
2a)
2b)
Schematic negating k
J
N
Q(t+1)
0
0
0
0
1
Q(t)
1
0
Q(t)
1
1
1
J
N
Q(t+1)
0
0
0
0
1→
0
0
1
0→
1
1
1
1
1
When you connect the two inputs (J ⇐
> N) meaning that If J = 0, N = 0, and if J = 1,
N = 1. This would create the table above.
This mimics a D type, when J = 0 → Q(t+1) = 0 (clear)
when J = 1 → Q(t+1) = 1
2C)
S
R
D
= [ S+Q(t) ] • R
Q(t+1)
0
0
Q(t)
Q(t)
0
1
0
0
1
0
1
1
1
1
0
0
3)
AaronGoldsmith.pdf (PDF, 279.1 KB)
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