Solution : Problem I
I.
JK flip-flop next state table
D flip-flop excitation table
Q
0
0
1
1
Qnext
0
1
0
1
D
0
1
0
1
Excitation table
D = JQ’ + K’Q
Solution : Problem II
The state table is derived as below : ( T 1 , T0 values also shown here, but they don’t
form a part of the state table ) : 20 points
Q1(t)
0
0
0
0
1
1
1
1
Q0(t)
0
0
1
1
0
0
1
1
X(t)
Q1(t+1)
0
1
1
1
0
0
1
1
0
1
0
1
0
1
0
1
Q0(t+1)
0
1
0
1
1
0
1
1
y
0
1
1
1
0
1
0
1
T1
T0
0
1
1
1
1
1
0
0
0
1
1
0
1
0
0
0
Excitation Table for a T Flip Flop,
PS
0
1
T
0
0
1
1
1
0
NS
T0, T1 values , K-Maps and correct T1 and T0 expressions : 10 points
K-Maps for T0 and T1
T0
Q1
Q0X 00
0
1
0
1
01
1
0
11
0
0
10
1
0
T1
Q1
0
1
Q0X 00
0
1
01
1
1
11
1
0
10
1
0
Implies
T0 = Q1’Q0’X + Q1Q0’X’ + Q1’Q0 X’
T1 = Q1Q0’ + Q1’X + Q1’Q0
QED.
Solution : Problem III
1. Tc ≥ Tpcq + Tpd + Tsetup
Longest path:
Tc ≥ Tpcq + 3*Tpd + Tsetup
Tc ≥ 40 + 3*35 + 50 = 195 ps
Max Frequency = 1/Tc = 5.12 GHz
2. Tccq + Tcd≥Thold + Tskew
Shortest Path:
Tccq + Tcd≥Thold + Tskew
30 + 25 ≥ 60 + Tskew
Tskew≤ -5
Solution : Problem IV
IV.1.
IV.2.
IV.3.