3. Design a counter with the following repeated binary sequence: 0,1,2,4,6. Use D flip-flop.
4. Design a counter to count with T flip-flops that goes through the following binary repeated sequence: 0,1,3,7,6,4. Find out the counter response towards the unused state. Illustrate the response with a state diagram.
5. Design a mod-7 counter (repeat binary sequence: 0,1,2,3,4,5,6) use JK flip-flop.
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3)
Given sequence is
0 --- 1 --- 2 ---4 ---6,
Step1:State Transition table
we can make the state diagram as follows.
Present State | Next State |
---|---|
0 | 1 |
1 | 2 |
2 | 4 |
4 | 6 |
6 | 0 |
Step2:State Assignment,
The max number in the sequence is 6,so we need a minimum of 3 binary bits to represent each state.
State | Assigned Binary Bits |
---|---|
0 | 000 |
1 | 001 |
2 | 010 |
4 | 100 |
6 | 110 |
Step3:
writing the state Transition table by using State assignment,we get
Present State | Next State | ||||
Q2 | Q1 | Q0 | Q2+ | Q1+ | Q0+ |
0 | 0 | 0 | 0 | 0 | 1 |
0 | 0 | 1 | 0 | 1 | 0 |
0 | 1 | 0 | 1 | 0 | 0 |
1 | 0 | 0 | 1 | 1 | 0 |
1 | 1 | 0 | 0 | 0 | 0 |
Step4: Circuit excitation table:
TO build circuit excitation table, we need the excitation table of D flip flop,
Present State | Next State | D flipflop Input(D) |
---|---|---|
0 | 0 | 0 |
0 | 1 | 1 |
1 | 0 | 0 |
1 | 1 | 1 |
Now, by considering the above excitation table, let us make our circuit excitation table, we get
Since we used three bits to represent each state, we need three flipflops.
present state | Next state | D ff inputs | ||||||
Q2 | Q1 | Q0 | Q2+ | Q1+ | Q0+ | D2 | D1 | D0 |
0 | 0 | 0 | 0 | 0 | 1 | 0 | 0 | 1 |
0 | 0 | 1 | 0 | 1 | 0 | 0 | 1 | 0 |
0 | 1 | 0 | 1 | 0 | 0 | 1 | 0 | 0 |
1 | 0 | 0 | 1 | 1 | 0 | 1 | 1 | 0 |
1 | 1 | 0 | 0 | 0 | 0 | 0 | 0 | 0 |
Step5:
Minimizing the above table by using any reduction techniques like kmap,
By using Kmap,
D2 with respect to Q2,Q2 and Q0,
D1 with respect to Q2,Q2 and Q0,
D0 with respect to Q2,Q2 and Q0,
4)
Given sequence is
0 --- 1 --- 3 ---7 ---6 ---4
Step1:State Transition table
we can make the state diagram as follows.
Present State | Next State |
---|---|
0 | 1 |
1 | 3 |
3 | 7 |
7 | 6 |
6 | 4 |
4 | 0 |
Step2:State Assignment,
The max number in the sequence is 6,so we need a minimum of 3 binary bits to represent each state.
State | Assigned Binary Bits |
---|---|
0 | 000 |
1 | 001 |
3 | 011 |
7 | 111 |
6 | 110 |
4 | 100 |
Step3:
writing the state Transition table by using State assignment,we get
Present State | Next State | ||||
Q2 | Q1 | Q0 | Q2+ | Q1+ | Q0+ |
0 | 0 | 0 | 0 | 0 | 1 |
0 | 0 | 1 | 0 | 1 | 1 |
0 | 1 | 1 | 1 | 1 | 1 |
1 | 1 | 1 | 1 | 1 | 0 |
1 | 1 | 0 | 1 | 0 | 0 |
1 | 0 | 0 | 0 | 0 | 0 |
Step4: Circuit excitation table:
TO build circuit excitation table, we need the excitation table of T flip flop,
Present State | Next State | T flipflop Input(T) |
---|---|---|
0 | 0 | 0 |
0 | 1 | 1 |
1 | 0 | 1 |
1 | 1 | 0 |
Now, by considering the above excitation table, let us make our circuit excitation table, we get
Since we used three bits to represent each state, we need three flipflops.
present state | Next state | T ff inputs | ||||||
Q2 | Q1 | Q0 | Q2+ | Q1+ | Q0+ | T2 | T1 | T0 |
0 | 0 | 0 | 0 | 0 | 1 | 0 | 0 | 1 |
0 | 0 | 1 | 0 | 1 | 1 | 0 | 1 | 0 |
0 | 1 | 1 | 1 | 1 | 1 | 1 | 0 | 0 |
1 | 1 | 1 | 1 | 1 | 0 | 0 | 1 | 1 |
1 | 1 | 0 | 1 | 0 | 0 | 0 | 1 | 0 |
1 | 0 | 0 | 0 | 0 | 0 | 1 | 0 | 0 |
Step5:
Minimizing the above table by using any reduction techniques like K-map,
By using K-map,
T2 with respect to Q2,Q2 and Q0,
T1 with respect to Q2,Q2 and Q0,
T0 with respect to Q2,Q2 and Q0,
5)Answer
Given sequence is
0 --- 1 --- 2 ---3 ---4 ---5---6
Step1:State Transition table
we can make the state diagram as follows.
Present State | Next State |
---|---|
0 | 1 |
1 | 2 |
2 | 3 |
3 | 4 |
4 | 5 |
5 | 6 |
6 | 0 |
Step2:State Assignment,
The max number in the sequence is 6,so we need a minimum of 3 binary bits to represent each state.
State | Assigned Binary Bits |
---|---|
0 | 000 |
1 | 001 |
2 | 010 |
3 | 011 |
4 | 100 |
5 | 101 |
6 | 110 |
Step3:
writing the state Transition table by using State assignment,we get
Present state | Next State | ||||
Q2 | Q1 | Q0 | Q2+ | Q1+ | Q0+ |
0 | 0 | 0 | 0 | 0 | 1 |
0 | 0 | 1 | 0 | 1 | 0 |
0 | 1 | 0 | 0 | 1 | 1 |
0 | 1 | 1 | 1 | 0 | 0 |
1 | 0 | 0 | 1 | 0 | 1 |
1 | 0 | 1 | 1 | 1 | 0 |
1 | 1 | 0 | 0 | 0 | 0 |
Step4:Circuit excitation table:
TO build circuit excitation table,we need the excitation table of JK flip flop,
Present State | Next State | J | K |
---|---|---|---|
0 | 0 | 0 | X |
0 | 1 | 1 | X |
1 | 0 | X | 1 |
1 | 1 | X | 0 |
Now,by considering the above excitation table,let us make our circuit excitation table,we get
Since,we used three bits to represent each state,we need three flipflops.
Present state | next State | JK ff inputs | |||||||||
Q2 | Q1 | Q0 | Q2+ | Q1+ | Q0+ | J2 | K2 | J1 | K1 | J0 | K0 |
0 | 0 | 0 | 0 | 0 | 1 | 0 | X | 0 | X | 1 | X |
0 | 0 | 1 | 0 | 1 | 0 | 0 | X | 1 | X | X | 1 |
0 | 1 | 0 | 0 | 1 | 1 | 0 | X | X | 0 | 1 | X |
0 | 1 | 1 | 1 | 0 | 0 | 1 | X | X | 1 | X | 1 |
1 | 0 | 0 | 1 | 0 | 1 | X | 0 | 0 | X | 1 | X |
1 | 0 | 1 | 1 | 1 | 0 | X | 0 | 1 | X | X | 1 |
1 | 1 | 0 | 0 | 0 | 0 | X | 1 | X | 1 | 0 | X |
Step5:
Minimizing the above table by using any reduction techniques like kmap,
By using Kmap,
D2 = Q2Q1Q0 +Q2Q1Q0
D2 = Q2Q1Q0 + Q2Q1Q0
Do = Q2Q1Qc
T2 = Q2Q1Q0 +Q22100
T1 = Q20100 +0201
T0 = Q22100 + Q2Q100
J2 = Q2Q100
K2 = Q1Q0
Jl = Q1Q0 K1 = Q200 + Q2Q0 JO=Q2+Q1 KO Q2 +01
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