Question

 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.

Answer 1

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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

Answer 2

Design a binary counter with the following repeated binary sequence: Use JK-type Flip-Flops. 0, 1, 2, 3, 4, 5, 6, 7