00,01,10,11 as output using 2 JK flip flops

Question

I want to know how the bit pattern 00, 01 ,10,11 may be obtained using a counter (2 JK flip flops, synchronous).

Truth table for JK :

clock	J	K	next state
0	dc	dc	old value
1	0	0	old value
1	0	1	0
1	1	0	1
1	1	1	toggle

Answer 1

 

Present State		Next State		Flipflop Inputs*
$A$	$B$	$A^+$	$B^+$	$J_A$	$K_A$	$J_B$	$K_B$
$0$	$0$	$0$	$1$	$0$	$X$	$1$	$X$
$0$	$1$	$1$	$0$	$1$	$X$	$X$	$1$
$1$	$0$	$1$	$1$	$X$	$0$	$1$	$X$
$1$	$1$	$0$	$0$	$X$	$1$	$X$	$1$

*use excitation table of JK flipflop 

Solve for $J_A,K_A,J_B\;\text{and}\;K_B$ using K-map 

$J_B=1\quad K_B=1\quad J_A=B\quad K_A=B$

Now draw the counter using these equations 

Comments

Unfortunately i am not able to connect/ paste the diagram.

I drew 2 JK FFs.

Initially :

J1=0  K1=1

J2=0 (clock providing input to it, basically clock is low)

Q1 serves as input to K2 (so, initially K2=0)

This is how, Q1 = Q2 = 0 initially.

Now, i worked it out using a combination of Truth table and excitation table as follows (It seems fine. I want to know if this approach is ok...)

clock	J1	K1	Q1	J2	K2	Q2
H	0	1	0(TT)	1	0	1(TT)
L	1	1	1(ET	0	0 retained from previous Q1	1(TT)
H	1	1	0(TT)	1	1 retained from previous Q1	0(TT)

---

initially $AB = 00$

$J_B=1 \;, K_B=1;\;\text{Toggle, or}\; B^+= J_BB'+K_B'B=1$

$J_A=B=0 \;, K_A=B=0;\;\text{No Change, or}\; A^+= J_AA'+K_A'A=0$

Next State $AB = 01$

$J_B=1 \;, K_B=1;\;\text{Toggle, or}\; B^+= J_BB'+K_B'B=0$

$J_A=B=1 \;, K_A=B=1;\;\text{Toggle, or}\; A^+= J_AA'+K_A'A=1$

Next State $AB =10$

$J_B=1 \;, K_B=1;\;\text{Toggle, or}\; B^+= J_BB'+K_B'B=1$

$J_A=B=0 \;, K_A=B=0;\;\text{No Change, or}\; A^+= J_AA'+K_A'A=1$

Next state $AB=11$

$J_B=1 \;, K_B=1;\;\text{Toggle, or}\; B^+= J_BB'+K_B'B=0$

$J_A=B=1 \;, K_A=B=1;\;\text{Toggle, or}\; A^+= J_AA'+K_A'A=0$

Next state $AB=00$ .... so on it will repeat the sequence

Answer 2

Well according to your question, assume values of q0=0 and q1=0.

 DESIRED OUTPUT IS : 00 01 10 11

 J1=K1=1, J1=Q0 exor Q1 , k0=0